Disha Resource Book
Disha Resource Book
Disha Resource Book
Resource Book
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D. P. Gupta [ Mathematics ]
Preetima Bajpai [ Chemistry ]
Sanjeev Kumar Jha [ Physics ]
DISHA PUBLICATION
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publisher do not take any legal responsibility for any errors or misrepresentations that might have crept in. We have
tried and made our best efforts to provide accurate up-to-date information in this book.
(ii)
Contents
• JEE Main - 2016 Solved Paper (with solutions) 2016-1-19
2. Physical Chemistry 1 C-5-10 12. AIEEE - 2012 PAPER (with solutions) 2012-1-20
3. Physical Chemistry 2 C-11-16 13. JEE Main - 2013 PAPER (with solutions) 2013-1-16
4. Physical Chemistry 3 C-17-20 14. JEE Main - 2014 PAPER (with solutions) 2014-1-20
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PART TEST SYLLABUS
JEE MAIN
Resource Book with CD
JEE Main - 2016
(Held on 3rd April, 2016)
Time : 3 Hours • Each correct answer has + 4 marks • Each wrong answer has – 1 mark. Max. Marks : 360
19. Identify the semiconductor devices whose characteristics 22. A screw gauge with a pitch of 0.5 mm and a circular scale
are given below, in the order (a), (b), (c), (d) : with 50 divisions is used to measure the thickness of a thin
sheet of Aluminium. Before starting the measurement, it is
I I found that wen the two jaws of the screw gauge are brought
in contact, the 45th division coincides with the main scale
line and the zero of the main scale is barely visible. What is
V V the thickness of the sheet if the main scale reading is 0.5 mm
and the 25th division coincides with the main scale line?
(1) 0.70 mm (2) 0.50 mm
(3) 0.75 mm (4) 0.80 mm
(a) (b) 23. A roller is made by joining together two cones at their
vertices O. It is kept on two rails AB and CD, which are
placed asymmetrically (see figure), with its axis perpendicular
I dark Resistance to CD and its centre O at the centre of line joining AB and Cd
(see figure). It is given a light push so that it starts rolling
with its centre O moving parallel to CD in the direction shown.
V As it moves, the roller will tend to :
Intensity
B D
Illuminated of light
(c) (d)
2mF
(A) (B)
+ –
These materials are used to make magnets for elecric
8V generators, transformer core and electromagnet core. Then
it is proper to use :
(1) 420 N/C (2) 480 N/C (1) A for transformers and B for electric generators.
(3) 240 N/C (4) 360 N/C (2) B for electromagnets and transformers.
21. A satellite is revolving in a circular orbit at a height 'h' from (3) A for electric generators and trasformers.
the earth's surface (radius of earth R; h < < R). The minimum (4) A for electromagnets and B for electric generators.
increase in its orbital velocity required, so that the satellite 25. The box of a pin hole camera, of length L, has a hole of
could escape from the earth's gravitational field, is close to : radius a. It is assumed that when the hole is illuminated by a
(Neglect the effect of atmosphere.) parallel beam of light of wavelength l the spread of the
spot (obtained on the opposite wall of the camera) is the
(1) gR / 2 (2) gR ( )
2 -1
sum of its geometrical spread and the spread due to
(3) 2gR (4) gR diffraction. The spot would then have its minimum size (say
bmin) when :
2016-4 JEE MAIN 2016 Solved Paper
l2
(2) a= and bmin = 4lL
L
31. The area (in sq. units) of the region {(x, y) : y2 ³ 2x and
x2 + y2 £ 4x, x ³ 0, y ³ 0} is :
l2 æ 2l2 ö
(3) a= and bmin = ç
ç L ÷
÷
L è ø 4 2 p 2 2
(1) p- (2) -
3 2 3
æ 2l2 ö
a = ll and bmin = ç ÷ 4 8
(4) ç L ÷ (3) p- (4) p-
è ø 3 3
26. A uniform string of length 20 m is suspended from a rigid æ1 ö
support. A short wave pulse is introduced at its lowest end. 32. If f(x) + 2f ç
ç ÷÷ = 3x , x ¹ 0 and
èx ø
It starts moving up the string. The time taken to reach the
S = {x Î R : f(x) = f(–x)}; then S:
supports is :
(1) contains exactly two elements.
(take g = 10 ms–2) (2) contains more than two elements.
(1) 2 2s (2) 2s (3) is an empty set.
(4) contains exactly one element.
(3) 2p 2 s (4) 2 s
2x12 + 5x 9
27. An ideal gas undergoes a quasi static, reversible process in 33. The integral ò 3
dx is equal to :
x5 + x3 + 1
which its molar heat capacity C remains constant. If during
this process the relation of pressure P and volume V is given
( )
by PVn = constant, then n is given by (Here CP and CV are x5 - x10
molar specific heat at constant pressure and constant volume, (1) 2
+C (2) 2
+C
5 3 5 3
respectively) : (
2 x + x +1
) (
2 x + x +1
)
CP - C C - CV
(1) n= (2) n= -x5 x10
C - CV C - CP (3) 2
+C (4) 2
+C
x5 + x3 + 1 2 x5 + x3 +1
CP C – CP
( ) ( )
(3) n= (4) n= where C is an arbitrary constant.
CV C – CV
34. For x Î R, f(x) = |log2 – sinx| and g(x) = f(f(x)), then :
28. An observer looks at a distant tree of height 10 m with a (1) g'(0) = – cos(log2)
telescope of magnifying power of 20. To the observer the (2) g is differentiable at x = 0 and g'(0) = – sin(log2)
tree appears : (3) g is not differentiable at x = 0
(1) 20 times taller (2) 20 times nearer (4) g'(0) = cos(log2)
(3) 10 times taller (4) 10 times nearer 35. The centres of those circles which touch the circle,
29. In an experiment for determination of refractive index of glass x2 + y2 – 8x – 8y – 4 = 0, externally and also touch the x-axis,
of a prism by i – d, plot it was found thata ray incident at lie on:
angle 35°, suffers a deviation of 40° and that it emerges at (1) a hyperbola (2) a parabola
angle 79°. In that case which of the following is closest to (3) a circle (4) an ellipse which is not a circle
the maximum possible value of the refractive index? 36. The sum of all real values of x satisfying the equation
(1) 1.7 (2) 1.8 2
(3) 1.5 (4) 1.6 (x 2 - 5 x+ 5) x + 4x - 60
= 1 is :
30. A pendulum clock loses 12 s a day if the temperature is 40°C (1) 6 (2) 5
and gains 4 s a day if the temperature is 20° C. The (3) 3 (4) – 4
temperature at which the clock will show correct time, and nd th th
37. If the 2 , 5 and 9 terms of a non-constant A.P. are in
the co-efficient of linear expansion (a) of the metal of the G.P., then the common ratio of this G.P. is :
pendulum shaft are respectively :
7
(1) 30°C; a = 1.85 × 10–3/°C (1) 1 (2)
4
(2) 55°C; a = 1.85 × 10–2/°C
(3) 25°C; a = 1.85 × 10–5/°C 8 4
(3) (4)
(4) 60°C; a = 1.85 × 10–4/°C 5 3
JEE MAIN 2016 Solved Paper 2016-5
2 æ 3ö æ 1 ö
(1) (2) 3 (1) sin -1 ç ÷ (2) sin -1 çç ÷÷
3 ç 4 ÷
è ø è 3ø
4 4
(3) (4) p p
3 3 (3) (4)
3 6
æ 2 4 ön 46. If the sum of the first ten terms of the series
39. If the number of terms in the expansion of çç1 - + 2 ÷÷ ,
è x x ø æ 3 ö2 æ 2 ö2 æ 1 ö2 2 æ 4 ö2 16
x ¹ 0, is 28, then the sum of the coefficients of all the terms çç1 ÷÷ + çç2 ÷÷ + çç3 ÷÷ + 4 + çç4 ÷÷ + ......., is m,
è 5ø è 5ø è 5ø è 5ø 5
in this expansion, is :
(1) 243 (2) 729 then m is equal to :
(3) 64 (4) 2187 (1) 100 (2) 99
40. The Boolean Expression (p Ù : q) Ú qÚ (: p Ù q) is (3) 102 (4) 101
47. The system of linear equations
equivalent to:
x + ly – z = 0
(1) p Ú q (2) p Ú : q
lx – y – z = 0
(3) : p Ù q (4) p Ù q x + y – lz = 0
41. Consider has a non-trivial solution for:
æ 1 + sin x ö æ ö (1) exactly two values of l.
f (x) = tan -1 ç ÷, x Î ç0,p÷.
ç 1 - sin x ÷ ç
è 2ø
÷ (2) exactly three values of l.
è ø
(3) infinitely many values of l.
p (4) exactly one value of l.
A normal to y = f(x) at x = also passes through the point :
6 x -3 y+2 z+4
48. If the line, = = lies in the plane, lx + my – z = 9,
æp ö æp ö 2 -1 3
(1) çç , 0 ÷÷ (2) ç , 0÷
ç ÷
è6 ø è4 ø then l2 + m2 is equal to :
(1) 5 (2) 2
æ 2pö
(3) (0, 0) (4) çç0, ÷÷ (3) 26 (4) 18
è 3 ø 49. If all the words (with or without meaning) having five letters,
1 formed using the letters of the word SMALL and arranged
æ(n + 1) (n + 2)...3n ön as in a dictionary; then the position of the word SMALL is :
42. lim ç ÷ is equal to: (1) 52nd (2) 58th
n ®¥ ç
è n 2n
÷
ø th
(3) 46 (4) 59th
9 50. If the standard deviation of the numbers 2, 3, a and 11 is 3.5,
(1) (2) 3 log 3 – 2 then which of the following is true?
e2
(1) 3a2 – 34a + 91 = 0
18 27 (2) 3a2 – 23a + 44 = 0
(3) 4 (4) (3) 3a2 – 26a + 55 = 0
e e2
43. If one of the diameters of the circle, given by the equation, (4) 3a2 – 32a + 84 = 0
x2 + y2 – 4x + 6y – 12 = 0, is a chord of a circle S, whose centre 51. A wire of length 2 units is cut into two parts which are bent
is at (–3, 2), then the radius of S is: respectively to form a square of side = x units and a circle of
(1) 5 (2) 10 radius = r units. If the sum of the areas of the square and the
(3) 5 2 (4) 5 3 circle so formed is minimum, then:
(1) x = 2r (2) 2x = r
44. Let two fair six-faced dice A and B be thrown simultaneously.
If E1 is the event that die A shows up four, E2 is the event (3) 2x = (p + 4)r (4) (4 – p) x = pr
that die B shows up two and E3 is the event that the sum of 1
2
numbers on both dice is odd, then which of the following
statements is NOT true ?
52. (
Let p = lim+ 1 + tan x
x®0
) 2x then log p is equal to :
(1) E1 and E3 are independent.
(2) E1, E2 and E3 are independent. 1 1
(1) (2)
(3) E1 and E2 are independent. 2 4
(4) E2 and E3 are independent. (3) 2 (4) 1
2016-6 JEE MAIN 2016 Solved Paper
53. y2
Let P be the point on the parabola, = 8x which is at a 59. Two sides of a rhombus are along the lines, x – y + 1 = 0 and
minimum distance from the centre C of the circle, 7x – y – 5 = 0. If its diagonals intersect at (–1, –2), then which
x2 + (y + 6)2 = 1. Then the equation of the circle, passing one of the following is a vertex of this rhombus?
through C and having its centre at P is: æ1 8 ö æ 10 7 ö
x
(1) ç ,- ÷
ç ÷ (2) çç- , - ÷÷
x 2 + y2 - + 2y - 24 = 0
è3 3 ø è 3 3ø
(1)
4 (3) (–3, –9) (4) (–3, –8)
(2) x2 + y2 – 4x + 9y + 18 = 0 60. If 0 £ x < 2p, then the number of real values of x, which
(3) x2 + y2 – 4x + 8y + 12 = 0 satisfy the equation
(4) x2 + y2 – x + 4y – 12 = 0 cos x + cos 2x + cos 3x + cos 4x = 0 is:
54. If a curve y = f(x) passes through the point (1, –1) and satisfies (1) 7 (2) 9
æ 1ö (3) 3 (4) 5
the differential equation, y(1 + xy) dx = x dy, then f ç
ç- ÷
÷ is
è 2ø Section - 3
equal to :
2 4
(1) (2)
5 5 61. Two closed bulbs of equal volume (V) containing an ideal
2 4 gas initially at pressure pi and temperature T1 are connected
(3) - (4) - through a narrow tube of negligible volume as shown in the
5 5
figure below. The temperature of one of the bulbs is then
® ® ® raised to T2. The final pressure pf is :
55. Let a , b and c be three unit vectors such that
T1 T1 T1 T2
® æ® ® ö æ® ® ö ®
÷ = 3 ç b + c ÷. If b is not parallel to c , then
®
a ´ç b ´ c pf,V pf,V
ç ÷ 2 ç ÷
è ø è ø
® ®
the angle between a and b is: æ T ö æ T 1T2 ö
(1) 2 pi ç 2 ÷ (2) 2 pi ç T + T ÷
è T1 + T2 ø è 1 2ø
2p 5p
(1) (2)
3 6 æ T T ö æ T1 ö
(3) pi ç 1 2 ÷ (4) 2 pi ç T + T ÷
3p p T + T
è 1 2ø è 1 2ø
(3) (4)
4 2 62. Which one of the following statements about water is FALSE ?
(1) There is extensive intramolecular hydrogen bonding
é5a - b ù
56. If A = ê ú and A adj A = A AT, then 5a + b is equal in the condensed phase.
ë 3 2 úû
ê (2) Ice formed by heavy water sinks in normal water.
to : (3) Water is oxidized to oxygen during photosynthesis.
(1) 4 (2) 13 (4) Water can act both as an acid and as a base.
63. In the Hofmann bromamide degradation reaction, the number
(3) –1 (4) 5
57. A man is walking towards a vertical pillar in a straight path, of moles of NaOH and Br2 used per mole of amine produced
at a uniform speed. At a certain point A on the path, he are :
observes that the angle of elevation of the top of the pillar is (1) Two moles of NaOH and two moles of Br 2.
30°. After walking for 10 minutes from A in the same direction, (2) Four moles of NaOH and one mole of Br 2.
at a point B, he observes that the angle of elevation of the (3) One mole of NaOH and one mole of Br 2.
top of the pillar is 60°. Then the time taken (in minutes) by (4) Four moles of NaOH and two moles of Br 2.
him, from B to reach the pillar, is: 64. Which of the following atoms has the highest first ionization
(1) 20 (2) 5 energy?
(3) 6 (4) 10 (1) K (2) Sc
58. The distance of the point (1, –5, 9) from the plane x – y + z = 5 (3) Rb (4) Na
measured along the line x = y = z is : 65. The concentration of fluoride, lead, nitrate and iron in a water
sample from an underground lake was found to be 1000 ppb,
10 20
(1) (2) 40 ppb, 100 ppm and 0.2 ppm, respectively. This water is
3 3 unsuitable for drinking due to high concentration of :
(3) (4) 10 3 (1) Nitrate (2) Iron
3 10
(3) Fluoride (4) Lead
JEE MAIN 2016 Solved Paper 2016-7
66. The heats of combustion of carbon and carbon monoxide 75. Which of the following statements about low density
are –393.5 and –283.5 kJ mol–1, respectively. The heat of polythene is FALSE?
formation (in kJ) of carbon monoxide per mole is : (1) Its synthesis requires dioxygen or a peroxide initiator
(1) –676.5 (2) – 110.5 as a catalyst.
(3) 110.5 (4) 676.5 (2) It is used in the manufacture of buckets, dust-bins etc.
(3) Its synthesis requires high pressure.
67. The equilibrium constant at 298 K for a reaction A +B (4) It is a poor conductor of electricity.
C+D is 100. If the initial concentration of all the four species 76. Which of the following compounds is metallic and
were 1 M each, then equilibrium concentration of D (in mol ferromagnetic?
L–1) will be : (1) VO2 (2) MnO2
(1) 1.818 (2) 1.182 (3) TiO2 (4) CrO2
(3) 0.182 (4) 0.818 77. The product of the reaction given below is:
68. The absolute configuration of
1. NBS/hv
CO2H 2. H 2O/K 2CO3
X
H OH
H Cl
O CO2H
CH3
is : (1) (2)
(1) (2S, 3S) (2) (2R, 3R)
(3) (2R, 3S) (4) (2S, 3R)
69. For a linear plot of log (x/m) versus log p in a Freundlich OH
adsorption isotherm, which of the following statements is
correct? (k and n are constants) (3) (4)
(1) Only 1 n appears as the slope.
78. The hottest region of Bunsen flame shown in the figure
(2) log (1 n ) appears as the intercept. below is :
region 4
(3) Both k and 1 n appear in the slope term.
(4) 1 n appears as the intercept. region 3
70. The distillation technique most suited for separating glycerol region 2
from spent-lye in the soap industry is :
(1) Steam distillation. region 1
(2) Distillation under reduced pressure.
(3) Simple distillation
(4) Fractional distillation
71. Which of the following is an anionic detergent?
(1) Cetyltrimethyl ammonium bromide. (1) region 3 (2) region 4
(2) Glyceryl oleate. (3) region 1 (4) region 2
(3) Sodium stearate. 79. At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires
(4) Sodium lauryl sulphate. 375 mL air containing 20% O2 by volume for complete
72. The species in which the N atom is in a state of sp combustion. After combustion the gases occupy 330 mL.
hybridization is : Assuming that the water formed is in liquid form and the
volumes were measured at the same temperature and
(1) NO-3 (2) NO2
pressure, the formula of the hydrocarbon is:
(3) NO+2 (4) NO-2 (1) C4H8 (2) C4H10
73. Thiol group is present in : (3) C3H6 (4) C3H8
80. The pair in which phosphorous atoms have a formal
(1) Cysteine (2) Methionine
oxidation state of + 3 is :
(3) Cytosine (4) Cystine
(1) Orthophosphorous and hypophosphoric acids
74. Which one of the following ores is best concentrated by
(2) Pyrophosphorous and pyrophosphoric acids
froth floatation method?
(3) Orthophosphorous and pyrophosphorous acids
(1) Galena (2) Malachite (4) Pyrophosphorous and hypophosphoric acids
(3) Magnetite (4) Siderite
2016-8 JEE MAIN 2016 Solved Paper
81. The reaction of propene with HOCl (Cl 2 + H2O) proceeds 86. The reaction of zinc with dilute and concentrated nitric acid,
through the intermediate: respectively, produces:
(1) NO and N2O (2) NO2 and N2O
(1) CH 3 – CH ( OH ) - CH 2+ (3) N2O and NO2 (4) NO2 and NO
87. Decomposition of H2O2 follows a first order reaction. In
(2) CH3 – CHCl - CH 2+
fifty minutes the concentration of H2O2 decreases from 0.5
to 0.125 M in one such decomposition. When the
(3) CH3 – CH + - CH 2 – OH
concentration of H2O2 reaches 0.05 M, the rate of formation
(4) CH3 – CH + - CH 2 – Cl of O2 will be:
(1) 2.66 L min–1 at STP
82. 2-chloro-2-methylpentane on reaction with sodium methoxide
(2) 1.34 × 10–2 mol min–1
in methanol yields:
(3) 6.96 × 10–2 mol min–1
CH3 (4) 6.93 × 10–4 mol min–1
88. The pair having the same magnetic moment is:
(i) C2H5CH2C OCH3 (ii) C2H5CH2C = CH2
[At. No.: Cr = 24, Mn = 25, Fe = 26, Co = 27]
CH3 CH3 (1) [Mn(H2O)6]2+ and [Cr(H2O)6]2+
(2) [CoCl4]2– and [Fe(H2O)6]2+
(iii) C2H 5CH = C – CH3 (3) [Cr(H2O)6]2+ and [CoCl4]2–
(4) [Cr(H2O)6]2+ and [Fe(H2O)6]2+
CH 3
89. Galvanization is applying a coating of:
(1) (iii) only (2) (i) and (ii) (1) Cu (2) Zn
(3) All of these (4) (i) and (iii) (3) Pb (4) Cr
83. Which one of the following complexes shows optical 90. A stream of electrons from a heated filaments was passed
isomerism? two charged plates kept at a potential difference V esu. If e
(1) trans [Co(en)2Cl2]Cl (2) [Co(NH3)4Cl2]Cl and m are charge and mass of an electron, respectively, then
(3) [Co(NH3)3Cl3] (4) cis[Co(en)2Cl2]Cl the value of h/l (where l is wavelength associated with
(en = ethylenediamine) electron wave) is given by:
84. The main oxides formed on combustion of Li, Na and K in (1) (2)
meV 2meV
excess of air are, respectively:
(3) meV (4) 2meV
(1) Li2O2, Na2O2 and KO2 (2) Li2O, Na2O2 and KO2
(3) Li2O, Na2O and KO2 (4) LiO2, Na2O2 and K2O
85. 18 g glucose (C6H12O6) is added to 178.2 g water. The vapour
pressure of water (in torr) for this aqueous solution is:
(1) 752.4 (2) 759.0
(3) 7.6 (4) 76.0
JEE MAIN 2016 Solved Paper 2016-9
æ R ö
1. (2) We know that V = w
A2 - x 2 distance r^ is çè - a÷
2 ø
2
w A2 æ 2A ö 6. (3) In amplitude modulation, the amplitude of the high
Initially V= -ç ÷
è 3 ø frequency carrier wave made to vary in proportional to
the amplitude of audio signal.
2
w A' 2 æ 2A ö
Finally 3v = -ç ÷
è 3 ø Audio signal
2A
Where A'= final amplitude (Given at x = , velocity Carrier wave
3
to trebled)
2 Amplitude modulated wave
æ 2A ö
A' 2 - ç ÷
3 è 3 ø
On dividing we get =
1 æ 2A ö
2
A2 - ç ÷ 1
mv2
è 3 ø 7. (3) hn02 – hn0 =
2
é 2 4A 2 ù 4A 2 7A 4 1
9 ê A - 9 ú = A'2 – \ A' = \ hn 0 - hn 0 = mv ' 2
êë úû 9 3 3 2
Ic Ic
2. (2, 4)We know that a = and b = 4 4
Ie Ib n - n0 n - n0
v '2
3
Also Ie = Ib + Ic \ = \ v' = v 3
v2 n - n0 n - n0
Ic
Ic I b
\ a= = b = 4
I b + Ic Ic 1 + b \ v' > v
1+ 3
Ib
Option (2) and (4) are therefore correct.
| DT1 | + | DT2 | + | DT3 | + | DT4 | 8. (2) Case (a) :
3. (3) DT =
4
2 +1+ 3 + 0
= = 1.5 µ0 I µ I
4 bA = ´ 2p = 0 ´ 2p (2pR = l)
As the resolution of measuring clock is 1.5 therefore 4p R 4p l / 2p
the mean time should be 92 ± 1.5
4. (1) In case of an 'OR' gate the input is zero when all inputs µ0 I
= ´ (2p)2
are zero. If any one input is ' 1', then the output is '1'. 4p l
5. (1) We know that L = mvr^ Case (b) :
y
D V C 45°
a a a BB
V
a a V a/2
a
A V B
R/ 2
µ0 I
BB = 4 × [sin 45° + sin 45°]
4p a / 2
O a X
µ0 I 2 µ I
R = 4´ ´ ´ = 0 ´ 32 2 [4a = l]
4p l / 8 2 4p l
2
2016-10 JEE MAIN 2016 Solved Paper
f - P0
9. (2) P = V V + 3P
0
l l -P0
[slope = , c = 3P0]
V0
(a) (b) PV0 + P0V = 3P0V0 ...(i)
But pv = nRT
v nRT
The fundamental frequency in case (a) is f =
2l \p= ...(ii)
v
The fundamental frequency in case (b) is
nRT
v u From (i) & (ii) V0 + P0V = 3P0V0
f ' = 4(l / 2) = 2l = f v
\ nRT V0 + P0V2 = 3P0V0 ...(iii)
10. (3) Applying Gauss's law
dT
uur uur Q For temperature to be maximum =0
Ñò S E. ds = Î0 dv
Differentiating e.q. (iii) by 'v' we get
Q + 2par 2 - 2pAa 2 dT
\ E × 4pr2 = nRV0 + P0(2v) = 3P0V0
Î0 dv
dr r dr dT
r= \ nRV0 = 3P0V0 – 2 P0V
dv Q a Gaussiam dv
Q = r4pr2 surface
b dT 3P0 V0 - 2P0 V
.
A = =0
dv nRV0
ò r 4pr
2
Q= dr =2pA [r2 – a2]
a 3V0 3P0
V= \ p= [From (i)]
1 é Q - 2pAa 2 ù 2 2
E= ê + 2pA ú
4p Î0 êë r 2
úû 9Po Vo
\ Tmax = [From (iii)]
For E to be independent of 'r' 4nR
Q – 2pAa2 = 0 W mgh ´ 1000 10 ´ 9.8 ´ 1 ´ 1000
13. (2) n = = =
Q input input input
\ A=
2pa 2 98000
11. (2) Here Input = = 49 × 104J
0.2
e e e
i= = = 49 ´ 10 4
R 2
+ X 2L 2
R +w L 2 2 2 2 2 2
R + 4p v L Fat used = = 12.89 × 10–3kg.
3.8 ´ 107
220 V 80 14. (1) Loss in P.E. = Work done against friction from p ® Q
10 = [Q R = = = 8] + work done against friction from Q ® R
64 + 4p (50) L 2 2 I 10
mgh = m(mgcosq) PQ + mmg (QR)
On solving we get h = m cos q × PQ + m(QR)
L = 0.065 H
12. (3) The equation for the line is 3 2
2= m× × + mx
2 sin 30°
P
2 = 2 3 m + mx --- (i)
3Po
2
c [sin 30° = ]
PQ
2Po q
Also work done P ® Q = work done Q ® R
Po
q \ 2 3 m = mx
Po
Vo
\ x » 3.5m
V
Vo 2Vo
JEE MAIN 2016 Solved Paper 2016-11
From (i) 2 = 2 3 m + 2 3 m = 4 3 m 4
The voltage across CP is VP = ×8 = 2v
2 1 4 + 12
m= = = 0.29 \ Voltage across 9mF is also 2V
4 3 2 ´ 1.732
\ Charge on 9mF capacitor = 9 × 2 = 18mC
r P \ Total charge on 4 mF and 9mF = 42mc
15. (1)
KQ 42 ´ 10-6
\E = = 9 × 109 × = 420 Nc–1
r2 30 ´ 30
T T
Semiconductor 21. (2) For h << R, the orbital velocity is gR
Metal (for limited –Eg
range of temperature) KT
Escape velocity =
B
2gR
E, Decreases \ The minimum increase in its orbital velocity
16. (3)
g-rays X-rays uv-rays Visible rays IR rays Radio = 2gR – gR = gR ( 2 – 1)
VIBGYOR Microwaves waves
Radio wave < yellow light < blue light < X-rays 0.5
(Increasing order of energy) 22. (4) L.C. = = 0.01 mm
50
17. (3) Ig G = ( I – Ig)s
Zero error = 5 × 0.01 = 0.05 mm (Negative)
\ 10–3 × 100 = (10 – 10–3) × S
\ S » 0.01 W Reading = (0.5 + 25 × 0.01) + 0.05 = 0.80 mm
18. (2) For At½ = 20 min, t = 80 min, number of half lifes n = 4
23. (3) As shown in the diagram, the normal reaction of AB on
No
\ Nuclei remaining = . Therefore nuclei decayed roller will shift towards O.
24 This will lead to tending of the system of cones to turn
No left.
= N0 - B D
24
For Bt½ = 40 min., t = 80 min, number of half lifes n = 2
No
\ Nuclei remaining = . Therefore nuclei decayed
22 O
No
= N0 -
22
A C
No 1 24. (2) Graph [A] is for material used for making permanent
No - 4 1-
2 16 15 4 5 magnets (high coercivity)
\Required ratio = No = 1 = ´ = Graph [B] is for making electromagnets and
No - 2 1- 16 3 4
2 4 transformers.
19. (3) Graph (a) is for a simple diode. 25. (1) Given geometrical spread = a
Graph (b) is showing the V Break down used for zener l lL
diode. Diffraction spread = ´L=
Graph (c) is for solar cell which shows cut-off voltage a a
and open circuit current. lL
Graph (d) shows the variation of resistance h and hence The sum b = a +
a
current with intensity of light.
db d æ lL ö
3µF C1 = 4µF 12µF = CP For b to be minimum =0 ça + ÷=0
20. (1) 4µF da da è a ø
9µF a = lL
Þ
2µF b min =
lL + lL = 2 lL = 4lL
8v 26. (1) We know that velocity in string is given by
8v
T
é æ 12 ö ù v= ...(I)
Charge on C1 is q1 = ê ç
è ÷ø ´ 8 ú ´ 4 = 24mc m
ë 4 + 12 û
2016-12 JEE MAIN 2016 Solved Paper
m mass of string 1
where m = = 4 = a (q – 20) ´ 86400 ....(ii)
l length of string 2
40 – q
The tension T =
m
´ x´g On dividing we get, 3 =
..(II) q – 20
l
3q – 60 = 40 – q
4q = 100 Þ q = 25°C
From (1) and (2)
l
dx SECTION-2 : MATHEMATICS
= gx T
dt A (2, 2)
x
l l
x -1/2 dx = g dt \ ò x -1/2dx - g ò dt
0 0 31. (4)
O B
l 20
2 l = g´t \ t = 2 =2 =2 2
g 10
(2, –2)
27. (4) For a polytropic process
Points of intersection of the two curves are (0, 0), (2, 2)
R R and (2, –2)
C = Cv + \ C - Cv =
1- n 1- n Area = Area (OAB) – area under parabola (0 to 2)
2
\ 1- n =
R
\ 1-
R
=n p ´ (2) 2
= - ò 2 x dx
C - Cv C - Cv 4
0
C - C v - R C - C v - Cp + C v 8
\ n= = =p-
C - Cv C - Cv 3
C - Cp æ1ö
= (Q C p - C v = R ) 32. (1) f (x) + 2f ç ÷ = 3x .......(1)
C - Cv èxø
æ A + dm ö æ 74 + d m ö æ1ö 3
sin ç ÷ sin ç ÷ Substracting (1) from (2) Þ f (x) - f ç ÷ = - 3x
è 2 ø= è 2 ø èxø x
But m =
sinA / 2 74 On adding the above equations
sin
2
2
Þ f (x) = -x
5 æ d ö x
= sin ç 37° + m ÷
3 è 2 ø 2 -2 2
f (x) = f (- x) Þ -x= +xÞx =
x x x
5 5
m max can be . That is m max is less than = 1.67 x2 = 2 or x = 2, - 2 .
3 3
But dm will be less than 40° so 2x12 + 5x9
33. (4) ò (x 5 + x3 + 1)3 dx
5 5
m< sin 57° < sin 60° Þ m = 1.45 Dividing by x15 in numerator and denominator
3 3
2 5
1 + 6 dx
30. (3) Time lost/gained per day = µ Dq ´ 86400 second 3
2 ò æ x 1 x 1 ö3
1 ç1 + 2 + 5 ÷
12 = a (40 – q) ´ 86400 .... (i) è x x ø
2
JEE MAIN 2016 Solved Paper 2016-13
1 1 2
+ 4x - 60
Substitute 1+ 2 + 5 = t 36. (3) (x 2 - 5x + 5) x =1
x x Case I
æ –2 5 ö x2 – 5x + 5 = 1 and x2 + 4x – 60 can be any real number
Þ ç – dx = dt
è x 3 x 6 ÷ø Þ x = 1, 4
Case II
æ2 5 ö x2 – 5x + 5 = –1 and x2 + 4x – 60 has to be an even
Þ ç ç 3+ 6÷ ÷ dx = - dt number
èx x ø
This gives, Þ x = 2, 3
where 3 is rejected because for x = 3, x2 + 4x – 60 is odd.
2 5 Case III
3
+ 6 dx
– dt
ò x 1 x 1 3 = ò t3 x2 – 5x + 5 can be any real number and x2 + 4x – 60 = 0
Þ x = –10, 6
æ ö
çè1 + 2 + 5 ÷ø Þ Sum of all values of x = –10 + 6 + 2 + 1 + 4 = 3
x x
37. (4) Let the GP be a, ar and ar 2 then a = A + d; ar = A + 4d;
1 ar2 = A + 8d
= +C
2t 2 ar 2 - ar (A + 8d)-(A + 4d)
Þ =
1 ar - a (A + 4d)-(A + d)
= +C
æ 1 1 ö2 4
2 çç1+ + ÷÷ r=
è x 2 x5 ø 3
x10 2b 2 1
= +C 38. (1) =8 and 2b = (2ae)
5 3 2 a 2
2( x + x +1)
34. (4) g (x) = f (f (x)) Þ 4b2 = a 2 e2 Þ 4a 2 (e 2 - 1) = a 2e 2
In the neighbourhood of x = 0,
f(x) = | log2 – sin x| = (log 2 – sin x) 2
\ g (x) = |log 2 – sin| log 2 – sin x || Þ 3e2 = 4 Þ e =
3
= (log 2 – sin(log 2 – sin x))
39. (2) Total number of terms = n+2C2 = 28
\ g (x) is differentiable
(n + 2) (n + 1) = 56
and g'(x) = – cos(log 2 – sin x) (– cos x)
x= 6
Þ g'(0) = cos (log 2) Sum of coefficients = (1 – 2 + 4)n = 36 = 729
40. (1) (pÙ : q) Ú q Ú (: p Ù q)
35 (2)
Þ {(p Ú q) Ù (: q Ú q)} Ú (: p Ù q)
C(4, 4)
Þ {(p Ú q) Ù T} Ú (: p Ù q)
6 Þ (p Ú q) Ú (: p Ù q)
k Þ {(p Ú q) Ú : p} Ù (p Ú q Ú q)
P(h, k) Þ TÙ (p Ú q)
Þ pÚq
k
X' X æ 1 + sin x ö
f ( x ) = tan –1 ç
è 1– sin x ÷ø
41. (4)
For the given circle,
centre : (4, 4)
æ 2ö
radius = 6 æ x xö æ xö
ç çè sin + cos ÷ø ÷ 1 + tan
2 2 ÷ ç 2÷
6+ k =
2
+ (k - 4) 2 = tan –1 ç –1
2 ÷ = tan ç
(h - 4) ç æ x xö x÷
ç çè sin – cos ÷ø ÷ ç 1 – tan ÷
(h – 4)2 = 20k + 20 è è 2ø
x 2 ø
\ locus of (h, k) is
(x – 4)2 = 20(y + 1),
æ æ p xö ö
which is a parabola. = tan –1 ç tan ç + ÷ ÷
è è 4 2ø ø
2016-14 JEE MAIN 2016 Solved Paper
æp p pö 1 1 1
P(E1Ç E 2 ) =, P(E 2 Ç E3 ) = , P(E1Ç E3 ) =
At çè , + ÷ø 36 12 12
6 4 12
And P(E1 Ç E2 Ç E 3) = 0 ¹ P (E1) . P(E2) . P(E3)
æp pö æ pö Þ E1, E2, E3 are not independent.
y – ç + ÷ = –2 ç x – ÷
è 4 12 ø è 6ø 45. (2) Rationalizing the given expression
4p 2p (2 + 3isin q)(1 + 2isin q)
y– = –2 x +
12 6 1 + 4sin 2 q
p p For the given expression to be purely imaginary, real
y– = –2 x + part of the above expression should be equal to zero.
3 3
2p 2 - 6 sin 2 q 1
y = –2 x + Þ =0 Þ sin 2 q =
2 3
3 1 + 4 sin q
æ 2p ö 1
This equation is satisfied only by the point ç 0, ÷ Þ sin q = ±
è 3ø
3
1
æ (n + 1)(n + 2)...3n ön æ 8 ö2 æ12 ö2 æ16 ö2 æ 20 ö2 æ 44 ö2
42. (4) y = lim çç ÷
÷ 46. (4) çç ÷÷ + çç ÷÷ + çç ÷÷ + çç ÷÷ ... + çç ÷÷
n ®¥ è 2n
n ø è5 ø è 5 ø è 5 ø è 5 ø è5ø
16 2 2 2
ln y = lim
1 æ 1 öæ 2 ö æ 2n ö
ln ç
ç1 + ÷ç
n ®¥ n è n ø è
÷ç1 + n ÷
÷.... ç
ç1 + ÷
ø è n ø
÷
S=
25 (
2 + 3 + 4 + ... + 112 )
1 é æ 1ö æ 2ö æ 2n öù 16 æ11(11 + 1)(22 + 1) ö
ln y = lim êln çç1 + ÷÷ + ln çç1 + ÷÷ + .... + ln çç1 + ÷
÷ú = ç
ç - 1÷
÷
n ®¥ n ë è
ê nø è nø è n ø úû 25 è 6 ø
1 2n æ r ö 16 16
= ´505 = ´101
= lim å ln çç1 + n ÷÷ø =ò02 ln(1+ x)dx
n ®¥ n r =1 è
25 5
Let 1 + x = t Þ dx = dt 16 16
Þ m = ´101
when x = 0, t = 1 5 5
x = 2, t = 3
Þ m = 101.
3
æ 33 ö æ 27 ö 47. (2) For trivial solution,
3
ln y =ò 1
ln t d t = [t ln t – t]1 = ln ç ÷
ç 2 ÷ = ln çç 2 ÷÷
èe ø èe ø 1 l -1
27 l - 1 -1 = 0
Þ y=
e 2 1 1 -l
( 1)(l- 1) = 0
Þ -ll+
43. (4) S
(–3, 2) Þl= 0, +1, –1
O
48. (2) Line lies in the plane Þ (3, –2, –4) lie in the plane
Þ 3l – 2m + 4 = 9 or 3l – 2m = 5 ..... (1)
5 2 Also, l, m,–1 are dr's of line perpendicular to plane and
5
B A 2, –1, 3 are dr's of line lying in the plane
(2, –3) Þ 2l – m – 3 = 0 or 2l – m = 3 .....(2)
Solving (1) and (2) we get l = 1 and m = –1
Þ l2 + m2 = 2.
JEE MAIN 2016 Solved Paper 2016-15
0
Þ 5a + b = 5
5=0
+m=
h
57. (2) tan 30° = O (–1,–2)
–
x+a
7x – y
7x – y
1 h
Þ = Þ 3h = x + a ...(1)
3 x +a
h h
tan 60° = Þ 3 = A x–y +l=0 B
a a
Let other two sides of rhombus are
Þ h = 3a ...(2) x – y+ l=0
and 7x –y + m = 0
then O is equidistant from AB and DC and from AD and BC
\ -1 + 2 + 1 = -1 + 2 +l Þl= – 3
h
and -7 + 2 - 5 = -7 + 2 +m Þm= 15
30° 60°
A \ Other two sides are x – y – 3 = 0 and 7x – y + 15 = 0
x B a On solving the eqns of sides pairwise, we get
From (1) and (2) æ 1 -8 ö æ -7 -4 ö
ç , ÷÷ , (1, 2), çç , ÷÷ , (-3, -6)
the vertices as ç
è3 3 ø è3 3ø
3a = x + a Þ x = 2a 60. (1) cos x + cos 2x + cos 3x + cos 4x = 0
Þ 2 cos 2x cos x + 2 cos 3x cos x = 0
Here, the speed is uniform
æ 5x xö
Þ 2cos x ç
ç2 cos cos ÷
÷= 0
So, time taken to cover x = 2 (time taken to cover a) è 2 2ø
10 5x x
\ Time taken to cover a = minutes = 5 minutes cos x = 0, cos = 0 , cos = 0
2 2 2
58. (4) p 3p p 3p 7p 9p
x =p, , , , , ,
2 2 5 5 5 5
P(1, –5, 9)
SECTION-3 : CHEMISTRY
x=y=z
61. (1) For a given mass of an ideal gas, the volume and amount
(moles) of the gas are directly proportional if the
temperature and pressure are constant. i.e
Vµn
Hence in the given case.
Q Initial moles and final moles are equal (nT)i = (n T)f
O Pi V Pi V Pf V Pf V
+ = +
RT1 RT1 RT1 RT2
Pi Pf Pf
x -1 y + 5 z - 9 2 = +
eqn of PO : = = =l T1 T1 T2
1 1 1
2 Pi T2
Þ x =l+ 1; y =l- 5;z =l+ 9. Pf =
T1 + T2
Putting these in eqn of plane :-
62. (1) There is extensive intermolecular hydrogen bonding
l+ 1 -l+ 5 +l+ 9 = 5 in the condensed phase instead of intramolecular
Þl= -10 H-bonding.
Þ O is (–9, –15, –1) 63. (2) 4 moles of NaOH and one mole of Br2 is required during
production of one mole of amine during Hoffmann's
Þ distance OP = 10 3 bromamide degradation reaction.
O
||
R–C–NH2 + Br2 + 4NaOH ® R–NH2 + K2CO3 + 2NaBr + 2H2O
JEE MAIN 2016 Solved Paper 2016-17
64. (2) Alkali metals have the lowest ionization energy in each Thus if a graph is plotted between log(x/m) and log P, a
period on the other hand Sc is a d - block element. straight line will be obtained
Transition metals have smaller atomic radii and higher
nuclear charge leading to high ionisation energy.
n
65. (1) The maximum limit of nitrate in drinking water is 50 1/
=
ppm. Excess nitrate in drinking water can cause disease pe
s lo
log x/m
such as methemoglobinemia ('blue baby' syndrome).
66. (2) Given
C(s) + O2(g) ® CO2(g); DH = –393.5 kJ mol–1 …(i)
Intercept = log K
1
CO(g) + O2(g) ® CO2(g); DH = –283.5 kJ mol–1…(ii)
2 log P
\ Heat of formation of CO = eqn(i) – eqn(ii)
= –393.5 – (–283.5) The slope of the line is equal to 1/n and the intercept
= –110 kJ on log x/m axis will correspond to log K.
67. (1) Given, 70. (2) Spent-lye and glycerol are separated by distillation
under reduced pressure.
C + D
A + B Under the reduced pressure the liquid boil at low
temperature and the temperature of decomposition will
No. of moles initially 1 1 1 1
not r each. e.g. glycerol boils at 290°C with
At equilibrium 1–a 1–a 1+a 1+a
decomposition but at reduced pressure it boils at
2 180° C without decomposition.
æ 1+ a ö
\ Kc = ç ÷ = 100 71. (4) Sodium lauryl sulphate (C11H23CH2OSO3– Na+) is an
è 1- a ø
anionic detergent. Glyceryl oleate is a glyceryl ester of
1+ a oleic acid. Sodium stearate (C17H35COO–Na+) is a soap.
\ = 10
1- a Cetyltrimethyl ammonium bromide
On solving é + ù -
a = 0.81 ê CH3 (CH 2 )15 N(CH 3 )3 ú Br is a cationic detergent.
[D]At eq = 1 + a = 1 + 0.81 = 1.81 ë û
68. (4) CO2H 1
72. (3) Hybridization (H) = [no. of valence electrons of central
H 1 OH 2
atom + no. of Monovalent atoms attached to it + (–ve
2
H Cl charge if any) – (+ve charge if any)]
CH3 1
NO2+ = [5 + 0 + 0 - 1] = 2 i.e. sp hybridisation
2
At (1),
1
3 S 1 NO–2 = [5 + 0 + 1 - 0] = 3 i.e. sp2 hybridisation
2
4 1 2 3 1
NO–3 = [5 + 0 + 1 - 0] = 3 i.e. sp2 hybridisation
2
2 4 The lewis structure of NO2 shows a bent molecular
It is 'S'configurated geometry with trigonal planar electron pair geometry
hence the hybridization will be sp2
At. (2),
N
2 1 R
4 1 3 2 O O
73. (1) Among 20 naturally occuring amino acids "Cysteine"
3 4 has '– SH' or thiol functional group.
It is 'R' configurated. Þ General formula of amino acid ® R–CH–COOH
69. (1) According to Freundlich adsorption isotherm |
NH2
x 1
log = log K + log P Þ Value of R = –CH2–SH in Cysteine.
m n
2016-18 JEE MAIN 2016 Solved Paper
74. (1) Froth floatation method is mainly applicable for 80. (3) Phosphorous acid contain P in +3 oxidation state.
sulphide ores. Acid Formula Oxidation state of
(1) Malachite ore : Cu(OH)2 . CuCO3 Phosphorous
(2) Magnetite ore : Fe3O4 Pyrophosphorous acid H4P2O5 +3
(3) Siderite ore : FeCO3 Pyrophosphoric acid H4P2O7 +5
(4) Galena ore : PbS (Sulphide Ore) Orthophosphorous acid H3PO3 +3
75. (2) High density polythene is used in the manufacture of Hypophosphoric acid H4P2O6 +4
housewares like buckets, dustbins, bottles, pipes etc. 81. (4)
Low density polythene is used for insulating electric CH2 – CH – CH3
wires and in the manufacture of flexible pipes, toys, Å Cl More stable intermediate
coats, bottles etc. CH2 = CH – CH3 + Cl –
Å
76. (4) Out of all the four given metallic oxides CrO2 is attracted CH2 – CH – CH3
by magnetic field very strongly. The effect persists Cl
even when the magnetic field is removed. Thus CrO2 is
metallic and ferromagnetic in nature Å
6
77. (4) N – bromosuccinimide results into bromination at allylic CH2 – CH – CH3 OH CH2 – CH – CH3
and benzylic positions Cl Cl OH
82. (1) When tert -alkyl halides are used in Williamson
synthesis elimination occurs rather than substitution
resulting into formation of alkene. Here alkoxide ion
abstract one of the b-hydrogen atom along with acting
NBS/hv
as a nucleophile.
CH3
+– CH3OH
CH3 CH2 CH2 C CH3 + Na OCH3
More stable Cl
2-Chloro-2-methylpentane
Br HO
NBS H2O/K2CO3
H CH3
CH OH
CH3CH2 C C CH3 + CH3OH + NaBr
2-Methyl-pent-2-ene
Time : 3 Hours • Each correct answer has + 4 marks • Each wrong answer has – 1 mark. Max. Marks : 360
Section - 1 E PE
E
KE
PE KE
(c) d
(d) d
1. Distance of the centre of mass of a solid uniform cone from its
vertex is z0. If the radius of its base is R and its height is h 5. A train is moving on a straight track with speed 20 ms–1. It is
then z0 is equal to : blowing its whistle at the frequency of 1000 Hz. The
5h 3h 2 percentage change in the frequency heard by a person
(a) (b) standing near the track as the train passes him is (speed of
8 8R
sound = 320 ms–1) close to :
h2 3h
(c) (d) (a) 18% (b) 24%
4R 4 (c) 6% (d) 12%
2. A red LED emits light at 0.1 watt uniformly around it. The 6. When 5V potential difference is applied across a wire of length
amplitude of the electric field of the light at a distance of 1 m 0.1 m, the drift speed of electrons is 2.5 × 10–4 ms–1. If the
from the diode is : electron density in the wire is 8 × 1028 m–3, the resistivity of
(a) 5.48 V/m (b) 7.75 V/m the material is close to :
(c) 1.73 V/m (d) 2.45 V/m (a) 1.6 × 10–6 Wm (b) 1.6 × 10–5 Wm
–8
(c) 1.6 × 10 Wm (d) 1.6 × 10–7 Wm
3. A pendulum made of a uniform wire of cross sectional area
A has time period T. When an additional mass M is added to 7. Two long current carrying thin wires, both with current I, are
its bob, the time period changes to TM. If the Young's modulus held by insulating threads of length L and are in equilibrium
1 as shown in the figure, with threads making an angle 'q' with
of the material of the wire is Y then is equal to : the vertical. If wires have mass l per unit length then the
Y value of I is :
(g = gravitational acceleration)
(g = gravitational acceleration)
é æ T ö2 ù A é æ T ö2 ù A
(b) ê1 - ç T ÷ ú Mg
M
(a) ê1 - ç ÷ ú
ë è T ø û Mg ëê è M ø úû
éæ T ö2 ù A éæ T ö2 ù Mg
M
(c) êç ÷ - 1ú (d) êç M ÷ - 1ú L
ëè T ø û Mg ëè T ø û A
q
4. For a simple pendulum, a graph is plotted between its kinetic
energy (KE) and potential energy (PE) against its displacement d.
Which one of the following represents these correctly?
(graphs are schematic and not drawn to scale)
I I
E
E KE PE
pgL plgL
(a) 2 tan q (b) tan q
µ0 µ0
(a) d (b) KE
plgL plgL
(c) sin q (d) 2sin q
PE µ0 cos q µ0 cos q
2015-2 JEE MAIN 2015 Solved Paper
8. In the circuit shown, the current in the 1W resistor is :
L1 L2
6V P 2W QMax
2
Q
2
Q2 Q2
K2 (c) (d)
C C
1µF 3µF 1µF 3µF
15V K1
13. From a solid sphere of mass M and radius R a cube of maximum
possible volume is cut. Moment of inertia of cube about an
(a) 6.7 mA (b) 0.67 mA axis passing through its center and perpendicular to one of
(c) 100 mA (d) 67 mA its faces is :
11. An LCR circuit is equivalent to a damped pendulum. In an 4MR 2 4MR 2
(a) (b)
LCR circuit the capacitor is charged to Q0 and then connected 9 3p 3 3p
to the L and R as shown below : MR 2 MR 2
(c) (d)
L 32 2p 16 2p
R
L
14. The period of oscillation of a simple pendulum is T = 2p .
g
C Measured value of L is 20.0 cm known to 1 mm accuracy and
time for 100 oscillations of the pendulum is found to be 90 s
using a wrist watch of 1s resolution. The accuracy in the
If a student plots graphs of the square of maximum charge
determination of g is :
( QMax
2
) on the capacitor with time(t) for two different values (a) 1% (b) 5%
(c) 2% (d) 3%
L1 and L2 (L1 > L2) of L then which of the following represents
15. On a hot summer night, the refractive index of air is smallest
this graph correctly ? (plots are schematic and not drawn to
near the ground and increases with height from the ground.
scale)
When a light beam is directed horizontally, the Huygens'
principle leads us to conclude that as it travels, the light beam :
L1 (a) bends downwards
2 2
QMax QMax Q0 (For both L1 and L2) (b) bends upwards
(a) L2 (b) (c) becomes narrower
t t (d) goes horizontally without any deflection
JEE MAIN 2015 Solved Paper 2015-3
(y2 – y1) m é æ æ 1 öù
(a) q > cos -1 êµsin ç A + sin -1 ç ÷ ú
240 ëê è è µ ø ûú
(a)
é æ æ 1 öù
t(s) (b) q < cos-1 êµsin ç A + sin -1 ç ÷ ú
8 12 êë è è µ ø úû
-1 é æ -1 æ 1 ö ù
(y2 – y1 ) m (c) q > sin êµsin ç A - sin ç ÷ ú
240 êë è è µ ø úû
(b)
-1 é æ -1 æ 1 ö ù
(d) q < sin êµsin ç A - sin ç ÷ ú
t(s) ëê è è µ ø ûú
8 12
2015-4 JEE MAIN 2015 Solved Paper
22. A rectangular loop of sides 10 cm and 5 cm carrying a current 25. Consider an ideal gas confined in an isolated closed chamber.
1 of 12 A is placed in different orientations as shown in the As the gas undergoes an adiabatic expansion, the average
figures below : time of collision between molecules increases as Vq, where V
z æ Cp ö
is the volume of the gas. The value of q is : ç g = ÷
I è Cv ø
B
I I g +1 g -1
(A) y (a) (b)
I 2 2
x
3g + 5 3g - 5
(c) (d)
z 6 6
26. From a solid sphere of mass M and radius R, a spherical
B portion of radius R/2 is removed, as shown in the figure.
(B) I Taking gravitational potential V = 0 at r = ¥, the potential at
I y the centre of the cavity thus formed is :
I
x I (G = gravitational constant)
I B
I
(C) I y
I
x
-2GM -2GM
(a) (b)
z 3R R
B -GM -GM
(c) (d)
2R R
(D) I
I y 27. Given in the figure are two blocks A and B of weight 20 N and
I 100 N, respectively. These are being pressed against a wall
x I
by a force F as shown. If the coefficient of friction between
If there is a uniform magnetic field of 0.3 T in the positive z the blocks is 0.1 and between block B and the wall is 0.15, the
direction, in which orientations the loop would be in (i) stable frictional force applied by the wall on block B is:
equilibrium and (ii) unstable equilibrium ?
(a) (B) and (D), respectively
(b) (B) and (C), respectively
(c) (A) and (B), respectively F
(d) (A) and (C), respectively A B
23. Two coaxial solenoids of different radius carry current I in the
uur
same direction. F1 be the magnetic force on the inner solenoid
uur
due to the outer one and F2 be the magnetic force on the
outer solenoid due to the inner one. Then :
uur uur
(a) F1 is radially inwards and F2 = 0
uur uur
(b) F1 is radially outwards and F2 = 0 (a) 120 N (b) 150 N
uur uur (c) 100 N (d) 80 N
(c) F1 = F2 = 0
uur uur 28. A long cylindrical shell carries positive surface charge s in
(d) F1 is radially inwards and F2 is radially outwards the upper half and negative surface charge - s in the lower
24. A particle of mass m moving in the x direction with speed 2v is half. The electric field lines around the cylinder will look like
hit by another particle of mass 2m moving in the y direction figure given in : (figures are schematic and not drawn to
with speed v. If the collision is perfectly inelastic, the percentage scale)
loss in the energy during the collision is close to :
(a) 56% (b) 62% (a) (b)
(c) 44% (d) 50%
JEE MAIN 2015 Solved Paper 2015-5
has its hydration enthalpy greater than its lattice enthalpy ? the product C is :
(a) BaSO4 (b) SrSO4 (a) C6H5CH2OH (b) C6H5CHO
(c) CaSO4 (d) BeSO4 (c) C6H5COOH (d) C6H5CH3
2015-6 JEE MAIN 2015 Solved Paper
42. Higher order (>3) reactions are rare due to : 50. The color of KMnO4 is due to :
(a) shifting of equilibrium towards reactants due to elastic (a) L ® M charge transfer transition
collisions (b) s - s* transition
(b) loss of active species on collision (c) M ® L charge transfer transition
(c) low probability of simultaneous collision of all the
(d) d – d transition
reacting species
51. The synthesis of alkyl fluorides is best accomplished by :
(d) increase in entropy and activation energy as more
molecules are involved (a) Finkelstein reaction (b) Swarts reaction
43. Which of the following compounds will exhibit geometrical (c) Free radical fluorination (d) Sandmeyer's reaction
isomerism ? 52. 3 g of activated charcoal was added to 50 mL of acetic acid
(a) 2 - Phenyl -1 - butene solution (0.06N) in a flask. After an hour it was filtered and the
(b) 1, 1 - Diphenyl - 1 - propene strength of the filtrate was found to be 0.042 N. The amount
(c) 1 - Phenyl - 2 - butene of acetic acid adsorbed (per gram of charcoal) is :
(d) 3 - Phenyl -1 - butene (a) 42 mg (b) 54 mg
44. Match the catalysts to the correct processes : (c) 18 mg (d) 36 mg
Catalyst Process 53. The vapour pressure of acetone at 20°C is 185 torr. When 1.2
(A) TiCl4 (i) Wacker process g of a non-volatile substance was dissolved in 100 g of acetone
(B) PdCl2 (ii) Ziegler - Natta at 20°C, its vapour pressure was 183 torr. The molar mass (g
polymerization mol–1) of the substance is :
(C) CuCl2 (iii) Contact process (a) 128 (b) 488
(D) V2O5 (iv) Deacon's process (c) 32 (d) 64
(a) (A) - (ii), (B) - (iii), (C) - (iv), (D) - (i) 54. Which among the following is the most reactive ?
(b) (A) - (iii), (B) - (i), (C) - (ii), (D) - (iv) (a) I2 (b) IC1
(c) (A) - (iii), (B) - (ii), (C) - (iv), (D) - (i) (c) Cl2 (d) Br2
(d) (A) - (ii), (B) - (i), (C) - (iv), (D) - (iii) 55. The standard Gibbs energy change at 300 K for the reaction
45. The intermolecular interaction that is dependent on the inverse
cube of distance between the molecules is : B + C is 2494.2 J. At a given time, the composition
2A
(a) London force (b) hydrogen bond
(c) ion - ion interaction (d) ion - dipole interaction 1 1
of the reaction mixture is [A] = , [B] = 2 and [C] = . The
46. The molecular formula of a commercial resin used for 2 2
exchanging ions in water softening is C8H7SO3– Na+ (Mol. reaction proceeds in the : [R = 8.314 J/K/mol, e = 2.718]
wt. 206. What would be the maximum uptake of Ca2 + ions by (a) forward direction because Q < Kc
the resin when expressed in mole per gram resin ? (b) reverse direction because Q < Kc
2 1 (c) forward direction because Q > Kc
(a) (b)
309 412 (d) reverse direction because Q > Kc
56. Assertion: Nitrogen and oxygen are the main components in
1 1
(c) (d) the atmosphere but these do not react to form oxides of
103 206 nitrogen.
47. Two Faraday of electricity is passed through a solution of Reason: The reaction between nitrogen and oxygen requires
CuSO4. The mass of copper deposited at the cathode is high temperature.
(at. mass of Cu = 63.5 amu) (a) The assertion is incorrect, but the reason is correct
(a) 2g (b) 127 g
(b) Both the assertion and reason are incorrect
(c) 0 g (d) 63.5 g
(c) Both assertion and reason are correct, and the reason is
48. The number of geometric isomers that can exist for square
the correct explanation for the assertion
plan ar complex [Pt (Cl) (py) (NH3 ) (NH2 OH)] + is
(py = pyridine) : (d) Both assertion and reason are correct, but the reason is
(a) 4 (b) 6 not the correct explanation for the assertion
(c) 2 (d) 3 57. Which one has the highest boiling point ?
49. In Carius method of estimation of halogens, 250 mg of an (a) Kr (b) Xe
organic compound gave 141 mg of AgBr. The percentage of (c) He (d) Ne
bromine in the compound is : 58. Which polymer is used in the manufacture of paints and
(at. mass Ag =108; Br = 80) lacquers ?
(a) 48 (b) 60 (a) Polypropene (b) Polyvinyl chloride
(c) 24 (d) 36 (c) Bakelite (d) Glyptal
JEE MAIN 2015 Solved Paper 2015-7
59. The following reaction is performed at 298 K. 66. Let A and B be two sets containing four and two elements
respectively. Then the number of subsets of the set A × B,
2NO (g)
2NO(g) + O2(g) each having at least three elements is :
2
The standard free energy of formation of NO(g) is 86.6 kj/mol (a) 275 (b) 510
at 298 K. What is the standard free energy of formation of (c) 219 (d) 256
NO2(g) at 298 K? (Kp = 1.6 × 1012) 67. Locus of the image of the point (2, 3) in the line (2x – 3y + 4) +
k (x – 2y + 3) = 0, k Î R, is a :
ln (1.6 ´ 1012 ) (a) circle of radius 2.
(a) 86600 –
R (298)
(b) circle of radius 3.
(b) 0.5[2 × 86,600 – R(298) ln(1.6 × 1012)] (c) straight line parallel to x-axis
(c) R(298) ln(1.6 × 1012) – 86600 (d) straight line parallel to y-axis
(d) 86600 + R(298) ln(1.6 × l012)
(1 - cos 2x)(3 + cosx)
60. From the following statements regarding H2O2, choose the 68. lim is equal to :
incorrect statement :
x ®0 x tan 4x
1
(a) It has to be stored in plastic or wax lined glass bottles in (a) 2 (b)
dark 2
(b) It has to be kept away from dust (c) 4 (d) 3
(c) It can act only as an oxidizing agent 69. The distance of the point (1, 0, 2) from the point of intersection
(d) It decomposes on exposure to light x - 2 y +1 z - 2
of the line = = and the plane x – y + z = 16,
3 4 12
Section - 3 is
(a) 3 21 (b) 13
® ® ® (c) 2 14 (d) 8
61. Let a , b and c be three non-zero vectors such that no two of
70. The sum of coefficients of integral power of x in the binomial
® ® ® 1® ®®
them are collinear and (a ´ b) ´ c = b c a . If q is the angle
( )
50
® ® 3 expansion 1 - 2 x is :
between vectors b and c , then a value of sin q is :
1 50 1 50
(a)
2
(b)
-2 3 (a)
2
(3 -1 ) (b)
2
(
2 +1 )
3 3
1 50 1 50
2 2 - 2 (c)
2
(
3 +1 ) (d)
2
3 ( )
(c) (d)
3 3 71. The sum of first 9 terms of the series.
62. Let O be the vertex and Q be any point on the parabola,
x2 = 8y. If the point P divides the line segment OQ internally in 13 13 + 23 13 + 23 + 33
+ + + ....
the ratio 1 : 3, then locus of P is : 1 1+ 3 1+ 3 + 5
(a) y2 = 2x (b) x2 = 2y (a) 142 (b) 192
2
(c) x = y (d) y2 = x (c) 71 (d) 96
63. If the angles of elevation of the top of a tower from three 72. The area (in sq. units) of the region described by
collinear points A, B and C, on a line leading to the foot of the {(x, y) : y2 £ 2x and y ³ 4x – 1} is
tower, are 30°, 45° and 60° respectively, then the ratio, AB : 15 9
(a) (b)
BC, is : 64 32
(a) 1: 3 (b) 2 : 3 7 5
(c) (d)
32 64
(c) 3 :1 (d) 3: 2 73. The set of all values of l for which the system of linear
64. The number of points, having both co-ordinates as integers, equations :
that lie in the interior of the triangle with vertices (0, 0), 2x1 – 2x2 + x3 = lx1
(0, 41) and (41, 0) is : 2x1 – 3x2 + 2x3 = lx2
(a) 820 (b) 780 –x1 + 2x2 = lx3
(c) 901 (d) 861 has a non-trivial solution,
65. The equation of the plane containing the line 2x – 5y + z = 3; (a) contains two elements.
x + y + 4z = 5, and parallel to the plane, x + 3y + 6z = 1, is : (b) contains more than two elements
(a) x + 3y + 6z = 7 (b) 2x + 6y + 12z = – 13 (c) is an empty set.
(c) 2x + 6y + 12z = 13 (d) x + 3y + 6z = –7 (d) is a singleton
2015-8 JEE MAIN 2015 Solved Paper
74. A complex number z is said to be unimodular if |z| = 1. Suppose 1
z1 - 2z 2 where or x < . Then a value of y is :
z 1 and z 2 are complex numbers such that 2 - z z is 3
1 2
unimodular and z2 is not unimodular. Then the point z1 lies 3x - x 3 3x + x 3
on a: (a) (b)
1 + 3x 2 1 + 3x 2
(a) circle of radius 2.
(b) circle of radius 2. 3x – x 3 3x + x 3
(c) (d)
(c) straight line parallel to x-axis 1 – 3x 2
1 - 3x 2
(d) straight line parallel to y-axis. 84. If the function.
75. The number of common tangents to the circles x2 + y2 – 4x
ïì k x + 1, 0 £ x £ 3
– 6x – 12 = 0 and x2 + y2 + 6x + 18y + 26 = 0, is : g(x) = í is differentiable, then the value
(a) 3 (b) 4 ïî m x + 2, 3 < x £ 5
(c) 1 (d) 2 of k + m is :
76. The number of integers greater than 6,000 that can be formed,
10
using the digits 3, 5, 6, 7 and 8, without repetition, is : (a) (b) 4
(a) 120 (b) 72 3
16
(c) 216 (d) 192 (c) 2 (d)
5
77. Let y(x) be the solution of the differential equation 85. The mean of the data set comprising of 16 observations is 16.
dy If one of the observation valued 16 is deleted and three new
(x log x) + y = 2x log x, (x ³ 1). Then y (e) is equal to: observations valued 3, 4 and 5 are added to the data, then the
dx
mean of the resultant data, is:
(a) 2 (b) 2e
(a) 15.8 (b) 14.0
(c) e (d) 0
(c) 16.8 (d) 16.0
é1 2 2 ù 86. The integral
ê ú
78. If A = ê 2 1 -2ú is a matrix satisfying the equation
4
log x 2
êë a 2 b úû ò log x 2 + log(36 - 12 x + x 2 ) dx is equal to :
2
T
AA = 9I, where I is 3 × 3 identity matrix, then the ordered pair (a) 1 (b) 6
(a, b) is equal to: (c) 2 (d) 4
(a) (2, 1) (b) (–2, – 1) 87. Let a and b be the roots of equation x2 – 6x – 2 = 0. If an = an
(c) (2, – 1) (d) (–2, 1)
79. If m is the A.M. of two distinct real numbers l and n(l, n > 1) a10 - 2a 8
– bn, for n ³ 1, then the value of 2a 9 is equal to :
and G1, G2 and G3 are three geometric means between l and n,
then G14 + 2G 42 + G 34 equals. (a) 3 (b) – 3
(c) 6 (d) – 6
(a) 4 lmn2 (b) 4 l2m2n2 88. Let f(x) be a polynomial of degree four having extreme values
(c) 4 l2 mn (d) 4 lm2n
80. The negation of ~ s Ú (~ r Ù s) is equivalent to : é f (x) ù
at x = 1 and x = 2. If xlim 1 + 2 ú = 3, then f(2) is equal to :
®0 ê
ë x û
(a) s Ú (r Ú ~ s) (b) sÙ r
(a) 0 (b) 4
(c) s Ù ~ r (d) s Ù (r Ù ~ s) (c) – 8 (d) – 4
89. The area (in sq. units) of the quadrilateral formed by the
dx
81. The integral ò x 2 (x 4 + 1)3/4 equals :
tangents at the end points of the latera recta to the ellipse
x 2 y2
1 + = 1, is :
1 æ x 4 + 1ö 4 9 5
(a) - (x 4 + 1) 4 + c (b) -ç 4 ÷ + c 27
è x ø (a) (b) 27
2
1
æ x 4 + 1ö 4 1 27
(c) ç 4 ÷ + c (d) (x 4 + 1) 4 + c (c) (d) 18
4
è x ø 90. If 12 identical balls are to be placed in 3 identical boxes, then
82. The normal to the curve, x2 + 2xy – 3y2 = 0, at (1, 1) the probability that one of the boxes contains exactly 3 balls
(a) meets the curve again in the third quadrant. is :
(b) meets the curve again in the fourth quadrant. 12 11
(c) does not meet the curve again. æ 1ö æ 1ö
(a) 220 ç ÷ (b) 22 ç ÷
(d) meets the curve again in the second quadrant. è 3ø è 3ø
83. Let
11 10
æ 2x ö 55 æ 2 ö æ 2ö
tan –1 y = tan –1 x + tan –1 ç (c) ç ÷ (d) 55 ç ÷
è 1 - x 2 ÷ø , 3 è 3ø è 3ø
JEE MAIN 2015 Solved Paper 2015-9
ANSWER K EY
PHYS ICS
1 (d ) 6 (b ) 11 (c) 16 (a) 21 (c) 26 (d)
2 (d ) 7 (d ) 12 (d ) 17 (d) 22 (a) 27 (a)
3 (c) 8 (a) 13 (a) 18 (a) 23 (c) 28 (c)
4 (d ) 9 (d ) 14 (d ) 19 (b) 24 (a) 29 (c)
5 (d ) 10 (b ) 15 (b ) 20 (a,b ) 25 (a) 30 (a)
CHEMIS TRY
31 (d ) 36 (c) 41 (b ) 46 (b) 51 (b) 56 (c)
32 (c) 37 (a) 42 (c) 47 (d) 52 (c) 57 (b)
33 (d ) 38 (a) 43 (c) 48 (d) 53 (d) 58 (d)
34 (a) 39 (a) 44 (d ) 49 (c) 54 (b) 59 (b)
35 (c) 40 (b ) 45 (b ) 50 (a) 55 (d) 60 (c)
MATHEMATICS
61 (c) 66 (c) 71 (d ) 76 (d) 81 (b) 86 (a)
62 (b ) 67 (a) 72 (b ) 77 (a) 82 (c) 87 (a)
63 (c) 68 (a) 73 (a) 78 (b) 83 (c) 88 (a)
64 (b ) 69 (b ) 74 (a) 79 (d) 84 (c) 89 (b)
65 (a) 70 (c) 75 (a) 80 (b) 85 (b) 90 (c)
Lsin q Lsin q FB
T sin q
(ll)g 3W Q 3W
plgL 2
Therefore, I = 2 sin q \ x=
u 0 cos q 23
3W q 4W 0.25cm q
6 = 3 I 1 + I1 – I2
4I1 – I2 = 6 ...(1)
25cm
– 9 + 2I2 – (I1 – I2) + 3I2 = 0
– I1 + 6I2 = 9 ...(2) 1.22 l
Resolving power = = 30 mm.
On solving (1) and (2) 2 m sin q
I1 = 0.13A 15 ´ 100
Direction Q to P, since I1 > I2. 10. (b) I(0) = = 0.1A
0.15 ´ 103
Alternatively I (¥) = 0
p
3 –t
V I (t) = [I (0) – I (¥)] e L / R + i( ¥ )
6V 9V 8
1W 1W
–t R
3W 5W 15 W I I(t) = 0.1 e L / R = 0.1 e L
8
0.15 ´1000
d
I(t) = 0.1 e 0.03 = 0.67mA
JEE MAIN 2015 Solved Paper 2015-11
4 3
pR
3 3
+ – = 3
= p. a
æ 2 ö 2
q c çè R ÷ø
q di 3
- iR - L = 0
c dt 2M
M¢ =
dq q dq Ld 2q 3p
i=- Þ + R+ 2 =0 Moment of inertia of the cube about the given axis,
dt c dt dt
M¢ a 2
d 2q R dq q I=
2
+ + =0 6
dt L dt Lc
2
From damped harmonic oscillator, the amplitude is 2M æ 2 ö
´ç R÷
dt 3p è 3 ø 4MR 2
given by A = Ao e - = =
2m 6 9 3p
d2x b dx k l
Double differential equation 2
+ + x=0 14. (d) As, g = 4 p
2
dt m dt m T2
Rt Rt Dg Dl DT
Qmax = Q oe - Þ Q 2max = Qo2e - So, ´ 100 = ´ 100 + 2 ´ 100
2L L g l T
Hence damping will be faster for lesser self inductance. 0.1 1
= ´ 100 + 2 ´ ´ 100 = 2.72 ; 3%
Q1 1mF 20 90
12. (d)
Q Plane WF
15. (b) (Light bends µ increases
C Vel decreases
upwards)
Q2 2mF Refracted
WF
16. (a) Amplitude modulated wave consists of three
2 2 frequencies are wc + wm, w,wc – wm
From figure, Q2 = Q = Q
2 +1 3 i.e. 2005 KHz, 2000KHz, 1995 KHz
17. (d) The entropy change of the body in the two cases is
æC ´ 3ö same as entropy is a state function.
Q = E çè ÷
C + 3ø
1æUö
18. (a) As, P = ç ÷
2 æ 3CE ö 2CE 3èVø
\ Q2 = çè ÷ =
3 C + 3ø C + 3 U
But = KT 4
Therefore graph d correctly dipicts. V
1 4
Charge So, P = KT
3
uRT 1
or = KT 4 [As PV = u RT]
V 3
4
pR 3T 3 = constant
3
C 1
1mF 3mF Therefore, Tµ
R
2015-12 JEE MAIN 2015 Solved Paper
19. (b) y1 = 10t – 5t2 ; y2 = 40t – 5t2 By eq. (iii) and (iv)
for y1 = – 240m, t = 8s 1
\ y2 – y1 = 30t for t < 8s. sin q = µsin A 1 - - cos A
µ2
for t > 8s, on further solving we can show for ray not to transmitted
1 through face AC
y2 – y1 = 240 – 40t – gt2 é
2 –1 æ 1 ö ù
Kq q = sin–1 ê u sin(A – sin çè µ ÷ø ú
20. (a,b) We know, V0 = = Vsurface ë û
R So, for transmission through face AC
Kq
Now, Vi = (3R 2 – r 2 ) [For r < R] é –1 æ 1 ö ù
2R 3
q > sin–1 ê u sin(A – sin çè µ ÷ø ú
At the centre of sphare r = 0. Here ë û
r r
3 22. (a) For stable equilibrium M || B
V = V0 r r
2 For unstable equilibrium M || (–B)
Kq r r
5 Kq 23. (c) F1 = F2 = 0
Now, = (3R 2 – r 2 )
4 R 2R 3 because of action and reaction pair
R Y
R2 =
2 pf = 3 m V
3 Kq Kq m pi
= 3 2v 45°
4 R R 24. (a) X
1 Kq Kq
= v
4 R R4 2m
R4 = 4R Initial momentum of the system
Also, R1 = 0 and R2 < (R4 – R3)
pi = [m(2V)2 ´ m(2V)2 ]
21. (c) When r2 = C, ÐN2Rc = 90°
= 2m ´ 2V
Where C = critical angle
Final momentum of the system = 3mV
1 By the law of conservation of momentum
As sin C = = sin r2
v
2 2mv = 3mV
A
2 2v
Þ = Vcombined
N1 3
Loss in energy
q N2
1 1 1
Q r1 DE = m1V12 + m2V22 - (m1 + m2 )Vcombined
2
r2 R 2 2 2
P
4 5
DE = 3mv2 - mv 2 = mv 2 = 55.55%
3 3
B C
Percentage loss in energy during the collision ; 56%
1
Applying snell's law at ‘R’ 25. (a) t =
æ N ö 3RT
µ sin r2 = 1 sin90° ...(i) 2pd2 ç ÷
èVø M
Applying snell's law at ‘Q’ V
1 × sin q = µ sin r1 ...(ii) tµ
T
But r1 = A – r2 As, TVg–1 = K
So, sin q = µ sin (A – r 2) So, t µ Vg + 1/2
g +1
sin q = µ sin A cos r 2 – cos A ...(iii) [using (i)] Therefore, q =
2
From (1) 26. (d) Due to complete solid sphere, potential at point P
- GM é 2 æ R ö
2 1 2ù
cos r2 = 1 – sin r2 = 1– 2 ...(iv) Vsphere = ê3R - ç ÷ ú
µ 2R 3 êë è2ø úû
JEE MAIN 2015 Solved Paper 2015-13
SECTION-2 : CHEMISTRY
-GM æ 11R 2 ö GM
= = -11 31. (d) When 1, 3-dimethylcyclopentene is heated with ozone
3 ç 4 ÷ 8R
2R è ø and then with zinc and acetic acid, oxidative cleavage
leads to keto - aldehyde.
Solid
sphere CH3 CH3
O
1O - 78° C
¾ ¾ ¾3¾ ¾ ¾ ¾ ® O C–H
2 - Zn- CH 3 COOH
P
CH3 CH3
Cavity
O O
|| ||
Due to cavity part potential at point P CH3 — C— CH 2 — CH 2 — CH— C— H
|
GM CH3
3 8 3GM
Vcavity = - =- 5- keto – 2 – methylhexanal
2 R 8R 32. (c) Water-soluble vitamins dissolve in water and are not
2 stored by the body. The water soluble vitamins include
So potential at the centre of cavity the vitamin B-complex group and vitamin C.
33. (d) In alkaline earth metals, ionic size increases down the
11GM æ 3 GM ö -GM
= Vsphere - Vcavity = - - ç- = group. The lattice energy remains constant because
8R è 8 R ÷ø R sulphate ion is so large, so that small change in cationic
f1 f2 size does not make any difference. On moving down
the group the degree of hydration of metal ions
decreases very much leading to decrease in solubility.
27. (a) F N \ BeSO 4 > MgSO 4 > CaSO 4 > SrSO 4 > BaSO 4
A B
34. (a)
+ –
f1 NH2 N = NCl CN
20N 100N
Assuming both the blocks are stationary
N= F ¾ NaNO
¾ ¾ 2¾/HCl
¾¾ ® ¾ CuCN/KCN
¾ ¾ ¾ ¾®
0 - 5° C D
f1 = 20 N
f2 = 100 + 20 = 120N CH3 CH3 CH3
f 35. (c) In bcc the atoms touch along body diagonal
\ 2r + 2r = 3a
3a 3 ´ 4.29
\ r= = = 1.857Å
4 4
36. (c)
–13.6Z2
37. (a) Total energy = eV
120N n2
Considering the two blocks as one system and due to where n = 2, 3, 4 ....
equilibrium f = 120N Putting n = 2
28. (c) Field lines originate perpendicular from positive charge -13.6
and terminate perpendicular at negative charge. Further ET = = -3.4eV
4
this system can be treated as an electric dipole. 38. (a) Phenelzine is an antidepressant, while others are
ze2
k ze 2 antacids.
29. (c) U = –K ; T.E = – 39. (a) For isoelectronic species, size of anion increases as
r 2 r
negative charge increases. Thus the correct order is
k ze 2 N 3- > O 2 - > F -
K.E = . Here r decreases
2 r (1.71) (1.40) (1.36)
30. (a) Frank-Hertz experiment - Discrete energy levels of atom 40. (b) In the metallurgy of aluminium, purified Al2O3 is mixed
Photoelectric effect - Particle nature of light with Na3AlF6 or CaF2 which lowers the melting point
Davison - Germer experiment - wave nature of electron. of the mix and brings conductivity.
2015-14 JEE MAIN 2015 Solved Paper
41. (b) Mass of AgBr = 141 mg = 0.141 g
1 mole of AgBr = 1 g atom of Br
CH3 COOH COCl CHO 188 g of AgBr = 80 g of Br
\ 188 g of AgBr contain bromine = 80 g
KMnO4 SOCl2 H2/Pd
80
BaSO 4 0.141 g of AgBr contain bromine = ´ 0.141
188
(A) (B) (C) This much amount of bromine present in 0.250 g of
organic compound
Rosenmund’s 80 0.414
reaction \ % of bromine = ´ ´100 = 24%
188 0.250
42. (c) Reactions of higher order (>3) are very rare due to very 50. (a) L ® M charge transfer spectra. KMnO4 is colored
less chances of many molecules to undergo effective because it absorbs light in the visible range of
collisions. electromagnetic radiation. The permanganate ion is the
source of color, as a ligand to metal, (L ® M) charge
H transfer takes place between oxygen's p orbitals and
|
H3C — C = CH — CH 2 the empty d-orbitals on the metal. This charge transfer
43. (c)
| takes place when a photon of light is absorbed, which
Ph leads to the purple color of the compound.
51. (b) Alkyl fluorides are more conveniently prepared by
1- Phenyl-2-butene the two groups around each of the heating suitable chloro – or bromo-alkanes with organic
doubly bonded carbon fluorides such as AsF3, SbF3, CoF2, AgF, Hg2F2 etc.
Because, all are different. This compound can show This reaction is called Swarts reaction.
cis-and trans-isomerism. CH 3Br + AgF ¾¾ ® CH 3 F + AgBr
44. (d) (A) - (ii), (B) - (i), (C) - (iv), (D) - (iii)
2CH3CH 2 Cl + Hg 2 F2 ¾¾ ® 2CH 3CH 2 F + Hg 2Cl 2
45. (b) Hydrogen bond is a type of strong electrostatic dipole-
dipole interaction and dependent on the inverse cube 52. (c) Let the weight of acetic acid initially be w1 in 50 ml of
of distance between the molecular ion-dipole interaction 0.060 N solution.
w ´ 1000
1 Let the N = 1 (Normality = 0.06 N)
µ . M.wt . ´ 50
r3
w1 ´ 1000
46. (b) 2 mole of water softner require 1 mole of Ca 2+ ion 0.06 =
60 ´ 50
1
So, 1 mole of water softner require mole of Ca2+ ion 0.06 ´ 60 ´ 50
2 Þ w1 = = 0.18 g = 180 mg.
1000
1 1 After an hour, the strength of acetic acid = 0.042 N
Thus, = mol / g will be maximum uptake
2 ´ 206 412 so, let the weight of acetic acid be w2
w ´1000
47. (d) Cu 2+ + 2e – ¾¾ ® Cu N= 2
60 ´ 50
2F i.e. 2 × 96500 C deposit Cu = 1 mol = 63.5 g
48. (d) Square planar complexes of type M[ABCD] form three w ´1000
0.042 = 2
isomers. Their position may be obtained by fixing the 3000
position of one ligand and placing at the trans position Þ w2 = 0.126 g = 126 mg
any one of the remaining three ligands one by one. So amount of acetic acid adsorbed per 3g
= 180 – 126 mg = 54 mg
HOH2N Cl HOH2N NH3 54
Amount of acetic acid adsorbed per g = = 18 mg
3
Pt Pt 53. (d) Using relation,
p° - ps w 2 M1
=
py NH3 py Cl ps w1M 2
trans cis where w1, M1 = mass in g and mol. mass of solvent
HOH2N w2, M2 = mass in g and mol. mass of solute
NH3 Let M2 = x
p° = 185 torr
Pt ps = 183 torr
185 - 183 1.2 ´ 58
= (Mol. mass of acetone = 58)
Cl py 183 100 x
cis x = 64
49. (c) Mass of substance = 250 mg = 0.250 g \ Molar mass of substance = 64
JEE MAIN 2015 Solved Paper 2015-15
r r r rr r 1 r rr 64. (b) Total number of integral points inside the square OABC
Þ ( )
– b c cos qa + c.a b = b c a
3
= 40 × 40 = 1600
No. of integral points on AC
2015-16 JEE MAIN 2015 Solved Paper
P
(1, 2)
O A B(a, b)
(0, 0) (41, 0)
Since, P is the fixed point for given family of lines
= No. of integral points on OB
So, PB = PA
= 40 [namely (1, 1), (2, 2) ... (40, 40)]
(a – 1)2 + (b – 2)2 = (2 – 1)2 + (3 – 2)2
\ No. of integral points inside the DOAC
(a – 1)2 + (b –2)2 = 1 + 1 = 2
1600 – 40
= = 780
2 (x – 1)2 + (y – 2)2 = ( 2)2
65. (a) Equation of the plane containing the lines (x – a)2 + (y – b)2 = r2
2x – 5y + z = 3 and x + y + 4z = 5 is Therefore, given locus is a circle with centre (1, 2) and
2x – 5y + z – 3 + l (x + y + 4z – 5) = 0 radius2.
Þ (2 + l) x + (–5 + l) y + (1 + 4l)z + (–3 – 5l) = 0 68. (a) Multiply and divide by x in the given expression, we
get
...(i)
(1 - cos 2 x ) (3 + cos x) x
Since the plane (i) parallel to the given plane x + 3y + 6z lim ·
=1 x®0 x2 1 tan 4 x
2 + l -5 + l 1 + 4l 2sin 2 x 3 + cos x x
\ = = = lim · ·
1 3 6 x®0 x2 1 tan 4 x
11 sin 2 x x
Þ l=- = 2 lim · lim 3 + cos x · lim
2 x®0 x 2 x®0 x ®0 tan 4 x
Hence equation of the required plane is
1 4x 1
= 2.4 lim = 2.4. = 2
æ 11ö æ 11ö æ 44 ö æ 55 ö 4 x®0 tan 4 x 4
çè 2 - ÷ø x + çè -5 - ÷ø y + çè1 - ÷ø z + çè -3 + ÷ø = 0
2 2 2 2 69. (b) General point on given line º P(3r + 2, 4r – 1, 12r + 2)
Þ (4 - 11)x + (-10 - 11)y + (2 - 44)z + ( -6 + 55) = 0 Point P must satisfy equation of plane
(3r + 2) – (4r – 1) + (12r + 2) = 16
Þ -7x - 21y - 42z + 49 = 0
11r + 5 = 16
Þ x + 3y + 6z - 7 = 0 r=1
Þ x + 3y + 6z = 7 P(3 × 1 + 2, 4 × 1 – 1, 12 × 1 + 2) = P(5, 3, 14)
(1 - 2 50 50 D=0
x) + (1 + 2 x )
2 2
B æ 1 ,1ö Þ z1 + 4 z 2 = 4 + z1 2 z 2 2
çè ÷
2 ø
2 2 2 2
Þ z1 + 4 z 2 – 4 – z1 z2 =0
O C æ 1 - 1ö
çè , ÷ø
8 2
(z 1
2
)(
- 4 1 - z2
2
)=0
Q z2 ¹ 1
1 1
y +1 y2 2
= ò 4
dy - ò 2
dy \ z1 =4
-1/ 2 -1/ 2
1 1
Þ z1 = 2
1 é y2 ù 1 é y3 ù
= ê + yú - ê ú Þ Point z1 lies on circle of radius 2.
4 ëê 2 ûú -1/ 2 2 ëê 3 ûú -1/ 2
75. (a) x2 + y2 – 4x – 6y – 12 = 0 ...(i)
1 é 3 3 ù 9 15 9 27 9 Centre, C1 = (2, 3)
= + - = - = =
4 êë 2 8 úû 48 32 48 96 32 Radius, r1 = 5 units
73. (a) 2x1 - 2x 2 + x 3 = lx1ü x2 + y2 + 6x + 18y + 26 = 0 ...(ii)
ï
2x1 - 3x 2 + 2x 3 = lx 2 ý Centre, C2 = (–3, –9)
- x1 + 2x 2 = lx3 ïþ Radius, r2 = 8 units
Þ (2 – l)x1 – 2x2 + x3 = 0
C1C2 = (2 + 3)2 + (3 + 9)2 = 13 units
2x1 – (3 + l) x2 + 2x3 = 0
r1 + r2 = 5 + 8 = 13
– x1 + 2x2 – lx3 = 0
\ C1 C2 = r1 + r2
2015-18 JEE MAIN 2015 Solved Paper
and a = – 2
(a, b) = (–2, –1)
1
l+n æ nö 4
C1 79. (d) m= and common ratio of G.P. = r = çè ÷ø
2 l
C2 \ G1 = l3/4n1/4, G2 = l1/2n1/2, G3 = l1/4 n3/4
3 2 2 3
G14 + 2G 4 + G34 = l n + 2l n + ln
Therefore there are three common tangents. = ln (l + n)2
76. (d) Four digits number can be arranged in 3 × 4! ways. = ln × 2m2
Five digits number can be arranged in 5! ways. = 4lm2n
Number of integers = 3 × 4! + 5! = 192. 80. (b) :[:sÚ(:r Ù s)]
= sÙ:(:r Ù s)
dy æ 1 ö
+ y=2 = sÙ(r Ú:s)
dx çè x log x ÷ø
77. (a) Given,
= (s Ù r) Ú (s Ù: s)
1 = (s Ù r) Ú 0
I.F. = ò x log x dx =sÙr
e
dx dx
= elog(log x) = log x 81. (b) I=ò = ò x 3 (1 + x -4 )3/ 4
2 4 3/ 4
x (x + 1)
y. logx = ò 2 log xdx + c Let x–4 = y
y logx = 2[x log x – x] + c Þ –4x–3 dx = dy
Put x = 1, y.0 = –2 + c -1 3
Þ dx = x dy
c=2 4
Put x = e
-1 x 3dy -1 dy
y loge = 2e(log e – 1) + c \I= ò 3
4 x (1 + y) 3 / 4
= ò
4 (1 + y)3/ 4
y(e) = c = 2
-1
= ´ 4(1 + y)1/ 4 = -(1 + x -4 )1/ 4 + C
é 1 2 2 ù é 1 2 a ù é 9 0 0ù 4
ê 2 1 -2ú ê2 1 2ú = ê0 9 0ú
78. (b) ê úê ú ê ú 1/ 4
êë a 2 b úû êë2 -2 b úû êë0 0 9 úû æ x 4 + 1ö
= –ç 4 ÷ +C
è x ø
é 1+ 4 + 4 2+ 2- 4 a + 4 + 2b ù é9 0 0ù
ê ú 82. (c) Given curve is
2+2-4 4 + 1 + 4 2a + 2 - 2bú = êê0 9 0úú
Þê x2 + 2xy – 3y2 = 0 ...(1)
ê 2 2 ú ê0 0 9ûú
ëa + 4 + 2b 2a + 2 - 2b a + 4 + b û ë dy dy
Differentiate w.r.t. x, 2x + 2x + 2y - 6y =0
dx dx
Þ a + 4 + 2b = 0 Þ a + 2b = – 4 ...(i)
2a + 2 – 2b = 0 Þ 2a – 2b = – 2 æ dy ö
ç ÷ =1
Þ a – b = –1 ...(ii) è dx ø(1, 1)
é 2x ù 6 ± 36 + 8
83. (c) tan–1 y = tan -1 x + tan -1 ê ú 87. (a) a, b = = 3 ± 11
ë1 - x 2 û 2
a = 3 + 11 , b = 3 - 11
= tan -1 x + 2 tan -1 x
= 3 tan–1x
( ) – (3 – 11)
n n
\ an = 3 + 11
é
-1 3x - x
3ù
tan–1 y = tan ê 2ú a10 – 2a 8
ëê 1 - 3x ûú 2a 9
3x - x 3
y =
1 - 3x 2 ( 3 + 11)10 – ( 3 – 11)10 – 2 ( 3 + 11)8 + 2 (3 - 11)8
=
2 ê( 3 + 11) - ( 3 - 11 ) ú
é 9 9ù
84. (c) Since g (x) is differentiable, it will be continuous at x = 3
ë û
\ xlim g(x) = lim+ g(x)
®3- x®3
k= 4m ...(2) ë û
Solving (1) and (2), we get
( ) – 6 (3 - 11) 6
9 9
2 8 6 3 + 11
m= , k= = = =3
2 ê( 3 + 11 ) - ( 3 - 11 ) ú
5 5 é 9 9ù 2
k+m =2 ë û
85. (b) Sum of 16 observations = 16 × 16 = 256
Sum of resultant 18 observations = 256 – 16 + (3 + 4+5) é f (x) ù
88. (a) lim ê1 + 2 ú = 3
x ®0 ë x û
= 252
252 f (x)
Mean of observations = = 14 Þ lim =2
18 x ®0 x2
4 So, f(x) contain terms in x2, x3 and x4.
log x2
86. (a) I=ò dx
2
2 log x + log(36 -12x + x 2 ) Let f(x) = a1x2 + a2x3 + a3x4
4 f (x)
log x2 Since lim = 2 Þ a1 = 2
I= ò dx . `..(i) x ®0 x2
2
2 log x + log(6 - x)2
Hence, f(x) = 2x2 + a2x3 + a3x4
4
log(6 - x)2 f ¢(x) = 4x + 3a2x2 + 4a3x3
I= ò dx ...(ii)
2
2 log(6 - x) + log x 2
As given : f ¢ (1) = 0 adn f ¢(2) = 0
Adding (i) and (ii)
Hence, 4 + 3a2 + 4a3 = 0 ...(1)
4
2I = ò dx = [ x ] = 2
4
and 8 + 12a2 + 32a3 = 0 ...(2)
2
2
By 4x(eq1) – eq(2), we get
I=1
2015-20 JEE MAIN 2015 Solved Paper
16 + 12a2 + 16a3 – (8 + 12a2 + 32a3) = 0 It meets x-axis at A (9/2, 0) and y-axis at B (0, 3)
Þ 8 – 16a3 = 0 Þ a3 = 1/2 1 9 27
\ Area of DOAB = ´ ´3=
2 2 4
and by eqn. (1), 4 + 3a2 + 4/2 = 0 Þ a2 = –2
Y
1 B
Þ f(x) = 2x2 – 2x3 + x4
2 (0, 3)
1 L(2,5/3)
f(2) = 2 × 4 – 2 × 8 + × 16 = 0
2
89. (b) The end point of latus rectum of ellipse C O S A X
(9/2, 0)
x2 y2 æ b2 ö
+ = 1 in first quadrant is ç ae, ÷ and the
a2 b2 è aø
D
2x y
Equation of tangent at L is + =1
9 3
PHYSICS
PART MECHANICS - I
TEST 1
Time : 1 hour
Max. Marks : 120
INSTRUCTIONS
Ø Crack this section in just 1 hour.
Ø This part test contains 30 questions.
Ø For every correct answer you will be awarded +4 marks. There is no negative marking.
Ø Please fill the correct bubbles in the Response Sheet provided at the end of the test.
Ø Do not look at the solutions before attempting the complete test.
. . . Best of Luck
1. If the displacement of a particle is given by x = 2t + 5 , 6. A person climbs up a stalled escalator in 60 sec. When he
stands on it then he is carried in the moving escalator in 30
what is the nature of its motion? sec. How much time will he take in climbing up in a moving
(a) Uniform (b) Accelerated escalator?
(c) Retarded (d) At rest (a) 10 sec (b) 30 sec (c) 20 sec (d) 40 sec
2. A bus is moving with a velocity of 10m/s on a straight road. 7. A projectile is thrown with an initial velocity of (xi + yj) m/
A scooterist wishes to overtake the bus in 100 seconds. If sec. If the range of projectile is doubled the maximum height
reached by it is
the bus is at a distance of 1 km from the scooterist, at what
(a) x = 2y (b) y = 2x (c) x = y (d) y = 4x
velocity should the scooterist chase the bus?
8. A heavy uniform chain lies on horizontal table top. If the
(a) 50 m/sec (b) 40 m/sec coefficient of friction between the chain and the table surface
(c) 30 m/sec (d) 20 m/sec is 0.25 then the maximum fraction of length of chain that can
3. A particle of mass m describes a circle of radius r. The hang over on edge of table is
centripetal acceleration on the particle is 4/r2. What will be (a) 20% (b) 35% (c) 25% (d) 15%
the momentum of the particle ? 9. Drops of water fall from the roof of a building 9m high at
regular intervals of time. The first drop reaching the ground
(a) 4 m/r (b) 2 m/r at the same instant fourth drop starts its fall. What are the
(c) (d) 2m / r distances of second and third drops from roof ?
4m / r
(a) 6m & 3m (b) 4m & 2m
4. If percentage errors in the measurement of mass and velocity
(c) 6m & 2 m (d) 4m & 1m
are 1% and 2% respectively, the ratio of the percentage 10. Wind is blowing east to west along parallel tracks. Two
errors in the measurement of linear momentum and kinetic trains moving with same speed in opposite direction have
energy is steam track of one double than other. The speed of each
(a) 3/5 (b) 5/3 (c) 1/2 (d) 2/1 train is
5. A 55 gm tennis ball strikes the ground and rebounds back. If (a) half that of wind
its velocities are 5 m/s and 4 m/s before and after the collision (b) three times that of wind
respectively then the impulse imparted will be (c) double that of wind
(d) equal to that of wind
(a) –0.495 Ns (b) 0.495 Ns
11. A stone is projected from a horizontal plane. It attains
(c) 495 Ns (d) –495 Ns maximum height ‘H’ & strikes a stationary smooth wall &
falls on the ground vertically below the maximum height.
P-2 Mechanics - I
Assume the collision to be elastic, the height of the point What will be the kinetic energy of the body when the
on the wall where ball will strike is – distance covered by it is 2m ?
20 B
F in Newton
H
h A C
0
1 2 3 4 5
d in meter
(a) H/2 (b) H/4 (a) 40 J (b) 10 J (c) 50 J (d) 30 J
(c) 3H/4 (d) None of these 18. A stone, tied to one end of a string, is revolved in the
12. A body is moved along a straight line by a machine delivering horizontal plane. The tension in the string is T. If the length
constant power. The distance moved by body in time t is of the string is reduced to one third of its initial value, then
proportional to the tension will increase by
(a) t 1/2 (b) t 3/4 (c) t 3/2 (d) t 2 (a) T (b) 26T (c) 37T (d) 8T
13. If E, m, J and G denote energy, mass and angular momentum 19. A pendulum consists of a wooden bob of mass m and length
and gravitational constant the dimensions of EJ2/m5G2 ‘l’. A bullet of mass m1 is fired towards the pendulum with a
should be the same as those of speed v1. The bullet emerges out of the bob with a speed
(a) angle (b) length v1/3, and the bob just completes motion along a vertical
(c) mass (d) time circle. Then v1 is
14. A block of mass M is placed on an inclined plane of inclination æ m ö 3æ m ö
(a) çç ÷÷ 5gl (b) ç ÷ 5gl
45°. See figure. The inclined plane moves horizontally with
è m1 ø 2 çè m1 ÷ø
an acceleration of 10 ms–2 as shown in the figure. If
acceleration due to gravity is 10 ms–2 the block will : 2 æ m1 ö æm ö
1
(c) ç ÷ 5gl (d) çç m ÷÷ gl
3 çè m ÷ø è ø
20. A rod of length L is placed on x – axis between
x = 0 and x = L. The linear density i.e., mass per unit length
denoted by r, of this rod, varies as, r = a + bx. What should be
the dimensions of b?
(a) M2L1T 0 (b) M1 L–2 T 0
–1
(c) M L T 3 1 (d) M–1 L2T 3
(a) move upwards along the plane
21. A river is flowing from west to east at the speed of 5 m/
(b) move downwards along the plane minute. A man who can swim at 10 m/minute in still water is
(c) be thrown away from the plane on the south bank of the river. He wants to go across the
(d) be at rest on the plane river in shortest time. How much time he will take if the
15. The time taken by a body in sliding down a rough inclined width of the river is 200 m?
plane of angle of inclination 45° is n times the time taken by (a) 20 min. (b) 30 min.
the same body in slipping down a similar frictionless plane. (c) 15 min. (d) 60 min.
The coefficient of dynamic friction between the body and 22. A stuntman plans to run across a roof top and horizontally
the plane will be jumps on to another roof 4.9 m below the first one and at a
(a) 1/(1–n2) (b) 1 – (1/n2) distance of 6.2 m away. What is the minimum velocity, he
2
(c) Ö{1– (1/n )} (d) Ö{1/(1–n2)} must have before the jump ?
16. A particle of mass m 1 collides head-on with another (a) 3.1 m/s (b) 4.0 m/s
stationary particle of mass m2 (m2 > m1). The collision is (c) 4.9 m/s (d) 6.2 m/s
perfectly inelastic. The fraction of kinetic energy which is
23. A bullet comes out of a gun with a velocity of 600 m/s. At
converted into heat in this collision is
what height from the target should the gun be aimed in
(a) m2/(m1+m2) (b) m1/(m1+m2)
order to hit the target situated at a distance of 200 m from
(c) m1/(m1–m2) (d) m2/(m1–m2)
the gun ?
17. The retarding force (F) acting on a body and the distance
(d) covered by it are shown in the following figure. The (a) 0.55 m (b) 55 m
mass of the body is 25 kg and its initial velocity is 2m/s. (c) 5.5 m (d) 55´10–4m
Mechanics - I P-3
24. A ball with velocity 9m/s collides with another similar 28. A small ball of mass m is allowed to slip down from the top
stationary ball. After the collision the two balls move in of a hemispherical dome of radius R. At what height h from
directions making an angle of 30° with the initial direction. the lower end the ball will leave contact with the dome ?
The ratio of the speeds of balls after the collision will be m
R
30°
30°
(a) 2R / 3 (b) R / 2 (c) R / 4 (d) R / 6
29. The masses of the blocks A and B are 0.5 kg and 1 kg respectively.
(a) v1/v2 = 1 (b) v1/v2 > 1 These are arranged as shown in the figure and are connected
(c) v1/v2 < 1 (d) v2/v1 = 2 by a massless string. The coefficient of friction between all
25. The potential energy of a body of mass 1.25 kg is given by contact surfaces is 0.4. The force, necessary to move the block
U = 3x – 4y where x, y are the co-ordinates of the body. The B with constant velocity, will be (g = 10m/s2)
acceleration of the body is
(a) 3 units (b) 4 units A
Response Sheet
ANSWER KEY
1 (b) 7 (b) 13 (a) 19 (b) 25 (b)
2 (d) 8 (a) 14 (d) 20 (b) 26 (d)
3 (d) 9 (d) 15 (b) 21 (a) 27 (a)
4 (a) 10 (b) 16 (a) 22 (d) 28 (a)
5 (a) 11 (c) 17 (a) 23 (a) 29 (b)
6 (c) 12 (c) 18 (b) 24 (a) 30 (a)
P-4 Mechanics - I
v2 v2 4 2 2x (v + w ) ´ t
3. (d) Centripetal acc. = ; = , v= Þ = Þ 2 v - 2 w = v + 2 Þ v = 3w
r r r 2 r x (v - w) ´ t
m´ 2 1
Momentum = m × v = 11. (c) H= g(2t) 2 = 2gt 2 .......... (1)
r 2
4. (a) We have p = mv
1
\ D p/p = Dm/m + Dv/v = 1 + 2 = 3% and K.E. = E = ½ h = H - gt 2 .......... (2)
mv2 2
\ D E/E = Dm/m + 2 Dv/v = 1 + 4 = 5%
hence (D p/p) /(D E/E) = 3/5
5. (a) Impulse = Change in momentum t
= Final momentum – Initial momentum
= [ 0.055 ´ (–4)] – [ 0.055 ´ 5] 2t
= –0.055 ´ 9 = –0.495 N/sec
6. (c) Let the person covers a distance x on escalator in time t. H t
h
x
Velocity of escalator v1 = m / sec . , Velocity of
30
x H 3H
person v2 = m / sec . ;
60 By (1) and (2), h = H - =
4 4
x x 12. (c) P=Fv=const. or ma.at=const. or a2t = const.
t= = = 20 sec .
v1 + v 2 x x
+ 1 2 1
30 60 Þ s= at But a µ ; s µ at2
2 t
7. (b) u sinq = y, u cosq = x.
u 2 sin 2 q y 2 t2
Therefore H = = ; Þsµ = t3 / 2 .
2g 2g t
13. (a) We get
u 2 2 sin q cos q 2xy
R= = ;
g g EJ 2
=
[ E][ J ]2 =
ML2 T -2 ´ (ML2 T -1 ) 2
=
2
m5 G 2 [ m ]5 [ G ]2 M5 ´ é M -1L3T -2 ù
2
2xy 2y ë û
As R = 2H Þ = Þ y = 2x
g 2g M0L0T0
8. (a) Force of Friction on chain = weight of hanging chain. which are same as dimension of angle.
µ(l – x)rg = xrg or m (l– x) = x; 14. (d) Newton’s third law of motion
m l = (m + l)x or x = m l /(l + m). According to question, g sin q = a cos q
.25l .25l x g sin 45° = a cos 45° Þ g = a
x= = = .2l Þ = .2 = 20%. Hence, the block remains at rest.
.25 + 1 1.25 l
Mechanics - I P-5
[ ]
2m1v1 2 m1v1 3 m
= mv Þ v = or v1 = 5gl d
3 3 m 2 m1 or ( H - h )1 / 2 + h1 / 2 = 0 .
dh
P-6 Mechanics - I
T T
a a
2m1g 2m2g
PART MECHANICS - II
TEST 2
Time : 1 hour
Max. Marks : 120
INSTRUCTIONS
Ø Crack this section in just 1 hour.
Ø This part test contains 30 questions.
Ø For every correct answer you will be awarded +4 marks. There is no negative marking.
Ø Please fill the correct bubbles in the Response Sheet provided at the end of the test.
Ø Do not look at the solutions before attempting the complete test.
. . . Best of Luck
10. From a vertical water jet placed on the ground, water drops
come out with same velocity at different angles. On landing 2Gr 2Gr
(a) (b)
if they jump with 1/4th of the original velocity and stop, the M+m M+m
maximum distance upto which water can reach is
(a) v2/g (b) v4 /g2 2G ( M + m ) 2G (M + m)
(c) (d)
2
(c) (17/16) v /g (d) None r r
11. A person of mass m is standing on one end of a plank of 18. A metallic wire of density d is lying horizontal on the surface
mass M and length L and floating in water. The person of water. The maximum length of wire so that it may not sink
moves from one end to another and stops. The displacement will be
of the plank is 2Tg 2pT
(a) Lm/(m+M) (b) L(M+m) (a) (b) dg
pd
(c) (M+m)/Lm (d) LM/(m+M)
12. The escape velocity from spherical satellite is Ve. The escape 2T
(c) pdg (d) any length
velocity from another satellite of double radius and half the
mean density will be
19. A wheel is rolling on a plane road. The linear velocity of
(a) Ö2Ve (b) Ve/2
centre of mass is v. Then velocities of the points A & B on
(c) 2Ve (d) 4Ve
circumference of wheel relative to road will be
13. The motion of a planet around sun in an elliptical orbit is
shown in the following figure. Sun is situated on one focus. B
The shaded areas are equal. If the planet takes time t1 and t2
v
in moving from A to B and from C to D respectively then
D
A
C
(a) vA = v, vB = 0 (b) vA = vB = 0
(c) vA = 0, vB = v (d) vA = 0, vB = 2v
S
20. Two men are carrying a uniform wire on their shoulders. The
A B bar is of length L and the younger man gets 1/4th of load
that happens to be at one extreme end. What is the distance
(a) t1 > t2 (b) t1 < t2 of the other man from him ?
(c) t1 = t2 (d) information incomplete (a) L / 3 (b) L / 2
14. A thin circular ring of mass M and radius R is rotating about (c) 2 L / 3 (d) 3L / 4
its axis with a constant angular speed w . Two blocks, each 21. A wooden block of volume 1000 cc is suspended from a
of mass m are attached gently to the opposite ends of a spring balance. It weighs 12 N in air. It is then held suspended
diameter of the ring. The angular speed of the ring, will be in water with half of it inside water. What would be the
(a) [(2M)/(M+2m)] w (b) [(M-2m)/(M+2m)] w reading in spring balance now?
(c) [M/(M+2m)] w (d) [(M+2m)/M] w (a) 10 N (b) 9 N
15. If a wire is stretched by applying forces at one of its ends (c) 8 N (d) 7 N
then the elastic potential energy density in terms of Young’s 22. Three-point masses m each are located at the vertices of an
modulus Y and linear strain a will be equilateral triangle. What is the moment of inertia of this
(a) a Y2/2 (b) Y a /2 system about an altitude of the triangle passing through
(c) a 2Y/2 (d) 2 a 2Y the vertex, if ‘a’ is the size of each side of the triangle ?
16. An incompressible liquid is continuously flowing through
a cylindrical pipe whose radius is 2R at point A. The radius mA
at point B, in the direction of flow, is R. If the velocity of
liquid at point A is V then its velocity at point B will be a a
(a) V (b) 4V
a
(c) 2V (d) V/2 mC
mB D
17. Two masses M & m situated at infinite distance apart are at Axis
rest. Due to mutual interaction they start approaching each
other, the relative velocity of the system when the distance (a) m a2/2 (b) 3m a2
between them is r, will be (c) m a2 (d) 3/4ma2
Mechanics - II P-9
23. Two points of a rod move with velocities 3 v and v (c) the gravitational potential is same at all points of the
perpendicular to the rod and in the same direction, separated circle y2 + z2 = 36
by a distance r. Then the angular velocity of the rod is – (d) both (a) and (c)
(a) 3v/r (b) 4v/r 27. Three particles each of mass ‘m’ are situated at the vertices
of an equilateral triangle of side length ‘a’. If they are kept in
(c) 5v/r (d) 2v/r
circular path due to their own mutual force of attraction, the
24. A square wire frame of size L is dropped in a liquid. On
time period of the motion will be
taking out, a membrane is formed. If the surface tension of
the liquid is T, force acting on the frame will be a
(a) 2 T. L (b) 4 T. L (a) zero (b) 2p
g
(c) 8 T. L (d) 10 T. L
25. A wheel is rolling straight on ground without slipping. If the
axis of the wheel has speed v, the instantenous velocity of 1 g a3
(c) (d) 2p
a point P on the rim, defined by angle q, relative to the 2p a 3Gm
ground will be 28. A tank is filled with water to a depth of h metre. A hole of
cross section a at the bottom allows the water to drain out
P (as in fig.), then at what distance below the hole, the cross
q sectional area of the stream is half the area of the hole when
H will be
æ1 ö æ1 ö
(a) v cosç q ÷ (b) 2v cosç q ÷
è2 ø è2 ø
B
(c) v(1 + sin q) (d) v(1 + cos q) a
Y (a) 3 h (b) 6 h
(c) 9 h (d) 1 h
29. A solid cylinder of mass M and radius R rolls down an
A B
inclined plane without slipping. The speed of its centre of
X mass when it reaches the bottom is
O
(a) (2gh) (b) [(4 3)gh]
Z
(c) [(3 4)gh] (d) [(3 4)gh]
Two spheres of equal radii 1 unit, with their centres at A (–2, 30. Suppose the gravitational force varies inversely as the nth
0, 0) and B (2, 0, 0) respectively, are taken out of the solid, power of the distance. Then the time period of a planet in
leaving behind spherical cavities as shown in figure, then circular orbit of radius R around the sun will be proportional
(a) the gravitational force due to this object at the origin is to
zero (a) Rn (b) R-n
(b) the gravitational force at point B(2, 0, 0) is zero (n+1)/2
(c) R (d) R (n-1)/2
Response Sheet
ANSWER KEY
1 (c) 7 (b) 13 (c) 19 (d) 25 (b)
2 (d) 8 (b) 14 (c) 20 (c) 26 (d)
3 (a) 9 (c) 15 (c) 21 (d) 27 (d)
4 (d) 10 (c) 16 (b) 22 (a) 28 (a)
5 (c) 11 (a) 17 (d) 23 (a) 29 (b)
6 (c) 12 (a) 18 (d) 24 (c) 30 (c)
F Ve 2 R 2 d2 2R1 1
A m
m Þ = = ´ = 2
D B Ve1 R1 d1 R1 2
13. (c) According to Kepler’s second law the line joining the
\ mv2/r = Ö (F2 + F2 + 2F2cos60°) = Ö3 F planet to the sun covers equal area in equal time. As
But r = (Ö3 L/2) ´ (2/3) = 1 / Ö3 shaded areas are equal hence time to cover these areas
\ mv2Ö3 / L = Ö3GM2 / L2 Þ v =Ö(GM/L) must also be same. i.e. t1=t2
Mechanics - II P-11
GMm 1 mM 2 2G(M + m )
W= = v or v =
r 2 m+M r
q
18. (d) Mg = 2TL Þ pr2 Ldg = 2TL Þ pr2 dg = 2T.. = 2v2 (1 + cos q) = 2v cos
This relation is independent of L. 2
19. (d) Point A is at rest w.r.t. motion hence v at A = 0. At point 26. (d) Sin ce, the two spheres are of same size and
B there are two horizontal velocities. Hence vB = 2V. symmetrically placed w.r.t. O, their gravitational force
on O will be equal and opposite therefore the net force
W at O is zero.
20. (c) For the balance of forces, y + = W , therefore y
4 y2 + z2 = 36 is a circle of radius 6 cm. This circle lies
3W outside the sphere. Thus potential at each point of it is
= . same. Hence, the correct option is (d).
4
m
Further for a balance of torque about CG we get,
W L 3W L
´ = ´ x therefore x = . a
4 2 4 6 a a
27. (d) 3
y O
W/4
x m a m
C.G. ìï 2 2 3 a üï
íQ ´ h = ´ = ý
ïî 3 3 2 3 ïþ
L/2 W The net force acting on m towards O is
Thus distance from the left end of the rod Gm 2
2F cos 30º where F =
=
L L 4 L 2L
+ = = a2
2 6 6 3
21. (d) Apparent wt. 2 æ a ö÷
Also, 2 F cos 30º = mw çç ÷
= Real wt.–Upthrust è 3ø
1
= 12 – (1000 ´ 10- 6 ) ´103 ´10 = 12 - 5 = 7 N 2Gm 2 3 a
2 Þ ´ = mw2
2 2
a 3
22. (a) We have
I = I1(m at A) + I2 (m at B) + I3(m at C)
= m ´ 02 + m ´ (a/2)2 + m (a/2)2 3Gm a3
or w = Þ T = 2p
= ma2/2 a3 3Gm
P-12 Mechanics - II
1 1 R n -1
mv 2 + Iw 2 = mgh
2 2
Therefore T µ R (n +1) / 2
PART HEAT
TEST 3
Time : 1 hour
Max. Marks : 120
INSTRUCTIONS
Ø Crack this section in just 1 hour.
Ø This part test contains 30 questions.
Ø For every correct answer you will be awarded +4 marks. There is no negative marking.
Ø Please fill the correct bubbles in the Response Sheet provided at the end of the test.
Ø Do not look at the solutions before attempting the complete test.
. . . Best of Luck
1. Kinetic energy of molecules of 1 gm He gas at 100oC will be 6. How many degree of freedom the gas molecules have if
(a) 6 × 10–21 joule (b) 1200 joule under STP the gas density r=1.3 kg/m3 and the velocity of
(c) 6 × 103 joule (d) 4 × 10 –23 joule sound propagation in it is 330 m/sec
2. For hydrogen gas Cp – Cv = a and for oxygen gas Cp – Cv=b. (a) 3 (b) 5
So, the relation between a and b is given by (c) 7 (d) 8
(a) a = 16 b (b) 16a = b 7. At a specific temperature, the energy densities for a black
body for three different wavelengths are 10, 19 and 7 units.
(c) a = 4b (d) a = b
For these wavelengths if the absorption coefficient of a
3. Heat required to melt 1 gm of ice is 80cal. A man melts 60 body is respectively 0.8, 0.3 and 0.9 then the emissive powers
gms of ice by chewing in 1 min. His power is of this body for these wavelengths are in the ratio
(a) 4800 Watt (b) 336 Watt (a) 10:19:7 (b) 8:3:9
(c) 1.33 Watt (d) 0.75 Watt (c) 8:5.7:6.3 (d) 8:9.5:4.3
4. 1/2 mole of helium gas is contained in a container at S.T.P. 8. 10g of ice at –20oC is dropped into a calorimeter containing
The heat energy needed to double the pressure of the gas, 10g of water at 10oC. The specific heat of water is twice that
keeping the volume constant (heat capacity of the gas of ice. When equilibrium is reached, the calorimeter will
= 3 Jg–1 K–1) is contain
(a) 3276 J (b) 1638 J (a) 10 g ice and 10g water (b) 20 g water
(c) 819 J (d) 409.5 J (c) 5 g ice and 15g water (d) 20 g ice
5. The graph shown in the figure is a plot of the temperature of 9. For a certain gas, the ratio of specific heats is given to be
a body in oC and oF. The value of sin q = g = 1.5. For this gas
(a) Cv=3R/J (b) Cp=3R/J
(c) Cp=5R/J (d) Cv=5R/J
ºC 10. A glass tube of length 133 cm and of uniform cross-section
must be filled with mercury so that the volume of the tube
unoccupied by mercury remains the same at all temperatures.
q If cubical coefficients for glass and mercury are respectively
ºF 0.000026 per oC and 0.000182 per oC, then the length of
mercury column is
(a) 5/9 (b) 5/Ö86
(a) 19 cm (b) 133 cm
(c) 5/Ö106 (d) 9/Ö86
(c) 7 cm (d) None
P-14 Heat
11. A nonconducting partition divides a container into two equal (a) 2 (b) 2Ö2
compartments. One is filled with helium gas at 200 K and (c) 4 (d) 8
the other is filled with oxygen gas at 400 K. The number of 18. A system changes from the state (P 1 , V 1 ) to
molecules in each gas is the same. If the partition is removed (P2, V2) as shown in the figure below. What is the work
to allow the gases to mix, the final temperature will be done by the system
(a) 350 K (b) 325 K
(c) 300 K (d) 275 K
(P,22V)
12. The length of the rail is 10 m. The coefficient of linear
5 ´ 105 B
expansion of steel is 1.1 x 10-5 per oC. If the temperature
Pressure in N/ m2
5
variations are 50oC, how much space must be left between 4 ´ 10
the two rails? 3 ´ 105
(a) 5.5 mm (b) 2.75 mm
2 ´ 105
(c) 5.5 cm (d) 2.75 cm A
13. A sphere, a cube and a thin circular plate all made of the 1´ 105 (P,1V) 1
same material and having the same mass, are initially heated
to a temperature of 200oC. Which of these objects will cool
slowest when left in air at room temperature? D
1 2 3 4 5
(a) the sphere (b) the cube
Volume in metre 3
(c) the circular plate (d) all will cool at same rate
14. A rigid container of negligible heat capacity contains one (a) 7.5 × 105 joule (b) 7.5 × 105 ergs
mole of an ideal gas. The temperature of the gas increases 5
(c) 12 × 10 joule (d) 6 × 105 joule
by 1oC if 3.0 cal of heat is added to it. The gas may be 19. The molar heat capacity of a certain substance varies with
(a) helium (b) nitrogen temperature according to the given equation, C = 27.2 +
(c) oxygen (d) carbon dioxide (4 × 10–3) T. The heat necessary to change the temperature
of 2 mol. of the substance from 300K to 700K is
15. P–V plots for two gases during adiabatic processes are
(a) 3.46 × 104J (b) 2.33 × 103 J
shown in the figure. Plots 1 and 2 should correspond
(c) 3.46 × 103 J (d) 2.33 × 104 J
respectively to
20. An electric heater, assumed to be a black body has a
temperature of 727oC. If its temperature is raised to 1727oC,
the amount of energy radiated per unit time now as compared
with that in the first case will be
(a) twice (b) 4 times
P
1 (c) 16 times (d) 100 times
21. An ideal refrigerator has a freezer at a temperature of –13oC.
2 The coefficient of performance of the engine is 5. The
temperature of the air (to which heat is rejected) is.
V
(a) 320oC (b) 39oC
(a) He and Ar (b) He and O2 (c) 325 K (d) 325oC
22. Certain perfect gas is found to obey PV3/2 = constant during
(c) O2 and N2 (d) O2 and He
adiabatic process. If such a gas at initial temperature T is
16. Inside a cylinder closed at both ends is a movable piston. adiabatically compressed to half the initial volume, its final
On one side of the piston is a mass m of a gas, and on the temperature will be
other side a mass 2m of the same gas. What fraction of (a) 2.T (b) 2T
volume of the cylinder will be occupied by the larger mass
of the gas when the piston is in equilibrium? The temperature (c) 2 2 . T (d) 4T
is the same throughout. 23. A bucket full of hot water is kept in a room and it cools from
75oC to 70oC in T1 minutes, from 70oC to 65oC in T2 minutes
(a) 1/4 (b) 1/2
and from 65oC to 60oC in T3 minutes. Then
(c) 2/3 (d) 1/3 (a) T1= T2 = T3 (b) T1 < T2 < T3
17. A mixture of two gases X and Y is enclosed, at constant (c) T1 > T2 > T3 (d) T1< T3< T2
temperature. The relative molecular mass of X, which is 24. The figure shows the cooling curve of pure wax material after
diatomic, is 8 times that of Y which is monoatomic. What is heating. It cools from A to B and solidifies along BD. If L and
the ratio of the r.m.s speed of molecules of Y to that of C are respective values of the specific latent heat of fusion
molecules of X? and specific heat capacity of the liquid wax, the ratio L/C is
Heat P-15
(a) C = 0 (b) C = Cv
(c) C > Cv (d) C < Cv
28. The accompanying graph shows the variation of
temperature (T) of one kilogram material with heat (Q)
supplied to it. At O, the substance is in solid state.Which of
the following interpretations from the graph is correct?
(Q3, T2)
C g
D
Temperature T
(a) 40 (b) 80 (Q1, T1) (Q4, T2)
(c) 100 (d) 20 A b
B
25. Two metallic rods are connected in series. Both are of same (Q2, T1)
material of same length and same area of cross-section. If a
the conductivity of each rod be K, then what will be the Heat added Q
conductivity of the combination?
(a) 4 K (b) 2 K (a) T2 is the melting point of the solid
(c) K (d) K/2 (b) B C represents the change of state from solid to liquid
26. Two spheres of the same material have radii 1m and 4 m and (c) (Q1-Q2) represents the latent heat of fusion of the
temperatures 4000 K and 2000 K respectively. The energy substance
radiated per second by the first sphere is
(d) (Q3-Q1) represents the latent heat of vaporisation of
(a) greater than that by the second the liquid
(b) less than that by the second 29. A cylinder of radius R made of a material of thermal
(c) equal in both cases conductivity K1 is surrounded by a cylindrical shell of inner
(d) the information is incomplete to draw any conclusion radius R & outer radius 2R made of a material of thermal
27. The figure shows a process on a gas in which pressure and conductivity K2. The two ends of the combined system are
volume both change. The molar heat capacity for this process maintained at two different temperatures. There is no loss
is C. Then of heat across the cylindrical surface and the system is in
steady state. The effective thermal conductivity of the
system is
(a) K1+ K2 (b) K1K2/ (K1+ K2)
(c) (K1+ 3K2)/4 (d) (3K1+K2)/4
30. A Carnot’s engine works as a refrigerator between 250 K
and 300 K. If it receives 750 calories of heat from the reservoir
at the lower temperature, the amount of heat rejected at the
higher temperature is
(a) 900 cal. (b) 625 cal.
V (c) 750 cal. (d) 1000 cal.
Response Sheet
ANSWER KEY
1 (b) 7 (c) 13 (a) 19 (d) 25 (c)
2 (d) 8 (d) 14 (a) 20 (c) 26 (c)
3 (b) 9 (b) 15 (d) 21 (b) 27 (c)
4 (b) 10 (a) 16 (c) 22 (a) 28 (c)
5 (c) 11 (c) 17 (b) 23 (b) 29 (c)
6 (b) 12 (a) 18 (c) 24 (d) 30 (a)
3 3 1 gP g ´ 105
1. (b) K.E = nRT = ´ ´ 8.314 ´ 273 =» 1200J 6. (b) V= , 330 × 330 = = 1.4
2 2 4 d 1.3
g = 1.4 for diatomic gas which has 5 degrees of freedom.
2. (d) Both are diatomic gases and CP – CV = R for all gases. 7. (c) Emissive power = aÎ
Q a1 Î1 : a2 Î2 : a3Î3
3. (b) P=
t
8 3 9
From this we get, Þ 10 ´ : ´ 19 : 7 ´
10 10 10
Q ML 60 ´ 80 ´ 4.2
P= = = = 336 W Þ 8 : 5.7 : 6.3
t t 60 8. (d) Heat lost by water to cool from 10°C to 0°C
4. (b) T1 = 273 K, T2 = 2×273 ( on doubling P at const V, T = mSDT = 10 × 1 × 10 = 100 cal
doubles) Heat gained by ice to rise its temperature from –20°C to
DT = 273 K, Q = nCvdt, ms = 3J/g/K, nCv = 3 0°C
1 1
Þ ´ CV = 3 = m × S × Dt = 10 ´ ´ 20 = 100 cal.
4 2
1 Since heat lost = heat gained, therefore the equilibrium
\ CV = 12 J/mol/K, n' Cv = ´12 = 6 temperataure is 0°C. Also at 0°C there will be ice only.
2
Hence the calorimeter will contain (10 + 10)g i.e 20g ice.
Q = n'Cvdt = 6×273 = 1638J
CP CP R
5F 32 ´ 5 9. (b) = g = 1.5; CV = ; CP – C V =
5. (c) We know that, C = F – 32 Þ C = - CV 1.5 J
100 180 9 9
Compare this equation with standard equation for p C RP .5C R
straight line y = mx ± c Þ CP– 1.5 = J Þ 1.5 = J
5 3R
tan q = m = Þ CP=
9 J
bm bg
10. (a) According to question, lm ´ D t = lg ´ ´ Dt
3 3
Þ l × .000182 = .000026× 133
133´ 26
Þ l= 182
= 19 cm
11. (c) Here the number of molecules is same. Hence
5 T1 + T2 200 + 400
\ sin q = Tfinal = = = 300 K
106 2 2
Heat P-17
28. (c) Melting occurs at temp T1 between A and B. Hence, Let K be the equivalent thermal conductivity of the
Q1 – Q2 represents the latent heat of fusion. system.
29. (c) Heat flowing per sec. through cylinder of radius R,
æ T1 – T2 ö
æ T1 – T2 ö Then Q = Kp (2R)2 ç ÷ ....(2)
è l ø
Q1 = K1 (pR2) çç l ÷÷
è ø From eqs. (1) and (2), we have (K1 + 3K2) = 4K
Heat flowing per sec through outer shell of radius 2R,
K1 + 3K 2
or K =
æ T1 – T2 ö 4
Q2 = K2 (p (2R)2 – pR2) çç ÷÷
è l ø T2 300 - 250 50 1
30. (a) We know that h = 1 - = = =
T1 300 300 6
æ T1 – T2 ö
Total Q = Q1 + Q2 = (K1 + 3K2) pR2 çç l ÷÷ ....(1)
è ø Q2 1 Q - 750
or h = 1 - Þ = 1
Q1 6 Q1
2R
R Q1 = 6Q1 –4500 Þ –5Q1 = -4500 Þ Q1 = 900 Cal
Cylinder
T1 l T2
PART ELECTRICITY
TEST 4
Time : 1 hour
Max. Marks : 120
INSTRUCTIONS
Ø Crack this section in just 1 hour.
Ø This part test contains 30 questions.
Ø For every correct answer you will be awarded +4 marks. There is no negative marking.
Ø Please fill the correct bubbles in the Response Sheet provided at the end of the test.
Ø Do not look at the solutions before attempting the complete test.
. . . Best of Luck
1. The equivalent capacity of the network, (with all capacitors 4. A charge Q is distributed over two concentric hollow spheres
having the same C) of radii r and R (>r) such that the surface densities are equal
and placed on the same axial points. Then the potential at
A
the common centre is
B Q(R 2 + r 2 ) Q
(a) (b)
4pe 0 (R + r ) R+r
(a) ¥ (b) zero
(c) C[(Ö3-1)/2] (d) C[(Ö3+1)/2] Q(R + r)
(c) Zero (d)
2. The temperature of cold junction of a thermocouple is 4pe 0 (R 2 + r 2 )
–20oC and the temperature of inversion is 560oC. The neutral
5. Three charges of (+ 2q), (– q) and (– q) are placed at the
temperature is
corners A, B and C of the equilateral triangle of side a as
(a) 270oC (b) 560oC
shown in the given figure. Then the dipole moment of this
(c) 1120 C o (d) 280oC combination is
3. Figure shows a closed surface which intersects a conducting
sphere. If a positive charged is placed at the point P, the flux
of the electric field through the closed surface A + 2q
a a
P D
–q B a C –q
conducting
closed sphere
surface (a) qa
(b) zero
(a) will remain zero
(c) qa 3
(b) will become positive
(c) will become negative 2
(d) qa
(d) will become undefined 3
P-20 Electricity
6. The current I and voltage V graphs for a given metallic wire (a) 2 ampere (b) 4 ampere
at two different temperature T1 and T2 are shown in the (c) 1 ampere (d) 6 ampere
figure. It is concluded that 11. If a battery is connected across series combination of a
capacitor and a resistor, at t = 0. If at an instant t potential
T1
difference across the capacitor be ‘V’ and energy stored in
it be U. Then which of the following graph is correct.
T2
I U U
(a) (b)
V
O V O V
(a) T1 > T2 (b) T1 < T2
(c) T1 = T2 (d) T1 = 2T2
U U
7. A dipole consisting of two charges +q and –q joined by a
massless rod of length l, is seen oscillating with a small
amplitude in a uniform electric field of magnitude E. The
period of oscillation (c) (d)
(a) is proportional to E O O
V V
(b) is proportional to 1/E
12. The adjoining figure shows two identical parallel plate
ml capacitors connected to a battery with switch S close.
(c) is p
3qE The switch is now opened and the plates are filled with a
dielectric of dielectric constant 3. The ratio of the total
1 ml electrostatic energy stored in both the capacitors before
(d) is proportional to but ¹ p
E 3qE and after the introduction of the dielectric is
S
8. If a tiny test charge, is placed in a cavity, a region in between
the molecules of a dielectric, will experience a force equal to B
A
its charge times C
V C
(a) Electric Field E
(b) Electric displacement D
(a) 2 : 5 (b) 3 : 5
(c) E + (4K/3) P
(c) 5 : 2 (d) 5 : 3
(d) A combination of E and P depending on the shape of 13. The field between two plates of a parallel plate capacitor of
the cavity separation d is 1NC-1. A charge +Q, of mass m, is placed at
9. A positive point charge q is carried from a point B to a point distance l from +ve plate, with a velocity v towards it. The
A in the electric field of a point charge + Q at O. If the charge.
permitivity of free space is e0, the work done in the process (a) will migrate to the –ve plate always
is given by (b) will migrate to the +ve plate
qQ æ1 1 ö qQ æ1 1 ö (c) will oscillate about a mean position
(a) ç + ÷ (b) ç - ÷
4 p eo è a b ø 4 p eo è a b ø (d) will migrate to the +ve plate if v exceeds critical values.
14. An ideal ammeter (zero resistance) and an ideal voltmeter
qQ æ 1 1 ö qQ æ 1 1 ö (infinite resistance) are connected as shown. The ammeter
(c) çç - ÷÷ (d) çç + ÷÷
4 p eo èa 2
b2 ø 4 p eo èa 2
b2 ø and voltmeter reading for R1 = 5 W, R2 = 15 W, R3 = 1.25 W
and E = 20 V are given as
10. In the circuit shown, the total current supplied by the battery is
R1=5W
R2 =5W
R3= 1.25W
E = 20 V
(a) 6.25 A, 3.75 V (b) 3.00 A, 5 V
(c) 3.75 A, 3.75 V (d) 6.25 A, 6.25 V
Electricity P-21
15. Parallel plate capacitors C1 & C2 have the same area with C1 19. In the circuit shown the effective resistance between B and
having half the plate separation as C2. C1 is inserted wholly C is
inside C2 as shown with their plates joined as shown. The
capacitance across terminals AB will be
C1
A B
C2
C1C 2
(a) C1 +C2 (b)
C1 + C 2
(c) C1 (d) C2
16. We have a parallel plate capacitor of capacitance C, potential
difference V and plate separation d. Someone pulls the plate
apart such that the plate separation becomes 2d. The amount (a) 3 W (b) 4 W
of work done is (given m = mass of either plate) (c) 4/3 W (d) 3/4 W
20. The electric potential V is given as a function of distance x
1
(a) mgd (b) CV2 by V = (5x2 + 10x – 9) volt. Value of electric field at x = 1m is
4
(a) –20 V/m (b) 6 V/m
æ 3ö 2 (c) 11 V/m (d) +20 V/m
(c) ç 4 ÷ CV (d) (1/9)CV2
è ø 21. A cell of e.m.f. E is connected across a reistance R. The
potential difference between the terminals of the cell is found
17. In the figure, the battery has potential difference of 20 V.
to be V. The internal resistance of the cell must be
Find the equivalent capacitance of all the capacitors.
(a) [2(E–V)V]/R (b) [2(E–V)/V]R
(c) [(E–V)/V]R (d) (E–V)R
22. In the circuit diagram shown below, the magnitude and
direction of flow of current respectively would be
a b
e
Q1 Q2
d2
l2
R
2
q d1
A l1
B
R
d
R d
E1 = 18V 15V=E2
5W 4W (a) 1 mF (b) 2 mF
r1 = 1W 2W = r2
(c) 3 mF (d) 4 mF
2m F 30. A galvanometer with 50 divisions on the scale has a
resistance of 25 W. A current of 2 x 10-4 A gives a deflection
(a) 30 mC (b) 20 mC of one scale division. The additional series resistance
(c) 25 mC (d) 48 mC required to convert it into a voltmeter reading up to 25 V is
27. A parallel plate capacitor of separation 'd' is deformed as (a) 1200 W (b) 1000 W
shown, with l1, l2 >> d2 or d1 or d for the capacitance to (c) 2475 W (d) 2500 W
remain unchanged.
Response Sheet
ANSWER KEY
1 (c) 7 (c) 13 (d) 19 (c) 25 (b)
2 (a) 8 (d) 14 (b) 20 (a) 26 (a)
3 (b) 9 (b) 15 (c) 21 (c) 27 (c)
4 (d) 10 (b) 16 (b) 22 (d) 28 (a)
5 (c) 11 (a) 17 (d) 23 (a) 29 (b)
6 (b) 12 (b) 18 (c) 24 (b) 30 (c)
1 1 2 3C + 2C a q1 Q
Q R+r
1. (c) = + = Þ = ( ;V= )
C ¥ C ¥ +C C Ca ( C a + C) r 2
R +r 2 2
4pe 0 R 2 + r 2
æ -1 + 3 ö 5. (c) This figure can be shown as
2C ¥ 2 +2CC ¥ -C 2 = 0 ® C ¥= Cç
è 2 ÷ø
2. (a) qi – qN = q N – qC
Using this, we get 560 –qN = qN – ( – 20)
Þ 560 = 2qN
\ qN = 2700 C
3. (b) The positive & negative charges developed on
different parts of conducting sphere due to induction.
Hence the flux through the closed surface is positive. Hence
Positive 3
charge q × r1 = 2q × a
2
Negative
+ – charge = 3qa
+ –
+ –
–
–
6. (b) R with increasing temp, V = IR
+
conducting
I 1 1
Slope of graph = = ; Slope of T1 is more i.e R is
closed sphere V R 1
surface more, hence R1 is less. This concludes that T1 will be
4. (d) Let the charges on spheres of radii r and R be q1 and q2 less than T2 as R1 is less than R2
respectively. Then Q = q1+q2. The potential at the ml 2
centre will be : 7. (c) I= ;
12
1 q1 q 2
V= ( + ) d 2q
4pe 0 r R Restoring troque = qElsin q » qEl q =
dt 2
As surface densities are equal, hence
qEl qEl ´12 2p
q1 q2 æ R2 ö \w = = =
s= = or q 2 = q ç
1ç 2 ÷
÷ I ml 2 T
4pr 2 4p R 2 è r ø
ml
1 q1 R 1 q \T = p
V= ( + q1 ) = ´ 1 (r + R ) 3qE
4pe o r r 2 4pe o r 2
4pP
8. (d) It is E + 4 p P for a long cavity || to E ; it is E +for
R 2 3
Now q1 + q2 = Q or q1 + q1 =Q a spherical cavity and different for other shapes.
r2
P-24 Electricity
A 2W 6W 1.5W 4A 4 – I1
6V 3W 6V 3W 6V 3W
1.5W 1.5W 3W
E
6´2 We know that I = = 20 = 4 A
Hence, R AB = = 1.5W R eq 5
6+2
Potential difference across R1 and R2 are same
60 3´3 9 (Parallel combination) I1R1 = (4 – I1)R2
V = IR Þ =I Þ I = 4A [Q R eq = =
15 3+ 3 6
Þ 5I1 = (4 – I1) ´ 15 Þ I1 = 12–3I1 Þ I1 = 3A
= 1.5W]
Thus reading of ammeter = 3A
1 Voltage across 1.25W = I ´ R = 4 ´ 1.25 = 5V (Reading
11. (a) U = CV2 of voltmeter)
2
From this, we get U µ V2 15. (c) Since the plates are joined.
C D
U
U C1
A B
2
V V
12. (b) When the switch is closed, each capacitor has energy
Potential of A is equal to potential of C and potential of
1
CV 2 . B is equal to potential of D.
2
Hence potential difference between AB is equal to
1 1 potential difference between CD.
So total energy = CV 2 + CV 2 = CV 2 \ Capacitance across AB is equal to capacitance across
2 2
When switch is opened, capacitance of A becomes CD which is C1.
KC, when dielectric material was introduced its energy 1 e0A
16. (b) W= CV2 Þ C =
2 d
1 Q2
becomes ( KC )V 2 . Energy of B becomes
2 2( KC ) 1 Î0 A 2
W' = V
where Q is the charge on it. 2 2d
C2 V 2 CV 2
Þ = 20V 20V 6mF
2( KC ) 2K 2mF 2mF 2mF
1 1 CV 2 5
Total energy = KCV 2 + = CV 2 3mF 3mF 6mF
2 2 K 3
\ Ratios of the energies in the two cases
1 1
Þ CV 2 :
5
CV 2 = 3 : 5 Þ CV2 = CV2
3 2´2 4
17. (d) The capacitors of 4mF and 4mF are in series.
1
13. (d) If mv2 >qEl,it will reach +ve plate. Otherwise to the -ve plate. 4´ 4
2 Þ C1 = = 2mF
4+ 4
R 1R 2 Now C1 is parallel to 2mF Þ C2 = C1 + 2 = 4mF
14. (b) Req = ( Parallel combination);
R1 + R 2 Again, C2 is parallel to 2mF Þ C3 = 4 + 2 = 6mF
Electricity P-25
The capacitors 3mF and 3mF are in parallel. 23. (a) The workdone in moving a charge along an
Þ C4 = 3 + 3 = 6mF equipotential surface is zero.
Now, C3 and C4 are in series
1 é Q1 Q ù
6´ 6 36 24. (b) VA = ê + 2 ú;
Hence, Capacitance = = = 3mF 4p Î0 ë R 2R û
6+6 12
18. (c) Both the capacitors are in series, hence total voltage 1 é Q2 Q ù
(E) VB= ê + 1 ú
4p Î0 ëR 2R û
= E1 + E2
(As +ve terminal of one is connected to –ve terminal of
other). So, Q is constant. 2R 2R
C1C 2 R R
Q = Ceq × E Þ C2V2 = ×E
C1 + C 2
C1 A R B
V2 = (E1+E2)
C1 + C 2
\ P D across x and y = P.D across C2
R 1R 2 4´2 8 4
19. (c) C.I. Req = (Parallel) = = = W
R1 + R 2 4+ 2 6 3
1 é Q2 Q ù
VA– VB = êQ1 + – Q2 – 1 ú
4p Î0 R ë 2 2û
Work done = Q × V = q × (VA– VB)
A
q é Q2 Q ù
3W
3W = 4p Î R êQ1 + – Q2 – 1 ú
6W 6W 0 ë 2 2û
º 2W 2W
[ 2Q1 + Q 2 – ]
6W
q 1
= ´ 2Q 2 – Q1
4p Î0 R 2
3W B 2W C
q ( Q1 – Q2 ) ( 2 – 1)
4W =
( 2 4p Î0 R)
25. (b) As S2 is open, hence C1 & C3 are in series, also C2 &
º
C4 are in series combination
C
3 3
B 2W Q= ×V = × 12 = 9mC;
4 4
-dV
20. 1(a) E = V = 5x2 + 10x –9,
dx 1´ 3
= 3 / 4F
1+ 3
-d C1
E= (5x2 + 10x – 9) = – (10x + 10)
dx 2´ 4
= 3 / 4F
On putting value of x in it we get, 2+4
E = –(10 × 1 + 10) = –20 V/m
21. (c) Let r be the internal resistance of the cell.
Terminal potential difference (V) < EMF of a cell (E). C2
12V
E - V æ E - Vö
V = E – Ir or r = =ç R
I è V ÷ø Charge on C1' = 9mC;
22. (d) Total R of circuit = 1 + 2 + 3 = 6 W 4
Charge on C2'= ×12 = 16mC
E 10 – 4 3
Current in circuit, I == = 1 amp; The direction Hence charge on C1 and C3 is 9mC, as both are in series
R 6
of current would be from a to b via e. combination.
P-26 Electricity
1 1 1 2mF
= +
C C1 C 2
B
C is the capacitance for length l1 + l2, C1 for length l1
and C2 for l2.
A
d d1 d2
Þ = +
2K Î0 (l1 + l 2 ) 2K Î0 l1 2K Î0 l 2 1mF 2mF
l1 + l 2 l1 l 2 l1d 2 + l 2 d1
Þ = + = B
d d1 d 2 d1d 2 2mF
1 V
28. (a) 1. U = CV2; 30. (c) I =
2 R+G
1 Q2 Total current through galvanometer = 50 × 2 ×10–4
2. U = = 10–2A
2 C
V
In first case, We know that, I =
R+G
1 Q2
U1' = (As charge is constant) Þ 10 (25+R) = 25 Þ 25+R = 2500
–2
2 C \ R = 2475 W
PART MAGNETISM
TEST 5
Time : 1 hour
Max. Marks : 120
INSTRUCTIONS
Ø Crack this section in just 1 hour.
Ø This part test contains 30 questions.
Ø For every correct answer you will be awarded +4 marks. There is no negative marking.
Ø Please fill the correct bubbles in the Response Sheet provided at the end of the test.
Ø Do not look at the solutions before attempting the complete test.
. . . Best of Luck
1. A straight section PO of a circuit lies along the x-axis from 4. Torques t1 and t 2 are required for a magnetic needle to
x = –a/2 to x = +a/2, and carries a steady current ‘I’. The remain perpendicular to the magnetic fields at two different
magnitude of magnetic field due to the section PO at a point places. The magnetic fields at the those places are B1 and
y = + a will be
B1
(a) proportional to a (b) proportional to a2 B2 respectively; then ratio is
B2
(c) proportional to 1/a (d) equal to zero
2. The instantaneous values of current and voltage in an A.C. t2 t1
(a) (b)
æ pö t1 t2
circuit are I = 4 sin wt and E = 100 cos ç wt + ÷ respectively..
è 3ø
t1 + t 2 t1 - t 2
The phase difference between voltage and current is (c) (d)
t1 - t 2 t1 + t 2
p 2p
(a) (b) 5. Two straight long conductors AOB and COD are
3 3 perpendicular to each other and carry currents I 1 and I2
5p 7p respectively. The magnitude of the magnetic induction at a
(c) (d) point P at a distance ‘a’ from the point O in a direction
6 6
perpendicular to the plane ABCD is
3. The B – H curve (i) and (ii) shown in Fig associated with
m0 m0
(a) (I1 + I 2 ) (b) (I1 – I 2 )
2 pa 2 pa
(i)
m 0 é 2 2 ù 1/ 2 m 0 æ I1I 2 ö
(c) I1 + I 2 (d) ç ÷
2pa çè I1 + I 2 ÷
(ii)
2pa ë û ø
H 6. Two coils have a mutual inductance 0.005 H. The current
changes in the first coil according to equation I=I0 sin wt,
where I0 = 10A and w = 100p radian/sec. The maximum
value of e.m.f. in the second coil is
(a) 2p (b) 5p
(c) p (d) 4p
(a) (i) diamagnetic and (ii) paramagnetic substance 7. XY is a boundary region separating two regions as shown in
(b) (i) paramagnetic and (ii) ferromagnetic substance figure. There is no magnetic field in region I, but a constant
(c) (i) soft iron and (ii) steel respectively magnetic field in region II. ACD is a semi circular conductor
(d) (i) steel and (ii) soft iron respectively. of radius r lying in region I. It is rotated with a constant angular
P-28 Magnetism
speed w about an axis passing through O and perpendicular 14. A circuit has a self inductance of 1 henry and carries a current
to the plane of the page. Which one of the following graphs of 2A. To prevent sparking, when the circuit is switched off,
represents correctly the variation of induced e.m.f. with time? a capacitor which can withstand 400V is used.The least
capacitance of the capacitor connected across the switch
w II must be equal to
I X
D (a) 50 mF (b) 25 mF (c) 100 mF (d) 12.5 mF
15. A thin rectangular magnet suspended freely has a period of
oscillation equal to T. Now it is broken into two equal halves
O (each having half of the original length) and one piece is
made to oscillate freely in the same field. If its period of
r oscillation is T', the ratio T' / T is
C
A (a) 1/ 2 2 (b) 1 / 2
Y (c) 2 (d) 1 / 4
e e
16. In a series resonant circuit, having L, C and R as its elements,
the resonant current is i. The power dissipated in the circuit
(a) (b)
t t at resonance is
0 0
e e i 2R
(a) æ 1 ö (b) zero
ç wL - ÷
è wC ø
(c) 0 (d) 0 t
t (c) i2wL (d) i2R
17. A bar magnet is suspended by a thin wire in uniform magnetic
field. When upper end of the wire is twisted by 120° , then
8. An A.C. circuit has two sources V1 and V2 of frequencies the magnet gets deflected by 30° from its position. How
w1 and w2 ; when V1 = 0, the peak current through a Resistor much upper end of the wire be twisted so that magnet may
is I2 and when V2=0 the peak current is I1. When V1 and V2 rotate by 90° from its initial position
are present, the peak current (a) 270° (b) 360° (c) 180° (d) 45°
(a) is (I1 + I2) 18. The field normal to the plane of a wire of n turns and radius
(b) is higher of I1 and I2 r which carries a current i is measured on the axis of the coil
(c) lies between |I1 – I2| and | I1 + I2 | at a small distance h from the centre of the coil. This is
1 smaller than the field at the centre by the fraction
(d) is |I1 + I2|
2 3 h2 2 h2 3 r2 2 r2
9. The mutual inductance of a pair of coils, each of N turns, is (a) . (b) . (c) . (d) .
M henry. If a current of I ampere in one of the coils is 2 r2 3 r2 2 h2 3 h2
brought to zero in t second, the emf induced per turn in the 19. A thin magnetic needle vibrates in horizontal plane with period
other coil, in volt, will be of 4s. The needle is cut into two halves by a plane normal to
MI NMI MN MI the magnetic axis of the needle. The period of each half is
(a) (b) (c) (d) approximately
t t It Nt (a) 4s (b) 2.8 s (c) 8 s (d) 1.4 s
10. The mutual inductance of two solenoids, of length L1 and
L2 with number of turns equal to N1 and N2 respectively. 20. In fig, what is the magnetic field induction at point O
(a) depends only on (N1N2/L1L2)
(b) depends on N1, N2,L1, L2 & orientation i
(c) is proportional to (N1N2)2
(d) is dependent on N1, N2,L1 ,L2 but not on orientation
11. A direct current of 2A and an alternating current having a
maximum value of 2A flow through two identical resistances.
The ratio of heat produced in the two resistances will be r
(a) 1 : 1 (b) 1 : 2 (c) 2 : 1 (d) 4 : 1 O
12. A transformer is used to light a 140 W, 24 V bulb from a
240 V a.c. mains. The current in the main cable is 0.7 A. The
efficiency of the transformer is m0 i m 0 i m0 i
(a) 63.8 % (b) 83.3 % (a) (b) +
4pr 4r 2p r
(c) 16.7 % (d) 36.2 %
m 0 i m0 i
(d) m 0 i - m0 i
13. In an A.C. circuit with voltage V and current I the power
dissipated
(c) +
4r 4p r 4r 4p r
(a) depends on the phase between V and I
21. Earth’s magnetic induction at a certain point is 7 x 10–5
(b) 1 Wb/m2. This field is to be annuled by the magnetic induction
VI
2 at the centre of a circular conducting loop 5.0 cm in radius.
1 The required current is
(c) VI (a) 0.056A (b) 5.6A (c) 56A (d) 12.8A
2
(d) VI
Magnetism P-29
22. In the given circuit, the current drawn from the source is
(a) 3 3 :8 (b) 16 : 9 3
(c) 4 : 9 (d) 2 2 : 3
V = 100x sin(100pt )
27. Two circular coils X and Y, having equal number of turns, carry
X C = 20W
X L =10W
R = 20W equal currents in the same sense and subtend same solid angle
~ at point O. If the smaller coil X is midway between O and Y, then
if we represent the magnetic induction due to bigger coil Y at O
as BY and due to smaller coil X at O as BX then
Y d
(a) 20 A (b) 10 A d/2
(c) 5 A (d) 5 2 A 2r
O
23. A proton, a deutron and an a particle accelerated through r
the same potential difference enter a region of uniform X
magnetic field, moving at right angles to B. What is the ratio
of their K.E. BY BY
(a) =1 (b) =2
(a) 1 : 2 : 2 (b) 2 : 2 : 1 (c) 1 : 2 : 1 (d) 1 : 1 : 2 BX BX
24. The magnetic field at the centre of a circular coil of radius r
BY 1 BY 1
is p times that due to a long straight wire at a distance r from (c) = (d) =
BX 2 BX 4
it, for equal currents. The following diagram shows three
cases : in all cases the circular part has radius r and straight 28. Two short bar magnets of equal dipole moment ‘M’ are
ones are infinitely long. For the same current, the magnetic fastened perpendicularly at their centres. The magnitude of
field at centres P in case 1, 2, 3 have the ratio two magnetic fields at a distance ‘d’ from the centre on the
bisector of the right angle is
S
P
P m0 M m0 M 2
P (a) (b)
4p d 3 4p d 3 N S
m0 2 2 M m 0 2M d
æ – p ö æ p ö æ 3p 1 ö æ – p ö æ p ö æ 3p 1 ö (c) (d) 4p 3
(a) ç ÷ :ç ÷ :ç – ÷ (b) ç + 1÷ : ç + 1÷ : ç + ÷ 4p d 3 P
è 2 ø è2ø è 4 2ø è 2 ø è2 ø è 4 2ø d N
29. Two different coils have self inductances L1 = 8 mH and
– p p 3p æ – p ö æ p 1 ö æ 3p 1 ö L2 = 2 mH. The current in one coil is increased at a constant
(c) : : (d) ç – 1÷ : ç – ÷ : ç + ÷
2 2 4 è 2 ø è 2 4ø è 4 2ø rate. The current in the second coils is also increased at the
25. The magnetic field at the centre due to motion of electron in same constant rate. At a certain instant, the power given to
the first Bohr orbit is B. The magnetic field due to motion of the two coil is the same. At that instant of time, if W1 and
electron in second Bohr orbit at the centre will be W2 are the energies stored in the first and the second coil,
respectively, the W1 : W2 is
B B (a) 1 : 2 (b) 2 : 1
(a) (b)
4 8 (c) 4 : 1 (d) 1 : 4
B B 30. A long solenoid having 200 turns per cm carries a current of
(c) (d) 1.5 amp. At the centre of it is placed a coil of 100 turns of
32 64
26. A magnet is suspended in such a way that it oscillates in the cross-sectional area 3.14 × 10–4 m2 having its axis parallel to
horizontal plane. It makes 20 oscillations per minute at a the field produced by the solenoid. When the direction of
place where dip angle is 30° and 15 oscillations per minute current in the solenoid is reversed within
at a place where dip angle is 60°. Ratio of total earth’s 0.05 sec, the induced em.f. in the coil is
magnetic field at the two places is (a) 0.48 V (b) 0.048 V
(c) 0.0048 V (d) 48 V
Response Sheet
ANSWER KEY
1 (c) 7 (d) 13 (a) 19 (b) 25 (c)
2 (c) 8 (c) 14 (b) 20 (c) 26 (b)
3 (c) 9 (a) 15 (b) 21 (b) 27 (c)
4 (b) 10 (b) 16 (d) 22 (d) 28 (c)
5 (c) 11 (c) 17 (a) 23 (d) 29 (d)
6 (b) 12 (b) 18 (a) 24 (a) 30 (b)
B=
m 0 2 2 1/ 2
2pa
( I1 + I 2 ) ( ^ to plane ABCD) = I 2rms Rt = ( 2 )2 Rt = 2Rt ;
H 1 4Rt 2
= =
H 2 2Rt 1
12. (b) Power of source = EI = 240 × 0.7 = 166
140
Þ Efficiency = Þ h = 83.3%
166
13. (a) Power dissipated = Erms Irms = (Erms) (Irms) cosq
Hence power dissipated depends upon phase
di
6. (b) e = -M difference.
dt
Magnetism P-31
E –5 –2
Current through circuit i = m0I 7 ´ 10 ´ 2 ´ 5.0 ´ 10
R 21. (b) B0 = , I= = 5.6amp.
Power dissipated at Resonance = i2R 2r 4p ´ 10 –7
17. (a) Torque = MB Sinf = Cq;
We know that T = Cq V0 100 V0 100
22. (d) iR = = = 5 , iL = = = 10 and
q is net deflection. R 20 X L 10
MB Sin f = Cq
f = angle by which magnet is deflected. V0 100
iC = = =5
MBSin 30° C(120 - 30) 1 90 XC 20
= = = or q = 270°.
MBSin 90° C(q - 90) 2 q - 90
Current, i = i 2R + (iC - i L ) 2 = 52 + 52 = 5 2 amp.
18. (a) The magnetic field on the axis of a current i carrying coil of
turns n, radius r and at a distance h from the centre of the 23. (d) K.E. gained by charged particle of charge q when
coil accelerated under a pot. diff. V will be Ek = qV;
m0 2pnir 2 For a given V, E µ q.
B= ´ ...(1)
4p ( r 2 + h 2 ) 3 / 2 For proton, deutron and a-particle, the ratio of charges
is 1 : 1 : 2.
The field at the centre is given by
m 0 2pi ´ n µ 0i µ i µ0 i
24. (a) In figure (1) B1 = Ä+ 0 +
B centre = ´ ....(2) (Q At centre h = 0) 4pr 4r 4pr
4p r
µ 0i
B r3 r3 B1 =
= = 4r
Bcentre (r 2 + h 2 )3 / 2 é h2 ù
3/ 2
In fig. (2), No field at centre due to straight part.
r 3 ê1 + ú
2
ëê r ûú \ B2 =
µ 0i
Ä
4r
1 æ 3 h2 ö In fig. (3), field at P due to upper straight part = 0
= or B ç1 + ÷ = Bcentre \ Total field a P
æ 3 h2 ö ç 2 r2 ÷
ç1 + ÷ è ø
ç 2 r2 ÷ 3 æ µ 0i ö µ i µ 0 i é 3p 1 ù
è ø B3 = .ç ÷Ä+ 0 = - Ä
4 è 2r ø 4pr 2 pr êë 4 2 úû
3 h2 p p æ 3p 1 ö
\ (B centre - B) / B = or B1 : B 2 : B3 : - : :ç - ÷
2 r2 2 2 è 4 2ø
I/4 m0I
I 25. (c) We know that, B =
19. (b) T = 2p , T ' = 2p M / 2B ; 2a
MB H H
where a is the radius of circular path.
T 4 m 0 ef
= 2 , T' = = 2.8s. I = ef, Hence B = ;
T' 2 2a
P-32 Magnetism
1 m 0 2M
We also know that frequency µ 3 and 28. (c) B1 =
n 4p d 3
2
a µ n (n is no. of orbits) (Along axial point) B2
1
Using all these we get B µ 5 m 0 2M
n B2 =
4p d 3 B
I (Along axial point) B1
26. (b) T = 2p ;
MB H
m 0 2M 2
Let the total magnetic fields due to earth at two places B2 = B12 + B 22 ; B =
4p d 3
be B1 and B2. If the horizontal components be (BH)1
and (BH)2 respectively then At final stage, the inductance acts as connecting wire
(BH)1 = B1 Cos 30° and (BH)2 = B2 Cos 60° hence whole of the current passes through it and no
Here T1 = 3 sec. and T2 = 4 sec. current pass through 10 W resistance.
I dI 1
T1 = 3 = 2p 29. (d) E = -L ; W = LI 2
MB1cos30° , dt 2
I
T2 = 4 = 2p We know that E = - LdI As dI is same,
MB2cos60° dt dt
1/ 2 E1 L1
3 æ B2 cos60° ö B 16 cos60° Hence = =4
= Þ 1 = ´ E 2 L2
4 èç B1cos30° ø÷ B2 9 cos30°
Power is constant so E 1I1 = E 2 I 2 ;
B1 16 1 2 16
Þ = ´ ´ = or B1 : B2 = 16 : 9 3
B2 9 2 3 9 3 E1 I 2 1
= = 4 We know that W = LI 2 ;
27. (c) Magnetic induction at O due to coil Y is given by, E2 I1 2
m0 2p I(2 r) 2 W1 L I 2 1 1
BY = ´ ......(1) \ = 1 1 = 4´ =
4p é 3/ 2
(2r)2 + d2 ù W2 L 2 I 2 2 16 4
ë û
Similarly, the magnetic induction at O due to coil X is 30. (b) B = m 0 n i = ( 4 p ´ 10 - 7 ) ( 200 ´ 10 -2 ) ´ 1.5
given by
= 3.8 ´ 10 -2 W / m 2
2
m0 2 p Ir Magnetic flux through each turn of the coil
BX = ´ ......(2)
4p é 2 2 3/ 2
r + ( d / 2) ù f=BA = (3.8 ´ 10 -2 ) (3.14 ´ 10 -4 ) = 1.2 ´10 -5 weber
ë û
When the current in the solenoid is reversed, the
BY 1 change in magnetic flux
From eq. (1) & (2) =
BX 2
= 2 ´ (1.2 ´ 10-5 ) = 2.4 ´10 -5 weber
df 2.4 ´ 10 -5
Induced e.m.f. = N = 100 ´ = 0.048 V.
dt 0.05
WAVE MOTION &
PART OPTICS
TEST 6
Time : 1 hour
Max. Marks : 120
INSTRUCTIONS
Ø Crack this section in just 1 hour.
Ø This part test contains 30 questions.
Ø For every correct answer you will be awarded +4 marks. There is no negative marking.
Ø Please fill the correct bubbles in the Response Sheet provided at the end of the test.
Ø Do not look at the solutions before attempting the complete test.
. . . Best of Luck
1. A particle of mass m is under a potential 3. A block is placed on a frictionless horizontal table. The mass
v(x) = (1/2)k1x2 for x > 0 and v(x) = k1x2 for x < 0. When of the block is m and springs are attached on either side
disturbed a little from the position x = 0, it will
with force constants K1 and K2. If the block is displaced a
(a) not execute SHM
little and left to oscillate, then the angular frequency of
m oscillation will be
(b) execute SHM with T = 2p
3k1
1 1
(c) execute SHM with T2 = 2p2m/(k12) æ K1 + K 2 ö2 é K1K 2 ù 2
(a) çç ÷÷ (b) ê ú
p m è m ø ë m (K 1 + K 2 ) û
(d) execute SHM with T =
æ 1 ö
k1 çç1 + ÷÷
è 2ø 1 1
é K 1K 2 ù 2 é K12 + K 22 ù 2
2. A flat plate P of mass ‘M’ executes SHM in a horizontal (c) ê ú (d) ê ú
plane by sliding over a frictionless surface with a frequency ë (K 1 - K 2 )m û êë (K1 + K 2 )m úû
V. A block ‘B’ of mass ‘m’ rests on the plate as shown in
figure. Coefficient of friction between the surface of B and P 4. 56 tuning forks are so arranged in series that each fork gives
is m. What is the maximum amplitude of oscillation that the 4 beats per sec with the previous one. The frequency of the
plate block system can have if the block B is not to slip on last fork is 3 times that of the first. The frequency of the first
the plate :
fork is
(a) 110 (b) 56
(c) 60 (d) 52
5. A beam of light of intensity 12 watt/cm2 is incident on a
totally reflecting plane mirror of area 1.5 cm2, then the force
mg (in newton) acting on the mirror will be
mg
(a) (b) (a) 2.4 × 10–6
4p 2 V 2 4p 2 V
(b) 1.2 × 10–7
m mg
(c) (d) (c) 3.6 × 10–8
4p V 2 g
2
2p 2 V 2
(d) 5.6×10–5
P-34 Wave Motion & Optics
6. A simple pendulum with a bob of mass ‘m’ oscillates from A to 14. Microwaves from a transmitter are directed towards a plane
C and back to A such that PB is H. If the acceleration due to reflector. A detector moves along the normal to the reflector.
gravity is ‘g’, then the velocity of the bob as it passes through Between positions of 14 successive maxima, the detector
B is P travels a distance of 0.14 m. What is the frequency of
A C
H transmitter?
B (a) 1.5 × 1010 Hz (b) 3.0 × 1010 Hz
9
(c) 1.5 × 10 Hz (d) 3.0×109 Hz
(a) zero (b) 2 gH
15. If we lower the temperature then for the e.m. radiation emitted
(c) mgH (d) 2g H
(a) the total energy changes but not the distribution
7. Two periodic waves of intensities I1 and I2 pass through a (b) the distribution changes but not the total energy
region at the same time in the same direction. The sum of the (c) both the distribution and total energy changes
maximum and minimum intensities is (d) neither the distribution nor the total energy changes
(a) 2 (I1 + I2) (b) I1 + I2 16. Two slits separated by a distance of 1 mm are illuminated
(c) ( I1 + I 2 ) 2 (d) ( I1 - I 2 ) 2 with red light of wavelength 6.5 × 10–7 m. The interference
fringes are observed on a screen placed 1 m from the slits.
8. A pipe closed at one end produces a fundamental note of The distance between third dark fringe & the fifth bright
412 Hz. It is cut into two pieces of equal length. the fringe is equal to
fundamental frequencies produced by the two pieces are
(a) 0.65 mm (b) 1.63 mm
(a) 206 Hz, 412 Hz (b) 824 Hz, 1648 Hz
(c) 3.25 mm (d) 4.88 mm
(c) 412 Hz, 824 Hz (d) 206 Hz, 824 Hz
17. A far sighted person can see clearly object beyond 100 cm
9. A thin sheet of glass (m = 1.5) of thickness 6 microns
distance. If he wants to see clearly an object at 40 cm
introduced in the path of one of interfering beams in a double
distance, then the power of the lens he shall require is
slit experiment shifts the central fringe to a position
(a) + 1.5 D (b) –1.5D
previously occupied by fifth bright fringe. Then the wave
(c) +3.0 D (d) –3.0 D
length of the light used is
18. An object of length 2d<<f, lies along the axis of a concave
(a) 6000 Å (b) 3000 Å
mirror of focal length. The centre of the object is at a distance
(c) 4500 Å (d) 7500 Å
2f. The approximate size of the image is
10. A beam of photons passing through a strong magnetic field
(a) d/2 (b) d
would
(c) 3d/2 (d) 2d
(a) lose energy (b) gain energy
(c) remain unaffected (d) slow down 19. An object and its image by a mirror of focal length f have
11. Producing interference fringes by light from two electric distances from the focus in the ratio 1:4. The object distance
bulbs is is:
(a) possible if they are on the same branch (a) f/2 (b) 2f
(b) possible if there are no fluctuations (c) 3f/2 (d) 2f/3
(c) possible if they have the same wattage 20. A glass slab has the left half of refractive index n 1, and the
(d) not possible right half of n2=3n 1. The effective refractive index of the
12. A ray of light is incident at an angle of 60° on one face of a whole slab is
prism of angle 30°. The ray emerging out of the prism makes n1
an angle of 30° with the incident ray. The emergent ray is (a) (b) 2n
2
(a) Normal to the face through which it emerges
(b) Inclined at 30° to the face through which it emerges 3n 1 2n 1
(c) Inclined at 60° to the face through which it emerges (c) (d)
2 3
(d) Inclined at 90° to the normal at face through which it
21. At the point of minimum deviation, the incidence angle i1
emerges
and the emergent angle i 2 satisfy
13. The dispersive powers of the materials of the two lenses are
in the ratio 4/3. If the achromatic combination of these two sin i1 sin i 2
(a) =m (b) =m
thin lens in contact is a convex lens of focal length 60 cm, sin i 2 sin i1
then the focal length of the component lenses are
(a) – 20 cm and +25 cm (b) –20 cm and –25 cm A
(c ) i1= i2 (d) i1= i2 =
(c) –15 cm and +20 cm (d) +15 cm and –20 cm 2
Wave Motion & Optics P-35
22. A sphere is placed in front of a convex lens of focal length (a) 0.08° (b) 0.16°
f. The radius of the sphere is much smaller compared to f. (c) 0.20° (d) 0.32°
The image of the sphere would look spherical if the object 28. A ray incident on face AB of a prism comes out of AC at
distance is grazing angle. Angle A is 900 & m =2. The incidence angle is
3f approximately
(a) f (b)
2
A
f
(c) 2f (d)
2 R
S
23. The magnification produced by the objective lens and the
eye lens of a compound microscope are 25 and 6 respectively. Q P
The magnifying power of this microscope is B C
(a) 19 (b) 31
(c) 150 (d) p
150 (a) 10 deg (b) rad
9
24. When light rays are incident on a prism at an angle of 45°
the minimum deviation is obtained. If the refractive index of p
(c) rad (d) 4.5deg
18
the material of prism is 2 , then the angle of prism will be
29. A convex lens A of focal length 20 cm and a concave lens B
(a) 30° (b) 40°
of focal length 5 cm are kept along the same axis with a
(c) 50° (d) 60°
distance d between them. If a parallel beam of light falling
25. A lens is placed between a source of light and a wall. It
on A leaves B as a parallel beam, then the distance d in cm
forms images of area A1 and A2 on the wall, for its two
will be
different positions. The area of the source of light is
(a) 25 (b) 15
(a) (A1A 2 ) (b) (A1+A2)/2 (c) 30 (d) 50
30. In the arrangement shown L1,L2 are slits and S1, S2 two
(c)
( A1 + A 2 )2 (d)
é 1
ê +
1 ù
ú
–1
independent sources on the screen, interference fringes
2 A
ë 1 A 2û screen
26. A ray of light from a denser medium strikes a rarer medium. L1
The reflected and refracted rays make an angle of 90° with × ×
S1 S2
each other. The angles of reflection and refraction are r and
r’. The critical angle would be L2
(a) sin –1 (tan r) (b) tan –1 (sin r) (a) will not be there
–1
(c) sin (tan r’) (d) tan –1 (sin r’) (b) will not be there if the intensity of light reaching the
27. In Young’s double slit interference experiment the distance screen from S1 and S2 are equal.
between two sources is 0.1 mm. The distance of the screen
(c) will be there under all circumstances
from the sources is 20 cm. The wavelength of light used is
(d) we will have only the central fringe
5460 Å. Then the angular position of the first dark fringe is
Response Sheet
ANSWER KEY
1 (a) 7 (a) 13 (d) 19 (c) 25 (a)
2 (a) 8 (b) 14 (a) 20 (c) 26 (a)
3 (a) 9 (a) 15 (c) 21 (c) 27 (b)
4 (a) 10 (c) 16 (b) 22 (c) 28 (b)
5 (b) 11 (d). 17 (a) 23 (c) 29 (b)
6 (d) 12 (a) 18 (d) 24 (d) 30 (b)
2IA 2 ´ 12 ´ 1.5 ´ 10 –4 10. (c) Only charged particles in motion are affected by B .
ÞF= = Photons are not charged.
C 10 –4 ´ 3 ´ 108
11. (d) Light from two different bulbs is not coherent and hence
8 ´15 ´ 10 -8 no fringes will be observed.
= = 12 ´ 10 -8 = 1.2 ´ 10 -7 N 12. (a) A + d = i + i' Þ i' = A + d - i = 30 + 30 - 60 = 0° .
10
6. (d) At B, the velocity is maximum. Taking vertical downward W1 f 4 4
motion of body from A to B, we have, u = 0, a = g, s = H 13. (d) =- 1 = or f1 = - f 2 ......(1)
W2 f2 3 3
and v = ?
1 1 1
As v2 = u2 + 2 as ; so v2 = 0 + 2g H or v = 2g H Further + = ......(2)
f1 f 2 60
14. (a) Detector receives direct & reflected waves. Distance 23. (c) Magnifying Power = M0 × Me = 25×6 = 150
moved between 2 consecutive positions of maxima=l/2. sin i
24. (d) m= ; r = A/2,
p sin r
For 14 successive maxima = 14 ´ = .14m(given ) ;
2 From concept (1) we get,
l = 2 × 10–2 m; sin 45 1 1
2= Þ sin r = = r = 30°
sin r 2´ 2 2
c 3 ´ 10 8
v= = = 1.5 ´1010 Hz
l 2 ´10 -2 A
From concept (2) we get, 30 = Þ A = 60°
15. (c) Both distribution and total energy are governed by 2
Plank’s law dependent on T. 25. (a) By displacement method, the size of object is given by
l D 6.5 ´ 10 –7 O= I1 ´ I 2 ;
16. (b) b= = ´ 1 = 0.65 × 10–3 = 0.65 mm
2d 10-3 \ A = A1 ´ A 2
The distance between fifth bright fringe from third dark
26. (a) From figure i = r and r’ = 90 - r
fringe = 2.5 b = 2.5 × .65 = 1.63 mm.
17. (a) The person suffers from hypermetropia. He should use sin r ' sin(90 - r )
Now =
m =
convex lens which may form image of 40 cm distant sin r sin r
object at 100 cm. u = -40 cm, v = –100 cm
cos r 1
1 1 1 1 1 Þ=
m =
sin r tan r
Now, f = v - u = - 100 - (-40)
1 60
Þ = = 1 .5 D = P
f 100 ´ 40
u 1,2 f
18. (d) u 1,2 = 2f + d ; \ v1,2 = » v1, 2
u 1, 2 - f
\ Size » 2d
19. (c) u = x + f , v = 4 ´ +f
1 1 1
\ = + ;
f x + f 4x + f
1
We know that m =
4x 2 sin C
\ f + 5x + = 5 x + 2f
f 1 1
where C is critical angle. = ;
f 3f sin C tan r
\ 4x 2 = f 2 , \ x = or u = sin C = tan r, C = sin–1 (tan r) = sin–1 (tan r)
2 2
20. (c) Total time taken to travel distance d is : b lD l
27. (b) The angular position, q = = =
D 2dD 2d
d d æ n +n2 ö d
+ = dçç 1 ÷=
÷ n ; The first dark fringe will be at half of the fringe width
2n 1 2n 2 è 2n 1 n 2 ø eff
from midpoint of central maximum. Thus the angular
3 q 1 l
n 2 = 3n 1 Þ n eff =n1 position of first dark fringe will be a = = ´ ;
2 2 2 2d
Here, 2d = .1 mm, l = 5460Å;
21. (c) By symmetry i1 = i2 since there is only one angle for
minimum deviation. 1 5460 ´10 -10
22. (c) Lateral magnitude = v/u; Hence, a = ´ = 2750 ´ 10 -6 radians
2 .1´ 10-3
dv u 2
Mag. along axis = = = 1 if v = u , \ u = 2 180°
du v 2 = 2750 ´ 10 -6 ´ = 0.16°
p
P-38 Wave Motion & Optics
é iù
ÐSRP = sin -1 êsin ú = q 2 ;
ë m û
\ Ð RPS = p - q1 - q 2 = p - A; F2
æ1ö sin i
\ A = sin -1 çç ÷÷ + sin -1
m
è ø m
sin i
\ 40 0 = 30 0 + sin -1 ;
m 30. (b) On the screen, we have four amplitudes pair wise
coherent.
( ) æpö p
\ sin i = 2 sin 10 0 » 2 sin ç ÷ »
è 18 ø 9
(A1 + A 2 ) + (A 3 + A 4 ) º A12 + A 34
However, If A12 and A34 have equal magnitude because
29. (b) In the absence of concave lens the parallel beam will of random phase of A12 and A34, no fringes will be
be focussed at F2 i.e. at a distance 20 cm. from lens A. seen.
PART MODERN PHYSICS
TEST 7
Time : 1 hour
Max. Marks : 120
INSTRUCTIONS
Ø Crack this section in just 1 hour.
Ø This part test contains 30 questions.
Ø For every correct answer you will be awarded +4 marks. There is no negative marking.
Ø Please fill the correct bubbles in the Response Sheet provided at the end of the test.
Ø Do not look at the solutions before attempting the complete test.
. . . Best of Luck
1. In short wave communication waves of which of the 6. What should be the maximum acceptance angle at the air
folowing frequencies will be reflected back by the core interface of an optical fibre if n1 and n2 are the refractive
ionospheric layer, having electron density 1011 per m3 indices of the core and the cladding, respectively
(a) 2 MHz (b) 10 MHz (a) sin -1 (n 2 / n1 )
(c) 12 MHz (d) 18 MHz
2. If radium and chlorine combine to form radium chloride, the (b) sin -1 n12 - n 22
compound is
(a) no longer radioactive é -1 n 2 ù
(b) half as radioactive as the radium content (c) ê tan ú
ë n1 û
(c) as radioactive as the radium content
(d) twice as radioactive as the radium content
é -1 n1 ù
3. Range of frequencies alloted for commercial FM radio (d) ê tan ú
ë n2 û
broadcast is
(a) 88 to 108 MHz 7. Four equal resistors, each of resistance 10 ohms are
(b) 88 to 108 kHz connected as shown in the circuit diagram. Then the
(c) 8 to 108 MHz equivalent resistance between points A and B is
(d) 88 to 108 GHz
4. Two electrons of K.E. 2.5 eV fall on a metal plate, which has
work function of 4.0 eV. Number of electrons ejected from
the metal surface is
(a) One
(b) Two
(c) No (a) 40 ohms (b) 20 ohms
(d) More than two (c) 10 ohms (d) 5 ohms
5. An oscillator is producing FM waves of frequency 2 kHz 8. A hydrogen-like atom has one electron revolving around a
stationary nucleus. The energy required to excite the electron
with a variation of 10 kHz. What is the modulating index
from the second orbit to the third orbit is 47.2 eV. The atomic
(a) 0.20 (b) 5.0
number of the atom is
(c) 0.67 (d) 1.5
(a) 3 (b) 4 (c) 5 (d) 6
P-40 Modern Physics
9. What is the ratio of the circumference of the first Bohr orbit (a) 2.2 MeV (b) 23.6 MeV
for the electron in the hydrogen atom to the de Brogile (c) 28.0 MeV (d) 30.2MeV
wavelength of electrons having the same velocity as the 19. A sky wave with a frequency 55 MHz is incident on D-
electron in the first Bohr orbit of the hydrogen atom? region of earth's atmosphere at 45°. The angle of refraction
(a) 1 : 1 (b) 1 : 2 (c) 1 : 4 (d) 2 : 1 is (electron density for D-region is 400 electron/cm3)
10. The counting rate observed from a radioactive source at t = (a) 60° (b) 45° (c) 30° (d) 15°
0 second was 1600 counts per second and at t = 8 seconds
20. The radioactivity of a sample is R1 at a time T1 and R2 at a
it was 100 counts per second. The counting rate observed,
time T2. If the half life of the specimen is T, the number of
as counts per second at t = 6 seconds will be
atoms that have disintegrated in the time (T 2 –T 1 ) is
(a) 400 (b) 300 (c) 200 (d) 150
proportional to
11. A photon materializes into an electron-positron pair. The
kinetic energy of the electron is found to be 0.19 MeV. What (a) (R1T1 – R2T2) (b) (R1 – R2)
was the energy of the photon? (c) (R1 – R2)/T (d) (R1–R2) ×T
(a) 0.38 MeV (b) 0.70 MeV 21. Consider an optical communication system operating at l ~
(c) 1.40 MeV (d) None 800 nm. Suppose, only 1% of the optical source frequency is
12. A transistor has b = 40. A change in base current of 100 mA, the available channel bandwidth for optical communication.
produces change in collector current How many channels can be accommodated for transmitting
(a) 40 × 100 microampere (b) (100 – 40) microampere audio signals requiring a bandwidth of 8 kHz ?
(c) (100 + 40) microampere (d) 100/40 microampere (a) 4.8 × 108 (b) 48
13. If 10% of a radioactive material decays in 5 days, then the (c) 6.2 × 10 8 (d) 4.8 × 105
amount of the original material left after 20 days is 22. An electron and a proton fired with the same momentum into
approximately a magnetic field perpendicular to the field. What is the nature
(a) 60% (b) 65% (c) 70% (d) 75% of their trajectories?
14. Lights of two different frequencies, whose photons have (a) Electron trajectory is less curved
energies 1 eV and 2.5 eV respectively, successively illuminate
(b) Proton trajectory is less curved
a metal whose work function is 0.5 eV. The ratio of the
(c) Both trajectories are equally curved
maximum speeds of the emitted electrons will be
(d) Both trajectories are straight lines
(a) 1 : 5 (b) 1 : 4 (c) 1 : 2 (d) 1 : 1
23. The work function of a metallic surface is 5.01 eV,
15. In frequency modulation
(a) The amplitude of modulated wave varies as frequency photoelectrons are emitted when light of wavelength 2000Å
of carrier wave falls on it. The minimum potential difference required to
(b) The frequency of modulated wave varies as amplitude stop the fastest photoelectrons (h = 4.14 x 10–15 eV–s)
of modulating wave (a) 1.2 volts (b) 2.4 volt
(c) The amplitude of modulated wave varies as amplitude (c) 3.6 volt (d) 4.8 volt
of carrier wave 24. The activity of a radioactive element decreases to one-third
(d) The frequency of modulated wave varies as frequency of the original activity I0 in a period of nine years. After a
of modulating wave further lapse of nine years its activity will be
16. The energy levels of a certain atom for 1st, 2nd and 3rd levels
2 I0 I0
are E, 4E/3 and 2E respectively. A photon of wavelength l is (a) I0 (b) I0 (c) (d)
3 9 6
emitted for a transition 3 ®1. What will be the wavelength
of emission for transition 2 ® 1? 25. If the total binding energies of ( H), ( He ), ( Fe) and
2
1
4
2
56
26
(a) l /3 (b) 4l /3 (c) 3 l /4 (d) 3l
17. In the X-rays tube before striking the target we accelerate ( 235
92 U )nuclei are 2.22, 28.3, 492 and 1786 MeV respectively..
the electrons through a potential difference of U volt. For
Identify the most stable nucleus out of the following
which of the following value of U, we will have X-rays of
(a) Fe (b) H (c) He (d) U
largest wavelength?
26. In Millikan oil drop experiment, a charged drop of mass 1.8 x
(a) 10 kV (b) 20 kV (c) 30kV (d) 40 kV
10–14 kg is stationary between its plates. The distance
18. The binding energy per nucleon for deuteron ( H) and
2
1 between its plates is 0.90 cm and potential difference is 2.0
( )
helium 42 He are 1.1 MeV and 7.0 MeV. The energy released kilovolts. The number of electrons on the drop is
when two deuterons fuse to form a helium nucleus is (a) 500 (b) 50 (c) 5 (d) 0
Modern Physics P-41
27. A hydrogen atom emits green light when it changes from electron, an alpha particle and a proton each having the
n = 4 energy level to the n = 2 level. Which colour of light would same de-Broglie wavelength then
the atom emit when it changes from n = 5 level to the n=2 level? (a) E1> E3> E2 (b) E2> E3> E1
(a) red (b) yellow (c) green (d) violet (c) E1> E2> E3 (d) E1= E2= E3
28. In a npn transistor 1010 electrons enter the emitter in 10–6 s. 30. Laser beams are used to measure long distances because
4% of the electrons are lost in the base. The current transfer (a) They are monochromatic
ratio will be (b) They are highly polarised
(a) 0.98 (b) 0.97 (c) 0.96 (d) 0.94 (c) They are coherent
29. If E1, E2 and E3 are the respective kinetic energies of an (d) They have high degree of parallelism
Response Sheet
ANSWER KEY
1 (a) 7 (c) 13 (b) 19 (b) 25 (a)
2 (c) 8 (c) 14 (c) 20 (d) 26 (c)
3 (a) 9 (a) 15 (b) 21 (a) 27 (d)
4 (c) 10 (c) 16 (d) 22 (c) 28 (c)
5 (b) 11 (c) 17 (a) 23 (a) 29 (a)
6 (b) 12 (a) 18 (b) 24 (c) 30 (d)
14. (c) Einstein equation KEmax = E - Work function; Bandwidth of channel (1% of above) = 3.8 × 1012Hz
1 Number of channels = (Total bandwidth of channel)
mv2 = E - W / (Bandwidth needed per channel)
2
Using this concept, (a) Number of channels for audio signal
= (3.8 ´ 1012 ) /(8 ´ 103 ) ~ 4.8 ´108
1
mV12 max
2 1 - .5 1 V max 1 mv
= = or 1 =
1 2 2.5 - .5 4 V 22. (c) r = qB as mv(momentum) is same for both. Thus r is
V2 max 2 max 2
2
same for both proton and electron. (q is also same for
15. (b) The process of changing the frequency of a carrier both of them.)
wave (modulated wave) in accordance with the audio hc
frequency signal (modulating wave) is known as 23. (a) eV = –W
l
frequency modulation (FM).
According to question,
hc
16. (d) Concept involved DE = hc
l eV = –5.01
l
hc hc
Acc. to question E3 – E1 = i.e. 2E – E = ....(1) 12400
l l Þ eV = – 5.01,
2000
E2 – E1 =
hc
or 4 E – E =
hc E
; =
hc
....(2) Þ eV = 6.2 –5.01 eV = 1.2 eV = 1.2 Volt
l' 3 l ' 3 l'
Initial amount
24. (c) Left amount = ; T = n×t½
hc hc
= (2)n
Put the value of E in eq 2 we get or l ' = 3 l
3l l '
9 = n 1 × t½ ......(1)
12345 1 18 = n2 × t½ ......(2)
17. (a) Concept involved l = Å ; From above l¥
V V IO IO
In case (1) = ......(3);
Hence X rays of largest wavelength can be generated 3 9 / t1 / 2
by applying lowest potential i.e.10kV. (2)
2 4 IO IO
18. (b) (1) Reaction involved 2 H ® He In case (2), I= = ....(4)
1 2 18 / t1 / 2 2
(2) æ 9 / t1 / 2 ö
ç (2 ) ÷
(2) Conservation of energy è ø
B.E of Deuteron = 1.1 ´ 2 = 2.2 MeV
B.E of Helium = 7 ´ 4 = 28 MeV I0 I
From eq. (3) & (4), we get I = = 0
2 9
According to reaction, 2 × 2.2 = 28MeV + E (3)
\ E = 23.6MeV
B.E
25. (a) More , more the stability
æ 80.5N ö 80.5 ´ (400 ´ 10 ) 6 Nucleon
19. (b) n eff = n 0 1 - ç = 1 1- =1
2 ÷
è v ø (55 ´ 106 ) 2 56 492
· For Fe = = 8.78MeV
26 56
sin i
Also n eff =
sin r Þ sin r = sin r Þ r = i = 45°
2 2.22
· For H= = 1.11MeV
1 2
.693
20. (d) 1. l = 2. R = lN t 4 28.3
t1 / 2 · For He = = 7.07MeV
2 4
Radioactivity at T1 is R1 = l N1,
235 1786
Radioactivity at T2 is R2 = l N2 · For U= = 7.6MeV ;
92 235
\ Number of atoms decayed in time Hence Fe is most stable and H is least stable nucleus.
R 1 - R 2 ( R 1 - R 2 )T 26. (c) ne ´ E = m ´ g
(T1–T2) = (N1–N2) or =
l .693
n ´1.6 ´ 10 -19 ´ 2 ´10 3
ie ¥ (R1–R2)T or = 1.8 ´10 -14 ´10
21. (a) Optical source frequency .9 ´ 10 - 2
f =
c
= 3 ´ 108 /(800 ´ 10 -9 ) = 3.8 ´1014 Hz
1.8 ´10 -14 ´10 ´ .9 ´ 10 -2
n= or n » 5
l
1.6 ´ 10 -19 ´ 2 ´10 3
P-44 Modern Physics
1 1 1 4 -1 nC ´ e
27. (d) In first case : f1 = = RZ 2 ( - ) = RZ 2 ( ) Collector current, I C =
l1 4 16 16 t
\ Current transfer ratio,
3
= RZ 2 = .18RZ 2 (green light)
16
a=
IC n C
=
0.96 ´ 1010
= = 0.96
IE nE 1010
1 1 1
In second case : f 2 = = RZ 2 ( - ) or
l2 4 25 1 h h
29. (a) K.E. = Mv2; Now as l is same l = ( is
2 mv mv
21
RZ 2 ´ = .21RZ 2 constant)
100
For alpha mass is more hence its velocity is less leading
Hence f2 > f1 Among the options, only violet is one to lowest k.E. In the same way mass of electron is less,
which has frequency more than green. Hence answer hence its velocity is more leading to max k.E. Thus
is violet. EElectron >E Proton>E alpha
28. (c) No. of electrons reaching the collector, 30. (d) Laser beams are perfectly parallel. So that they are very
96 narrow and can travel a long distance without
nC = ´ 1010 = 0.96 ´ 1010
100 spreading. This is the feature of loser while they are
monochromatic and coherent these are characteristics
nE ´e only
Emitter current, I E =
t
EXPERIMENTAL
PART SKILLS IN PHYSICS
TEST 8
Time : 1 hour
Max. Marks : 120
INSTRUCTIONS
Ø Crack this section in just 1 hour.
Ø This part test contains 30 questions.
Ø For every correct answer you will be awarded +4 marks. There is no negative marking.
Ø Please fill the correct bubbles in the Response Sheet provided at the end of the test.
Ø Do not look at the solutions before attempting the complete test.
. . . Best of Luck
1. A potentiometer works on the principle that 3. A current of 4A produces a deflection of 30° in the
(a) when a current flows through a wire of iniform thickness galvanometer. The figure of merit is
and material, potential difference between its two points (a) 6.5 A/rad (b) 7.6 A/rad
is directly proportional to the length of the wire between (c) 7.5 A/rad (d) 8.0 A/rad
the two points 4. For the determination of the focal length of a convex mirror,
(b) when a current flows through a wire of uniform a convex lens is required because
thickness and material, potential difference between (a) it is not possible to obtain the image produced by a
its two points is inversely proportional to the length of convex mirror on the screen
the wire between the points (b) a convex lens has high resolving power so it helps to
(c) when a current flows through a wire of uniform measure the focal length correctly
thickness and material, potential difference between (c) a convex mirror always forms a real image which is
its two points doesn't depend on the length of the wire diminished by the convex lens
between the points (d) none of these
5. Which is the correct form of the graph of angle of incidence
(d) none of these
vs angle of deviation for a triangular prism during the parallax
2. The figure of merit of a galvanometer is defined as
method ?
(a) the voltage required to produce unit deflection in the
Y
galvanometer (a)
Angle of deviation
(b) Y
Reverse-bias
Angle of deviation
Reverse bias
current (IR)
Dm
O Angle of incidence X
Y'
(c) Y
Y
Angle of deviation
(VR)
(d)
O Angle of incidence X
O Reverse-bias current X
(IR)
(d) Y
7. Choose the wrong statement regarding the experiment to
Angle of deviation
(b) Vernier B
(c) Accuracy in measurement does not depend on vernier
constant
O Reverse-bias current X (d) Both A and B are equally accurate.
(IR) 9. A cooling curve is plotted between the temperature of a hot
body and time. Which of the following is not true for the
cooling curve?
O
(IR) (a) Cooling is faster from calorimeter having a large surface
(b) Reverse-bias current X
area than a smaller one.
Reverse-bias
voltage (VR)
10. The speed of sound is measured by a resonance tube at the calculated in the experiment. Which of the following is true?
room temperature once by filling the tube with water and (a) v1 = v2
then with glycerine as v1 and v2. Which of the following (b) v1 > v2
relations is/are true in this context? (c) v1 < v2
(a) v1 = 2v2 (d) v1 and v2 are independent of r 1 and r 2
(b) v1 > v2 19. Specific heat of a solid does not depend on which of the
(c) v1 < v2 following factors?
(d) v1 and v2 does not depend on the nature of the liquid (a) Mass of the solid
taken in the tube. (b) Temperature increase of solid
11. The null point should be obtained on the metre bridge wire (c) Nature of the solid
to get maximum accuracy at (d) Volume of the solid.
(a) the middle of the wire 20. The refractive index of the material of a prism does not
(b) the left end of the wire depend on which of the following factor?
(c) the right end of the wire (a) Nature of the material
(d) the 1/4th distance from the left end (b) Wavelength or colour of light
12. What is the null point in a potentiometer (c) Temperature
(a) flowing of no current in the cell circuit. (d) Angle of the prism.
(b) flowing of equal and opposite currents in the cell circuit. 21. The transfer characteristics of a transistor means a plot of
(c) flowing of more current from the driver cell than the (a) input voltage versus input current
unknown cell (b) output voltage versus output current.
(d) flowing of more current from the unknown cell than the (c) input voltage versus output voltage
driver cell. (d) input current versus output current.
13. To measure the resistance of a galvanometer by half 22. Current gain is maximum in the following configuration of a
deflection method, a shunt is connected to the galvanometer. transistor
The shunt is– (a) common emitter configuration
(a) low resistance connected in parallel (b) common base configuration
(b) low resistance connected in series (c) common collector configuration
(c) high resistance connected in parallel (d) equal in both common emitter and common base
(d) high resistance connected in series configuration
14. The u-v graph for a convex lens is a 23. How are the currents flowing in the emitter, base and the
(a) straight line passing through the origin. collector related to each other?
(b) straight line having an y-intercept. (a) Ic =Ib + Ie (b) Ie =Ib + Ic
(c) parabola. (c) Ib =Ie + Ic (d) Ie =Ic – Ib
(d) hyperbola. 24. Which of the following operations will not increase the
15. An LED operates under which biasing condition? sensitivity of a potentiometer?
(a) Forward bias (a) Increase in the number of wires of the potentiometer.
(b) Reverse bias (b) Reducing the potential gradient.
(c) Can operate both in forward and reverse bias (c) Increasing the current through the potentiometer
(d) No biasing is required. (d) Increasing the sensitivity of the galvanometer.
16. Which of the following is not a two legged device? 25. The potential gradient of a potentiometer can be increased
(a) Resistor (b) Capacitor by which of the following operation?
(c) p-n junction diode (d) Integrated circuit (a) By increasing the area of cross-section of the
17. Zener diode can be used as potentiometer wire.
(a) half-wave rectifier (b) oscillator (b) By decreasing the area of cross-section of the
(c) voltage regulator (d) transformer potentiometer wire.
18. Two spheres of radii r 1 and r2 (r1 > r2) is dropped through a (c) By decreasing the current through it.
tube full of glycerine. Their terminal velocities v1 and v2 are (d) By using a wire of material of low specific resistance.
P-48 Experimental Skills in Physics
26. If a wire of certain length is stretched to twice its length, (a) + 0.03 mm (b) – 0.03 mm
how will its specific resistance change? (c) + 0.03 cm (d) – 0.03 cm
(a) it increases 29. What happens when the applied load increases and upto
(b) it decreases breaking stress in the experiment to determine the Young's
(c) first decreases and then increases. modulus of elesticity ?
(d) remains same. (a) The area of wire goes on decreasing and wire extends
27. The reading of given vernier callipers is and breaks.
(b) The area of wire goes on decreasing and wire breaks.
0 5 M 10
(c) The wire extends and area remains constant.
0 v 5 (d) The area remains same and wire length is also same.
(a) 5.06 cm (b) 5.6 cm 30. What is effect on temperature vs time graph when the
(c) 5.06 mm (d) 5.60 mm temperature of enclosure is large in the experiment to plot
28. What is the zero error in the diagram? the cooling curve for the relationship between the
Circular scale temperature of a hot body and time ?
0 95 (a) graph becomes more flattend
Refrence line (b) graph becomes a straight line
N (c) graph become more curved
(d) none of these
Response Sheet
ANSWER KEY
1 (a) 7 (c) 13 (a) 19 (d) 25 (b)
2 (b) 8 (b) 14 (c) 20 (d) 26 (d)
3 (b) 9 (d) 15 (a) 21 (c) 27 (b)
4 (a) 10 (d) 16 (d) 22 (a) 28 (a)
5 (a) 11 (a) 17 (c) 23 (b) 29 (a)
6 (c) 12 (a) 18 (b) 24 (c) 30 (a)
1. (a) A potentiometer is device which is used to compare 10. (d) The velocity depends on the length of the resonating
e.m.f.'s of two cells as well as to determine the internal air column, not on the nature of the liquid.
resistance of a cell. It is based on the principle that when 11. (a) Null point at the middle gives exact balance to the bridge.
a current flows through a wire of uniform thickness and 12. (a) When the null point is obtained on the potentiometer
material, potential difference between its two points is e.m.f. is balanced by the potential drop across the wire,
directly proportional to the length of the wire between so no current flows.
the two points. 1c. (a) Shunt is a low resistance connected in parallel to
2. (b) The figure of merit of a galvanometer is the current galvanometer so that most of the current passes
required to produce unit deflection in the galvanometer. through the shunt.
Mathematically, 14. (c) Since for a lens
K = I/q, where k is the figure of merit of the
1 1 1
galvanometer. , < = constant.
v u f
3. (b) Here I = 4A
\ u vs. v graph is a parabola.
c
æ 30 ´ p ö pc 15. (a) LED is a p-n junction diode which always operates on
q =30° = ç ÷ =
è 180 ø 6 forward bias.
16. (d) An integrated circuit (IC) has more than 3 legs.
I 4 4´6´7 2´6´7 17. (c) Since Zener diode operates at break-down voltage
Now, k = = = =
q æpö 22 11 region so output voltage across it remains constant
ç ÷ even if input current changes.
è ø
6
2 r 2 (r - s)g
84 18. (b) vT = ; vT µ r 2
= = 7.6 A / rad 9 h
11
4. (a) A convex mirror always forms a virtual image which \ v1 > v2
can't be cast on a screen. Hence a convex lens is used Q
19. (d) Specific heat = C =
to find the focal length of a convex mirror. mDt
5. (a) The correct form of the graph between the angle of so it depends on mass and temperature difference but
incidence and the angle of deviation is represented by not on the volume.
the graph what is given in (a). 20. (d) Refractive index is the property of the material, hence it
6. (c) In reverse-bias mode, a reverse current flows. does not depend on angle of the prism.
Therefore, (c) represents the form. 21. (c) In a transfer characteristics Vi is plotted along x-axis
7. (c) For a resistor the pointer moves in both ways when and V0 along y-axis.
voltage is applied.
8. (b) Lower the vernier constant, more accurate measurement Χ IC
22. (a) Current gain in CE configuration is b = .
is possible by it. ΧI B
9. (d) The rate of cooling is faster at the beginning and then IC >> IB, hence it is maximum.
decreases and becomes constant.
P-50 Experimental Skills in Physics
23. (b) Emitter current is the sum of base and collector current 27. (b) Main scale reading = 5 cm
by Kirchhoff's 1st law. Vernier division coinciding (n) = 3
rl V Ir Vernier scale value, n × V.C = 3 × 0.2 = 0.6 cm
24. (c) V = IR = I \ < k < , Total reading = 5 + 0.6 = 5.6 cm.
A l A
increase in I, will increase k, so it will decrease 28. (b) 0 of the circular scale is up to the refrence line, so, it is '–
sensitivity. ve' and 3rd division coincides with refrence line,
Irl so zero error is = –0.03 mm.
25. (b) V = IR = \ potential gradient = k 29. (a) First, the length of wire goes on increasing i.e., area
A
when A is decreased, k will increase. decreases and finally at breaking stress the wire breaks.
26. (d) Specific resistance is the property of material hence it 30. (a) Because if the temperature of enclosure is large then q
does not change with change in its length. – q0 is very small i.e. the graph becomes flattened.
CHEMISTRY
GENERAL
PART CHEMISTRY
TEST 1
Time : 1 hour
Max. Marks : 120
INSTRUCTIONS
Ø Crack this section in just 1 hour.
Ø This part test contains 30 questions.
Ø For every correct answer you will be awarded +4. There is no negative marking.
Ø Please fill the correct bubbles in the Response Sheet provided at the end of the test.
Ø Do not look at the solutions before attempting the complete test.
. . . Best of Luck
1. For a given value of principal quantum number n, the number (a) Gaseous, metallic (b) Liquid, metallic
of allowed values of azimuthal quantum number l, is given by (c) Solid, metallic (d) Solid, nonmetallic
(a) n – 2 (b) n 8. Two containers P and Q of equal volume (1 litre each) contain
(c) n – 1 (d) n 2 6 g of O2 and SO2 respectively at 300 K and 1 atmosphere.
2. An atom X belongs to 4th period of the periodic table and then
has highest number of unpaired electrons in comparison to (a) Number of molecules in P is less than that
the other elements of the period. The atomic number of X is in Q
(a) 23 (b) 25 (b) Number of molecules in P and Q is same
(c) 24 (d) 33 (c) Number of molecules in Q is less than that in P
3. Which of the following species would be least likely to act (d) Either (a) or (b)
as Lewis base? 9. A sample was weighed using two different balances. The
(a) PCl3 (b) CN– results were
(c) SCl2 (d) I+ (i) 3.929 g (ii) 4.0 g
4. The wave number of first line of Balmer series of hydrogen
How would the weight of the sample be reported?
is 15200 cm–1.The wave number of first Balmer line of Li2+
(a) 3.93 g (b) 3g
ion is
(c) 3.9 g (d) 3.929 g
(a) 15200 cm–1 (b) 60800 cm–1
(c) 76000 cm –1 (d) 136800 cm–1 10. Which of the following is not a disproportionation reaction?
5. Three isotopes of an element have mass numbers M, (M+1) (a) 4 KClO3 ¾¾ ® 3KClO4 + KCl
and (M+2). The mean atomic mass is (M + 0.5). Then which (b) 2Cu + ¾¾® Cu2+ + Cu
of the following ratio may be accepted for M, (M+1), (M+2) (c) 5X + XO3 + 6H+ ¾¾
– – ® 3X2 + 3H2O
respectively (d) 4P + 3NaOH + 3H2O ® PH3 + 3NaH2PO2
(a) 1:1:1 (b) 4:1:1 11. What would happen when a small quantity of H2O2 is added
(c) 3:2:1 (d) 2:1:1 to a solution of FeSO4
6. ‘a’ gm of element A reacts with ‘b’ gm of element B. Also ‘b’ (a) Colour disappears
gm of element B combines with 2c gm of C. If one gm (b) H2 is evolved
equivalent of B weighs ‘b’ gm. Then the mass ratio in which (c) An electron is added to Fe2+
A and C combine is (d) An electron is lost by Fe2+
(a) a : c (b) a : 2c 12. Which of the following reaction involves neither oxidation
(c) 2a : c (d) 3a : c nor reduction
7. An element belongs to Group 13 and the second period of
(a) CrO42– ® Cr2O72– (b) Cr ® CrCl3
the periodic table. Which of the following properties will be
shown by the element? (c) Na ® Na+ (d) 2 S2O32– ® S4O62–
C-2 General Chemistry
13. N2 and O2 are converted to mono cations N2+ and O2+ (a) 2.0 (b) 2.5
respectively, which of the following is wrong? (c) 3.4 (d) 4.0
(a) In N2+, the N – N bond weakens 22. Why are strong acids generally used as standard solutions
(b) In O2+, the O – O bond order increases in acid base titration?
(c) In O2+, paramagnetism decreases (a) The pH at the equivalent point will always be 7
(d) N2+ becomes diamagnetic (b) They can be used to titrate both strong and weak bases
14. If the electronegativity difference between two atoms A and (c) Strong acids form more stable solutions than weak acids
B is 2.0 then the percentage of covalent character in the (d) The salts of strong acids do not hydrolyse
molecule is 23. In which reaction, there is change in oxidation number of N
(a) 54% (b) 46% (a) 2NO2 ® N2O4
(c) 23% (d) 72% (b) NH4OH ® NH4+ + OH–
15. The reaction in which hydrogen peroxide acts as a reducing
(c) N2O5 + H2O ® 2HNO3
agent is
(d) 2NO2 + H2O ® HNO3 + HNO2
(a) PbS + 4H 2 O 2 ® PbSO 4 + 4 H 2 O
24. The % of oxalate ion in a given sample of oxalate salt of
(b) 2 Kl + H 2 O2 ® 2KOH + I 2 which 0.3 g dissolved in 100 ml of water required 90 ml of
N/20 KMnO4 for complete oxidation would be.
(c) 2 FeSO 4 + H 2SO 4 + H 2 O 2 ® Fe 2 (SO 4 ) 3 + 2H 2 O
(a) 11% (b) 33%
(d) Ag 2 O + H 2 O 2 ® 2Ag + H 2 O + O 2 (c) 66% (d) 99%
16. In which of the following molecules, the vander Waals force 25. The threshold frequency of a metal is 1 ´ 1015 s–1. The ratio
is likely to be most important in determining mp and bp? of maximum kinetic energies of the photoelectrons when
(a) Br2 (b) CO the metal is irradiated with radiations of frequencies
(c) H2S (d) HCl 1.5 ´ 1015 s–1 and 2 ´ 1015 s–1 respectively would be
17. A wedding ring presented to a bride contains 788 mg of (a) 3 : 4 (b) 1 : 2
gold and rest diamond. If the ring weighs 1 g the bride (c) 2 : 1 (d) 4 : 3
receives (At wt : Au = 197, C = 12) 26. If Ka = 10–5 for a weak acid, pKb value of its conjugate base is
(a) more number of gold atoms (a) 5 (b) 6
(b) more number of carbon atoms (c) 7 (d) 9
(c) equal number of gold and carbon atoms 27. In PO43– ion, the formal charge on each O atom and P–O
(d) gold and carbon atoms in ratio of 4:1 approximately bond order are respectively
18. How many electrons are used in bonding in the Lewis (a) –0.75, 0.6 (b) –0.75, 1.0
structure of C2O42– ion (c) –0.75, 1.25 (d) –3, 1.25
(a) 22 (b) 20 28. In which of the following structure, the total number of sigma
(c) 18 (d) 14 bonds are equal to pi bonds?
19. The de Broglie wave length of a particle with mass 1 g and (a) 1, 2-propadiene (b) CO2
velocity 100 m/s is (c) Dicyanoethene (d) None
(a) 6.63 ´ 10–33 m (b) 6.63 ´ 10–34 m 29. The molecular species having highest bond order is
(c) 6.63 ´ 10–35 m (d) 6.65 ´ 10–35 m (a) O2 (b) N2 –
20. An aqueous solution of 6.3 g of oxalic acid dihydrate is (c) N2 (d) O2 2–
made up to 250 ml. The volume of 0.1 N NaOH required to 30. The lower electron affinity of flourine than that of chlorine
completely neutralise 10 ml of this solution is is due to
(a) 20 ml (b) 40 ml (a) Smaller size
(c) 10 ml (d) 4 ml (b) Smaller nuclear charge
21. pKa values of four acids are given below at 25°C. Which (c) Difference in their electronic configurations
value corresponds to the strongest acid? (d) Its highest reactivity
Response Sheet
ANSWER KEY
1 (b) 7 (d) 13 (d) 19 (a) 25 (b)
2 (c) 8 (c) 14 (b) 20 (b) 26 (d)
3 (d) 9 (a) 15 (d) 21 (a) 27 (c)
4 (d) 10 (c) 16 (a) 22 (b) 28 (b)
5 (b) 11 (d) 17 (b) 23 (d) 29 (c)
6 (b) 12 (a) 18 (d) 24 (c) 30 (a)
\ % of C2O4– – in sample
6.3 ´ 1000
20. (b) Normality of oxalic acid = = 0. 4 N
63 ´ 250 0.198 ´100
= = 66%
Using N1V1 = N2V2, we get 0.3
Þ 10 × 0.4 = V × 0.1 = 40 ml.
25. (b) ( K .E.)1 n - n0 (1.5 ´ 1015 - 1 ´ 1015 ) 1
21. (a) Smaller the pKa value, stronger is the acid. = 1 = =
( K .E.)2 n 2 - n0 (2.0 ´ 1015 - 1 ´ 1015 ) 2
22. (b) Strong acids can be used to titrate both strong and
weak bases. 26. (d) Ka = 10–5 = pKa = –log(Ka) = +5
23. (d) In reaction (d) oxidation number changes from + 4 in pKa + pKb = pKw = 14 Þ pKb = 14 – 5 = 9
NO2 to + 3 in HNO2. 27. (c) PO43– is a hybrid of following structures.
24. (c) Redox changes are O O
7+ - ++
Mn + 5e ¾¾
® Mn
P P
2– 4+ O O O O
C2 ¾¾
® 2C + 2e
O O
\ Meq of oxalate ion = Meq. of KMnO4
O O
w 1
´1000 = 90 ´
E 20 P P
88 O O O O
Where E = = 44 O O
--
C 2O 4 2
P-O bond order = 5/4 = 1.25;
w
´ 1000 =
90 Charge on each O = -3/4 = -0.75
\
44 20 28. (b) In CO2 (O=C=O), there are 2 sigma and 2 pi bonds.
90 44 29. (c) Bond orders are O2 = 2, N2 = 3, N2– = 2.5, O22– = 1.0
w= ´ = 0.198 30. (a) Since F has small size and its relatively compact
20 1000
2p-subshell’s electron-electron repulsion, do not allow
Q 0.3 g C2O4– – sample has oxalate ion the addition of an extra electron whereas Cl has bigger
= 0.198 g size than F, allows the addition of an extra electron
more easily.
PART PHYSICAL CHEMISTRY-1
TEST 2
Time : 1 hour
Max. Marks : 120
INSTRUCTIONS
Ø Crack this section in just 1 hour.
Ø This part test contains 30 questions.
Ø For every correct answer you will be awarded +4. There is no negative marking.
Ø Please fill the correct bubbles in the Response Sheet provided at the end of the test.
Ø Do not look at the solutions before attempting the complete test.
. . . Best of Luck
1. Which of the following figures represents the cross-section 4. A liquid can exist only
of an octahedral site ? (a) between boiling and melting point
(b) between triple point and critical temperature.
(c) between melting point and critical temperature
(d) at any temperature above the melting point.
5. The root mean square speed of gas molecules at 25K &
(a) (b) 1.5 ´ 105 Nm–2 is 100.5 ms–1. If the temperature is raised to
100K & pressure to 6.0 ´ 105 Nm–2 , the root mean square
speed becomes
(a) 100.5 ms–1
(b) 201.0 ms–1
(c) 402 ms–1
(d) 1608 ms–1
(c) (d) 6. 4.0 g of argon in a bulb at temperature T K had a pressure of
p. When the bulb was placed in a hotter bath at temperature
of 50° more than the first, 0.8 g of the gas had to be removed
2. The pure crystalline substance on being heated gradually to get the original pressure p. The original temperature is
first forms a turbid liquid at constant temperature and still at (a) 73 K
higher temperature turbidity completely disappears. The (b) 100K
behaviour is a characteristic of substance forming. (c) 200 K
(d) 570K
(a) Allotropic crystals
7. A sample of gas occupies 300dm3 at 27°C and 750mm
(b) Liquid crystals
pressure. What contraction in volume takes place when the
(c) Isomeric crystals
gas is cooled to -33°C at 750mm pressure?
(d) Isomorphous crystals. (a) 240 dm3
3. A crystal of ZnO is heated evolve O2 and forms defect; the (b) 60 dm3
type of defects formed will be (c) 120 dm3
(a) Metal excess (b) Metal deficiency (d) no change in volume
(c) Impurities (d) None of these
C-6 Physical Chemistry-1
8. Consider a 0.1M solution of two solutes A and B. A behaves the partial pressures of A and B are 3.0 and 1.0 atm
as a non-electrolyte while 80% of B dimerises. Which of the respectively and if ‘C’ has mol. wt of 2.0 what is the weight
following statement is correct regarding these solutions? of ‘C’ in g present in the mixture.
(a) The b.pt of A will be less than B (a) 8 (b) 12
(b) The osmotic pressure of B will be more than that of A (c) 3 (d) 6
(c) The freezing point of solution A will be less than that 17. The molecular velocities of two gases at same temperature
of B are u1 and u2, their masses are m1 and m2 respectively.
(d) Boiling points of both solutions will be same. Which of the following expression is correct?
9. The molality of H2SO4 in which mole fraction of water is (a) m1/u12 = m2/u22 (b) m1u1 =m2u2
0.85, is (c) m1/u1 = m2/u2 (d) m1u12 =m2u22
(a) 8.9 (b) 9.8 18. Which of the following statement is correct
(c) 0.98 (d) 0.89 (a) The volume of a given mass of the gas at 0°C and one
10. One litre solution of N/2 HCl is heated in a beaker. It was atm. pressure will be twice its volume at 273° C.
observed that when the volume of solution is reduced to (b) The volume of a given mass of a gas at 0°C and one
600ml, 3.25 g of HCl is lost. Calculate the normality of the atm. pressure will be half of its volume at–136.5°C.
new solution. (c) The mass ratio of the equal volumes of NH3 and H2S
(a) 6.85 (b) 0.685 under similar conditions of temperature and pressure
(c) 0.1043 (d) 6.50 is 1:2
11. The relationship between the values of osmotic pressure of (d) All of these
solutions obtained by dissolving 6.0 gL-1 of CH3COOH 19. If a 0.1 M solution of glucose (Mol. wt 180) and 0.1 molar
(p1) and 7.45 g L-1 of KCl (p2) is solution of urea (Mol. wt. 60) are placed on two sided
(a) p1 < p2 semipermeable membrane to equal heights, then it will be
(b) p1 > p2 correct to say that
(c) p1 = p2 (a) there will be no net movement across the membrane
(d) p1/(p1 +p2 )= p2 /(p1 +p2 ) (b) glucose will flow across the membrane into urea solution
12. A 5% solution of glucose (MW = 180) is isotonic with 3% of (c) urea will flow across the membrane into glucose
X. The molecular weight of substance X is solution
(a) Half of mol. wt. of glucose (d) water will flow from urea solution to glucose solution
(b) 3/5 of mol. wt. of glucose. 20. An element (atomic mass =100g/mol) having bcc structure
(c) 1.66 times of mol. wt. of glucose. has unit cell edge 400pm. The density (in g/cm3) of the
(d) 3/8 times of mol. wt. of glucose. element is
13. If an equimolar solutions of CaCl2 and AlCl3 in water have (a) 10.376 (b) 5.19
boiling point of t1 and t2 respectively then (c) 7.289 (d) 2.144
(a) t1 > t2 (b) t2 + t1 £ 0 21. The edge length of face centered unit cubic cells is 508 pm.
(c) t1 = t2 (d) t1 < t2 If the radius of the cation is 110 pm., the radius of anion is
14. The expression relating degree of dissociation of the weak
(a) 110 pm. (b) 144 pm.
electrolyte AxBy with its van’t Hoff factor is
(c) 618 pm. (d) 398 pm.
i -1 i +1 22. 0.15 g of a substance dissolved in 15 g of a solvent boiled at
(a) a= (b) a=
x + y -1 x + y -1 a temperature higher by 0.216ºC than that of the pure solvent,
find out the molecular weight of the substance (Kb for
x + y -1 x + y -1 solvent is 2.16ºC).
(c) a= (d) a=
i -1 i +1
(a) 1.01 (b) 10.1
15. The molarity of concentrated sulphuric acid (r=1.834
(c) 100 (d) 10
gcm–3) containing 95% of H2SO4 by mass is
(a) 4.44 M (b) 8.88 M 23. Which of the following salt will have same value of van't
(c) 13.32 M (d) 17.78 M Hoff’s factor as that of K4[Fe (CN)6]?
16. A gaseous mixture of three gases A, B and C has a pressure (a) Al2(SO4)3 (b) NaCl
of 10 atm. The total number of moles of all the gases is 10. If (c) Al(NO3)3 (d) Na2SO4
Physical Chemistry-1 C-7
(a) (u rms ) H 2 > (u rms ) CH4 > ( u rms ) NH3 > ( u rms ) CO2
(b) (u rms ) H 2 < ( u rms ) CH 4 < ( u rms ) NH3 < (u rms ) CO2 + a/2
Li
(c) (u rms ) H 2 < ( u rms ) CH 4 > (u rms ) NH3 > (u rms ) CO2 (a) 100pm (b) 200 pm
(c) 141.4pm (d) 282.8pm
(d) (u rms ) H 2 > (u rms ) CH4 < (u rms ) NH3 < ( u rms ) CO2
28. 0.70 gms of sample of Na2CO3.xH2O were dissolved in water
25. Which of the following solution pairs can be separated into and the volume made to 100 ml. 20 ml of this solution required
its pure components by fractional distillation? 19.8 ml of N/10 HCl for complete neutralization. The value of
(a) Water – HCl x is
(a) ~2 (b) ~3
(b) Water – C2H5OH
(c) ~4 (d) ~1
(c) Water – HNO3
29. Heptane and octane form ideal solution. At 373 K, the vapour
(d) Benzene – Toluene pressures of the two liquid components are 105.2kPa and
26. The circulation of blood in the human body supplies oxygen 46.8 kPa, respectively. What will be the vapour pressure, in
and removes CO2 . The concentration of oxygen and carbon bar, of a mixture of 25.0 g of heptane and 35.0 g of octane ?
dioxide is variable but on an average 100 ml of blood contains (a) 1.728 bar (b) 0.218 bar
0.02 g of O2 and 0.08 g of CO2. Calculate the volume of (c) 0.728 bar (d) 0.518 bar
oxygen and carbon dioxide at 1 atm and body temperature 30. The vapour pressure of benzene at a certain temperature is
37°C assuming that there is 10L of blood in the human body. 640 mm of Hg. A non volatile and non electrolyte solid
(a) 2L, 4L (b) 1.5L , 4.5L weighing 2.175 g is added to 39.08 of benzene. If the vapour
pressure of the solution is 600mm of Hg, what is the
(c) 1.6L, 4.58L (d) 3.82L, 4.62L
molecular weight of solid substance?
27. Lithium chloride has a cubic structure as shown below. If
(a) 49.50 (b) 59.60
the edge length is 400 pm., then the radii of Cl– ions is
(c) 69.60 (d) 79.87
Response Sheet
ANSWER KEY
1 (d) 7 (b) 13 (d) 19 (a) 25 (d)
2 (b) 8 (b) 14 (a) 20 (b) 26 (c)
3 (a) 9 (b) 15 (d). 21 (b) 27 (c)
4 (a) 10 (b) 16 (b) 22 (c) 28 (a)
5 (a) 11 (a) 17 (d) 23 (a) 29 (c)
6 (c) 12 (b) 18 (d) 24 (a) 30 (c)
1. (d) Figure (d) represents cross section of an octahedral 9. (b) X1 = 0.85, X2 = 1 – 0.85 = 0.15
site. The interstitial void formed by combination of two 1 Molal solution is the number of moles dissolved per
triangular voids of the first and second layer is called thousand grams of solvent
octahedral layer. \ no. of moles of solvent, n 1 = 1000/ 18 = 55.55 mol
2. (b) Liquid crystals on heating first become turbid and then X2 = n2/(n1 + n2) or 0.15 = n 2/ (55.55 + n2)
clear. On solving, we get n 2 = 9.8
10. (b) 1 L – 1 N HCl = 36.5 g
heat 1
3. (a) ZnO ¾¾¾
® Zn2+ + O2 + 2e - \1L – N/2 HCl = 18.25 g
2
Wt. of HCl lost = 3.25 g
Excess Zn2+ ions are trapped by vacant interstitial sites \Wt. of HCl left = 15.00 g
and the electrons are in neighbouring interstitial sites. Vol. of sol. left = 600 ml.
4. (a) A substance exists as a liquid above its melting point Normality = (15/ 36.5) ´ 1000/ 600 = 0.685 N
and below its boiling point. 11. (a) Effective conc. of KCl in solution > acetic acid conc. It
5. (a) In such a case there is no change in velocity is because KCl is a strong electrolyte and ionises in
u= (3RT/M)= (3PV/M) solution.
12. (b) W1/M1 = W2/M2 or M2= (W2M1)/ W1 = 3 ´ 180/ 5
The increase in temperature = 4 times &
13. (d) Conc. of AlCl3 is more as it gives 4 ions whereas CaCl 2
also the increase in pressure = 4 times. Both of these
gives 3 ions
reinforce each other
14. (a) AxBy xAy+ + yBx–
6. (c) According to ideal gas equation
1– a ax ay
PV = nRT = (W/M) RT
Total amount of species = 1+ (x+y–1)a
For given wt of gas at a particular temp;
(W1/M) RT1 = (W2/M) RT2 or W1T1 = W2T2 1 + (x + y - 1)a
Van’t Hoff factor, i =
Here T1 = TK and T2 = (T + 50)K 1
4T = 3.2 (T + 50) or 0.8T = 160 or T = 200 K
i -1
7. (b) At constant pressure, or a =
x + y -1
V1/T1 = V2/T2; V1 = 300 dm3;
V2 = ?, T1 = 300K, T2 = 273 – 33 = 240 K;
\V2 = (V1/T1).T2 = (300/300) ´ 240 = 240 dm3 (95 g / 98 g mol -1 )
15. (d) Molarity =
\ decrease in volume = 300 – 240 = 60 dm3 (100 g / 1.834 g cm -3 )
8. (b) The concentration of electrolyte, which ionises in water
= 0.01778 mol cm–3 = 17.78 mole dm–3 .
shall be more although 80% of it dimerises
Physical Chemistry-1 C-9
16. (b) Total pressure of mixture of gases A,B & C = 10atm. VO2 = 0.066 ´ 0.0821 ´ 310/ 1 = 1.6L ;
Partial pressure of A = (No. of moles of A ´ 10)/ 10
or (No. of moles of A ´ 10)/ 10 = 3 VCO 2 = 0.18 ´ 0.0821 ´ 310 / 1 = 4.58L
Hence, no. of moles of A = 3
\ Partial pressure of B = (No. of moles of B ´ 10)/ 10 27. (c) For a cube as given, the Cl – ions are at the corners and
(since partial pressure of B = 1atm) one each in the face centre i.e. it is a ccp structure.
\ No.of moles of B = 1 For a ccp structure 4r– = Ö2 a, The face diagonal = Ö2a
Now the no. of moles of C = 10 – ( 3+1) = 6; On the face diagonal there are only Cl – ions
1 mole of C weighs = 2 gm. \ 4r– = Ö2 a or r– = 1.414 ´ 400/4 = 141.4pm
\ 6 mole of C will weigh = 2 ´ 6 = 12
17. (d) u = (3RT/m)1/2 or mu2 = 3RT
at same temperature
m1u12 < m2u22 a/2
18. (d) For (a) and (b) apply V1T2 = V2T1 for (c) 1 mole of NH3
= 17g and 1 mole of H2S = 34 g. Thus mass ratio = 17/34 a/2
= 1:2; For (d) moles ratio of CH4: SO2 = 1/16 :1/64 = 4:1
19. (a) The two solutions are isotonic hence there will be no 28. (a) 100 ml of Sol. contains 0.7 gm of Na 2 CO 3
movement of H2O.
20. (b) For bcc lattice, number of atoms per unit cell = 2 0.7 ´ 1000
1000 ml of Sol. contains =
n´M 2 ´ 100
100
Now d = 3
=
-8
a ´ No (4 ´ 10 cm) 3 ´ 6.02 ´ 10 23
= 7 gm of Na 2 CO 3 xH2O
= 200/38.528 = 5.19 g/cc
21. (b) Edge length = 2r + + 2r–; Strength of Na 2 CO 3 . x H2O = 7 g/litre.
508/2 = r+ + r–: r–= 254 – 110 = 144pm
N
22. (c) M = Kb.WB.1000/ DTb. Now 20 ml of Na 2 CO 3 x H2O Sol. = 19.8 ml of HCl
10
WA = 2.16 ´ 0.15 ´ 1000 / 0.216 ´ 15 = 100
23. (a) K4[Fe(CN)6] dissociates as 4K+ & [Fe(CN)6]4–. Thus 19.8 19.8
Normality of Na 2 CO 3 xH2O Sol. = = N
1 molecule dissociates into five particles. In the same 20 ´10 200
way Al2 (SO4)3 also dissociates as
Al2(SO4)3 ® 2Al3+ + 3SO42– 7 ´ 200
\ Eq.Wt of Na 2 CO 3 xH2O = 19.8 = 70.70
24. (a) u rms = 3RT / M , i.e. u rms a 1 / M
M H 2 < M CH 4 < M NH3 < M CO2 But Eq. Wt. of Na 2 CO 3 xH2O = Mol .Wt .
2
25. (d) Others form azeotropic mixtures
2 ´ 23 + 12 + 3 ´ 16 + x ´ 18
26. (c) Volume of blood in human body = 10 L =
2
Wt. of oxygen in blood = 0.02 ´ 10 ´ 1000 / 100 = 2g
Wt. of CO2 in blood = 0.08 ´ 10 ´ 1000 / 100 = 8 g = 53 + 9x
TEST 3
Time : 1 hour
Max. Marks : 120
INSTRUCTIONS
Ø Crack this section in just 1 hour.
Ø This part test contains 30 questions.
Ø For every correct answer you will be awarded +4. There is no negative marking.
Ø Please fill the correct bubbles in the Response Sheet provided at the end of the test.
Ø Do not look at the solutions before attempting the complete test.
. . . Best of Luck
1. The tendency of a reaction to take place spontaneously is (b) (HCl, CH3COOH2+ ) & (CH3COOH,Cl–)
greatest when the reaction is (c) (CH3COOH2+, HCl) & (Cl–, CH3COOH)
(a) Exothermic and randomness increases (d) (HCl, Cl–) & (CH3COOH2+, CH3COOH)
(b) Endothermic & randomness increases 6. Normality of a mixed solution of sulphuric acid and
(c) Exothermic and randomness decreases hydrochloric acid is 0.6 N. 20 ml of this solution gives 0.4305
(d) Endothermic & randomness decreases g of AgCl on reacting with AgNO 3 solution. The strength
2. A solution of 0.1 m CuSO4 solution is
(a) acidic in nature of H 2SO 4 in g/ l in the mixed solution is
(b) alkaline in nature (a) 42.05 g/ l (b) 22.05 g/ l
(c) neutral in nature (c) 28.56 g/ l (d) 37.05 g/ l
(d) acidic at low temperature, neutral at room temperature 7. Consider the reaction
and alkaline at high temperature
CaCO3(s) CaO(s) + CO2(g) in a closed container at
3. For the expression, rate = k(2a)2 when the concentration of
equilibrium. What would be the effect of addition of CaCO3
reactants is doubled the rate of reaction would undergo
on the equilibrium concentration of CO2 ?
(a) An eight fold increase
(a) Increases (b) Decreases
(b) Six fold increase
(c) Data insufficient (d) Remains unaffected
(c) Two fold increase
8. On increasing the pressure three fold the rate of reaction of
(d) Four fold increase
2H2S + O2 ® Product, would increase
4. In a buffer solution, pH = pKa when the solution contains
(a) [salt] = [acid] (a) 27 times (b) 9 times
(b) [salt > [acid] (c) 3 times (d) 12 times
(c) [salt] < [acid] 9. A reaction rate constant is given by
(d) [salt] + [acid] has a maximum value K = 1.2 ´ 1010 e-2500/RT . It means
5. The following equilibrium is established when hydrogen (a) log K vs T will give a straight line
chloride is dissolved in acetic acid (b) log K vs 1/T gives a straight line with a slope
HCl + CH3COOH Cl- + CH3COOH2+ – 2500/2.303 R
The set that characterises the conjugate acid-base pairs is (c) half life of reaction will be more at higher temperature
(a) (HCl, CH3COOH) & (CH3COOH2+,Cl–) (d) log K vs 1/T gives a straight line with a slope 2500/R
C-12 Physical Chemistry-2
10. A reaction is found to be second order w.r.t. one of the (a) oxidise iodide ion to iodine
reactants & has rate constant of 0.5 mol–1 dm3 min–1. If (b) reduce iodine to iodide ion
initial concentration is 0.2 mol dm–3 then t1/2 of reaction is (c) form a soluble blue complex
(a) 5 min (b) 10 min (d) induce the reaction rate.
(c) 15 min (d) 20 min 20. For a chemical reaction X ® Y, the rate of reaction increases
11. Which of the following expressions is not true ? by a factor of 1.837 when the concentration of X is increased
by 1.5 times. The order of the reaction with respect to X is
(a) For a netural solution [H+] = [OH–] = Kw (a) 1 (b) 1.5
(b) For an acidic solution (c) 2 (d) 2.5
21. The half life of a first order reaction is 10 min. If initial amount
[H+] > K w and [OH - ] < K w is 0.08 mol L-1 and concentration at some instant is 0.01 mol
(c) For an alkaline solution L-1. What is the time elapsed ?
(a) 10 min (b) 20 min
[H+] < K w and[OH - ] > K w (c) 30 min (d) 40 min
(d) For a neutral solution [H+] = [OH–] = 10–7 M at all 22. The dissociation constants of two weak acids are K1 and
temperatures K2. The relative strength of the two acids is given by
12. The solubility of AgCl is 4 × 10–10 at 298K. The solubility of K1 + K 2
AgCl in 0.04 m CaCl2 will be (a) K1/K2 (b)
K 2 + K1
(a) 2 × 10–5 m (b) 5 × 10–9 m
–4 (d) 2.2×10–4 m
(c) 10 m (c) (K1/K2)½ (d) (K1/K2)3/2
13. Taking Ba(OH)2 to be completely ionised. The pH of its 23. At 100°C the gaseous reaction A ® 2B + C is found to be
0.001 M solution is first order. On starting from pure A it is found that at the end
(a) 11.3 (b) 2.7 of 10 min the total pressure of system is 176 mm of Hg. After
(c) 11 (d) 3 a long time, pressure becomes 270 mm Hg. The partial pressure
14. Which of the following is most soluble ? of A at the end of 10 min is (in mm of Hg)
(a) Bi2S3 (Ksp = 1 ´ 10–70) (b) MnS (Ksp = 7 ´ 10–16) (a) 43 (b) 47
(c) CuS (Ksp = 8 ´ 10–37) (d) Ag2S (Ksp = 6 ´ 10–51) (c) 94 (d) 86
15. Concentration of NH4Cl and NH4OH in a buffer solution is 24. The racemisation of pinene is first order reaction. In the gas
in the ratio of 1 : 1, Kb for NH4OH is 10-10. The pH of the phase the specific reaction rate was found to be 2.2. ´ 10–5
buffer is min–1 at 457.6 K and 3.07 ´ 10–3 min–1 at 510.1K. The activation
energy is
(a) 4 (b) 5
(a) ln 2.2 ´ 10 –5 ´ 457.6
(c) 9 (d) 11
(b) 3.048 ´ 10 –3
16. Initially, 0.8 mole of PCl5 and 0.2 mole of PCl3 are mixed in
(c) (510.1 ´ 457.6/52.5) R ln (307/2.2)
one litre vessel. At equilibrium, 0.4 mole of PCl3 is present.
(d) R (510.1 – 457.6) ln (3.07 ´10-3/2.2 ´ 10-5)
The value of Kc for the reaction
25. The volume in ml of 0.1 N HCl solution required to react
PCl5 (g) PCl3 (g) + Cl2 (g) would be
completely with 1.0 g of a mixture of Na 2CO 3 and NaHCO 3
(a) 0.13 mol 1–1 (b) 0.05 mol 1–1
(c) 0.013 mol 1–1 (d) 0.66mol 1–1 containing equimolar amounts of the two compounds
17. If the equilibrium concentration of the components in a (a) 119 ml (b) 158 ml
(c) 122 ml (d) 58 ml
reaction, A + B C + D are 3, 5, 10 and 15 mol L-1 26. The Ksp of CuS, Ag2S and HgS are 10-31, 10-44 and 10-54
respectively then what is DG0 for the reaction at 300 K? respectively. The solubility of the three sulphides are in the
(a) – 1.381 kcal (b) – 600 cal order
(c) – 1140 cal (d) –300 cal (a) Ag2S > CuS > HgS (b) Ag2S > HgS > CuS
18. If Kp for the reaction, (c) HgS > Ag2S > CuS (d) CuS > Ag2S > HgS
A(g) + 2B (g) 3C (g) + D(g), 27. 1.245 g of CuSO 4 . xH 2 O was dissolved in water and H 2S
is 0.05 atm at 1000 K. Its Kc in terms of R will be
was passed till CuS was completely precipitated. The
(a) 20000 R (b) 0.02 R
(c) 5 ´ 10-5 R (d) 5´10-5/R H 2SO 4 produced in the filtrate required 10 ml of N NaOH.
19. In the kinetic study of reaction of iodide ion with hydrogen The value of x is
peroxide, a known volume of sodium thiosulphate solution (a) 7 (b) 5
is added to (c) 8 (d) 6
Physical Chemistry-2 C-13
28. Two HCl solutions A and B have concentrations of 0.5 N of concentration of bicarbonate ion to carbon dioxide?
and 0.1 N respectively. The volume of solutions A and B (a) 3.8 ´ 10–12 (b) 3.8
needed to make 2 L of 0.2 N HCl are (c) 6 (d) 13.4
(a) 1.5 L of A + 0.5 L of B (b) 1.0 L of A + 1.0 L of B 30. The rate constant for the reaction,
(c) 0.75 L of A + 1.25 L of B (d) 0.5 L of A + 1.5L of B
2N2O5 ¾ ¾® 4NO2 + O2,
29. When CO2 dissolves in water, the following equilibrium is
established is 3.0 ´ 10–5sec. If the rate of reaction is 2.40 ´ 10–5
mol sec , then the concentration of N2O5 (in mol lit–1) is
–1 –1
CO2 + 2H2O H3O+ + HCO3–; for which the equilibrium
(a) 1.4 (b) 1.2
constant is 3.8 ´ 10 and pH = 6.0. What would be the ratio
–6
(c) 0.04 (d) 0.8
Response Sheet
ANSWER KEY
1 (a) 7 (d) 13 (a) 19 (d) 25 (b)
2 (a) 8 (a) 14 (b) 20 (b) 26 (a)
3 (d) 9 (b) 15 (a) 21 (c) 27 (b)
4 (a) 10 (b) 16 (a) 22 (c) 28 (d)
5 (d) 11 (d) 17 (a) 23 (b) 29 (b)
6 (b) 12 (b) 18 (d) 24 (c) 30 (d)
1. (a) For exothermic reaction DH = -ve and if randomness 8. (a) Rate = Kp2H 2S ´ pO2 = x(say); on increasing the
increases it means DS becomes positive. Thus pressure three fold,
according to D G = DH - TDS; DG becomes –ve, so the
reaction is spontaneous. Rate = K(3p H2S )2 (3pO2 ) = K ´ 9p 2H 2S ´ 3pO 2
2. (a) Solution is acidic. As per the hydrolysis reaction
Cu2+ + H2O Cu(OH)+ + H+
= K ´ 27 ´ pH 2S ´ pO2 = 27
[salt] 1
pH = pKa + log 10. (b) In general t1/ 2 of reaction µ
[acid] (a 0 )n -1
Obviously, pH = pKa if [salt] = [acid] For a second order reaction,
5. (d) HCl + CH3COOH Cl– + CH3COOH2+ 1 1 1
Acid1 Base1 Conj base Conj acid
t1/ 2 = n -1
= = = 10 min
K(a 0 ) K(a 0 ) 0.5 ´ 0.2
6. (b) Q 143.5 g of AgCl = 36.5 HCl
11. (d) The statement (d) is applicable only at 25º C and is not
36.5 ´ 0.4305 independent of temperature.
0.4305 g AgCl = g HCl
143.5
12. (b) [Cl–] due to CaCl2 = 2 ´ .04 = .08molL -1
36.5 ´ 0.4305 0.4305
geq of HCl = =
143.5 ´ 36.5 143.5 (AgCl ) 4 ´ 10 -10
S AgCl = [Ag + ] = K sp = = 5 × 10–9
[Cl - ] .08
0.4305 1000
Normality of HCl = × = 0.15
143.5 20
13. (a) [OH - ] = 2[Ba(OH) 2 ] = 2 ´ .001 = 2 ´ 10 -3
Normality of H 2SO 4 = 0.6 – 0.15 = 0.45
pOH = - log 2 ´ 10 -3 = 2.7
Strength of H 2SO 4 = 49 × 0.45 = 22.05 g/ l
7. (d) For the given r eaction K = [CO 2 ] because pH = 14 - pOH = 14 - 2.7 = 11.3
concentrations of solid CaCO3 and CaO have been 14. (b) For Bi2S3, S = (1 ´ 10–70/108)1/5 = 3.9 ´ 10–15 M;
taken as unity, since [CaCO3] or [CaO] do not come in For MnS, S = [7 ´ 10–16]1/2 = 2.6 ´ 10–8 M
the expression there will be no effect of addition of For CuS, S = [ 8 ´ 10–37] 1/2 = 8.9 ´ 10–19 M;
CaCO3.
For Ag2O, S = [6 ´10–51/4] = 1.4 ´ 10–17 M
Physical Chemistry-2 C-15
10 ´ 15 2.2 ´ 10-5 Ea é 1 1 ù
K= ; DGo =– 2.303 RT log K or ln = -
17. (a)
3´ 5 3.07 ´ 10 -3 R ë 510.1 457.6 úû
ê
(2 – x) × 0.1 = 0.2×2
its conc. remains constant)
Þ 0.5x + 0.2 – 0.1x = 0.4
Þ 0.4x = 0.2 Þ x = 0.5 éHCO 3- ù
ë û K 3.8 ´ 10 -6
= = =3.8
Vol. of A is 0.5 litres, Vol. of B is 2 – 0.5 = 1.5 litres. [CO 2 ] é H 3 O+ ù 10-6
ë û
é H 3 O+ ù é HCO 3- ù
29. (b) K=ë ûë û As pH = 6.0[H O]+=10–6
3
30. (d) The unit of K (sec-1) indicates that the reaction is of
[CO 2 ][ H 2 O]2 first order
2.4 ´ 10 -5
r = K (N2O5); [N 2 O 5 ] = = 0.8
3 ´ 10 -5
PART PHYSICAL CHEMISTRY-3
TEST 4
Time : 1 hour
Max. Marks : 120
INSTRUCTIONS
Ø Crack this section in just 1 hour.
Ø This part test contains 30 questions.
Ø For every correct answer you will be awarded +4. There is no negative marking.
Ø Please fill the correct bubbles in the Response Sheet provided at the end of the test.
Ø Do not look at the solutions before attempting the complete test.
. . . Best of Luck
1. Which of the following is an example of an isolated system? 8. Which of the following reactions corresponds to the
(a) cold water in a stoppered thermos flask definition of enthalphy of formation ?
(a) C(diamond) + O2(g) ® CO2(g)
(b) cold water in an open thermos flask
(b) C(graphite) + O2(l) ® CO2(g)
(c) a tree in a garden (c) C(graphite) + O2(g) ® CO2(g)
(d) a cup of hot coffee (d) C(g) + O2(g) ® CO2(g)
2. Which solution has highest specific conductance? 9. State which of the following statements is true ?
(a) 0.001N (b) 0.0001 N (a) First law of thermodynamics is not adequate in
(c) 0.1 N (d) 1.0 N predicting the direction of the process .
3. During the preparation of egg albumin sol. (lyophilic sol.) (b) In an exothermic reaction ,the total enthalpy of the
water taken should be cold because products is greater than that of reactants .
(c) The standard enthalpy of diamond is zero at 298K and
(a) in cold water precipitation of egg albumin takes place 1 atm pressure .
which helps in preparation of sol. (d) It is possible to calculate the value of D H for the
(b) in cold water precipitation of egg albumin does not reaction H2(g) + Br2(l) ® 2HBr(g) from the bond
take place. enthalpy data.
(c) in cold water egg albumin mix well whereas yolk get 10. The molar specific heat of air at room temperature and 1atm
separated. pressure is 25 J K–1mol–1. How much heat is required to
(d) cold water is a purest form of water. heat the room through 10 Kelvin at room temperature if 144g
4. The standard reduction potential of of air is present in room. Vapour density of air = 14.4.
Li+/Li, Ba2+/Ba, Na+/Na and Mg2+/Mg are –3.05, –2.73, – (a) 1250 kJ (b) 25 J
2.71 and –2.37 volts respectively. Which one of the following (c) 50 J (d) 1250 J
is strongest oxidising agent? 11. Heat of dissociation of benzene to elements is 5535 kJ mol–1.
(a) Na+ (b) Li+ The bond enthalpies of C–C, C=C, and C–H are 347.3, 615.0
(c) Ba 2+ (d) Mg2+ and 416.2 kJ respectively. Resonance energy of benzene is
(a) 1.51 kJ (b) 15.1 kJ
5. How many coulombs are required for the oxidation of 1 mol. (c) 151 kJ (d) 1511 kJ
of H2O2 to O2 ?
12. Lattice energy of certain univalent solid AX is 180kJmol-1.
(a) 93000C (b) 1.93 × 105C
4 The dissolution of solid is endothermic to the extent of 1 kJ
(c) 9.65 × 10 C (d) 19.3×102C
6. The phenomenon of syneresis is mole. If DHHydration of A+ and X- are in the ratio of
(a) separation of the dispersed phase from the gel 6 : 5; DHHyd of A+ ion is
(b) formation of a sol from a gel (a) + 98 kJ mol-1 (b) –98 kJmol-1
(c) + 100 kJ mol -1 (d) –80 kJmol-1
(c) migration of colloid in an electric field
(d) process of converting gel into a true solution. 13. A gas expands from 1.5 to 6.5 L against a constant pressure
7. At CMC the surfactant molecule of 0.5 atm and during this process the gas also absorbs 100J
(a) decomposes (b) becomes completelysoluble of heat. The change in the internal energy of the gas is
(c) associates (d) dissociate (a) 153.3 J (b) 353.3 J
(c) – 153.3 J (d) –353.3 J
C-18 Physical Chemistry-3
Response Sheet
1. (a) Isolated system cannot exchange heat and matter. 13. (c) DU = q + w = 100 J – (0.5atm) (6.5 L –1.5L)
2. (d) Specific conductance increases on increasing the
æ 8.314J ö
concentration. = 100 J – (2.5 atm L) çç 0.082 atm L ÷÷ = – 153.3 J
3. (b) During the formation of egg albumin sol hot water is è ø
14. (d) 2+ –
Fe + 2e ® Fe
not used because in hot water precipitation of egg
albumin takes place whereas in cold water formation of DG1° = –nFE° = – 2F (– 0.44) = 0.88F
precipitate does not occur. Fe+3 + 3e– ® Fe
4. (d) The strongest oxidising agent is one which has D G °2 = –nF E°2 = –3F (–0.036) = 0.108F
maximum tendency to gain electrons, i.e. whose E°Red Subtracting (ii) from (i) we get Fe3+ + e– ® Fe2+;
is maximum DG° = –nFE° = – 0.772 or E° = –0.772/1 = –0.772V
5. (b) H2O2 ® O20 + 2e– 15. (b) Smaller the atom, larger will be the bond dissociation
Oxidation of 1 mol of H2O2 to O2 requires 2F of energy because there would be more effective overlap.
electricity. 16. (b) x/m = kp1/n, Taking log
6. (a) Syneresis is defined as the shrinkage of gels on Log x/m = log k + (1/n) log p. \ slope = 1/n
standing by exudation of solvent. 17. (a) Fe(OH)3 is positive sol. K3[Fe(CN)6] will provide
[Fe(CN)6]4– for coagulation.
7. (c) CMC is critical micellization concentration, the
18. (c) DS Total = DS system + DS surrounding
surfactants at this concentration associate to form
micelles. For soaps, the CMC value is 10-3 mole/litre. 12 12
=– + = – 0.0375 + 0.400 = + 0.0025JK–1
8. (c) C(graphite) is the stable state of aggregation of carbon. 320 300
9. (a) The statements given under (b), (c) are not correct 19. (c) The reduction potential of A is more than B. Hence A
because statement (b) refers to an endothermic reaction will not be able to displace B from its solution. Hence
and the standard state for carbon is graphite under (c). no reaction occurs.
For the calculation of DH of the reaction under (d), 20. (b) The ions given by electrolytic impurities will run quickly
additional data in the heat of vapourisation of Br 2 (l) is under the applied electric field and hence will be remove
necessary. easily.
10. (d) Cp = q/T ; 21. (a) The equivalent conductance increase with dilution,
Heat required = No. of moles × Cp×T since inter ionic alterations decrease with dilution
= (144 × 25 × 10)/(14.4 × 2) = 1250J 22. (c) 1000 ml of solution contains .1 mole of CO2;
11. (c) Experimental heat of atomization of benzene 0.1
= 5535 kJ / mole. 200 ml of solution contains = ´ 200
1000
Theoretical heat of atomization (breaking of 3 C–C
bonds, 3C = C bonds and 6C – 4 bonds) = 0.02 mole CO2
= 3× 347.3 + 3 ´ 615.0 + 6 ´ 416.2 = 5384.1; We know that,
Hence resonance energy = 5535 – 5384.1= 151kJ Vol. of 1mole of CO2 at STP = 22.4 litres;
12. (b) Let DH hydration of A+ is –a kJ. Vol of 0.02 mole of CO2at STP = 0.448 litres
Now DHsol = DHlattice + DHhyd of A+ + DHhyd of X– 23. (a) Lesser the gold number better the protective colloid
or 1 kJ = 180 kJ – a–(5a)/6 24. (a) L eq = k × 1000/N Now, k = 1.5/250,
or –11a/6 = –179, a = 97.63
L
\ eq = (1.5/250) ×1000/0.1= 60W–1 cm2 eq–1
\ D Hhydration of A+ = –97.63 @ –98 kJ mol-1
C-20 Physical Chemistry-3
TEST 5
Time : 1 hour
Max. Marks : 120
INSTRUCTIONS
Ø Crack this section in just 1 hour.
Ø This part test contains 30 questions.
Ø For every correct answer you will be awarded +4. There is no negative marking.
Ø Please fill the correct bubbles in the Response Sheet provided at the end of the test.
Ø Do not look at the solutions before attempting the complete test.
. . . Best of Luck
1. What is not true for ozone? 6. A mixture of which pair of species react with water to produce
(a) The two O–O bond lengths are not equal a pure colourless gas that gives white fumes with conc. HCl :
(b) O–O bond order is between 1 and 2 (a) Calcium hydride, calcium carbide
(c) O–O–O angle is approximately 116º (b) Calcium carbide and aluminium carbide
(d) It is light blue gas with pungent odour
(c) Magnesium nitride and calcium nitride
2. Tin plague means-
(d) Calcium phosphide and calcium cyanamide
(a) Tin plating
(b) Conversion of white tin to grey tin 7. Identify the feasible reaction among the following:
(c) Emission of sound while bending a tin plate D
(a) K 2 CO3 ¾¾
® K 2 O + CO 2
(d) Conversion of stannous salts to stannic salts.
D
3. When FeCr2O4 (chromite) is reduced with ‘C’ in an electric (b) Na 2 CO 3 ¾¾® Na 2 O + CO 2
arc furnace : (c) D
Li 2 CO 3 ¾¾
® Li 2 O + CO 2
(a) Cr and Fe2O3 are formed
(b) Fe and Cr 2O3 are formed (d) Rb 2 CO 3 ¾¾
® Rb 2 O + CO 2
(c) Fe and Cr (Ferrochrome) are formed 8. Which of the following shows the tendency to form
(d) FeCrO4 is formed peroxide?
4. Removal of Fe, Cu, W from Sn metal after smelting is by (a) Lithium (b) Magnesium
............... because .............
(c) Beryllium (d) Radium
(a) Poling; of more affinity towards oxygen for impurities
Z LiH
(b) Selective oxidation; of more affinity towards oxygen 9. B X Y + LiBF4
for impurities
Heat
(c) Electrolytic refining; impurities undissolved in electrolyte
(d) Liquation; Sn having lowmelting point compared to impurities. Which of the statement is true for the above sequence of
reactions?
5. Aluminothermy used for on the spot welding of large iron
(a) Z is hydrogen
structure is based on the fact that-
(b) X is B2H6
(a) As compared to iron, aluminium has greater affinity for (c) Z and Y are F2 and B2H6 respectively
oxygen.
(d) Z is potassium hydroxide
(b) As compared to aluminium, iron has greater affinity for 10. Chlorine bleaches by
oxygen. (a) Oxidation (b) Reduction
(c) Reaction between aluminium and oxygen is endothermic. (c) Neutralization (d) Both (a)&(b)
(d) Reaction between iron and oxygen is endothermic.
C-22 Inorganic Chemistry-1
11. Among KO 2 , AlO-2 , BaO2 and NO +2 , unpaired 22. Which factor is responsible for the greater reducing
character of lithium ?
electron is present in (a) Its greater hydration energy
(a) NO +2 and BaO 2 (b) KO 2 and AlO-2 (b) Its greater ionisation energy
(c) KO 2 only (d) BaO 2 only (c) Its greater sublimation energy
12. Which pair of elements do not react? (d) Its greater electron affinity.
(a) P4 and S8 (b) P4 and Cl2 23. Among the reactions given below for B2H6, the one which
(c) P4 and N2 (d) P4 & O2 does not take place is
13. F2 reacts with cold water to give
(a) O2 (b) O3 (a) B2H6 + HCl ¾¾ ® B2H5Cl + H2
(c) OF2 (d) HOF (b) B2H6 + 2NH3 ¾¾D
® B3N3H6 (borazine)
14. How many P–H and O–H bonds respectively, are present in (c) B2H6 + 2N(CH3)3 ¾¾® 2(CH3)3 NBH3
H4P2O7 molecule?
(a) 1 and 3 (b) 0 and 4 (d) B2H6 + 6C2H4 ¾¾¾® H3O+ 3C2H5OH + 2B(OH)3
(c) 4 and 0 (d) 2 and 3 24. Which of the following does not occur in free state ?
15. Point out the incorrect statment among the following : (a) Pt (b) Os
(a) The oxidation state of oxygen is +2 in OF2. (c) Au (d) Cu
(b) Acidic character follows the order
H2O < H2S < H2Se < H2Te. 25. Which of the metal is extracted by Hall-Heroult process?
(c) The tendency to form multiple bonds increases in (a) Al (b) Cu
moving down the group from sulphur to tellurium (c) Ni (d) Zn
(towards C and N) 26. Basicity of orthoboric acid (H3BO3) is
(d) Sulphur has a strong tendency to catenate while (a) One (b) Two
oxygen shows this tendency to a limited extent. (c) Three (d) Zero
16. NaOCl on heating gives 27. The correct statement among the following is :
(a) NaClO2 (b) NaClO3 (a) The alkali metals when strongly heated in oxygen form
(c) NaClO4 (d) Cl2O superoxides.
17. A metal X on heating in nitrogen gas gives Y.Y on treatment (b) Caesium is used in photoelectric cells.
with H2O gives a colourless gas which when passed through
CuSO4 solution gives a blue colour. Y is (c) NaHCO3 is more soluble in water than KHCO3.
(a) Mg(NO3)2 (b) Mg3 N2 (d) The size of hydrated ions of alkali metals increases
(c) NH3 (d) MgO from top to bottom.
28. Which of the following statements, about the advantage of
18. Sodium hydroxide solution reacts with phosphorus to give
roasting of sulphide ore before reduction is not true?
phosphine. To bring about this reaction we need
(a) white phosphorus and dil. NaOH (a) The DG of of the sulphide is greater than those for CS2
(b) white phosphorus and conc. NaOH and H2S.
(c) red phosphorus and dil. NaOH (b) The DG of is negative for roasting of sulphide ore to oxide.
(d) red phosphorus and conc. NaOH (c) Roasting of the sulphide to the oxide is
19. Pure NaCl is prepared by saturating a solution of common thermodynamically feasible.
salt in water by bubbling HCl gas. The principle used is (d) Carbon and hydrogen are suitable reducing agents for
metal sulphides.
(a) Lechatlier principle (b) Displacement law 29. Diborane has
(c) Fractional distillation (d) Common ion effect (a) 2 two - centre -two - electron bonds and 4 three - centre
20. When NaOH pallets are left-open in air, they acquire a fluid - three - electron bonds.
layer around each pallet because (b) 4 two - centre - two - electron bonds and 2 three -
(a) they start melting centre - three - electron bonds
(b) they absorb moisture from air (c) 3 two - centre - two - electron bonds and 3 three -
(c) they absorb CO2 from air centre - three - electron bonds.
(d) they react with air to form a liquid compound (d) all six identical bonds.
30. Arrange CCl4, MgCl2, AlCl3, PCl5 and SiCl4 in increasing
21. Which of the following statements is (are) true ? extent of hydrolysis.
(a) Claude’s apparatus can be used for the isolation of (a) CCl4 < MgCl2 < AlCl3 < SiCl4 < PCl5
helium and argon
(b) SiCl4<PCl5<AlCl3<MgCl2<CCl4
(b) Helium cannot be adsorbed by coconut charcoal
(c) CaC2 can remove both Nitrogen & Oxygen (c) MgCl2 < AlCl3 < CCl4 < SiCl4 < PCl5
(d) All of these (d) AlCl3<SiCl4<CCl4<PCl5<MgCl2
Response Sheet
ANSWER KEY
1 (a) 7 (c) 13 (a) 19 (d) 25 (a)
2 (b) 8 (d) 14 (b) 20 (b) 26 (a)
3 (c) 9 (c) 15 (c) 21 (d) 27 (b)
4 (d) 10 (a) 16 (b) 22 (a) 28 (d)
5 (a) 11 (c) 17 (b) 23 (d) 29 (b)
6 (c) 12 (c) 18 (b) 24 (d) 30 (a)
10. (a) Moist chlorine gives oxygen hence it can act as an 19. (d) When HCl gas is passed through a common salt
oxidising and bleaching agent. solution pure NaCl precipitates due to common ion
effect.
Cl 2 + H 2 O ¾
¾® HCl + HClO
NaCl (aq.) Na+ + Cl–
HClO ¾
¾® HCl + [O] HCl H+ + Cl–
Hypochloro us acid
Common ion
11. (c) In NO2 + odd (unpaired) electron is removed. In
[Na+] × [Cl–] > Ksp of NaCl, hence precipitates out.
peroxides (O22–) no unpaired electrons are persent as
20. (b) NaOH is deliquescent.
the antibonding pi M.O.’s acquired one more electron
21. (d) All the three are true. A & B need no explanation. For C,
each for pairing. AlO-2 containing Al 3+ (2s 2 p6 ) CaC 2 + N 2 ¾
¾® CaCN 2 +C
configuration and 2 oxides (O2–) ions each of which Calcium cyanamide
does not contain unpaired electron. Superoxide O2– ¾® 2 CO ; 2CO + O 2 ¾
2C + O 2 ¾ ¾® 2CO 2 ;
has one unpaired electron in pi antibonding M.O.
C + O2 ¾
¾® CO 2
12. (c) Phosphorus and nitrogen do not react.
22. (a) Alkali metals are strong reducing agents
1
13. (a) F2 + H 2 O ¾
¾® 2HF + O2 because of
2
C-24 Inorganic Chemistry-1
(i) Low heat of atomization • Cesium used in photoelectricells due to its low
(ii) Low ionisation potential I.E. Hence statements (b) is the only choice correct.
(iii) Large atomic size 28. (d) The sulphide ore is roasted to oxide before reduction
The reducing power of alkali metals increases from Li because the DG of of most of the sulphides are greater
to Cs in gaseous state.
than those of CS2 and H2S, therefore neither C nor H
Anomalous Behaviour of Li : Although Li has the
can reduce metal sulphide to metal. Further, the standard
highest I.E., yet in aqueous solution, Li is the strongest free energies of formation of oxide are much less than
reducing agent because of large hydration energy of those of SO2. Hence oxidation of metal sulphides to
its cation. metal oxide is thermodynamically favourable.
23. (d) Reaction between diborane and alkene are carried out 29. (b) Structure of B2H6 is as follows:
in dry ether under an atmosphere of N2 because B2H6
and the products are very reactive. The products Hb
further treated with alkaline H2O2 to convert into Ht •• Ht
alcohols.
B B
alkaline
B2 H6 + 6C2 H 4 ¾¾
® B(C2 H4 )3 ¾¾¾¾
®
reactive
H 2O 2 Ht •• Ht
Hb
3CH3CH 2 OH + H3BO3
24. (d) Cu is only metal among the given options which does Ht Ht
Hb
not found in free state. It is found in the form of oxide
97°
ores (Cu2 O), carbonate ores (Melachite green), B B 121.5°
or
sulphide ores (copper glance) and chalcopyrites 1.33Å
1.19 Å
(CuFeS2). Ht Hb
Ht
25. (a) Hall-Heroult process is used in extraction of Al from 1.77Å
alumina. Thus the diborane molecule has four two-centre-two-
26. (a) H 3 BO 3 + H 2 O ¾¾®[ B(OH ) 4 ] - + H + electron bonds (2c–2e bonds) also called usual bonds
and two three-centre-two-electron bonds (3c–2e) also
H3BO3 gives only one H+ ion per molecule. called banana bonds. Hydrogen attached to usual and
27. (b) • Li does not form peroxide or superoxide due to it 8 banana bonds are called Ht (terminal H) and Hb
small size. (bridged H) respectively.
• Solubility of carbonates and biocarbonates 30. (a) CCl 4 having no vacant d-orbital hence does not
increases on moving down the group. undergo hydrolysis.
• The increasing order of size of hydrated ions of In a particular period, on going left to right the order of
alkalimetals is Li+ > Na+ > K+ > Rb+ Cs+ extent of hydrolysis increases.
PART INORGANIC CHEMISTRY-2
TEST 6
Time : 1 hour
Max. Marks : 120
INSTRUCTIONS
Ø Crack this section in just 1 hour.
Ø This part test containfs 30 questions.
Ø For every correct answer you will be awarded +4. There is no negative marking.
Ø Please fill the correct bubbles in the Response Sheet provided at the end of the test.
Ø Do not look at the solutions before attempting the complete test.
. . . Best of Luck
1. Which reagent can be used to identify nickel ion? (a) MX4 (b) Ma 3 b
(a) Resorcinol (b) Dimethyl glyoxime (c) M(aa’)2 (d) M(aa)2
(c) Diphenyl benzidine (d) Potassium ferrocyanide 11. The net bonding in a metal carbonyl is
2. Which of the following is a high-spin (spin-free) complex? M = C = O. This is due to -
(a) [Co(NH3)6]3+ (b) [Fe(CN)6]4– (a) The overlap of px orbital of the metal with the 2py
(c) [CoF6] 3– (d) [Zn(NH3)6]2+ orbital of carbon
3. In which of the following cases, the stability of two oxidation (b) Back bonding
states is correctly represented (c) The original M ® C s bond is very strong
(a) Ti3+ > Ti4+ (b) Mn2+ > Mn3+ (d) The C – O bond order remaining unaltered
2+
(c) Fe > Fe 3+ (d) Cu+ > Cu2+ 12. A white sodium salt dissolves readily in water to give a solution
4. Amongst the following the compound that is both which is neutral to litmus. When silver nitrate solution is added
paramagnetic and coloured is to the solution, a white precipitate is obtained which does not
(a) K2Cr2O7 (b) (NH4)2[TiCl6] dissolve in dil. HNO3. The anion could be
(c) CoSO4 (d) K3[Cu(CN)4] (a) CO32– (b) Cl– (c) SO42– (d) S2–
5. Which one has been tried as anticancer drug? 13. Of Cr (VI) as Cr2 O72 - and Cr O 24 - , which is better oxidising
(a) cis-dichlorodiammine platinum (II)
agent?
(b) K4[Fe(CN)6]
(a) CrO42–, basic medium (b) Cr2O72–, basic medium
(c) tris-(ethylenediamine) cobalt (III)
(d) [Ag(NH3)2](NH3)2] (c) Cr2O72–, acidic medium (d) CrO3, basic medium
6. The donor atoms in EDTA are 14. Which statement is correct?
(a) Two N and two O (b) Two N and four O (a) Fe2+ gives brown colour with ammonium thiocyanate
(c) Four N and two O (d) Three N and three O (b) Fe2+ gives blue precipitate with potassium ferricyanide
7. Which of the following can be termed as a mixed complex? (c) Fe3+ gives brown colour with potassium ferricyanide
(a) K4 [Fe(CN)6] (b) [Cu(NH3)4] SO4 (d) Fe3+ gives red colour with potassium ferrocyanide
(c) [Co(NH3)4NO2Cl] Cl (d) K2FeO4 15. The complexes of Nickel (II) can be
8. Which of the following reacts with AgNO3 to form a white (a) Square planar, tetrahedral and octahedral
ppt, which then changes to black colour? (b) Square planar and octahedral
(a) NaCNS (b) Na2S2O3(c) NaCl (d) Na2S (c) Tetrahedral and octahedral
9. The co-ordination number of copper in the complex formed (d) Square planar only
by adding excess of NH3 to CuSO4 solution is 16. Which of the following reagents can one use to distinguish
(a) 4 (b) 2 (c) 6 (d) 5 between Na2CO3 and Na2SO3?
10. Which one of the following square planar complexes will be (a) Lime-water (b) Baryta water
able to show geometrical isomerism? (c) Acidified K2Cr2O7 solution (d) H2SO4 solution
C-26 Inorganic Chemistry-2
17. The reagent that can distinguish between silver and lead salt is 23. Schwitzer’s reagent has the composition of
(a) H2S gas (a) [Ag (NH3)2] Cl (b) Fe4 [Fe(CN)6]3
(b) Hot dilute HCl solution (c) [Cu (NH3)4] SO4 (d) K2 [Hgl4]
(c) NH4Cl (solid) + NH4OH (solution) 24. Potassium permanganate acts as an oxidant in neutral,
(d) NH4Cl (solid) + (NH4)2CO3 solution alkaline as well as acidic media. The final products obtained
18. Which one of the following statements concerning from it in the three conditions are, respectively
lanthanide elements is false? (a) MnO42–, Mn3+ and Mn2+
(a) Lanthanides are seperated from one another by ion (b) MnO2, MnO2 and Mn2+
exchange methods. (c) MnO2, MnO2+ and Mn3+
(b) The ionic radii of trivalent lanthanides steadily increase (d) MnO, MnO2 and Mn2+
with increase in atomic number. 25. Which of the following statements is not true?
(c) All lanthanides are highly dense metals. (a) An acidified solution of K2Cr2O7 liberates iodine from
(d) Most characteristic oxidation state of lanthanides iodides
elements is + 3. (b) In acidic solution dichromate ions are converted to
19. CuSO4 reacts with an excess of hypo solution to produce a chromate ions
CuI complex of (c) (NH4 ) 2 Cr 2 O 7 on heating undergo exothermic
(a) Cu2S2O3 (b) Na3[Cu(S2O3)2] decomposition to give Cr2O3
(c) Na4[Cu6(S2O3)5] (d) Cu4[Na6(S2O3)5] (d) Potassium dichromate is used as a titrant for estimation
20. Which of the following is not an actinide? of Fe2+ ions
(a) Curium (b) Californium 26. The hybridization states of the central atoms in the
(c) Uranium (d) Terbium complexes [Fe(CN)6]3–, [Fe(CN)6]4– and [Co(NO2)6]3– are
21. Which of the following ions are optically active? (a) d2sp3, sp3 and d4s2 respectively
(b) d2sp3, sp3d and sp3d2 respectively
en en (c) d2sp3, sp3d2 and dsp2 respectively
Cl + Cl +
(d) all d2sp3
27. Which of the following has a shape different from others?
Co Co (a) [Zn(NH3)4]+2 (b) Ni(CO)4
(c) [Cd(CN)4]-2 (d) [Cu(NH3)4]+2
Cl Cl 28. Which one of the following is an example of co-ordination
en en isomerism ?
I II (a) [Co (NH3)5 Br] SO4 and [Co (NH3)5 SO4] Br
en (b) [Co (NH3)5 NO2] Cl2 and [Co (NH3)5 ONO]Cl2
3+ en + (c) [Cr (H2O)6] Cl3 and [Cr (H2O)5 Cl] Cl2 H2O
Cl
(d) [Co (NH3)6] [Cr (CN)6] and (Cr (NH3)6] [Co (CN)6]
en Co Co 29. A solution containing As3+, Cd2+, Ni2+ and Zn2+ is made
alkaline with dilute NH4OH and treated with H2S. The
precipitate obtained will consist of
Cl
en en (a) As2S3 and CdS (b) CdS, NiS and ZnS
III IV (c) NiS and ZnS (d) Sulphide of all ions
(a) I only (b) II only 30. Predict the order of Do for the following compounds :
(c) II and III (d) IV only I. [ Fe ( H 2 O ) 6 ] 2 + II. [Fe (CN ) 2 (H 2 O) 4 ]
22. Aqueous solution of a gas ‘X’ is treated with hydrogen 2-
III. [Fe(CN) 4 ( H 2 O) 2 ]
peroxide and thereafter allowed to react with barium chloride
solution. A white precipitate appears which is insoluble in (a) D o ( I) < D o ( II ) < D o ( III )
conc. HNO3. The gas ‘X’ is (b) D o ( II) < D o ( I ) < D o ( III )
(a) H2S (b) CO (c) D o (III) < D o (II) < D o (I)
(c) SO2 (d) CO2 (d) D o (II) < D o (III) < D o (I)
Response Sheet
ANSWER KEY
1 (b) 7 (c) 13 (c) 19 (c) 25 (b)
2 (c) 8 (b) 14 (b) 20 (d) 26 (d)
3 (b) 9 (a) 15 (a) 21 (c) 27 (d)
4 (c) 10 (c) 16 (c) 22 (c) 28 (d)
5 (a) 11 (b) 17 (b) 23 (c) 29 (d)
6 (b) 12 (b) 18 (b) 24 (b) 30 (a)
1. (b) Nickel salts reacts with dimethyl glyoxime in presence 7. (c) By definition, a mixed complex contains more than one
of NH4OH to give red ppt. of nickel dimethyl glyoxime. type of ligands.
CH3 – C = NOH 8. (b) 2AgNO 3 + Na 2S2 O 3 ¾
¾® Ag 2S2 O 3 + 2NaNO 3
2 + NiCl2 + 2NH4OH white
CH3 – C = NOH
Ag 2S2O 3 + H 2 O ¾
¾® Ag 2S + H 2SO 4
OH O black
INSTRUCTIONS
Ø Crack this section in just 1 hour.
Ø This part test contains 30 questions.
Ø For every correct answer you will be awarded +4. There is no negative marking.
Ø Please fill the correct bubbles in the Response Sheet provided at the end of the test.
Ø Do not look at the solutions before attempting the complete test.
. . . Best of Luck
1. The I.U.P.A.C. name of CH 2 - CH - C H 2 is 6. An organic compound is treated with NaNO 2 and dil. HCl
| | |
CN CN CN at 0°C. The resulting solution is added to an alkaline solution
(a) 1, 2, 3-propanetrinitrile of b -naphthol where by a brilliant red dye is produced. It
(b) 1, 2, 3 – tricyanopropane shows the presence of
(c) 3-cyano-1, 5-dinitrilepentane (a) – NO 2 group (b) aromatic – NH 2 group
(d) 3-cyano – 1, 5-pentanedinitrile
2. Which of the following compounds exhibit tautomerism? (c) – CONH 2 group (d) aliphatic – NH 2 group
(a) Chloroethane (b) Ethanol 7. Positive Beilstein test for halogens shows that
(c) Ethoxyethane (d) Nitreothane (a) a halogen is definitely present
3. Which of the following will not give lassaigne test for (b) a halogen may be present
nitrogen? (c) a halogen is absent
(a) NH2–NH2 (b) C6H5–N=N–C6H5 (d) none of these.
(c) CH3CONH2 (d) CH3C º N
4. During electrophilic substitution in benzene, the CºCH
intermediate species involved is
(a) carbocation (b) carbanion 8. H SO
+ H 2O ¾¾2 ¾¾
4 ®' X' , ' X'
(c) free radical (d) none of these HgSO 4
5. During steam distillation of a mixture of o-nitrophenol and
p-nitrophenol (a) Acetophenone (b) Benzophenone
(a) vapours of o-nitrophenol are carried by steam because (c) Benzaldehyde (d) Phenyl acetaldehyde
of its lower boiling point due to chelation. 9. Which of the following structures does not contain any
(b) vapours of o-nitrophenols are carried by steam because chiral C atom but represent the chirality in the structure.
of its lower boiling point and solubility in steam. (a) 2 – Ethyl – 3 – hexene (b) 2, 3-Pentadiene
(c) vapours of o-nitrophenol are carried by steam because (c) 1,3 – Butadiene (d) Pent – 3 – en – 1 – yne
its boiling point is reduced by steam. 10. Sodium adipate on electrolysis gives
(d) vapours of p-nitrophenol are carried by steam because (a) cyclobutane (b) 2-butene
of its lower boiling point. (c) propene (d) cyclopropane
C-30 Organic Chemistry-1
11. An organic compound ‘X’ on ozonolysis followed by (a) C6H5CH2Br (b) CH3Br
O (c) (CH3)2CHBr (d) (CH3)3CBr
||
reduction with Zn/H2O gives 2 moles of H - C - H and 19. 2-methyl-2-butene on oxidation with acidic KMnO4 gives
(a) acetone (b) acetic acid
O O (c) both (a)& (b) (d) CO2 & H2O
|| ||
H - C - CH 2 - C - H . ‘X’ is
(a) CH2 = CH – CH2 – CH = CH2 1.O / H O
¾¾ 3¾ ¾2¾® A
20.
(b) CH2 = CH–CH2–CH2–CH = CH2 2.D
(c) H – C º C – C º C – H
(d) CH2 = CH – CH = CH2 (a) CH3COOH (b) CHCOOH
12. Which of the following reagents will not yield alkane ? ||
CHCOOH
HI / Red P
(a) C 2 H 5 OH ¾¾ ¾ ¾
¾® (c) C6H5COOH (d) C6H4(COOH)2
150º C
(b) Refluxing butanone with Zn/Hg–HCl 21. A and B in the reaction,
(c) Hydroxylation of ethyne
Mg 1.HCHO
(d) Electrolysis of sodium propionate solution CH 3 CH 2 CH 2 Cl ¾¾¾® A ¾¾ ¾¾® B
ether
13. To detect iodine in presence of bromine, the sodium extract 2.H 3O +
is treated with NaNO 2 + glacial acetic acid + CCl 4 . Iodine (a) CH3CH2CH2MgCl, CH3CH2CH2CH2OH
is detected by the appearance of CH3CH2
(a) yellow colour of CCl4 layer (b) CH3CHMgCH2Cl, CHOH
H3C
(b) purple colour of CCl4
(c) brown colour in the organic layer of CCl4 H3C
(d) deep blue colour in CCl4
14. Dehalogenation of vicinal dihalides with Zn/alc. mainly produces (c) CH3CH2CH2MgCl, H3C – C – OH
(a) alcohol (b) alkene (c) alkyne (d) alkane H3C
15. Addition of HCl to vinyl ch lor ide gives
1, 1-dichloroethane because of (d) CH3CH2CH2MgCl, CH3CH2OCH2CH3
(a) inductive effect of Cl (b) mesomeric effect of Cl 22. Which one of the following is likely to give a precipitate
(c) restricted rotation around double bond with AgNO3 solution?
(d) none of these (a) CCl4 (b) CH3–CH2–Cl
(c) (CH3)3CCl (d) CHCl3
CH2 – CH3 23. The intermediate during the addition of HCl to propene in
the presence of peroxide is
16. Sun light
+ Br2 ¾¾ ¾¾® X ¾¾¾® Y .
KCN · Å
(a) CH 3 CHCH 2 Cl (b) CH 3 CHCH 3
Y is · Å
(c) CH 3 CH 2 CH 2 (d) CH3CH 2 CH 2
CN
CH2 – CH3 CH2 – CH2CN
24. The reaction of CH3 CH = CH OH with HBr
C º CH C º CCH2CH3 (c) H 3C - CH 2 :- = H 2 C = CH :- = H - C º C :-
(d) None of these
(b) A = ;B= 29. During the preparation of acetanilide from aniline a small
amount of zinc is added to the reaction mixture because.
(a) zinc induces the precipitation.
C º CH CH2CH2Cº CH
(b) zinc prevents the reduction of aniline during the reaction.
(c) zinc reduces the coloured impurities in the aniline and
(c) A= ;B= also prevents its ocidation during the reaction.
(d) zinc form a white crytstlline complex with aniline.
(d) None of these 30. Which reaction sequence would be best to prepare
26. An analysis of organic compound gave 74.0% C, 8.65% of 3-chloroaniline from benzene ?
H and 17.3% N. What is the empirical formula of compound? (a) Chlorination, nitration, reduction
(a) C5H8N (b) C10H12N (c) C5H7N (d) C10H14 N (b) Nitration, chlorination, reduction
27. An organic compound A (C4H10O) has two enantiomeric (c) Nitration, reduction, chlorination
forms and on dehydration it gives B(major product) and C
(d) Nitration, reduction, acylation, chlorination, hydrolysis
(minor product). B and C are treated with HBr/ Peroxide and
Response Sheet
ANSWER KEY
1 (d) 7 (b) 13 (b) 19 (c) 25 (b)
2 (d) 8 (a) 14 (b) 20 (a) 26 (c)
3 (c) 9 (b) 15 (b) 21 (a) 27 (b)
4 (a) 10 (a) 16 (a) 22 (c). 28 (a).
5 (a) 11 (a) 17 (a) 23 (b) 29 (c)
6 (b) 12 (c) 18 (a) 24 (b) 30 (b)
Rearrangement
9. (b) The molecule 2,3 - pentadiene does not have any chiral
3. (c) Hydrazine (NH 2 —NH 2 ) does not respond to C but at the same time it does not have any mirror
Lassaigne’s test because they donot have any carbon plane which makes the molecule chiral.
and hence NaCN is not formed. 10. (a) Kolbe’s electrolysis
+
X CH–CH–COONa
2 CH–2 CH2
2
4. (a) +X + CO2 + NaOH
H
CH–CH–COONa
2 2 CH–2 CH 2
Cyclobutane
5. (a) Only o-nitrophenol is capable of forming intramolecular
H-bonding (chelation) which leads to its lower b.p. (less
11. (a) CH2=CH – CH2– CH = CH2 O 3 HOOC– CH– COOH
than 100°C) and also lower solubility in water (steam). 2
+ 2HCOOH
Zn / H2O
OH 2HCHO+OHC–CH–
2 CHO Reduction
6. (b) N 2 Cl + ¾
¾®
12. (c) Hydroxylation of ethyne produces ethanal
H SO
OH CH º CH + H2O ¾¾¾¾
2 4 ® CH CHO
3
2
Hg +
N=N
+
CH3CHCH2 OH ......(I)
As the corresponding intermediate (free radical) is more
stable than the other.
+
17. (a) To prepare unsymmetrical alkanes, CH3 CH2CH OH ......(II)
Corey – House reaction is most important.
hence bromination occurs on 3rd C atom
dry
R - X + R ¢2 CuLi ¾¾¾
® R - R ¢ + R ¢Cu + LiX
ether
Br
18. (a) Because of the formation of the most stable
+
CHCH2Br Cº CH
carbonium ion, C6 H5 - C H 2
alc. KOH
25. (b)
NaNH2
CH2 CH2
26. (c) Relative number of atoms,
CH CH CH CH O O
20. (a) O3/H2O
|| || || || + || ||
CH CH O O CH CH 74
For C = = 6 .1 ,
CH2 12
CH2
8.65 17.3
For H = = 8.65 , For N = = 1.2
COOH– CH2– COOH + COOH– CH2– COOH 1 14
D
Simplest ratio :
2CH3COOH
(Two Carboxylic group on one 6 .1 8.65 1 .2
carbon are not stable) For C = = 5, H = =7,N= = 1;
1 .2 1 .2 1 .2
27. (b) Entiomers of C4H10O are electronegative than sp3 hybridized carbon. More
electronegative atom accommodates the negative
OH HO charge more easily.
| |
H3CH 2C - C - CH3 H 3C - C - CH 2 CH 3 29. (c) Zinc reduces the coloured impurites in the aniline
| |
H and also prevents its oxidation during the reduction.
H
NO2
OH
|
H SO
30. (b) conc. HNO3
¾conc
¾¾ ¾¾® ¾Cl / Fe
¾2 ¾¾®
CH 3 CH 2- C H - CH 3 ¾¾2¾¾
4® . H SO
2 4
A
¾Sn
¾/ ¾
HCl
¾®
28. (a) sp-Hybridized carbon atom is more electronegaive than
sp 2 hybridized carbon which in turn is more Cl Cl
PART ORGANIC CHEMISTRY-2
TEST 8 Time : 1 hour
Max. Marks : 120
INSTRUCTIONS
Ø Crack this section in just 1 hour.
Ø This part test contains 30 questions.
Ø For every correct answer you will be awarded +4. There is no negative marking.
Ø Please fill the correct bubbles in the Response Sheet provided at the end of the test.
Ø Do not look at the solutions before attempting the complete test.
. . . Best of Luck
1. Polarization of electrons in acrolein may be written as: 6. Alcohols can be obtained from all methods except
d- d+
(a) hydroboration-oxidation
(a) C H 2 = CH — CH = O (b) oxymercuration-demercuration
(c) reduction of aldehyde/ketones with Zn–Hg/HCl
d+ d-
(b) C H 2 = C H — CH = O (d) fermentation of starch
7. To prepare 3-ethylpentane-3-ol the reagents needed are
d+ d- (a) CH3CH2 MgBr + CH3COCH2CH3
(c) C H 2 = CH — CH = O
(b) CH3MgBr + CH3CH2CH2COCH2CH3
d- d+ (c) CH3CH2MgBr + CH3CH2COCH2CH3
(d) C H2 = C H — C H = O (d) CH3CH2CH2MgBr + CH3COCH2CH3
2. How many different dipeptides can be formed by two 8. The correct decreasing order of ease of acid catalysed
different amino acids ? esterification is-
(a) 4 (b) 1 (a) CH3CH2CH2CH2OH > CH3 CH2 CH (OH) CH3 > (CH3)2
(c) 3 (d) 2 C(OH)CH3
3. An example of zwitter ion is (b) CH3CH2CH(OH)CH3 > CH3 CH2 CH2CH2 OH > (CH3)2
C(OH) CH3
(a) alanine
(c) (CH3)2C(OH)CH3 > CH3CH2CH (OH) CH3 > CH3 CH2
(b) glycine hydrochloride CH2 CH2 OH
(c) urea (d) None of the above
(d) ammonium acetate 9. Aniline when diazotized in cold and when treated with
4. Degree of sweetness is minimum in dimethyl aniline gives a coloured product. Its structure
(a) glucose (b) fructose would be
(c) sucrose (d) invert suga (a) CH3NH N=N NHCH3
5. Breathalyzer tubes used to detect intoxication have a coating of
(b) CH3 N=N NH2
(a) blue litmus solution
(b) red litmus solution (c) (CH3)2N N=N
(c) sodium bicarbonate solution
(d) (CH3)2N NH
(d) sodium dichromate solution
C-36 Organic Chemistry-2
O (a) C + F2 + Cl2 ¾
¾® (b) CH3Cl + F2 ¾
¾®
||
SeO 2
10. CH 3 - C - CH 3 ¾¾¾® P + Se + H 2 O . SbCl 5
(c) CCl4+HF ¾¾¾ ® (d) CCl4 + F2 ¾
¾®
Here P is
O O
O
|| ||
(a) CH 3 - C - CHO (b) CH 3 - C - OCH 3 17. NH OH H SO
¾¾ ¾
2
¾® A ¾¾2 ¾
¾4
® B. The product B
O
||
is
(c) CH 3 - C - CHOH (d) None
(a) Caprolactam (b) Lactone
11. Which one of the following on treatment with 50% aqueous
(c) Perlone (d) Nylon-6
sodium hydroxide yields the corresponding alcohol and
acid? 18. Which of the following is the least basic ?
HO Cl NO2
(c) (d)
O
(c) (d)
22. The repeating units of acrilan are
O
O H H
| |
15. Isopropyl alcohol when heated with iodine and sodium (a) H 2 C = C - COOCH 3 (b) H 2 C = C - COOC 2 H 5
hydroxide, which of the following products will be formed?
(a) Isopropyl chloride (b) Tri-iodomethane H CH3
| |
(c) Propene (d) Ethanol (c) H 2 C = C - CN (d) H 2 C = C - COOCH3
16. Which of the following reagents will produce CCl2F2 ,
commonly called Freon-12?
Organic Chemistry-2 C-37
23. Which of the following gives n-nitrosoamine on reaction 27. In the following compounds,
with nitrous acid?
O
CH3 CH3
NH2 N N N N N
H H H
(a) (b) I II III IV
O
|| Br
(c) R - C- N (d) Both (a) & (b)
Br
Response Sheet
ANSWER KEY
1 (c) 7 (c) 13 (b) 19 (a) 25 (b)
2 (a) 8 (a) 14 (d) 20 (c) 26 (d)
3 (a). 9 (c) 15 (b) 21 (c) 27 (d)
4 (a) 10 (a) 16 (c) 22 (c) 28 (a)
5 (d) 11 (a) 17 (a) 23 (c) 29 (c)
6 (c) 12 (a) 18 (a) 24 (d) 30 (c)
13. (b) Hoffman bromamide reaction 22. (c) Polyacrylonitrile (PAN), acrilan or orlon
CH3 CH3 CN CN
| |
C–CONH2 Br2 / NaOH C –NH2 n CH 2 = CH ¾¾
® (CH 2 - CH)-n
CH3 CH3 Acrylonitrile orlon
14. (d) This is an example of aldol type condensation. It is hard used in preparing cloths, carpets
23. (c) Only secondary amines give nitroso amines on reaction
O O with nitrous acid. This reaction is known as Libermann’s
H nitroso reaction.
H 24. (d) –NO2 group is electron withdrawing, and hence III and
IV are strong acids whereas –CH3 is electron releasing
group, hence I is less acidic than the others.
O O 25. (b) Aldol condensation
H
Heat
O H O
(–H2O) || | ||
OH Ba (OH )
CH 3 - C H + C H 2 - C - H ¾¾¾¾
¾2®
D H
HO O
15. (b) CH3 CHCH3 + 4I 2 + 6NaOH ¾¾®
| | ||
|
( - H 2 O)
OH CH3 - HC - CH - C - H ¾¾¾¾ ®
heat
O NOH H CH 3 - CH = CH - CH 2 OH
O
N
17. (a) H2NOH H2SO4 26. (d) RCONH 2 + Br2 + 4 KOH ¾
¾®
RNH 2 + 2KBr + K 2CO 3 + 2H 2 O
Caprolactum
Mechanism :
18. (a) In case of pyridine the lone pair of electrons are
delocalised in the benzene nucleus. Br
R NH2 R N
Basicity order is Br2 H KOH
C C
Dimethyl amine > Methyl amine > Piperidine > –H
Pyridine O O
19. (a) sp-Hybrid C is more electronegative than sp2 hybrid C
which is more electronegative than sp3 hybrid C. As
the hybrid state of a-C changes from sp3 to sp, the R N Br R N
+ –KBr
acid strength tends to increase from X to Z. C K C
20. (c) Diallyl ether has double bond which shows addition
reaction with KMnO4 (change its colour) while di-n- O O
propyl ether does not react with KMnO4. H2O
21. (c) Sandmeyer’s reaction R–N=C=O RNH2+ CO 2
NH2 O
27. (d) ..
NaNO2 .. .. .. N
HCl N N N
H H
+ –
H
NH2 Cl Cl I II III IV
lp belongs to lp belongs to lpisnsp2 Although N is
CuCl sp3 orbital sp3 but disperse orbital, hence sp2 hybridised,
to same extent not a part of lp on N is in
towards O aromatic sextet p-orbital,
hence forms a
part of aromatic
sextet
C-40 Organic Chemistry-2
PART ALGEBRA
TEST 1
Time : 1 hour
Max. Marks : 120
INSTRUCTIONS
Ø Crack this section in just 1 hour.
Ø This part test contains 30 questions.
Ø For every correct answer you will be awarded +4. There is no negative marking.
Ø Please fill the correct bubbles in the Response Sheet provided at the end of the test.
Ø Do not look at the solutions before attempting the complete test.
. . . Best of Luck
1. ( 1+z+ z2 )8 = C0 + C1z +C2z2.......+C16z16, where z is real or and the sum of their squares is 35 . The sum to n terms of the
complex number then series is
C0 – C2 + C4 – C6 +.....+C16 is equal to (a) 3n2 (b) 2n2
(a) –1 (b) 1 (c) 6n – 2n 2 (d) 6n – n2
(c) 0 (d) none 6. n+4C – nC – 3.nC n n
r r r–1 – 3. Cr–2 – Cr–3 is equal to
2. The point on the curve |z – 5i | = 3 having the least positive (a) n+1C (b) n+2 C
r–1 r–1
argument, is n+3C n+4C
(c) r–1 (d) r–1
æ 12 16 ö 7. If a, b, c, d and p are distinct non zero real numbers such that
(a) ç , ÷ (b) (0,3)
è 5 5ø (a2 + b2 + c2) p2 –2(ab + bc + cd)p + (b2 + c2+ d2) £ 0 then
a,b,c,d are in
(c) (0,8) (d) (3,4)
(a) A.P. (b) G.P.
3. 5 boys of class VI, 6 boys of class VII and 7 boys of class
(c) H.P. (d) satisfy ab = cd
VIII sit in a row. The number of ways they can sit so that
8. The total no. of ways of distributing n identical balls among
boys of the same class sit together, is
k different boxes if each box can hold any no. of balls is
(a) (5!)(6!)(7!) (b) (3!)(5!)(6!) (7!)
(a) kn (b) nk
(c) 18!/(5!6!7!) (d) (4!)(5!)(6!)
(c) (k +n –1)!/[(k –1)!n!] (d) (k –n +1 )!/[k! (n–1)!]
4. If p is not equal to q then for the equations
9. The equation sinx (sinx + cosx) = p has real solution if p lies
x2 – px + q = 0 and x2 – qx + p = 0 which of the following is not
in the interval
correct?
(a) Both can simultaneously have a zero root é1 - 2 1 + 2 ù
(b) If the roots of the first one are both positive, so are the (a) ê , ú (b) [–1–Ö5, 1+Ö5]
roots of the second. ëê 2 2 úû
(c) If the roots of the first are both negative, those of the (c) [Ö5–1, Ö5+1] (d) [–1–Ö5, Ö5–1]
second have opposite sign.
10. If the complex numbers Z1 and Z1 represents the adjacent
(d) Both can have roots belonging to the set {1,2,3,5}
5. The sum of the first three consecutive terms of an A.P. is 9 vertices of a regular polygon of side n and if Im Z1/ Re Z1
M-2 Algebra
= 2 – Ö3 then the value of n is equal to 20. Let a and b be the roots of (x – a) (x – b) + c = 0, c ¹ 0. Then
(a) 8 (b) 12 the roots of (ab – c) x2 + (a + b) x + 1 = 0 are
(c) 16 (d) 24 (a) 1/a, 1/b (b) –1/a, 1/b
11. 2
If the inequality (k + 4) x – 2kx + 2k – 6< 0 is satisfied for all (c) 1/a, –1/b (d) –1/a, –1/b
real x, then the largest integral value of k is 21. The integer just greater than (3 +Ö5)(2n) is divisible by,
(a) –10 (b) 4 (n Î N),
(c) –4 (d) –7 (a) 2(n–1) (b) 2(n+1)
12. Find the maximum distance of points moving on following (c) 2 (n+2) (d) not divisible by 2
22. 2
The roots of the equation ax + 2bx + c = 0 are equal. Then
circles | z |= 2 & | z - 3 - 3i |= 1
ax2 + 2bx + c + k is positive for every real value of x if
(a) 3(1 + 2 ) (b) 3 (a) k > 0 (b) a > 0, k > 0
(c) k < 0 (d) c > 0, k > 0
(c) 3 2 (d) None
23. The most general value of q satisfying the equation
13. The greatest possible number of points of intersection of 8 (cosq + i sinq) (cos 22q + i sin 22q) …..(cos n2q + i sin n2q) = –i,
straight lines and 4 circles in a plane is is given by
(a) 32 (b) 64 (c) 76 (d) 104
(a) [3(4k–1)p] / [n(n+1)(2n+1)]
14. If a,b,c,d are four unequal positive numbers which are in
A.P. then (b) [3(4k+1)p] / [n(n+1)(2n+1)]
(a) 1/a +1/d = 1/c + 1/b (b) 1/a +1/d < 1/b + 1/c (c) [6(2k–1)p] / [n(n+1)(2n+1)]
(c) 1/a + 1/c > 1/b + 1/d (d) 1/b + 1/c > 4/(a + d)
(d) [6(2k+1)] / [n(n+1)(2n+1)]
15. If a + b + c = 3 and a > 0, b > 0, c > 0, then the greatest value
of a2 b3 c2 is 24. Let (1 – x – 2x2)6 = 1+ a1x + a2x2 + …. + a12 x12. Then
a2 a4 a6 a12
310.24 39.2 4 38.2 4 + + + ...... + is equal to
2 4 6
(a) 7 (b) 7 (c) 7 (d) None 2 2 2 212
7 7 7
(a) –1 (b) –1/2
16. For any complex number z, the minimum value of
(c) 0 (d) 1/2
5 |z + 1| + |(3 + 4i) (z – 2)| is
(a) 3 (b) 5
(c) 10 (d) 15 ASSERTION & REASON
17. Let a/(a–1) and b/(b–1) be the roots of x2 + ax + b = 0. Then
DIRECTIONS (Qs. 25 - 30) : Each of these questions
1/a and 1/b are the roots of
contains two statements: Statement-1(Assertion) and
(a) bx2 + ax + 1 = 0
Statement-2 (Reason). Choose the correct answer (ONLY
(b) bx2 – ax + 1 = 0
ONE option is correct) from the following -
(c) bx2 + (a + 2b)x + a + b + 1 = 0
(a) Statement-1 is false, Statement-2 is true.
(d) bx2 – (a + 2b) x + a + b + 1 = 0
(b) Statement-1 is true, Statement-2 is true; Statement-2
18. Let a and b be the roots of x2 + (2b – a2) x + b2 = 0, and a' and
is a correct explanation for Statement-1.
b' be those of x2 + (2a – b2)x + a2 = 0. If a - b = a ' - b¢ , (c) Statement-1 is true, Statement-2 is true; Statement-2
then a + b is equal to is not a correct explanation for Statement-1.
(a) –4 (b) –2 (d) Statement-1 is true, Statement-2 is false.
(c) 2 (d) 4
19. Let a, b be the roots of ax 2 + bx + c = 0 and 25. Statement–1 : The coefficient of x203 in the expression
a + k, b + k, g be the roots of px3 + qx2 + rx = 0. If a, b, k are (x – 1)(x2 – 2) (x2 – 3) . . . (x20 – 20) must be 13.
Statement–2 : The coefficient of x8 in the expression
(b 2 – 4ac) (2 + x)2 (3 + x)3 (4 + x)4 is equal to 30.
positive, then is equal to
(q2 – 4pr) 26. Statement–1: If n is an odd prime, then integral part of
27. Statement–1 : If w is complex cube root of unity then 29. Let z1 and z2 be complex number such that
(x – y) (xw – y) (xw2 – y) is equal to x3 + y2
z1 + z2 =| z1 | + | z 2 |
Statement–2 : If w is complex cube root of unity then
1 + w + w2 = 0 and w3 = 1
æ z1 ö
28. Statement–1 : If 0 < a < p/4, then the equation Statement–1 : arg ç z ÷ = 0
è 2ø
(x – sin a) (x – cos a) – 2 = 0 has both roots in (sin a, cos a)
Statement–2 : z1, z2 and origin are collinear and z1, z2 are on
Statement–2 : If f (a) and f (b) possess opposite signs then
there exist at least one solution of the equation f (x) = 0 in the same side of origin.
open interval (a, b). 30. Statement-1: Let the positive numbers a, b, c, d, e be in AP,
then abcd, abce, abde, acde, bcde are in HP.
Statement-2: If each term of an A.P. is divided by the same
number k, the resulting sequence is also.
Response Sheet
ANSWER KEY
1 (b) 7 (b) 13 (b) 19 (c) 25 (d)
2 (a) 8 (c) 14 (d) 20 (d) 26 (b)
3 (b) 9 (a) 15 (a) 21 (b) 27 (a)
4 (a) 10 (b) 16 (d) 22 (b) 28 (a)
5 (d) 11 (d) 17 (d) 23 (a) 29 (b)
6 (c) 12 (a) 18 (a) 24 (b) 30 (b)
a
= n + 4 C r - n + 2 C r + n + 2 C r -1
x
O = n + 4 C r - n +3 Cr = n +3 C r -1
3
Coordinates of P (4 cos a,4 sin a ) , a = cos -1 7. (b) (a 2 + b 2 + c 2 ) p 2 - 2(ab + bc + cd ) p
5
+ b2 + c2 + d 2 £ 0
æ 12 16 ö
\ Pç , ÷
è 5 5ø Þ (a 2 p 2 - 2abp + b 2 ) + (b 2 p 2 - 2 bcp + c 2 )
3. (b) Boys of class VI can sit among each other in 5! ways + (c 2 p 2 - 2cdp + d 2 ) £ 0
Boys of class VII can sit among each other in 6! ways
Boys of class VIII can sit among each other in 7! ways Þ (ap - b)2 + (bp - c) 2 + (cp - d) 2 £ 0
As boys of each class sit together, so we have 3 groups Þ ap - b = 0, bp - c = 0 & cp - d = 0
which can be organised in 3! ways.
Hence the number of ways = 3! 5! 6! 7! b c d
Þ = = Þ Hence in G.P
a b c
4. (a) Let f (x) = x 2 - px + q
Also ad = bc
g( x ) = x 2 - qx + p 8. (c) Consider the problem as dividing n identical balls into
f (0) = q k parts by introducing k–1 dividers. So, we have to
If 0 is root of f (x) (n + k – 1)!
arrange n + k – 1 items, that we can do in .
(k – 1)!n!
Þ q = 0 ¹ p, g(0) = p ¹ 0
Hence 0 is not the root of g(x) 1 - cos 2x + sin 2x
9. (a) P = sin 2 x + sin x cos x. =
5. (d) a - d + a + a + d = 9 Þ a = 3 2
(a - d) 2 + a 2 + (a + d) 2 = 35 Þ ( 2p - 1) 2 = 1 - sin 4 x Þ sin 4x = -4p 2 + 4p
Algebra M-5
-1 1 13. (b) One st. line can intersect 4 circles at maximum 8 pts. So
Þ -1 £ -4 ( p 2 - p ) £ 1 Þ £ p2 - p £ 8 st. lines can intersect at 64 pts. Note the language of
4 4
the question. It is asking only those pts and not pts
Let solution of inequality p = [a, b] created by intersection of 2 st lines or 2 circles.
where a, b are the roots of the equation 14. (d) b = a + r, c = a + 2r, d = a + 3r
1± 1+ 1
p 2 - p - 1 / 4 = 0 Þ a, b = option (a) ®
1
+
1
¹
1
+
1
not correct
2 a a + 3r a + 2 r a + r
2
(p –p)
option (b) ® 1 + 1 < 1 + 1 can’t say
a a + 3r a + 2r a + r
depend on a >r or a < r
1/4
option (c) ® 1 + 1 > 1 + 1 can’t say
1/2 a a + 2r a + r a + 3r
p depend on a > r or a < r
a 1 b option (d)
1/4 4 1 2 1 1 1
® = + = < +
2 - 1 2a + 4r a + 2r a + 2r a + 2r a + r a + 2r
Since p - p > (from graph) [since r > 0]
4
15. (a) Taking A.M. and G.M. of 7 numbers
æ1- 2 1 + 2 ö
Þ p =ç , ÷ a a b b b c c
, , , , , , , we get
ç 2 2 ÷
è ø 2 2 3 3 3 2 2
2p b
10. (b) = 2 tan -1 a b c 1
n a 2. + 3. + 2. ìïæ a ö 2 æ b ö3 æ c ö 2 üï 7
2 3 2 ³ ç ÷ ç ÷ ç ÷
b í ý
7 ïîè 2 ø è 3 ø è 2 ø ïþ
37 æ a 2 b3c 2 ö 37 a 2 b3 c 2
Þ 7 ³ çç 2 3 2 ÷÷ Þ 7 ³ 2 3 2
2p/n 7 è2 3 2 ø 7 2 .3 .2
a
310.2 4
Þ a2 b3 c2 £
77
310.2 4
2p \ greatest value of a2 b3 c2 =
= 2 tan -1 (2 - 3 ) = Þ n = 12 77
12 16. (d) 5 |z + 1| + |(3 + 4i) (z – 2)|
11. (d) f (x) = ax2 + bx + c = |5z + 5| + |5z – 10| > |(5z + 5) + (5z – 10)|
For f (x) < 0, a < 0 and D < 0 > |(5z – 5) – (5z – 10)| = 15
K + 4 < 0 Þ k < -4 & D < 0 17. (d) Let a' = a/(a – 1), b' = b/(b – 1)
Þ D = b2 - 4ac = 4k 2 - 4c 2 k - 6(k + 4) < 0 Þ a = a'/(a'– 1), b = b'/(b' – 1)
Þ 1/a = (a' – 1)/a', 1/b = (b' – 1)/b'
Þ K 2 - 2K 2 - 6 k + 6 k + 24 < 0 Then equation whose roots are 1/a, 1/b, is
Þ (4 - k)(6 + k ) < 0 x2 – (1/a + 1/b)x + 1/ab = 0
k = (-¥, -6) Þ x2 – [(a' – 1)/a' + (b' – 1)/b'] x + [(a' – 1) (b' –1)] / a'b' = 0
Integer k = – 6 –1= –7 Þ a'b'x2 – [2a'b' – (a' + b')]x + a'b' – (a' + b') + 1 = 0
12. (a) AB is the maximum distance Þ bx2 – (a + 2b)x + a + b + 1 = 0
18. (a) a + b = a2 – 2b, ab = b2, a' + b' = b2 – 2a, a'b' = a2,
2 2 Given: Öa – Öb = Öa'– Öb'
= 2 + 3 + 3 +1 = 3 + 3 2
B On squaring,
1 Þ a + b – 2Ö(ab) = a' + b' – 2Ö(a'b')
(3, 3) Þ a2 – 2b – 2b = b2 – 2a – 2a
Þ a2 – b2 = –4 (a – b) Þ a + b = –4
19. (c) It turns out that a and b are the roots of ax2 + bx + c = 0
x
while a' = a + k, b' = b + k are the roots of px2 + qx + r = 0
2 as x = 0 is one of the roots, which is g.
A
we have a '– b' = a – b
Þ (a' + b')2 – 4 a'b' = (a + b)2 – 4ab
M-6 Algebra
( )
n
Let [R] + 1 = I ( [ ] greatest integer function). Let, 5-2 = f¢ ; 0<f¢<1
Þ R + G = I (Q 0 < G < 1)
( ) -( )
n n
Þ 5+2 5-2 = Integer (Q n is odd)
(
Þ 3+ 5 ) + (3 - 5 )
2n 2n
=I é n -1 n-3 ù
( ) -( )
n n
Q 5+2 5-2 = 2 ê n c1 2.5 2 + n c3 23.5 2 + ... ú
Seeing the option put n = 1 êë úû
Þ I = 28 is divisible by 4 i.e 2n+1 Q ( f = f ¢)
22. (b) ax2 + 2bx + c = 0 has equal roots = > b2 = 4ac Þ I is divisible by 20n on using statement – II.
ax2 +2bx + c + k = a [{x + b/a}2 + (ac–b2)/a2 + k/a] 27. (a) (x – y) (xw – y) (xw2 – y)
= a [(x + b/a)2 + k/a] = x3 w2 – x2yw – x2yw2 + xy2 – x2yw + xy2w +
Which is +ve for every real x if k > 0, a > 0 xy2w2 – y3
(Alternatively) Let f(x) 3
=x –y 3
f (x) 28. (a) Let, f (x) = (x – sin a) (x – cos a) – 2
then, f (sin a) = – 2 < 0; f (cos a) = – 2 < 0
a >0 f (x)
a <0 p
Also as 0 < a < ;
4
equal roots \ sin a < cos a
= ax 2 + 2bx + c Therefore equation f(x) = 0
all a > 0 & k > 0 has one root in (- ¥, sin a)
23. (a) (cosq + isinq)(cos22q + sin22q) … and other in (cos a, ¥)
…(cos n2q + i sin n 2q) = –i
Þ cos (12 + 22 +… + n2)q + isin (12 + 22 + .. + n2) q sin a cos a
= cos(–p/2) + i sin(–p/2) –¥ ¥
INSTRUCTIONS
Ø Crack this section in just 1 hour.
Ø This part test contains 30 questions.
Ø For every correct answer you will be awarded +4. There is no negative marking.
Ø Please fill the correct bubbles in the Response Sheet provided at the end of the test.
Ø Do not look at the solutions before attempting the complete test.
. . . Best of Luck
1. If x =1/5, the value of cos (cos–1x + 2 sin–1 x) is 6. In an equilateral triangle, (circumradius) : (inradius) :
(exradius) is equal to
24 24 (a) 1 : 1 : 1
(a) - (b)
25 25 (b) 1 : 2 : 3
(c) 2 : 1 : 3
1 1
(c) - (d) (d) 3 : 2 : 4
5 5
7. For any real q, the maximum value of
cos2(cos q) + sin2(sin q) is
2. The sides of a D are sina, cosa and 1 + sin a cos a , for
(a) 1 (b) 1+sin21
p (c) 1+cos21 (d) 1/2
0<a< , then greatest angle of D is
2 8. The number of integral values of a for which the equation
(a) 60° (b) 150° cos 2x + a sin x = 2a – 7 possesses a solution is
(c) 120° (d) 90° (a) 2 (b) 3
3. The value of tan[sec–1(5/4) + cot–1(3/2)] is (c) 4 (d) 5
(a) 6/17 (b) 17/6 p
(c) 7/16 (d) 16/7 9. If 0 < a, b, g < p/2 such that a + b + g = and cot a, cot b,
2
4. The value of cos36 cos42 cos780 is
0 0
cot g are in arithmetic progression, then the value of cot a
é cot g is
5 -1 5 + 1ù
êGiven : sin 18 = and cos 36 = ú (a) 1 (b) 3
êë 2 2 úû
(c) cot b2 (d) cot a + cot g
(a) 1/4 (b) 1/8 10. The area of a circle is A1 and the area of a regular pentagon
(c) 1/16 (d) [(Ö5 – 1)/4]2 inscribed in the circle is A2. Then A1 : A2 is
5. The equation sin-1x = 3sin-1a has a solution for
p p 2p p
(a) –1 £ a £ 1 (a) cos (b) sec
5 10 5 10
(b) –Ö3/2 £ a £Ö3/2
(c) –1/Ö2 £ a £ 1/Ö2 2p p
(c) cos ec (d) None
(d) – 1/2 £ a £ 1/2 5 10
M-8 Trigonometry
A
11. If u = a 2 cos 2 q + b 2 sin 2 q + a 2 sin 2 q + b 2 cos 2 q 21. In any triangle ABC, sin is
2
then the difference between maximum & minimum value b+c a
(a) less than (b) less than or equal to
of u2 is given by a b+c
2a
(a) 2(a 2 + b 2 ) (b) (a - b ) 2 (c) greater than (d) None of these
a +b+c
22. In a D ABC, if a/cosA = b/cosB then
(c) (a + b ) 2 (d) 2 a 2 + b2 (a) 2 sinA cosB = sin C
12. If a > b > 0, the minimum value of a secq - b tanq is (b) 2 sinA sinB = sin C
(a) Ö(a2 +b2) (b) Ö(a2 - b2) (c) tan2 (A/2) + 2 tan (A/2) tan (C/2) – 1 = 0
2 2 (d) none of these
(c) 2Ö(a - b ) (d) a – b
¥
13. The maximum and minimum values of 6 sin x cos x + 4 cos 2x p
are
23. For 0 < f £
2
, if x = åcos 2n f ,
n =0
(a) 5, –5 (b) 4, –4 ¥ ¥
Response Sheet
ANSWER KEY
1 (c) 7 (b) 13 (a) 19 (b) 25 (c)
2 (c) 8 (d) 14 (b) 20 (a) 26 (b)
3 (b) 9 (b) 15 (c) 21 (b) 27 (d)
4 (b) 10 (b) 16 (c) 22 (a) 28 (d)
5 (d) 11 (b) 17 (a) 23 (c) 29 (b)
6 (c) 12 (b) 18 (b) 24 (c) 30 (b)
M-10 Trigonometry
1. (c) The given expression is equal to 8. (d) The given equation can be written as
cos(cos -1 x + sin -1 x + sin -1 x) 1 - 2 sin 2 x + a sin x = 2a - 7
æp ö 1 Þ 2 sin 2 x - a sin x + 2a - 8 = 0
= cos ç + sin -1 x ÷ = - sin(sin -1 x) = - x = -
è2 ø 5 a ± a 2 - 8(2a - 8) a ± (a - 8)
Þ sin x = =
-1 p -1 4 4
[Using cos x + sin x = ] Þ since sin x = 2 not possible
2
2. (c) Greatest angle opposite the greatest side a -4
Þ sin x =
i.e 1 + sin a cos a 2
a-4
b2 + c2 - a 2 which is possible if - 1 £ £ 1 or 2 £ a £ 6
Using, cos A = 2
2bc So the required values of a are 2,3,4,5,6 and hence
sin 2 a + cos 2 a - (1 + sin a cos a) 1 the required number is 5.
cos A = =- p p
2 sin a cos a 2 9. (b) a + b + g = Þ a + g = – b.
2 2
Þ A = 120°
æp ö
3. (b) tan[sec–1(5/4) + cot–1(3/2)] so that cot ( a + g ) = cot ç - b÷
= tan[tan–1(3/4) + tan–1(2/3)] è2 ø
= [(3/4) + (2/3)]/[1–(1/2)] = 17/6 cot a cot g - 1 1
Þ =
4. (b) cos36°cos42°cos78° cot a + cot g cot b
=cos36°cos(60°-18°)cos(60°+18°)
Þ cot a cot g - 1 = 2 Þ cot a cot g = 3.
=
5 +1
(cos 2
60° - sin 2 18° ) (since cot a + cot g = 2cot b)
4 10. (b) In the D OAB,
360°
æ 5 + 1 öé 1 æ 5 - 1 ö
2ù
=ç ÷ê - ç
ç 4 ÷ê 4 ç 4 ÷ ú =
÷ ú 5 +1 5 -1 1
=
( )( ) OA = OB = r and ÐAOB =
5
= 72°
è øë è ø û 1 1
32 8 \ area ( DOAB ) = . r. r. sin72° = r 2 cos18°;
–1 2 2
5. (d) Since –p/2 £ sin x £ p/2, it follows that
5 2
–p/6 £ sin–1a £ p/6 Þ –1/2 £ a £ 1/2 \ area ( pentagon ) = r cos18°
2
abc D r
6. (c) Use R = , r = and 2pr 2 2p p
4D s R \ A1 : A 2 =
2
= sec
5r cos 18° 5 10
a cos A + b cosB + c cosC D
=
a+b+c
A B C E O C
The ex-radius = 4R sin. . cos . cos r
2 2 2
A B
1 3 3 3
= 4R . . . = R
2 2 2 2 (a + b 4 ) cos2 q sin 2 q
4
11. (b) u 2 = a 2 + b2 + 2
7. (b) f (q) = cos 2 (cos q) + sin 2 (sin q) + a 2 b2 (cos4 q + sin 4 q )
Now put q = p /2 & 0
(a 4 + b4 ) cos 2 q sin 2 q + a 2 b 2
f (0) = cos 2 1 + 0 = cos 2 1 = a 2 + b2 + 2
[(cos2 q + sin 2 q) 2 - 2cos 2 q sin 2 q
2
f (p / 2) = 1 + sin 1
= a 2 + b 2 + 2 (a 2 - b 2 ) 2 cos 2 q sin 2 q + a 2 b 2
Since sin 2 1 > cos 2 1
p p p = a 2 + b 2 + (a 2 - b 2 ) 2 sin 2 2q + 4a 2 b 2
[Q sin q > cos q "
< q< &1> ]
4 2 4 u 2max = u 2min = a 2 + b 2 + (a 2 + b 2 ) - (a 2 + b 2 + 2ab)
Now seeing the options 1 + sin21 is greater than all
other options. = ( a - b) 2
Trigonometry M-11
12. (b) Let a secq – b tanq = x Þ a secq = x + b tanq PAB, PBC and PCA are equilateral and thus
Þ (a2 – b2)tan2q – 2bx tanq + a2 – x2 = 0 PA=PB=AB=a. In triangle ABC, OA is the bisector of
Using disc ³ 0, we have x2 ³ Ö(a2 – b2) OA a 2 a
13. (a) We have 6 sin x cos x + 4 cos 2x = 3 sin 2x + 4 cos 2x. angle A, so = sec 30° Þ OA = . =
Now the maximum and minimum values of a/2 2 3 3
Now from right angled triangle POA,
3 sin 2x + 4 cos 2x are 32 + 4 2 and - 32 + 4 2 i.e., PA2 = OP2 + OA2
5 and –5 respectively. a2 2a 2
14. (b) Let the point C be (h,k). Then h2 + (k+1)2 = 25 Þ a2 = h2 + Þ = h 2 Þ 2a 2 = 3h 2
Using slope of AC × AB = –1 3 3
2(k – 3) = –(h – 2), which are satisfied by (4,2) and (0,4). P
B
æ ö
ç 1+ x 2 ÷
15. (c) sin(cot -1 x) + cot ç sin -1 ÷
ç 2 + x 2 ÷ø O
è
æ 1 ö æ 1 ö 2
C A
= sinç sin -1 ÷ + cot ç cot -1 ÷= 20. (a) Here BD : DC = c : b, but BD +DC = a
ç ÷ ç ÷
è 1+ x 2 ø è 1+ x 2 ø 1+ x2 c
16. (c) tanA/2, tanB/2, tanC/2 are in A.P. \ BD = .a
Þ tanA/2 – tanB/2 = tanB/2 – tanC/2 b+c
sin A / 2 sin B / 2 sin B / 2 sin C / 2 BD AD
Þ - = - In DABD, =
cos A / 2 cos B / 2 cos B / 2 cos C / 2 A sin B
sin
Þ cos(C/2)sin(A/2 – B/2) 2
= cosA/2 sin (B/2 – C/2) ca sin B 2D A
Þ cos B - cos A = cosC - cosB \ AD = . = cosec
b+c A b+c 2
Þ 2cos B = cos A + cos C sin
2
Þ secA, secB, secC are in H.P.
17. (a) From the figure we have AI AB c b+c
Also = = =
æ4ö æ4ö ID BD ca /(b + c) a
BD = ç ÷BC = ç ÷a
11
è ø è 11 ø b+c D A
Þ AI = .AD = cosec
From (sine rule) triangle ABD a+b+c s 2
AD BD Similarly for BI and CI.
= A
sin B sin 60°
BD sinB 4a sin B
Þ AD = = I
sin60 ° 11 sin 60°
4 b sin A 4 7 sin120 ° 28 B D C
= = ´ =
11 sin 60° 11 sin60 ° 11 a sin A
21. (b) =
A b + c sin B + sin C
4 60º 60º 7 A
2 sin. cos
A
sin
A
= 2 2 = 2
B D C B+ C B-C B-C
2 sin . cos cos
2 2 2
18. (b) sin q + sin f = 3 (cos f - cos q)
A a B-C a
( q + f) (q - f) (q + f) q-f \ sin = cos £
Þ sin cos = 3 sin sin 2 b+c 2 b+c
2 2 2 2
Þ q + f = 2np or q - f = 60 ° + 2np 2a 2 sin A
=
a + b + c sin A + sin B + sin C
Now sin 3q + sin 3f
3 (q + f) ( q - f) A A A
= 2 sin cos 3 2.2 sin
cos sin
2 2 2 2 2 A
= = ³ sin
3(q + f) A B C B C 2
when q + f = 2np Þ sin =0 4 cos . cos . cos cos cos
2 2 2 2 2 2
22. (a) a/cosA = b/cosB, but we know that a/sinA = b/sinB;
q-f (q - f) \cosA sinB = sinA cosB
when = 30 + np Þ cos 3 =0
2 2 Þ sin(A–B) = 0 Þ A–B = np
19. (b) Let O be the center of the equilateral triangle ABC and but A < p, B < p for a D; \A – B = 0
OP be the tower of height h. Then each of the triangles Þ A = B Þ C = p – 2A
M-12 Trigonometry
INSTRUCTIONS
Ø Crack this section in just 1 hour.
Ø This part test contains 30 questions.
Ø For every correct answer you will be awarded +4. There is no negative marking.
Ø Please fill the correct bubbles in the Response Sheet provided at the end of the test.
Ø Do not look at the solutions before attempting the complete test.
. . . Best of Luck
14. The straight line passing through the point of intersection which P will move is bounded by parts of parabolas of which
of straight lines x – 3y + 1 = 0 and 2x + 5y – 9 = 0 and having one has the equation
infinite slope has the equation (a) y2 = a2 + 2ax (b) x2 = a2 + 2ay
(a) x = 2 (b) 3x + y – 1 = 0 2
(c) y + 2ax = a 2 (d) All
(c) y = 1 (d) None 23. The locus of the mid points of the chords of the ellipse
15. Two concentric circles of which smaller is x2 + y2 = 4, have x2/a2 + y2/b2 = k, k > 0, making equal intercepts on the
the difference in radius d, if line y = x + 1 cuts the outer circle coordinate axes, is
in two points, then d lies in the interval. (a) x = y (b) x + y = 0
æ
(c) x/a2 = y /b2 (d) x/a2 + y/b2 = 0
1 ö æ 1 ö
(a) çç – ¥,–2 – ÷÷ È çç – 2 + , ¥ ÷÷ 24. Two distinct chord drawn from the point (p, q) on the circle x2
è 2 ø è 2 ø + y2 = px + qy, where pq ¹ 0, are bisected by the x-axis then
æ 1 1 ö (a) | p | = | q | (b) | p | = 2 2 |q|
(b) çç – 2 + ,2 + ÷÷
è 2 2ø (c) | p | < 2 2 | q | (d) | p | > 2 2 | q |
æ 1 ö æ 1 ö 25. Two circles which can be drawn, to pass through (1, 0) and
(c) çç – ¥,1 – ÷÷ È çç1 + , ¥ ÷÷ (3, 0) and to touch the y-axis, intersect at an angle q, then
è 2ø è 2 ø cos q is equal;
æ 1 1 ö (a) 1/2 (b) –1/2 (c) 1/4 (d) –1/4
(d) çç1 – ,1 + ÷÷
è 2 2ø
16. The straight line 4x + 3y = 12 intersects the x-axis and the ASSERTION & REASON
y-axis at the points A and B respectively. Then the distance
BI where I is the centre of the incircle of the DOAB where O DIRECTIONS (Qs. 26 - 30) : Each of these questions
is the origin is equal to contains two statements: Statement-1(Assertion) and
Statement-2 (Reason). Choose the correct answer (ONLY
(a) 10 (b) 2 5 (c) 3 (d) 2 ONE option is correct) from the following -
17. 2 2
If tangents be drawn to the cricle x + y = 12 at its points of (a) Statement-1 is false, Statement-2 is true.
intersection with the circle x2 + y2 – 5x + 3y – 2 = 0, then the (b) Statement-1 is true, Statement-2 is true; Statement-2
tangents intersect at the point is a correct explanation for Statement-1.
æ 18 ö æ 18 ö æ 18 ö æ 18 ö (c) Statement-1 is true, Statement-2 is true; Statement-2
(a) ç – 6, ÷ (b) ç 6, ÷ (c) ç – 6,– ÷ (d) ç 6,– ÷ is not a correct explanation for Statement-1.
è 5ø è 5ø è 5ø è 5ø
(d) Statement-1 is true, Statement-2 is false.
18. One bisector of the angle between the lines given by
a(x – 1)2 + 2h (x – 1) y + by2 = 0 is 2x + y– 2 = 0. The other 26. Statement-1 : The incentre of the triangle formed by the
bisector is lines y = |x| and y = 1 is (0, 2 - 2) .
(a) x – 2y = 0 (b) x + 2y = 0 Statement-2 : The triangle is an isosceles right angled.
(c) x – 2y – 1 = 0 (d) x + 2y – 1 = 0 27. Statement 1 : Number of circles passing through (1, 4),
19. The orthocentre, circumcentre, centroid and incentre of the (2, 3), (– 1, 6) is 1
triangle formed by the line x + y = a with the coordinate axes Statement 2 : Though 3 non collinear points in a plane only
lie on one circle can be drawn
(a) x2 + y2 = 1 (b) y = x 28. Statement-1 : Slopes of tangents drawn from (1, 4) to the
(c) y = 2x (d) none of these parabola y2 = 12x are 1 and 3.
20. Two parabolas with the same axis, focus of each being Statement-2 : Two tangents can be drawn to parabola
exterior to the other and the latus rectum being 4a & 4b. The y2 = 4ax from the point (h, k) if 4ah – k2 > 0.
locus of the middle points of the intercepts between the 29. Let AB be a chord of the parabola y2 = 4ax, which is not
parabolas made on the lines parallel to the common axis is a parallel to the directrix of the parabola and subtends right
(a) straight line if a = b (b) parallel if a ¹ b angle at two other points C and D of the parabola.
(c) parabola for all a, b (d) ellipse if b > a Statement-1 : If parameters of the points A, B, C, D are
21. The equation of a common tangent to y2 = 4x and the curve respectively t1, t2, t3, t4 then |t1 + t2| = |t3 + t4|
x2 + 4y2 = 8 can be Statement-2 : t1, t2, t3, t4 are roots of the equation
(a) x – 2y + 2 = 0 (b) x + 2y + 4 = 0 at4 – t2 + 2at – 1 = 0.
(c) x – 2y = 4 (d) x + 2y = 4 30. Statement-1 : The straight line 2x + 3y = 4 intersects the
22. P is a point which moves in the x-y plane such that the point hyperbola 4x2 – 9y2 = 36 in exactly one point.
P is nearer to the center of a square than any of the sides. Statement-2 : The line is parallel to an asymptote of the
The four vertices of the square are (±a, ±a). The region in hyperbola.
Response Sheet
ANSWER KEY
1 (c) 7 (d) 13 (b) 19 (b) 25 (a)
2 (c) 8 (b) 14 (c) 20 (c) 26 (c)
3 (b) 9 (a) 15 (a) 21 (b) 27 (a)
4 (a) 10 (a) 16 (a) 22 (d) 28 (d)
5 (b) 11 (d) 17 (d) 23 (c) 29 (d)
6 (a) 12 (a) 18 (c) 24 (d) 30 (b)
æ m – m2 ö Þ mCC1 + mBB1 = 0
y=
Using tan q = ç 1 ÷
3
è 1 + m1m 2 ø C(1,6)
45°
B
B(b, b1) C(3b, b1)
M-16 Coordinate Geometry
6. (a) Circle x2 + y2 – x – y = 0
Þ 0 = -2ah - 2bk + a 2 + b 2 + 4
æ1 1ö 1
Centre: ç , ÷, radius = 1 Þ 2ax + 2by = a 2 + b 2 + 4
è2 2ø 2 B
2 12. (a) Let (h, k) be centroid of D
Maximum value of PA+PB 1 C
is when AB is diameter of 3h = 2 + (-2) + x1 & 3k = -3 + 1 + y1
2 æ 1 1ö
circle A çè , ÷ø
2 2 x y -2
Þ x1 = 3h Þ h = 1 Þ k = 1
Hence PA + PB = 26 P 3 3
(1, –2)
æ x ö æ y -2ö
7. (d) Find the point of intersection of lines 3x – 10y + 37 = 0, Þ 2ç 1 ÷ + 3ç 1 ÷ = 1 Þ 2x1 + 3y1 - 6 = 3
è 3 ø è 3 ø
9x + 2y – 17 = 0 Þ (1,4)
Check if it lies on the given diagonal 3x –2y = 14. We Þ 2x1 + 3y1 = 9
see that it does not lie on the diagonal. The situation is
13. (b) For a circle x 2 + y 2 = a 2 the locus of the point form
that the other diagonal will be ^ to given one that
passes through (1,4). So slope of diagonal will be = –23 which tangent to the above circle are ^ r , is known as
Þ y – y1=m (x – x1) will give the other equation of the
diagonal. Hence the diagonal is (y – 4) = (–2/3) (x–1) director circle and its locus given by x 2 + y 2 = 2a 2
which is equal to 2x + 3y = 14.
8. (b) In any triangle the centroid and the incentre lie within Equation of tangent, y = mx + a 2 m 2 + a 2
the triangle. If the triangle is obtuse angled then the
orthocentre and the circumcentre fall outside the Þ k = mh + a 2 m 2 + a 2 , m1m 2 = -1
triangle. Here, the angle between 7y = x and
Þ h 2 + k 2 = 2a 2 (equation of director circle)
3y + x = 0 is obtuse because the first line is inclined
2 2 5
at an angle less than 45° with the x-axis, while the second Now, Given circle is ( x - 1) + ( y - 2) =
2
line makes 150° with the x-axis.
9. (a) C1 (1, 0); C2 (0, –2) \ locus is ( x - 1) 2 + ( y - 2) 2 = 5
14. (c) Seeing the option only one line have the infinite slope
r1 = 1 + 15 = 4, r2 = 4 - 3 = 1
i.e., y = 1.
C1C 2 = 1 + 4 = 5 We only have to check. That this line passes through
the intersection of x - 3y + 1 = 0 & 2x + 5y = 9 which
r1 - r2 = 3 Þ C1C 2 < r1 - r2
is (2, 1), hence (c)
Hence, C2 lies inside C1. 15. (a) Difference in radius of circle is d;
10. (a) Let A and B have coordinates (x1,y1) and (x2,y2) Then Let the circles be x2 + y2 = 4 …. (1)
x1 + x2 = 3 and x1x2=1.Equation of the line AB is 2 2
And x + y = (2 + d) 2 ….(2)
2x + y = 0. This give, y1 + y2 = –6 and y1y2 = 4. Hence Line y = x + 1 cuts the circle in real points x 2 + (x + 1)2
the equation of the circle on AB as diameter. = (2 + d)2
(x – x1) (x – x2) + (y – y1) (y – y2) = 0 or 2x2+2x – (d2 + 4d + 3) = 0
Þ x2 + y2 – 3x + 6y + 5 = 0. For real points of cut D > 0
D= 4 – 4 × 2 × {– (d2 + 4d + 3)} > 0
11. (d) OC 2 = r12 + r2 2 (DOPC is right angle)
Þ 4 + 8(d2 + 4d + 3) > 0
(a, b)
or 2d2 + 8d + 7 > 0
P –8 ± 64 – 4 ´ 2 ´ 7 2 2 = –2 ± 1
(h, k) Þd= = –2 ±
C 2´ 2 4 2
16. (a) AB = 5
O é 4.3 + 3.0 + 5.0 4.0 + 3.4 + 5.0 ù
\I= ê , ú = [1, 1]
ë 12 12 û
\ BI = 10 units
Þ h 2 + k 2 = (h - a ) 2 + (k - b) 2 + 2 2
Coordinate Geometry M-17
h2 æ1 1ö
(0, 4) B \ 2a = ç - ÷,
4 èa bø
y 2 (b - a )
locus of P is 2x =
4 ab
A 2
O (3, 0) y = 4a(x – k)
B P A
17. (d) Let the point of contact be (h,k); equation of chord of
contact is T = 0
Þ xh – yk – 12 = 0 ………..(1) O
X
Equation of common chord C1 – C2 = 0 (0, 0)
Þ 5x – 3y –10 = 0 ………. (2)
(1) and (2) represent the same line 2
Þ h/k = k/(–3)= –12/(–10) y = –4b(x + k)
Þ h = 6.k = –18/5
x 2 y2
18. (c) The bisector of the angle between the lines are at right 21. (b) y 2 = 4x & + =1
8 2
angle. As such the other bisector will have slope 1 Equation of tangent to above curves are
2
and passes through (1, 0). respectively.
1
\ equation of line is y – 0 =
1
( x – 1) or x – 2y – 1 = 0 y 2 = mx + and y = mx + 8m 2 + 2
2 m
19. (b) The line x + y = a cuts the co-ordinate axes at A (a, 0), 1
B (0, a). Comparing = 8m 2 + 2
m
Y
Þ m 2 (8m 2 + 2) = 1
seeing the options
(0, a) B
1
m=± satisfy the equation
2
1
Þy=± x ± 2 Þ 2y = ± x ± 4
2
X i.e. 2y = x + 4 & x + 2y + 4 = 0
O A (a, 0)
22. (d) If P = (x,y) then
æa aö
Q circumcentre is the mid-point of AB, i.e. ç 2 , 2 ÷ x2 + y2 < a - x , x2 + y2 < a + x ,
è ø
æa aö
centroid is ç , ÷ , orthocentre is the origin. All lie on x2 + y2 < a - y , x 2 + y2 < a + y ,
è 3 3ø
y = x. Incentre lies on the angle bisector of ÐAOB , Therefore the region is bounded by the curves
which is also y = x. x 2 + y 2 = (a - x ) 2 , x 2 + y 2 = (a + x )2
20. (c) The equation of parabola can be taken as
y2 = 4a (x-k) & y2= –4b (x+k) x 2 + y 2 = (a - y ) 2 , x 2 + y 2 = (a + y) 2
A line parallel to the common axis is y = h
23. (c) The equation of chord having (x1,y1) as its mid-point
æ h2 ö æ h 2 ö÷
Then A = çç + k, h ÷ and B = ç - k - ,h is
÷ ç 4b ÷ø
è 4a ø è (xx1/a2)+ (yy1/b2 )– k = (x12/a2) + (y12/b2) – k
If P = (a, b) then Þ (xx1/a2) + (yy1/b2 ) = (x12/a2) + (y12/b2)...(1)
It makes equal intercepts on the axes when
1 æ h2 h 2 ö÷ (a2/x1) = (b2/y1) Þ (x1/a2) – (y1/b2 ) =0
a= ç +k -k - , b = h;
2 çè 4a 4b ÷ø Hence the locus of (x1, y1) is (x/a2) = (y/b2 )
M-18 Coordinate Geometry
24. (d) Let Q point be ( h , 0) 26. (c) The coordinates of the vertices of triangle OPQ are
respectively (0, 0), (1, 1), (–1, 1).
y
T Q P
Q(h, 0) I
P(p, q) x
O
Þ T (2h - p,-q) Also, PQ = 2, OP = OQ = 2 . So, the incentre is
T lies on circle
æ 0+ 2 + 2 ö
Þ ( 2 h - p ) 2 + q 2 = p( 2 h - p ) - q 2 çç 0, ÷÷ = (0, 2 - 2)
è 2+ 2 + 2 ø
Þ 4h 2 + p 2 - 4ph + q 2 = 2hp - p 2 - q 2 27. (a) Through 3 non collinear points a unique circle passes
through (1, 4), (2, 3) and (–1, 6).
Þ 4h 2 + 2(p 2 + q 2 ) - 6hp = 0
3- 4
Slope of (1, 4) (2, 3) = = -1 and
Since there are two distinct chords Þ D > 0 2 -1
Þ 36p 2 - 4 ´ 4(2)(p 2 + q 2 ) > 0 6-4
(1, 4) (–1, 6) = = -1 . Hence points are collinear..
-1-1
Þ 36p 2 - 32p 2 > 32q 2 Þ| p | > 2 2 | q | So no circle is drawn. Statement 1 is false Statement 2 is
25. (a) Let A = (1, 0), B = (3, 0) and C1, C2 be the center of true.
circles passing through A, B and touching the y-axis at 3
P1 and P2. If r be the radius of circles (note that radius 28. (d) A tangent to y2 = 12x is y = mx + , which passes
m
of both circles will be same), C1A = C2A = r = OD = 2
through (1, 4).
and C1 = (2, h).
Where, h2 = OC12 – AD2 = 4 – 1 = 3 3
Þ 4 =m+ Þ m 2 - 4m + 3 = 0 Þ m = 1, 3.
m
Þ C1 = (2, 3 ), C2 = (2, – 3 ) Two tangents can be drawn from (h, k) if k2 – 4ah > 0.
If ÐC1AC2 = d 29. (d) Let A (at12, 2at1) and B (at22, 2at2) subtend right angle
2at 1 - 2at 2at 2 - 2at
AC12 + AC 22 – C1C 22 1 at a point (at2, 2at) then at 2 - at 2 ´ at 2 - at 2 = -1
Þ cos a = =– 1 2
2AC1 .AC 2 2
Þ (t 1 + t)(t 2 + t) = -4
2p 1
Þa= Þ cos q = Þ t 2 + (t 1 + t 2 )t + t 1t 2 + 4 = 0
3 2
Clearly t3 and t4 are the roots of the above equation.
Hence t3 + t4 = – (t1 + t2) Þ |t1 + t2| = |t3 + t4|
x 2 y2
30. (b) The hyperbola is - = 1 . It asymptotes are
9 4
1 1
2
y= ± x . Clearly the given line is parallel to
3
2
y = - x , hence intersects the hyperbola at exactly
2 3
2
one point.
RELATIONS & FUNCTIONS
PART MATRICES & DETERMINANTS
TEST 4
Time : 1 hour
Max. Marks : 120
INSTRUCTIONS
Ø Crack this section in just 1 hour.
Ø This part test contains 30 questions.
Ø For every correct answer you will be awarded +4. There is no negative marking.
Ø Please fill the correct bubbles in the Response Sheet provided at the end of the test.
Ø Do not look at the solutions before attempting the complete test.
. . . Best of Luck
(c) If A and B are matrices of the same order, then (A + B) 2 (c) l =2 (d) l ¹2
= A2 + 2AB + B2 is possible if AB = I
(d) None of these écos a - sin a ù
6. If Aa = ê ú , then which of the following is
2. If A is a square matrix of order n, then adj (adj A) is equal to ë sin a cos a û
9. If g = {(1, 1), (2, 3), (3, 5), (4, 7),} is a function described by 17. For positive numbers x, y, z the numerical value of the
a b c (c) 1 (d) 3 3
is divisible by k, then the determinant l m n will 20. If A and B are two matrices such that A + B and AB are both
p q r defined, then
(a) A and B are two matrices not necessarily of same order
(a) be divisible by k (b) be divisible by k2
(b) A and B are square matrices of same order
(c) both of the above (d) none of the above
(c) Number of columns of A = Number of rows of B
p q-y r-z (d) None of these
p q r
14. If p - x q r - z = 0 , then the value of + + is 21. If A is a skew-symmetric matrix of order n and C is a column
x y z
p-x q-y r matrix of order n × 1, then C–1AC is
(a) A identity matrix of order n
(a) 0 (b) 1 (c) 2 (d) 4pqr
(b) A unit matrix of order one
15. In a upper triangular matrix n×n, minimum number of zeros is
(c) A zero matrix of order one
(a) n(n – 1)/2 (b) n(n + 1)/2
(d) None of these
(c) 2n(n – 1)/2 (d) None of these
22. Let l and a be real. The set of all values of x for which the
system of linear equations
(a x + a - x )2 (a x - a - x )2 1
l x + (sin a) y + (cos a) z = 0
16. (b x + b - x )2 (b x - b - x )2 1 =
(c x + c - x )2 (c x - c - x )2 1 x + (cos a) y + (sin a) z = 0
Relations & Functions Matrices & Determinants M-21
– x + (sin a) – (cos a) z = 0 has a non-trivial solution, is Statement-1 : The relation R is an equivalence relation.
Statement-2 : A relation R is an equivalence relation if it
(a) é 0, 2 ù (b) é - 2, 0 ù is reflexive, transitive and symmetric.
ë û ë û
27. Statement-1 : Every relation which is symmetric and
(c) é - 2, 2 ù (d) None of these transitive is also reflexive.
ë û
Statement-2 : If aRb then bRa as R is symmetric. Now
23. Let f(x) = | x – 1 |. Then aRb and bRa Þ aRa as R is transitive.
(a) f(x2) = (f(x))2 (b) f(x + y) = f(x) + f(y) 28. Statement 1 : The determinant of a matrix
(a) (I – A) (b) (I + A)
Statement 2 : The determinant of a skew symmetric
(c) (I – A)–1 (d) (I + A)–1 matrix of odd order is zero.
29. Statement-1 : If three lines
2
25. Let f (x) = , g(x) = cos x and h(x) = x + 3 then the L1 : a1x + b1y + c1 = 0,
x +1
L2 : a2x + b2y + c2 = 0 and
range of the composite function fogoh, is
L3 : a3x + b3y + c3 = 0 are concurrent,
(a) R+ (b) R – {0}
a1 b1 c1
(c) [1, ¥) (d) R+ – {1}
then a 2 b2 c2 = 0 .
ASSERTION & REASON a3 b3 c3
DIRECTIONS (Qs. 26 - 30) : Each of these questions
contains two statements: Statement-1(Assertion) and a1 b1 c1
Statement-2 (Reason). Choose the correct answer (ONLY
Statement-2 : If a 2 b2 c 2 = 0 , then the lines L1, L2,
ONE option is correct) from the following -
a3 b3 c3
(a) Statement-1 is false, Statement-2 is true.
(b) Statement-1 is true, Statement-2 is true; Statement-2 L3 must be concurrent.
is a correct explanation for Statement-1.
30. Statement 1 : The period of
(c) Statement-1 is true, Statement-2 is true; Statement-2
is not a correct explanation for Statement-1. 1
(d) Statement-1 is true, Statement-2 is false. f (x) = sin 2x cos [2x] – cos 2x sin [2x] is
2
that xRy Û x 2 + y2 = 1
Response Sheet
ANSWER KEY
1 (d) 7 (b) 13 (a) 19 (a) 25 (c)
2 (c) 8 (a) 14 (c) 20 (b) 26 (a)
3 (b) 9 (b) 15 (a) 21 (b) 27 (a)
4 (d) 10 (b) 16 (a) 22 (c) 28 (b)
5 (b) 11 (c) 17 (d) 23 (d) 29 (d)
6 (a) 12 (b) 18 (a) 24 (c) 30 (b)
1. (d) r r -1
(a) We have | AB | = | A | | B | 4. (d) det (Mr) = = 2r - 1
r -1 r
Also for a square matrix of order 3, | kA |= k 3 | A | 2007 2007
because each element of the matrix A is multiplied by k
and hence in this case we will have k3 common
å det (M r ) = 2 å r - 2007
r =1 r =1
\ | 3AB |= 33 | A || B |= 27( -1)(3) = -81 2007 ´ 2008
= 2´ - 2007 = (2007) 2
(b) Since A is invertible, therefore A –1 exists and 2
5. (b) For trivial solution,
AA -1 = 1 Þ det(AA -1 ) = det(I)
1 -2 1
Þ det(A) det(A -1 ) = 1 4
2 - 1 3 ¹ 0 Þ – 5l – 4 ¹ 0 or l ¹ -
5
1 l 1 -1
Þ det(A -1 ) =
det(A) écos a - sin aù é cos a sin a ù
6. (a) A a × A ( - a) = ê ú ê- sin a cos a ú
(c) 2
(A + B) = (A + B)(A + B) ë sin a cos a û ë û
é4 2 2ù l1 b c
ê T ú = k l2 m n
\ Adj A = C = ê- 5 0 5ú
ëê 1 - 2 3úû l3 q r
= k ( Integer) Þ D is divided by k
é4 2 2ù
-1 AdjA 1 ê ú p q-y r-z
\A = = - 5 0 5ú
| A | 10 ê 14. (c) p-x q r-z = 0
ëê 1 - 2 3úû p-x q-y r
Comparing , we get a = 5 Apply R1 ® R1 – R3 and R2 ® R2 – R3, we get
9. (b) (1, 1) satisfies g( x) = ax + b , \a + b = 1
x 0 -z
(2, 3) satisfies g( x) = ax + b , \ 2a + b = 3
0 y -z = 0
Solving the two equation, we get a = 2, b = –1.
p-x q-y r
10. (b) X Ç Y = {1,3,5}, (X - Y) È (Y - X)
= (2,4) È (7,9) = (2,4,7,9) Þ x[yr + z(q - y)] - z[0 - y(p - x)] = 0
We rewrite F as follows : [Expansion along first row]
Þ xyr + xzq - xzy + yzp - zyx = 0
Since 1 Î X and y = x + 2 = 1 + 2 = 3 Î Y,
\ (1,3) Î F p q r
Þ xyr + zxq + yzp = 2xyz Þ + + =2
x y z
Since 2 Î X but y = 2 + 2 = 4 Ï Y ,
15. (a) As we know, a square matrix A = | aij | is called an upper
we have (2,4) Î F. triangular matrix if aij = 0 for all i > j.
Since 3 Î X and y = 3 + 2 = 5 Î Y, é1 2 4 3ù
we have (3,5) Î F ê0 5 1 3úú
Such as A = ê
since 4 Î X and y = 4 + 2 = 6 Ï Y, ê0 0 2 9ú
ê ú
we have (4,6) Ï F ë0 0 0 5û 4´4
Finally since 5 Î X and y = 5 + 2 = 7 Î Y, we have
4(4 - 1)
(5,7) Î f. Hence F = {(1,3),(3,5),(5,7)} Number of zeros = =6
Here F is a relation from X to Y since F Ì XxY but it is 2
not a mapping since the elements 2 and 4 of the domain Hence number of zeros in a upper triangular matrix of
X have no image in Y under f. n(n - 1)
order n × n =
11. (c) Range of f is [1, ¥) Þ f is onto 2
For one-one f ( x1 ) = f (x 2 ) Þ x1 = x 2 16. (a) Put x = 0, which gives answer (a).
log a
Þ | x1 | +1 =| x 2 | +1 Þ | x1 | = | x 2 | 17. (d) Replace logba by
log b
Þ x1 = ± x 2 . Hence f is not one-one.
12. (b) Given f(x) = x–x2 + x3 – x4 + ....to ¥ log x log y log z
1
x \D = ´ log x 3log y log z
Þy= (Infinite G. P.) log x log y log z
1+ x log x log y 5log z
Þ y + xy = x Þ y = x (1 - y) Take log x, log y, log z common from C1, C2, C3
y y respectively.
Þ x= Þ f -1 ( y) =
1- y 1- y 1 1 1 1 0 0
x D = 1 3 1 = 1 2 0 = 1´ 2 ´ 4 = 8
Þ f -1 (x) = 1 1 5 1 0 4
1- x
a b c
N(N + 1)
13. (a) D = l m n , R1 ® 100 R1 + 10 R 2 + R 3 1 5
2
p q r N N(N + 1)(2N + 1)
18. (a) å Un = 2N + 1 2N + 1
abc b c kl1 b c n =1 6
{ }
2
D = lmn m n = kl 2 m n , N(N + 1)
3N 2 3N
pqr q r kl 3 q r 2
[where l1, l 2 , l3 are integers.]
M-24 Relations & Functions Matrices & Determinants
INSTRUCTIONS
Ø Crack this section in just 1 hour.
Ø This part test contains 30 questions.
Ø For every correct answer you will be awarded +4. There is no negative marking.
Ø Please fill the correct bubbles in the Response Sheet provided at the end of the test.
Ø Do not look at the solutions before attempting the complete test.
. . . Best of Luck
27. Let f (x) = 2 + cos x for all real x. Statement-2: L.H.D. at x = 2 = R.H.D. at x = 2
29. Statement–1 : Angle of intersects in between y = x2 and
Statement - 1 : For each real t, there exists a point c in
6y = 7 – x3 at (1, 1) is p/4.
[t, t + p] such that f '(c) = 0 because
Statement–2 : Angle of intersection between any two curve
Statement - 2 : f(t) = f(t + 2p) for each real t. is angle between the tangents at the point of intersection.
28. Statement-1: The function f(x) defined by 30. Statement-1: f : R ® R be a function such that
ì x, for x < 1 f (x) = x3 + x2 + 3x + sin x. Then f is one-one.
ï
f (x) = í2 - x, for 1 £ x £ 2 Statement-2: f (x) is neither increasing nor decreasing.
ïî -2 + 3x - x 2 , for x > 2
is differentiable at x = 2.
Response Sheet
ANSWER KEY
1 (c) 7 (b) 13 (c) 19 (a) 25 (b)
2 (d) 8 (d) 14 (c) 20 (c) 26 (b)
3 (c) 9 (b) 15 (b) 21 (d) 27 (c)
4 (d) 10 (d) 16 (c) 22 (d) 28 (b)
5 (c) 11 (a) 17 (a) 23 (c) 29 (a)
6 (c) 12 (c) 18 (c) 24 (b) 30 (d)
M-28 Differential Calculus
sin x -1
é ù tan x
Lim ê
{1 + (2 sin x - 1)(sin x - 1)}ú =e -1 / 2 é 1 ù
-
sin x
x® p / 6 ê 1 ú lim ê(1 + tan x ) tan x ú = e–1 ´ e–1 = e–2
x ®0 ê ú
ëê (2 sin x - 1)(sin x - 1) ûú
ë û
1
5. (c) f(x) = is discontinuous at x =1. 11. (a) Let f(x) = ax 2 + bx + c
x -1
( x - 1) 2 ax 3 bx 2
(gof) (x) = g(f(x)) = - , which is not defined g( x ) = f (x ) = + + cx
( 2x - 1)( x - 2) 3 2
at x = 1/2, 2. a b
Hence the set of points where (gof) (x) is discontinuous \ g(1) = + + c = 2a + 3b + c = 0
3 2
is {1/2, 1, 2}
6. (c) Let f(x) = ax2 + bx + c where a > 0, b2 < 4ac. g(0) = 0 = g(1)
Þ f(x) > 0 for all x \ g(x) is continuous apply Rolle’s theorem
Differential Calculus M-29
Þ g( x) = 0 for some value For local maxima, even derivative must be negative.
Since maximum value at x = a is b
x Î (0,1) Þ g' (x ) = f ( x) = ax 2 + bx + c = 0 \ f (x) = b – (x – a)2n + 2 (Q f2n + 2 (a) = –ve).
has a root in [0, 1] 17. (a) Since g(x) is decreasing,
\ g (x2) £ g (x1) when x2 ³ x1.
12. (c) f ( x ) = sin x + cos x + tan 2x + cot 2 x
Since f(x) is increasing,
sin x & cos x are cont. "x Î R But tan 2x is \ f [g(x1)] ³ f [g(x2)]
discontinuous at 2x = (2n + 1)p / 2 Þ h(x1) ³ h(x2) when x2 ³ x1.
Þ h(x) is a decreasing function of x and h(0) = 0.
2x = (2n + 1)p / 2
Also, domain of h = [0, ¥ ) and range of h = [0, ¥ ).
cot 2x is discontinuous. at 2x = np
\ h(x) = 0, " x Î [0, ¥ )
p 18. (c) x1/3 + y1/3 = a1/3 Þ a1/3
Þ x =n Þ f (x ) is discontinous at
2 1 2/3 1 –2/3 dy
Þ x + y =0
ì np pü ì np ü 3 3 dx
x Î í È ( 2n + 1) ý Þ x Î í "n Î Iý
î2 4þ î4 þ dy x -2/3 x 2/2
13. (c) At x =1, f (1 – 0) = 0, f (1 + 0) = 1 Þ = – -2/3 = - 2/3
dx y y
Þ f(x) is discontinuous as well as non-differentiable at
x = 1. æ dy ö
At x = 0, f (0 – 0) = f (0 + 0) = f (0) = 0 Þ çè dx ÷ø = –1
Þ f(x) is continuous at x = 0. (a / 8,a / 8)
f ( h ) - f ( 0)
The equation of the tangent at (a/8, a/8) is given by
f '(0–) = Lim-
h ®0 h a æ aö a
y- = -1 ç x - ÷ or x + y – = 0
-h - 0 8 è 8ø 4
= Lim- = –1 The x and y intercepts of this line on the coordinate
h ®0 h
axes are each equal to a/4, So we have
f ( h ) - f (0 )
f '(0+) = Lim+ =0 2
æ aö æ aö
2
h ®0 h çè ÷ø + çè ÷ø = 2 Þ a = 4
Thus f(x) is not differentiable at x = 0. 4 4
14. (c) Consider the function f( x ) = f (x ) - g(x ) on the 19. (a) By Rolle's theorem f(1) = f(3) gives
a + b + 11 – 6 = 27a + 9b + 33 – 6
interval [ x 0 , x ]. Then, f( x) satisfies all the conditions Þ 26a + 8b + 22 = 0
of Lagrange’s mean value theorem on [ x 0 , x ]. Þ 13a + 4b + 11 = 0
Again f ¢ (x) = 3ax2 + 2bx + 11
\ There exists at least one c Î (x 0 , x ) such that \ (1) or (3) are possible
f( x ) - f(x 0 ) = f¢(c)(x - x 0 ) æ 1 ö
Þ f(x ) = f¢(c)(x - x 0 ) (Q f( x 0 ) = 0) Since f ¢ çè 2 + ÷ =0
......(1) 3ø
Also f¢( x ) = f ¢( x) - g¢( x ) Þ f¢(c) = f ¢(c) - g¢(c) > 0 2
æ 1 ö æ 1 ö
(Q f ¢( x ) > g ¢( x ) for x > x0) \ 3a ç 2 +
è ÷ø + 2b çè 2 + 3 ÷ø + 11 = 0
3
\ From (1) , f(x ) > 0, for x > x0 Simplify it, we get b + 6a = 0 .....(2)
Þ f (x ) - g( x) > 0 for x > x 0 Clearly (a) satisfies it only and (c) does not.
Hence a = 1, b = – 6
or f ( x) > g(x ) for x > x 0 . 20. (c) Using Lagrange's Mean Value Theorem
15. (b) Let the point be ( x1, y1). Let f(x) be a function defined on [a, b]
Therefore y1 = (x1 – 3)2 ...(i) f (b) - f (a)
\ Now slope of the tangent at ( x1, y1) is then, f '(c) = ....(i) c Î [a, b]
b-a
2(x1 – 3), but it is equal to 1.
1
7 \ Given f(x) = logex \ f'(x) =
Therefore, 2(x1 – 3) = 1 Þ x1 = x
2
\ equation (i) become
2
æ7 ö 1 1 f (3) - f (1)
y1 = ç – 3÷ = . =
è2 ø 4 c 3 -1
æ 7 1ö 1 loge 3 - log e 1 log e3 2
Hence the point is ç , ÷ . Þ = = Þ c=
è 2 4ø c 2 2 loge 3
16. (c) For local maximum or local minimum odd derivative Þ c = 2 log3e
must be equal to zero.
M-30 Differential Calculus
21. (d) Let at any instant, the radius of the base and height of
the cone formed by the water in the filter be x and y æ1 1 1 ö
= lim+ ç + 2 + 3 + ...÷ = ¥ (infinite)
respectively x ®0 è x x 2x ! ø
\ volume of water in the filter at that time is
e1/ x
1 x 10 1 1 \ lim+ does not exist
V = px2y But = = \x= y x ®0 x
3 y 20 2 2
25. (b) f ¢(x) = ln (x + 1 + x 2 ) = – ln ( 1 + x 2 - x )
1 1 2 py3
\ V= p y . y= Þ f ¢ (x) > 0
3 4 12 Þ f (x) when x < 0
dV p 2 dy py 2 dy Þ f (x) is increasing when x > 0.
\ = 3y = Þ f (x) > f (0) Þ f (x) > 0.
dt 12 dt 4 dt Again f(x) is decreasing in (–¥, 0)
dV Þ f (x) > f (0) Þ f (x) > 0.
Given =5 26. (b) f(x) = sin x continuous in [–p/2, p/2]
dt
By intermediate value theorem
dy f (–p/2) = sin (–p/2) = –1
We are to find , when y = 15
dt æ pö p
f ç ÷ = sin = 1
(15) 2 dy è 2ø 2
\ 5=p
4 dt æ pö æ pö
dy 5´ 4 1 4 f ç - ÷ and f ç ÷ are of opposite sign
= . = è 2ø è 2ø
\ cm/sec.
dt 15 ´ 15 p 45p \ By intermediate value theorem,$ a point
22. (d) Let the speed of the train be v and distance to be covered c Î [–p/2, p/2] such that f(x) = 0
be s so that total time taken is s/v hours. Cost of fuel per $ a point x Î [–p/2, p/2] such that f (x) = 0
hour = kv2 (k is constant) i.e., sin x = 0
Also 48 = k. 162 by given condition
é p pù
3 thus sin x = 0 has at least one root between ê - , ú
\k= ë 2 2û
16 27. (c) Given that f (x) = 2 + cos x which is continuous and
3 2 differentiable every where.
\ Cost to fuel per hour v . Also f ' (x) = – sin x
16
\ f ' (x) = 0 Þ x = np
Other charges per hour are 300.
Total running cost , Þ There exists c Î [t, t + p] for t Î R
Such that f ' (c) = 0
æ 3 ös 3s 300 s \ Statements-1 is true.
C = ç v 2 + 300 ÷ = v +
è 16 ø v 16 v Also f (x) being periodic of period 2p, statement-2 is
true, but statement-2 is not a correct explanation of
dC 3s 300 s statement-1.
= - 2 = 0 Þ v = 40
dv 16 v 28. (b) L.H.D. at x = 2
d 2 C 600 s ìd ü
= >0 = í (2 - x) ý = -1
dv 2 v3 î dx þx = 2
\ v = 40 results in minimum running cost. R.H.D. at x = 2
ìd ü
21/ x 1 = í (-2 + 3x - x 2 )ý = -1
23. (c) lim = lim =1 î dx þ x =2
x ®0 1 + 21/ x x ® 0 1 + 2-1/ x
dy dy 1
cos -1 (1 - x ) q 29. (a) m1 = ; m2 = dx =-
lim = lim dx x =1 x =1 2
x ®0 x x ®0+ 1 - cos q
m1m2 = –1. Hence angle is p/2
(let, cos–1(1 – x) = q Þ 1 – x = cos q) 30. (d) Every increasing or decreasing function is one-one
q f ¢ (x) = 3x2 + 2x + 3 + cosx
lim = 2
x ®0 + æ qö æ 1ö 8
2
2 sin ç ÷
è 2ø = 3 ç x + ÷ + + cos x > 0
è 3ø 3
1 1 1 2
1+ + + ... æ 1ö 8 8
24. (b) e1/ x x 2! x 2 [Q | cosx | < 1 and 3 ç x + ÷ + ³ ]
lim = lim è 3 ø 3 2
x ® 0+ x x ® 0+ x
\ f (x) is strictly increasing.
PART INTEGRAL CALCULUS
TEST 6
Time : 1 hour
Max. Marks : 120
INSTRUCTIONS
Ø Crack this section in just 1 hour.
Ø This part test contains 30 questions.
Ø For every correct answer you will be awarded +4. There is no negative marking.
Ø Please fill the correct bubbles in the Response Sheet provided at the end of the test.
Ø Do not look at the solutions before attempting the complete test.
. . . Best of Luck
2 p 2m +1
ò 0 cos
æ d2 y ö 2 æ d2y ö
æ dy ö (c) x dx (d) none of these
1. The degree of the equation ç ÷ + çè ø÷ = x cos ç 2÷
è dx 2 ø dx è dx ø
is 1 tan -1 x p/2 x
(a) 1 (b) 2
6. If ò0 x
dx = k ò
0 sin x
dx, then the value of k is
(c) 3 (d) none
1
1
(a) 1 (b)
ò f (x )dx = log( f (x )) + C, then f(x) is
2. 2
4
(a) x + a (b) 2x + a 1
(c) 4 (d)
x 2
(c) +a (d) x2 + a
2 d e sin x
7. Let f (x) = ,x >0
f ( x) dx x
3. If f ¢( x) = f ( x ) and f (0) = 2, then ò 3 + 4f (x ) dx = sin x
4 2e
2
1
If ò1 x
dx = f (k) - f (1), then one of the possible
(a) x
log(3 + 8e ) + C (b) log(3 + 8e x ) + C
4 value of k is
(a) 4 (b) 16
1
(c) log(3 + 8e x ) + C (d) none of these (c) 8 (d) none of these
2 8. Find the area lying between the curves y = tan x, y = cot x
4. ò x | x | dx = and x-axis, x Î ê0, ú
é pù
2û ë
x3 x2 | x |
(a) (b) 1
3 3 (a) log 2 (b) log 2
2
x2 | x | æ 1 ö
(c) (d) none of these (c) 2 log ç (d) none of these.
2 è 2 ÷ø
p
5. Value of ò 0 sin
n
x cos 2m+1 x dx 9. Area bounded by the curve xy 2 = a 2 (a - x ) and y-axis is
( where m, n Î N) is pa 2
(a) (b) pa 2
(2m + 1)! (2m + 1)! 2
(a) (b)
n! n2 (c) 3pa 2 (d) none of these.
M-32 Integral Calculus
q q2 1 sin x - 1 1 2 sin x - 1
The point on the curve x = ò sin -1 -1 log - log +C
10. t dt, y = ò cos tdt , (a) 4 sin x + 1 2 2 sin x + 1
0 0
where the tangent is parallel to the x-axis is given by 1 cos x - 1 1 2 cos x - 1
(a) q = 0 (b) q = 1/2 (b) log - log +C
(c) q = 1 (d) q = –1 8 cos x + 1 2 2 2 cos x + 1
p/4
1 sin x - 1 1 2 sin x - 1
ò sin
4
11. The value of 2x cos 2 xd(x - [x ]) is (c) log - log +C
8 sin x + 1 4 2 2 sin x + 1
0
(a) 3p/64 + 1/20 (b) 3p/32 + 1/20 (d) none of these
(c) 3p/32 + 1/10 (d) 3p/16+1/20 1 x 1 x
If I1 = ò02 dx, I2 = ò02 dx,
3 4
17.
2
(log x + ax )
12. òe cos(bx 2 + c) dx is equal to
2 x 2 x
I3 = ò1 2 dx and I 4 = ò1 2 dx then
3 4
1 2 b
(a) e ax cos( bx 2 + c + tan -1 ) + A
2
a +b 2 a (a) I1 = I 2 (b) I 2 > I1
1 2 b (c) I3 > I 4 (d) I 4 > I 3 , I1 > I 2 .
(b) e ax cos( bx - c - tan -1 ) + A
2
2 a +b 2 a
18. If the function f ( x ) = Pe 2 x + Qe x + Rx satisfies the
1 2 b conditions f (0) = -1, f ¢(log 2) = 31 and
(c) e ax cos( bx + c - tan -1 ) + A
2
a +b 2 a
log 4 39
1 2 b ò0 (f (x) - Rx)dx =
2
, then
(d) e ax cos( bx + c - tan -1 ) + A
2 a 2 + b2 a (a) P = 5, Q = -6, R = 3 (b) P = 3, Q = 2, R = 3
13. The value of (c) P = -5, Q = 6, R = 3 (d) P = -5, Q = 6, R = 3
p/4 np+ t
ò (x | x | + sin3 x + x tan 2 x + 1) dx is
-p / 4
19. ò (| cos x | + | sin x |)dx
0
(a) 0 (b) 1 (a) cos t (b) n
(c) p/4 (d) p/2 (c) 2n + sin t + cos t (d) sin t - cos t + 4n + 1
x2 8a 3
2 20. Area included between y = and y = 2 is
xæ 1- x ö
14. òe çç
è 1+ x2
÷÷ dx =
ø
4a x + 4a 2
a2 a2
x x (a) ( 6p - 4) (b) (4 p + 3)
e 2xe 3 3
(a) +C (b) +C
1+ x 2 (1 + x 2 ) 2 a2
(c) (8p + 3) (d) none of these
3
e -x
(c) +C (d) none of these 4x -x d3y dy
1+ x2 21. If y = e + 2e satisfies the relation 3 + A + By = 0 ,
dx dx
2 then values of A and B respectively are
( x - 1)dx
15. If ò = log | tan -1 f ( x ) | + C, then
æ x2 +1 ö (a) –13, 14 (b) –13, –12
( x 4 + 3x 2 + 1) tan -1ç ÷ (c) –13, 12 (d) 12, – 13
ç x ÷
è ø 22. A steam boat is moving at velocity V when steam is shut off.
Given that the retardation at any subsequent time is equal
x 2 +1 to the magnitude of the velocity at that time. The velocity v
(a) f(x) = x2 + 1 (b) f ( x) =
2x in time t after steam is shut off is
(a) v = Vt (b) v = Vt – V
x2 +1 1 2
(c) v = Vet (d) v = Ve–t
(c) (d) f ( x) = (x + 1)
x 2 23. If the slope of the tangent at (x, y) to a curve passing
sin x æ pö y æyö
16. The primitive of
sin 4 x
is through ç 1, ÷ is given by - cos 2 ç ÷ , then the
è 4ø x èxø
equation of the curve is
Integral Calculus M-33
Response Sheet
ANSWER KEY
1 (d) 7 (b) 13 (d) 19 (d) 25 (d)
2 (c) 8 (a) 14 (a) 20 (a) 26 (b)
3 (b) 9 (b) 15 (c) 21 (b) 27 (d)
4 (b) 10 (c) 16 (c) 22 (d) 28 (b)
5 (d) 11 (a) 17 (d) 23 (c) 29 (a)
6 (d) 12 (d) 18 (a) 24 (a) 30 (b)
x
Since f (0) = 2, therefore f (x) = 2ex.
t an
y=
p/4 p/2
y=
ex ò0 tan x dx + ò cot x dx
co
=
\ I = 2ò
tx
dx p/4
x
3 + 8e
Put ex = t, \ dt = ex dx p/4 p/2
= [ log sec x ] 0 + [ log sin x ] p / 4 X
dt 1 O x = p/2
\ I = 2ò Þ I = log(3 + 8e x ) + C p/4
3 + 8t 4 = log 2 + log 2 = 2 log 2 = log 2
4. (b) ò ò
I = x. | x |= | x | .x dx ¥ ¥ a3
9. (b) Area = 2 ò xdy = 2ò dy
(Integrating by parts taking |x| as first function) 0 0 y2 + a 2
¥ A)(a, 0)
1é yù p
x2 | x | x2 x2 | x | 1 = 2a 3 . ê tan -1 ú = 2a 2 = pa 2
=| x | . -ò . dx = - ò | x | x dx aë aû0 2
2 x 2 2 2
2 q
æ 1ö x |x| 1 2 dx
Þ Iç1 + ÷ = ÞI= x |x| 10. (c) x = ò sin -1 tdt Þ = sin -1 q ;
è 2ø 2 3 dq
0
p
5. (d) I = ò sin n x cos2m +1 x dx q2
-1 dy
0 y= ò cos tdt Þ = 2q cos -1 q
p dq
= ò sin n (p - x).cos 2m+1 (p - x) dx 0
0
dy dy / dq pq
p = = = -2q;
= -ò sin n x cos2m+1 x dx = -I Þ I = 0 dx dx / dq sin -1 q
0
Similarly, one can show that dy p
= 0 Þ sin -1 q = Þ q = 1
p 2m+1 dx 2
ò 0 cos x dx = 0
p/4
(d) Put x = tanq Þ dx = sec2q dq
6.
-1
11. (a) ò sin 4 2x cos 2 x d(x - [x])
1 tan x p/4 q 0
\ ò0 x
dx = ò
0 tan q
sec 2 q dq
1
p/4
1
p/4
sin 4 2x dx + ò sin 4 2x cos 2x dx
2 ò
p/4 q =
=ò dq 0
2
0
sin q cos q
0
(Since d (x – [x] ) = d(x)– d[x] = d(x) – 0 as d[x] is constant)
p/4 q 1 p/2 t
= 2ò dq = ò dt where 2q = t 1
p/2
1 é 5 p/4 3p 1
0 sin 2q 2 0 sin t sin 4 t dt + sin 2x ù
1
=
4 ò 20 ë û 0
= +
64 20
Hence k = 0
2
Integral Calculus M-35
è ø 2
2
a +b 2 a
17. (d) For 0 < x < 1, x 4 < x 3
Put b = 0 & a = 1 Þ ò xe x cos c dx = Kex cos(c) + A
2 2
4 3 4 3
\ 2 x < 2 x and 2 x > 2 x
cos c x
e = Ke x cos c + A 1 4 1 3
Þ \ ò 0 2 dx < ò 0 2 dx
x x
2 2
2
2 1 2 x4 1 3
Comparing coeff. of e x cos c , we get K =
2 and ò1 2 dx > ò 2x dx
2
p/4 Þ I 2 < I1 and I 4 > I3 Þ (d) is correct.
13. (d) I = ò (x | x |+ sin 3 x + x tan 2 x + 1) dx
p/4 ¯ ¯ ¯ 18. (a) f ( x ) = Pe 2 x + Qe x + Rx
odd f odd f odd f
p/4 Þ f ¢( x ) = 2Pe 2 x + Qe x + R
p
I = ò dx =
-p / 4 2 Þ 31 = 2Pe 2 log 2 + Qelog 2 + R
a Þ 8P + 2Q + R = 31 ......... (i)
Also, 0 = P + Q ......... (ii)
ò
[Q f ( x )dx = 0, as f (x ) is an odd function ]
log 4 39
-a
& ò 0 (f (x) - Rx)dx =
2 ì 2 ü 2
æ 1- x ö x ï1 + x - 2x ï
14. (a) I = ò ex ç ÷ dx = ò e í 2 2 ý
dx log 4 39
è 1+ x2 ø îï (1 + x ) þï Þò (Pe2x + Qex )dx =
0 2
ìï 1 -2x üï æ 1 ö éP ù
log 4
39
= ò ex í + 2 2ý
dx = e x çç ÷÷ + C. Þ ê e 2x + Qe x ú =
îï1 + x
2
(1 + x ) þï è1+ x2 ø ë2 û0 2
(x 2 - 1) dx P 2log 4 P 39
15. (c) I=ò Þ e + Q elog 4 - - Q =
æ x + 1ö 2 2 2 2
(x 4 + 3x 2 + 1) tan -1 ç ÷ Þ 15P + 6Q = 39 ......... (iii)
è x ø Solving (i), (ii) and (iii), we get P = 5, Q = – 6, R = 3
(Dividing Num. and Den. by x2) p
æ 1ö 19. (d) Since |sin x| + |cos x| is periodic with period , \
2
çè1 - 2 ÷ø dx dt
x np+ t
I =ò =ò
æ 2 1ö -1 æ 1 ö 2
( )
t + 1 tan -1 t ò0 (| cos x | + |sin x |)dx
çè x + 3 + 2 ÷ø tan çè x + x ÷ø p/2 t
x = 2n ò (| cos x | + |sin x |)dx + ò (| cos x | + |sin x |) dx
0 0
æ 1 æ 1ö ö
çè where t = x + x Þ dt = çè1 - 2 ÷ø dx ÷ø
p/2 t
x = 2n ò (cos x + sin x) dx + ò (cos x + sin x) dx
0 0
æ x 2 + 1ö p/2
= 2n [ sin x - cos x ] + [ sin x - cos x ] 0
t
= log | tan -1 t | + C = log tan -1 ç ÷ +C 0
è x ø = 2 n.2 + sin t - cos t + 1 = ( 4n + 1) + sin t - cos t.
x2 + 1 20. (a) The curve of y (x2 + 4a2) = 8a3 is symmetrical about
Þ f (x) = y-axis and cuts it at A (0, 2a). Tangent at A is parallel to
x x-axis. x-axis is asymptote. This curve meets x2 = 4ay
sin x
16. (c) I = ò
sin 4x
dx
Where,
x2
=
8a 3
4a x + 4a 2
2
Þ x 4 + 4a 2 x 2 - 32a 4 = 0
sin x dx
I= ò
2 sin 2x cos 2x
dx = ò
4 cos x cos 2x
Þ (x 2 - 4a 2 )(x 2 + 8a 2 ) = 0 Þ x = ± 2a
é 2a 8a 3 2a x
2
cos x \ Required area = 2 ê ò dx - ò
=ò dx 2 2 0 4a
dx
4(1 - sin 2 x )(1 - 2 sin 2 x ) êë 0 x + 4a
Put sin x = t Þ cos x dx = dt
A(0, 2a)
æ ö
dt 1 ç 1 1 ÷
\I = ò = òç 2 - ÷ dt
(–2a, a) (2a, a)
4(1 - t 2 )(1 - 2 t 2 ) 4 2 1
ç t -1 t - ÷
è 2ø
O
M-36 Integral Calculus
a2 3p /2 2p
= (6p - 4).
3
- ò cos xdx + ò cos x dx
5p / 4 3p / 2
21. (b) Given, y = e 4 x + 2e - x
Differentiating, we get (4 2 - 1)
= sq.units
2 2
dy d y
= 4e 4 x - 2e - x Þ = 16e 4 x + 2e - x 25. (d) A = 2 × 1/2 = 1 sq. unit
dx dx 2
d3 y
Þ = 64e4x - 2e - x
dx 3
0 1 2
d3 y dy
Putting these values in +A + By = 0
dx 3 dx 2p p 2p
We have,
(64 + 4A + B)e 4 x + (-2 - 2A + 2B)e - x = 0
ò | sin x |dx = ò sin xdx - ò sin x dx
0 0 p
Þ 64 + 4A + B = 0, - 2 - 2A + 2B = 0 = 4 sq. unit.
Solving these eqs., we get A = –13, B = –12 26. (b) Statement – II is true as any member of the family will
dv
22. (d) The retardation at time t = -
dt
. Hence, the differential
have equation
x2
+
( y - b) 2 = 1 , where 0 < e < 1,
equation is a2 (
a 2 1 - e2 )
dv dv
- =vÞ = -dt ...(i) a > 0, b Î R and ae = 2
dt v Hence F is a two parameter family.
Integrating, we get log v = - t + c ...(ii) Statement – I is true, because of statement – II, because
When t = 0, v = V Þ C = log V order of a differential equation of a n parameter family
The equation (ii) becomes log v = – t + logV is n
v v c4
Þ log = - t Þ = e - t Þ v = Ve - t 27. (d) y = c1ex + (c2 + c3) ex × e
V V
= ex (c1 + (c2 + c3) ec 4 )
dy y æyö
23. (c) Given, = - cos 2 ç ÷
dx x èxø
y = cex … (1) {here c = c + (c 1 2 + c3 )e c4 }
dy dv dy dy / dx
Putting y = vx so that = v+x = ce x Þ c = Put in (1)
dx dx dx ex
dv dy / dx x
We get, v + x = v - cos 2 v y= ´e
dx ex
dv dx dx dy
Þ 2
=- Þ sec 2 v dv = - So = y and order is 1.
cos v x x dx
Integrating, we get dx
28. (b) ò =x
eò
Pdx
æyö =e x
tan v = – ln x + ln c tan ç ÷ = - ln x + ln c
èxø \ sol. is xy = ò x 2 dx + c
Here A = a2 + b2
p 5p/4
òe
x
O p/4 (sec 2 x + tan x) dx = ex + ax + c
3p/2 X
y = cos x 2p 4 4
INSTRUCTIONS
Ø Crack this section in just 1 hour.
Ø This part test contains 30 questions.
Ø For every correct answer you will be awarded +4. There is no negative marking.
Ø Please fill the correct bubbles in the Response Sheet provided at the end of the test.
Ø Do not look at the solutions before attempting the complete test.
. . . Best of Luck
r r
1. If a = ˆi – 2ˆj + 3kˆ and b = –3iˆ + ˆj - kˆ and (a - d ).( b - c ) = ( b - d ).( c - a ) = 0 .
r r r r r r r r Then for the DABC, D is its
r ´ a = b ´ a, r ´ b = a ´ b
r (a) incentre (b) circumcentre
then a unit vector in the direction of r is (c) orthocentre (d) centroid
1 1 6. Given a parallelogram OACB. The length of the vectors
(a) (–2 î + ˆj - k̂ ) (b) (–2 î - ĵ + 2 k̂ )
3 3 OA , OB & AB are a, b, & c respectively. The scalar
1
(c) (–2 î - ˆj - 2 k̂ ) (d) None of these product of the vectors OC & OB is
3
(a) (a2 – 3b2 + c2)/2 (b) (3a2 + b2 – c2)/2
2. For nonzero vectors a , b , c the equality (c) (3a2 – b2 + c2)/2 (d) (a2 + 3b2 – c2)/2
ur ur ur ur ur ur r r r r r r r r r r r r
| (a ´ b ).c | = | a || b || c | holds if and only if 7. If (a ´ b) ´ (c ´ d) + (a ´ c) ´ (d ´ b) + (a ´ d) ´ (b ´ c) = la,
where l is a scalar then the value of l is
(a) a .b = 0, b .c = 0 (b) b .c = 0, c .a = 0 rrr
(a) zero (b) –2 [b c d]
(c) c .a = 0, a .b = 0 (d) a .b = b .c = c .a = 0 rrr rrr
(c) 2[a b c] (d) [b c d]
3. Let A & B be two non-parallel unit vectors in a plane. If r r r r r r
8. For a non-zero vector r , ( r .î )( r ´ î ) + ( r . ĵ)( r ´ ĵ) + ( r .k̂ )( r ´ k̂ )
(a A + B ) bisects the internal angle between A and is always equals to
B , then a is equal to (a) a (b) 2 a
(a) 1/2 (b) 1 (c) 3 a (d) None
(c) 2 (d) 4 9. Let a, b, c be distinct non-zero numbers. If the vector
r
4. A vector x is coplanar with vectors
r ˆ ˆ ˆ a i – a j + b k , i + k and ci + c j + b k lie in a plane then
a = –i + j + k & = 2iˆ + kˆ and is orthogonal to the vector b is
r r r r (a) the AM of a and c (b) the GM of a and c
b . If x . a = 7 then the vector x =
(c) the HM of a and c (d) equal to zero
(a) (–3iˆ + 5jˆ + 6k)
ˆ (b) 1/2 (–3iˆ + 5jˆ + 6k)
ˆ
10. If the vector product of a constant vector OA with a variable
(c) (3iˆ – 5jˆ – 6k)
ˆ (d) 1/2 (3iˆ – 5jˆ + 6k)
ˆ
vector OB in a fixed plane OAB be a constant vector then
5. A, B, C and D are four points in a plane with position vectors locus of B is
a , b , c and d respectively such that
M-38 Vector and 3-D Geometry
(a) a 2 + b2 + c2 (b)
1 2
a + b2 + c2
(d) ( -8 14 + 1, 12 14 - 1, - a4 14 )
2
(c) a2 + b2 + c2 (d) None
Vector and 3-D Geometry M-39
Response Sheet
ANSWER KEY
1 (b) 7 (b) 13 (a) 19 (b) 25 (d)
2 (d) 8 (d) 14 (c) 20 (a) 26 (b)
3 (b) 9 (c) 15 (a) 21 (d) 27 (a)
4 (b) 10 (b) 16 (c) 22 (d) 28 (d)
5 (c) 11 (a) 17 (b) 23 (c) 29 (d)
6 (d) 12 (b) 18 (a) 24 (c) 30 (c)
M-40 Vector and 3-D Geometry
1. (b) We have
(b) B C(a + b)
r ´a + r ´b = a ´b + b ´a = 0 é ù
êFrom, | b - a |=| c | ú
Þ r ´ (a + b ) = 0 ê ú
b c êb 2 + a 2 = 2a .b = c 2 ú
ê ú
Þ r = l (a + b ) = l (–2 î – ˆj + 2 k̂ ) ê a 2 + b 2 - c2 ú
êëa .b = 2 úû
1
Þ r̂ = (–2 î – ĵ + 2k̂ ) O a A (a)
3
ur ur ur 7. (b)
2. ( )
(d) | a ´ b .c | = volume of the parallelepiped
(a ´ b ) ´ (c ´ d ) + ( a ´ c )
´ ( d ´ b ) + (a ´ d ) ´ ( b ´ c ) = c
= | a || b || c |
Þ It is a rectangular parallelopiped, i.e., concurrent Þ [c d a ]b – [c d b ]a + [d b a ]c
edges are perpendicular to each other. - [d b c ]a + [a d c ]b - [a d b ]b = 2[b c d ]a
So, a .b = 0, b .c = 0, c .a = 0
Hence, l = -2[ b c d ]
3. (b) Vector bisecting internal angle between A & B 8. (d) Let r = x î + yĵ + zk̂ Þ r .î = x
r r
A B r r ur
is = r + r = a A + B Þ a = 1 and r ´ ˆi = zjˆ – ykˆ
|A| |B| ur
The resultant of two unit vectors is the angle bisector Similarly r .jˆ = y , r ´ ĵ = xk̂ – zk̂ , r .k̂ = z and
vector between the two vectors.
r ´ k̂ = yî – xˆj . Thus the given expression becomes
x (zĵ – yk̂) + y( xk̂ – z î ) + z ( yî – xˆj) = 0
A
a -a b
9. (c) Here | a b c |= 0 . So, 1 0 1 = 0
B
r r
= (m - 1)q + nr ......(1) 1 2
Similarly radius a + b2 + c2 .
r r r r r r 2
= q ´ {(x - r) ´ q} = (n - 1)r + l p ......(2) 18. (a) Given that l – 5m + 3n = 0 …..(1)
r r r r r r & 7l2 + 5m2 – 3n2 = 0 ……..(2).
& = r ´ {x - p) ´ r} = (l - 1)p + mq ......(3)
Adding (1), (2), (3) Substituting the value of l from (1) in (2), we get
r r r 7(5m – 3n)2 + 5m2 – 3n2 = 0,
= (2l - 1)p + (2m - 1)q + (2n - 1)r = 0 or 180 m2 – 210 mn + 60n2 = 0
r r r
r (p + q + r) m 2 1
Þ l = m = n =1/ 2 \ x = or (3m – 2n) (2m – n) = 0 or = ,
2 n 3 2
r rr r r
13. (a) p r + ( r .b ) a = c ...(I)
rr rr rr rr If m = 2
p( r .b) + ( r.b)(a.b) = c.b n 3
rr
r r c.b rr Therefore, m = n = 5m - 3n = l (from (1))
Þ r.b = , since a.b = 0 , putting in (I), 2 3 5.2 - 3.3 1
p
r rr rr r
r
r c (c.b) r c b.c a
Þr= - a = –
( ) or,
l m n
= = =
l2 + m 2 + n 2
=
1
p p2 p p2 1 2 3 (1)2 + (2)2 + (3) 2 14
14. (c) If the given lines are coplanar, then a line which is Therefore direction cosines of one line are
normal to the plane in which they lie, will be
perpendicular to all the three given lines. Let the 1 2 3
direction cosines of the normal be l, m, u. – , . If m = 1 ,
,
14 14 n 2 14
\ ll1 + mm1 + un1 = 0, ll2 + mm2 + un2 = 0,
ll3 + mm3 + un3 = 0. Eliminating l, m, u from these m n 5m - 3n l [from (1)] or,,
equations, = = =
1 2 5.1 - 3.2 -1
l1 m1 n1
we get, l 2 m 2 n 2 = 0 . l m n l2 + m 2 + n 2 1
= = = = .
l3 m3 n 3 -1 1 2 2 2
( -1) + (1) + (2) 2 6
\ The direction cosines of other line are
15. (a) The equation of a line passing through A(2, –3, –1)
and B(8, –1, 2) is 1 1 2
- , ,
x - 2 y + 3 z -1 x - 2 y + 3 z -1 6 6 6
= = Þ = = 19. (b) Let OP be the normal from
6 2 3 6 2 3 O to the plane of DABC,
7 7 7 let l, m, n be the direction
The coordinates of points on this line at a distance r cosines of OP
Let s be the area of DABC.
æ 6r 2r 3r ö Therefore
from A are given by ç 2 ± ,-3 + ,1 ± ÷ .
è 7 7 7ø OAC = Projection of DABC on zx plane = s.m.
Putting r = 14, we get the required points as (4, 1, 5) and But OAC is a right angled triangle, so that area of
(–10, –7, –7). 1
16. (c) The equation of a plane through the line of inter-section DOAC = a.c.
2
of the planes ax + by + cz + d = 0
1 ac
& a'x + b'y+c'z+d' = 0 is \ ac = s.m. Þ m = . Similarly in DOBC,
(ax + by + cz + d) +l(a'x + b'y + c'z + d') =0 2 2s
or x(a + la') + y(b + lb') + z(c + lc') + d + ld' =0 …..(i). bc ab
This is parallel to x-axis i.e. y = 0, z = 0. l= and in D OAB, n =
2s 2s
\ x (a+ l a') + 0 (b+ l b') + 0(c+ l c') = 0 Since, l2 + m2 + n2 = 1
a 2 2 2
Þ l = - . Putting the value of l in (i), the required æ ac ö æ bc ö æ ab ö
a' \ ç ÷ +ç ÷ +ç ÷ =1
plane is è 2s ø è 2s ø è 2s ø
y(a'b – ab') + z(a'c – ac') + a'd–ad' = 0 1 2 2
Þ y(ab' – a'b) + z(ac' – a'c) + ad' + a'd = 0 Þs= b c + c 2a 2 + a 2 b 2
2
17. (b) Let P (a, b, g) be the foot of the perpendicular from the 20. (a) Let the equation of the required sphere be x2 + y2 +z2
origin O on the given plane. The plane passes through
+ 2ux+ 2vy + 2wz + l = 0. This passes through (1, 0, 0),
A (a, b, c). Therefore, AP ^ OP Þ AP .OP = 0 (0, 1, 0) and (0, 0, 1), therefore 1 + 2u + l = 0 …… (i),
1 + 2v + l = 0 …….(ii) and 1 + 2w + l = 0 …. (iii)
Þ [( a - a )î + (b - b ) ĵ + ( g - c)k̂ ].(a î + b ĵ + gk̂ ) = 0
æ l +1 ö æ l +1ö æ l +1ö
Þ a (a - a ) + b(b - b) + g ( g - c) = 0 . Hence, the locus \ u= -ç ÷, v = -ç ÷, w = - ç ÷.
of (a, b, g) is x(x – a) + y (y – b) + z (z –c) = 0 è 2 ø è 2 ø è 2 ø
or x2 + y2 + z2 –ax – by – cz = 0. Clearly, it is a sphere of Let R be the radius of the sphere.
M-42 Vector and 3-D Geometry
Then R2 = u2 + v2 +w2 – l 24. (c) The coordinates of any point on the given line are
2 (2r + 1, –3r – 1, r)
æ l +1 ö 1 2 The distance of this point from the point (1, –1, 0) is
or R2 = 3ç ÷ - l = (3l + 2l + 3)
è 2 ø 4
given to be 4 14
3 éæ 8ù
2
1ö
= êç l + ÷ + ú Þ (2r)2 + (–3r)2 + (r)2 = (4 14 )2
4 êè 3ø 9ú Þ 14 r2 = 16 × 14 Þ r = ± 4
ë û
1 So the coordinates of the required point are (9, –13, 4)
Þ R is minimum if l + = 0 i.e. l = –1/3 or (–7, 11, – 4)
3 Out of which nearer the origin is (–7, 11, – 4)
Therefore, u = v = w = –1 / 3. Hence, the required sphere 25. (d) P1 = P2 = 0, P2 = P3 = 0 and P3 = P1 = 0 are lines of
is 3 (x2 + y2 + z2) – 2 (x + y + z) – 1 = 0 intersection of the three planes P1, P2 and P3.
r r r
r r r
21. (d) If a, b, c are linearly independent vectors, then rc As n1 , n 2 and n3 are non-coplanar, planes P1, P2 and
r P3 will intersect at unique point. So the given lines will
should be a linear combination of ar and b . pass through a fixed point.
r r r 26. (b) Statement 1 :
Let c = pa + qb
r r r r (iˆ + 2ˆj - k).(i
ˆ ˆ + k)ˆ 1+ 0 -1
( ) (
i.e. ˆi + aˆj + bkˆ = p ˆi + ˆj + kˆ + q 4iˆ + 3jˆ + 4kˆ ) cos (a + b, a - b) =
6 2
=
12
= 0;
r r r r r r r
Equating the coefficient i , j, k , we get 1 = p + 4q, (a + b, a - b) = 90°
a = p + 3q and b = p + 4q. r r r r
(a + b).( a - b) 0
From first and third, b = 1. Statement 2 : r r = =0
|a-b| 2
Now | rc | = 3 Þ 1 + a 2 + b 2 = 3
a -1 a a +1 0 a a +1
Þ a 2 = 1, \ a = ± 1.
Hence , a = ± 1, b = 1. 27. (a) D = b -1 b b + 1 = 0 b b +1
r r r r r c -1 c c +1 0 c c +1
22. (d) We have, | (a ´ b).c | = | a | | b | | c |
r r (R 1 ® R 1 + R 3 - 2R 2 )
Þ ( ) r r
| a | | b | sin q .nˆ .c = | ar | | b | | rc | \ D = 0, which is not a natural number.
r r
= | | ar | | b | | rc | sin q cos a | = | ar | | b | | rc | r r r r r
28. (d) | v | = | a ´ b | = | a || b | sin q = sin q
r r r r r r r
Þ |sin q | |cos a | = 1 Þ q =
p
2
and a = 0 ( )
u = a - a. b b = a - b cos q
r r r r rr
r r Þ | u | = | a |2 + | b |2 cos 2 q - 2a.b cos q
Þ a ^ b and c is parallel to n̂
r r
r r r r r = 1 + cos2 q - 2 cos2 q = sin2 q Þ | u |=| v |
Þ a ^ b and c ^ both a and b
r r r If q is the angle between the vectors
Þ a, b, c are mutually perpendicular.. 1
Þ a.b rr rr rr ( 2i - 2 j + k ) and ( 2i - 4 j + 3k ) , then
= b.c = c.a = 0 . 3
r r 1
23. (c) If angle between a and c is q then ( 4 + 8 + 3) 15
a.c = | a | | c | cos q = 1.2 cos q = 2 cos q 3 æ 5 ö
cos q = = Þ q = cos -1 ç
But a × (a × c) + b = 0 29 3 29 è 29 ÷ø
Þ (a. c) a – (a.a) c + b = 0 29. (d) Two lines are coplanar if
Þ (2 cos q).a – 1. c = – b x 2 - x1 y 2 - y1 z 2 - z1
Þ [(2 cos q) a –c]2 = [–b]2 a1 b1 c1 = 0
Þ 4cos2 q | a |2 – 2. (2 cos q) a.c + | c |2 = | b |2
Þ 4cos2 q – 4 cosq (2 cos q) + 4 = 1 a2 b2 c2
uuur uuur uuur uuur
Þ 4 (1 – cos2 q) = 1 [Q | a | = 1, | b | = 1] 30. (c) In DABC AB + BC = AC = -CA
Þ sin q =1/2 Þ q = p/6 uuur uuur uuur
Þ AB + BC + CA = 0
uuur uuur uuur
OA + AB = OB is triangle law of addition, A is true R
is false
SETS, MATHEMATICAL
PART REASONING, STATISTICS &
TEST 8 PROBABILITY
Time : 1 hour
Max. Marks : 120
INSTRUCTIONS
Ø Crack this section in just 1 hour.
Ø This part test contains 30 questions.
Ø For every correct answer you will be awarded +4. There is no negative marking.
Ø Please fill the correct bubbles in the Response Sheet provided at the end of the test.
Ø Do not look at the solutions before attempting the complete test.
. . . Best of Luck
23. A is one of 6 horses entered for a race, and is to be ridden by 26. Consider the following argument :
one of two jockey B and C. It is 2 to 1 that B rides A, in which “If it is cloudy tonight then it will rain tomorrow and if it
case all the horses are equally likely to win, if C rides A, his
rains tomorrow, I shall be on leave tomorrow, and the
chance is trebled. The odds against his winning are
conclusion is if it is cloudy tonight then I shall be on leave
(a) 5 : 18 (b) 18 : 5 tomorrow.”
(c) 13: 5 (d) 5 : 13
Statement-1 : The given argument is a valid
24. A and B are two independent witnesses (i.e. there is no argument.
collusion between them) in a case. The probability that A
Statement-2 : For three statements p, q and r
will speak the truth is x and the probability that B will speak
the truth is y. A and B agree in a certain statement. The p Þ q and q Þ r then p Þ r
probability that the statement is true is
27. Statement-1 : Two sets A and B have 4 and 5
x–y xy elements respectively then A È B has
(a) (b)
x+y 1 + x + y + xy at most 9 elements.
Statement-2 : For two finite sets A and B
x–y xy
(c) (d)
1 – x – y + 2 xy 1 – x – y + 2 xy max {n(A), n(B)} £ n(A È B) £ n(A) + n(B)
28. Let A and B are two events such that P(A) = 3/5 and
ASSERTION & REASON P(B)= 2/3, then
DIRECTIONS (Qs. 25 - 30) : Each of these questions 4 3
contains two statements: Statement-1(Assertion) and Statement–1 : £ P ( A Ç B) £ .
15 5
Statement-2 (Reason). Choose the correct answer (ONLY
ONE option is correct) from the following - 2 æ Aö 9
Statement–2 : £ Pç ÷ £
(a) Statement-1 is false, Statement-2 is true. 5 è B ø 10
(b) Statement-1 is true, Statement-2 is true; Statement-2
29. Statement-1 : If P(A/B) ³ P(A) thenP(B/A) ³ P(B)
is a correct explanation for Statement-1.
(c) Statement-1 is true, Statement-2 is true; Statement-2 P(A Ç B)
Statement-2 : P(A/B) =
is not a correct explanation for Statement-1. P(B)
(d) Statement-1 is true, Statement-2 is false.
30. Let X1 and X2 are means of two distributions such that
25. Statement-1 : The statement (p Ú q) Ù ~ p and
~ p Ù q are logically equivalent. X1 < X 2 and X is the mean of the combined distribution.
Statement-2 : The end columns of the truth table of Statement-1 : X1 < X < X 2
both statements are identical.
X1 + X 2
Statement-2 : X=
2
Response Sheet
ANSWER KEY
1 (c) 7 (c) 13 (a) 19 (a) 25 (b)
2 (b) 8 (b) 14 (c) 20 (a) 26 (b)
3 (a) 9 (b) 15 (c) 21 (d) 27 (b)
4 (c) 10 (a) 16 (a) 22 (a) 28 (b)
5 (b) 11 (a) 17 (b) 23 (c) 29 (b)
6 (b) 12 (b) 18 (a) 24 (d) 30 (d)
1. (c) A Ç (A È B)' = A Ç (A'ÇB' ) 7. (c) If I do not stay at home, then it does not rain.
8. (b) P(exactly one event out of A and B occurs)
= (A Ç A' ) Ç B' = f Ç B = f
2. (b) A = {x : | x | < 1} = (–1, 1) = P[( A Ç B ') È ( A 'Ç B)]
Since | x | < 1 Þ –1< x < 1 = P( A È B ) – P( A Ç B ) = P(A) + P(B) – 2 P( A Ç B )
B = {x : | x–1 | ³ 1} = (– ¥, 0] È [2, ¥)
Since | x –1 | ³ 1 Þ x–1£ –1 or x – 1³ 1 \ P(A) + P(B) – 2 P( A Ç B ) = 1 – a ..... (1)
Þ x £ 0 or x ³ 2
\ A È B = (– ¥ , 0] È [2, ¥ ) È (–1, 1)
= (– ¥, 1) È [2, ¥ ) = R – [1, 2)
\ D = [1, 2) = { x : 1 £ x < 2}
3. (a) Let U be the set of consumers questioned X, the set of
consumers who liked the product A and Y, the set of
consumers who liked the product B. Then n (U) = 1000, Similarly, P(B) + P(C) – 2 P( B Ç C ) = 1 – 2a ..... (2)
n (X) = 720, n (Y) = 450
P(C) + P(A) – 2 P(C Ç A) = 1 – a ..... (3)
n (X È Y) = n (X) + n (Y) – n (X Ç Y) = 1170 – n (X Ç Y)
2
\ n (X Ç Y) = 1170 – n (X È Y) P( A Ç B Ç C ) = a ..... (4)
Clearly n (X Ç Y) is least Now P( A È B È C ) = P(A) + P(B) + P(C) –
When n (X È Y) is maximum.
P( A Ç B ) – P( B Ç C ) – P(C Ç A) + P( A Ç B Ç C )
Now, XÈY Ì U
\ n(XÈY) £ n (U) = 1000 1
= [P(A) + P(B) – 2 P( B Ç C ) + P(B)
\ the maximum value of n (XÈY) is 1000. 2
Thus the least value of n (XÇY) is 170 + P(C) – 2 P( B Ç C ) + P(C) + P(A) – 2 P(C Ç A) ]
4. (c) x + 2 = 6 Þ x = 4, Its converse is x = 4 Þ x + 2 = 6 i.e. if
+ P( A Ç B Ç C )
x = 4 then 4 + 2 = 6
5. (b) Its contra positive is (sum of digits of n is not divisible 1
by 9) Þ (n is not divisible by 9) = [1 – a + 1 – 2a + 1 – a ] + a 2
2
6. (b) We have N=10000, n(A) = 40% of 10000 = 4000, [Using (1), (2), (3) and (4)]
n(B) = 2000, n(C) = 1000, n (A Ç B) =500,
3 1 2 1
n (B Ç C) = 300, n(C Ç A) = 400, n (A Ç B Ç C) = 200 = – 2a + a 2 = + (a –1) >
2 2 2
We have to find
9. (b) We have, P È Q
n(A Ç B Ç C) = n (A Ç (B È C)' )
= {x Î R : f ( x ) = 0 or g (x ) = 0}
= n (A) - n (A Ç (B È C))
= {x Î R : f ( x) g (x ) = 0}
n (A) - n ((A Ç B) È (A Ç C))
(a) represents the set for which either f (x) = 0
= n (A) - {n (A Ç B) + n (A Ç C) - n (A Ç B Ç C)} and g (x) = 0 or f (x) = – g (x)
= 4000–(500+400–200)=3300 (c) represents the set for which f (x) and g (x) both zero.
Sets, Mathematical Reasoning, Statistics & Probability M-47
10. (a) p q ~ p ~ q p Ù ~ q ~ p Ú q (p Ù ~ q) Ù (~ p Ú q) 4 2 4 4 4 2
= ´ + ´ ´ ´ + .......
T T F F F T F 6 6 6 6 6 6
T F F T T F F
F T T F F T F é ù
F F F T F T F 2 4ê 1 ú 6 9 2
= ´ ê ú= ´ =
6 6 ê 4 ú 36 5 5
Clearly, (p Ù ~ q) Ù (pÚ ~ q) is a contradiction. 1-
êë 9 úû
11. (a) If A will be final winner
Þ (A beats B ) (B beats C)
16. (a) y
+ (A beats C) (C beats B)
2 2 æ 1 - 2 öæ 1 - 2 ö 4 1 5
= ´ +ç ÷ç ÷ = ´ =
3 3 è 3 øè 3 ø 9 9 9
5
12. (b) P(A) = where A is an event of getting 6 by A
36
x
6
P(B) = where B is an event of getting 7 by B
36 Since x1 < x 2 Þ f (x1) < f (x 2 ) if x1 - - - x n are in
P(A win) ascending order
= W + L L W + LLLLW + ....... Þ f (x1 ) - - - f ( x n ) also in ascending order
A win A Loose B loose A win If M is the median along x, then f (M) is the median
along y
2
5 31 30 5 æ 31 30 ö 5
= + ´ ´ +ç ´ ÷ +-- aX + b
36 36 36 36 è 36 36 ø 36 17. (b) Let Y =
c
5 1 5 36 ´ 36 30 1 a
= ´ = ´ = Then Y = (a X + b ) Þ Y - Y = ( X - X )
36 31´ 30 36 366 61 c c
1-
36 ´ 36
1 a2 1
13. (a) The variance of combined data
Þ
N å (Y - Y ) 2 =
c2 N
å (X - X ) 2
1 é n 1n 2 ù
s2 = 2 2
ê n 1 s1 + n 2 s 2 + (x1 - x 2 ) 2 ú Therefore S.D. of Y
n1 + n 2
ë n1 + n 2 û
1 3´5 a2 1 a2 a
=
3+5
[(3 ´ 24) + (5 ´ 18) +
3+5
2
(8 - 8) ] =
c2 N
å (X - X) 2 = c 2
s2 =
c
s
= 20.25
18. (a) Putting n = 99 and p = 1/2,
14. (c) Probability of getting a blue ball at any draw,
1
10 1 we have (n + 1)p = (100) ( ) = 50,
= = 2
20 2 so that the maximum value of P(X = r) occurs at
P [getting a blue ball 4th time in 7th draw] = P [getting r = (n + 1)p = 50 and at r = (n + 1)p–1 = 49
3 blue balls in 6 draws] × P [a blue ball in the 7th draw]. 19. (a) The required event occurs if two sixes are observed in
3 3 the first seven throws and a six is observed on the
6 æ1ö æ1ö 1
= C3 ç ÷ ç ÷ . eighth throw. If p is the probability that a six shows on
è 2ø è2ø 2 the die, the number of throws n is 7, and X is the number
7 of times a six is observed, then X ~ B(7,p). Therefore
6 ´ 5´ 4 æ 1 ö 1 5 the required probability equals P(X = 2) times the
= ç ÷ = 20 ´ =
1´ 2 ´ 3 è 2 ø 32 ´ 4 32 probability of getting a six on the eighth throw, i.e., it
equals
15. (c) 2W
4R 2 5
æ1ö æ5ö æ1ö
P (B draws white ball before A) ( 7 C2 p2 q5 )(p) = (7 C2 ) ç ÷ ç ÷ ç ÷
è6ø è6ø è6ø
= L W + LLLW + LLLLLW + ......
¯ ¯ 7
C2 (55 )
A Loose B win =
68
M-48 Sets, Mathematical Reasoning, Statistics & Probability
20. (a) Sx = 170, Sx2 = 2830 E2 = A and B both tell a lie Þ P (E2) = (1 – x ) (1 – y)
Increase in Sx = 10, then E = A and B agree in a certain statement
Sx' = 170 + 10 = 180
Clearly, P(E / E1 ) = 1 and P(E / E 2 ) = 1
Increase in Sx2 = 900 – 400 = 500, then
Sx'2 = 2830 + 500 = 3330 The required probability is P(E1 / E ) . Using Baye’ss
2 theorem
1 æ Sx ' ö
\ Variance = Sx '2 - ç ÷ P(E1 )P(E / E1 )
n è n ø P(E1 / E ) =
P(E1 )P(E / E1 ) + P(E 2 )P(E / E 2 )
2
3330 æ 180 ö
= -ç ÷ xy.1 xy
15 è 15 ø = =
xy.1 + (1 – x )(1 – y).1 1 – x – y + 2xy
= 222 - 144 = 78
25. (b) Truth table has been given below :
21. (d) Since, root mean square ³ arithmetic mean
p q ~ p p Ú q (p Ú q) Ù ~ p ~ p Ù q
n n
T T F T F F
å xi2 å x i T F F T F F
i =1 i=1 400 80
\ ³ = ³
n n n n F T T T T T
Þ n ³ 16 F F T F F F
Hence, possible value of n = 18 26. (b) Define the statements
22. (a) Total number of function from A to B = 57. p = It is cloudy tonight
Total number of onto functions from A to B is q = It will rain tomorrow
r = I shall be on leave tomorrow
7! 7! 1 7! ´ 20
n (E) = ´ 5! + ´ ´ 5! = The assumptions are p Þ q, q Þ r and the conclusion
3!4! 3!2! 2!2! 6
is p Þ r validity can be checked using truth table.
n(E) 7! ´ 2
\ P (E) = n(S) = 27. (b) Statement-2 is the correct reason.
3 ´ 56
28. (b) Q P ( A Ç B) = P ( A) + P ( B) - P ( A È B) ³ P(A) + P(A) – 1
23. (c) Let E1 be the event that B rides A.
E2 the event C rides A and E the event that A wins. 3 2 4
Then by the given \ P(A Ç B) ³ + - 1 Þ P(A Ç B) ³ .. . (i)
5 3 15
2 2 1 3
P ( E1 ) = , P (E 2 ) = 1 - = .
3 3 3 Q P(A Ç B) £ P(A) Þ P(A Ç B) £ 5
. . . (ii)
5 P(A Ç B)
Hence probability of A’s win is so that odds against Þ £ P(B) Þ P(B/A) £ P(B)
18 P(A)
A’s wins are as (18–5) : 5 that is 13 : 5. 30. (d) If n1 and n 2 are the numbers of items in the two
24. (d) A and B will agree in a certain statement if both speak
truth or both tell a lie. We define following events n1 X1 + n 2 X 2
distributions then X =
E1 = A and B both speak truth Þ P(E1) = xy n1 + n 2
AIEEE - 2002
Paper - I
Time : 2½ hours Max. Marks : 450
33. By increasing the temperature, the specific resistance of a 43. If mass-energy equivalence is taken into account, when
conductor and a semiconductor water is cooled to form ice, the mass of water should
(a) increases for both (b) decreases for both (a) increase
(c) increases, decreases (d) decreases, increases (b) remain unchanged
34. If there are n capacitors in parallel connected to V volt source, (c) decrease
then the energy stored is equal to (d) first increase then decrease
44. The energy band gap is maximum in
1 1
(a) CV (b) nCV2 (c) CV2 (d) CV2 (a) metals (b) superconductors
2 2n (c) insulators (d) semiconductors.
35. Which of the following is more close to a black body? 45. The part of a transistor which is most heavily doped to
(a) black board paint (b) green leaves produce large number of majority carriers is
(c) black holes (d) red roses (a) emmiter
36. The inductance between A and D is (b) base
(c) collector
(d) can be any of the above three.
46. Energy required to move a body of mass m from an orbit of
A 3H 3H 3H D
radius 2R to 3R is
(a) GMm/12R2 (b) GMm/3R2
(a) 3.66 H (b) 9 H (c) 0.66 H (d) 1 H. (c) GMm/8R (d) GMm/6R.
37. A ball whose kinetic energy is E, is projected at an angle of 47. If a spring has time period T, and is cut into n equal parts,
45° to the horizontal. The kinetic energy of the ball at the then the time period of each part will be
highest point of its flight will be
(a) T n (b) T / n
(a) E (b) E / 2 (c) E/2 (d) zero. (c) nT (d) T
38. From a building two balls A and B are thrown such that A is 48. A charged particle q is placed at the centre O of cube of
thrown upwards and B downwards (both vertically). If vA length L (A B C D E F G H). Another same charge q is placed
and v B are their respective velocities on reaching the at a distance L from O. Then the electric flux through ABCD
ground, then is
E F
(a) vB > vA D
(b) vA = vB c
O
(c) vA > vB H q q
(d) their velocities depend on their masses. G
39. If a body looses half of its velocity on penetrating 3 cm in a A
B
wooden block, then how much will it penetrate more before L
coming to rest? (a) q /4 0L (b) zero
(a) 1 cm (b) 2 cm (c) q/2 0L (d) q/3 0L
(c) 3 cm (d) 4 cm.
49. If in the circuit, power dissipation is 150 W, then R is
40. If suddenly the gravitational force of attraction between
Earth and a satellite revolving around it becomes zero, then R
the satellite will
(a) continue to move in its orbit with same velocity
(b) move tangentially to the original orbit in the same 2
velocity
15 V
(c) become stationary in its orbit
(d) move towards the earth (a) 2 (b) 6 (c) 5 (d) 4
41. Cooking gas containers are kept in a lorry moving with 50. Wavelength of light used in an optical instrument are
uniform speed. The temperature of the gas molecules inside
1 4000 A an d 2 5000 A , then ratio of their
will
(a) increase respective resolving powers (corresponding to and )
1 2
(b) decrease is
(c) remain same (a) 16 : 25 (b) 9 : 1 (c) 4 : 5 (d) 5 : 4.
(d) decrease for some, while increase for others 51. A child swinging on a swing in sitting position, stands up,
42. When temperature increases, the frequency of a tuning fork then the time period of the swing will
(a) increases (a) increase (b) decrease
(b) decreases (c) remains same
(c) remains same (d) increases of the child is long and decreases if the child
(d) increases or decreases depending on the material is short
S-4 The Pattern Target AIEEE
52. A lift is moving down with acceleration a. A man in the lift 64. If a charge q is placed at the centre of the line joining two
drops a ball inside the lift. The acceleration of the ball as equal charges Q such that the system is in equilibrium then
observed by the man in the lift and a man standing stationary the value of q is
on the ground are respectively (a) Q/2 (b) –Q/2
(a) g, g (b) g – a, g – a (c) Q/4 (d) –Q/4
(c) g – a, g (d) a, g 65. Capacitance (in F) of a spherical conductor with radius 1 m
53. The mass of product liberated on anode in an electrochemical is
cell depends on 10 6
(a) (It)1/2 (b) It (a) 1.1 10 (b) 10
(c) I/t (d) I2t (c) 9 10 9 (d) 10 3
(where t is the time period for which the current is passed). 66. A light string passing over a smooth light pulley connects
54. At what temperature is the r.m.s velocity of a hydrogen two blocks of masses m1 and m 2 (vertically). If the
molecule equal to that of an oxygen molecule at 47°C? acceleration of the system is g/8, then the ratio of the masses
(a) 80 K (b) –73 K is
(c) 3 K (d) 20 K. (a) 8 : 1 (b) 9 : 7
55. The time period of a charged particle undergoing a circular
(c) 4 : 3 (d) 5 : 3.
motion in a uniform magnetic field is independent of its
67. Two spheres of the same material have radii 1 m and 4 m and
(a) speed (b) mass
temperatures 4000 K and 2000 K respectively. The ratio of
(c) charge (d) magnetic induction
the energy radiated per second by the first sphere to that by
56. A solid sphere, a hollow sphere and a ring are released from
the second is
top of an inclined plane (frictionless) so that they slide down
(a) 1 : 1 (b) 16 : 1
the plane. Then maximum acceleration down the plane is for
(c) 4 : 1 (d) 1 : 9.
(no rolling)
(a) solid sphere (b) hollow sphere 68. Three identical blocks of masses m = 2 kg are drawn by a
(c) ring (d) all same. force F = 10. 2 N with an acceleration of 0. 6 ms-2 on a
57. In a transformer, number of turns in the primary coil are 140 frictions surface, then what is the tension (in N) in the string
and that in the secondary coil are 280. If current in primary between the blocks B and C?
coil is 4 A, then that in the secondary coil is C B A F
(a) 4 A (b) 2 A (c) 6 A (d) 10 A.
58. Even Carnot engine cannot give 100% efficiency because (a) 9.2 (b) 7.8
we cannot (c) 4 (d) 9.8
(a) prevent radiation 69. One end of a massless rope, which passes over a massless
(b) find ideal sources and frictionless pulley P is tied to a hook C while the other
(c) reach absolute zero temperature end is free. Maximum tension that the rope can bear is 360 N.
(d) eliminate friction. With what value of maximum safe acceleration (in ms-2) can a
59. Moment of inertia of a circular wire of mass M and radius R man of 60 kg climb on the rope?
about its diameter is
(a) MR2/2 (b) MR2 (c) 2MR2 (d) MR2/4. P
60. When forces F1, F2, F3 are acting on a particle of mass m
such that F2 and F3 are mutually perpendicular, then the C
particle remains stationary. If the force F1 is now removed
then the acceleration of the particle is
(a) F1/m (b) F2F3/mF1
(c) (F2 - F3)/m (d) F2/m. (a) 16 (b) 6
61. Two forces are such that the sum of their magnitudes is 18 N (c) 4 (d) 8.
and their resultant is 12 N which is perpendicular to the 70. A particle of mass m moves along line PC with velocity v as
smaller force. Then the magnitudes of the forces are shown. What is the angular momentum of the particle about
(a) 12 N, 6 N (b) 13 N, 5 N P?
(c) 10 N, 8 N (d) 16N, 2N.
62. Speeds of two identical cars are u and 4u at the specific C
instant. The ratio of the respective distances in which the
L
two cars are stopped from that instant is
(a) 1 : 1 (b) 1 : 4 (c) 1 : 8 (d) 1 : 16. P
63. 1 mole of a gas with = 7/5 is mixed with 1 mole of a gas l
with = 5/3, then the value of for the resulting mixture is O
(a) 7/5 (b) 2/5 (a) mvL (b) mvl
(c) 24/16 (d) 12/7. (c) mvr (d) zero.
AIEEE-2002 Solved Paper S-5
71. Which of the following is used in optical fibres? 82. CH3 – Mg – Br is an organo metallic compound due to
(a) total internal reflection (a) Mg – Br bond (b) C – Mg bond
(b) scattering (c) C – Br bond (d) C – H bond.
(c) diffraction 83. 1 M NaCl and 1 M HCl are present in an aqueous solution.
(d) refraction. The solution is
72. The escape velocity of a body depends upon mass as (a) not a buffer solution with pH < 7
(a) m0 (b) m1 (c) m2 (d) m3. (b) not a buffer solution with pH > 7
73. Which of the following are not electromagnetic waves? (c) a buffer solution with pH < 7
(a) cosmic rays (b) gamma rays (d) a buffer solution with pH > 7.
(c) -rays (d) X-rays. 84. Species acting as both Bronsted acid and base is
(a) (HSO4)–1 (b) Na2CO3
74. Identify the pair whose dimensions are equal.
(a) torque and work (b) stress and energy (c) NH3 (d) OH–1.
(c) force and stress (d) force and work. 85. Let the solubility of an aqueous solution of Mg(OH)2 be x
then its ksp is
75. If i , is the inversion temperature, n is the neutral (a) 4x3 (b) 108x5
temperature, is the temperature of the cold junction, (c) 27x 4 (d) 9x.
c
then 86. Units of rate constant of first and zero order reactions in
terms of molarity M unit are respectively
(a) i c n (b) i c 2 n (a) sec–1, Msec–1 (b) sec–1, M
–1
(c) Msec , sec –1 (d) M, sec–1.
i C
(c) n (d) c i 2 n 87. In XeF2, XeF4, XeF6 the number of lone pairs on Xe are
2
respectively
Section - 2 (a) 2, 3, 1 (b) 1, 2, 3
(c) 4, 1, 2 (d) 3, 2, 1.
88. In which of the following species the interatomic bond angle
is 109° 28’?
76. When H2S is passed through Hg2S we get (a) NH3, (BF4)–1 (b) (NH4)+, BF3
(a) HgS (b) HgS + Hg2S (c) NH3, BF4 (d) (NH2)–1, BF3.
(c) Hg2S + Hg (d) None of these. 89. For the reaction A + 2B C, rate is given by R = [A] [B]2
77. Alum helps in purifying water by then the order of the reaction is
(a) forming Si complex with clay partiles (a) 3 (b) 6
(b) sulphate part which combines with the dirt and removes it (c) 5 (d) 7.
(c) coagulaing the mud particles 90. RNA is different from DNA because RNA contains
(d) making mud water soluble. (a) ribose sugar and thymine
78. A square planar complex is formed by hybridisation of which (b) ribose sugar and uracil
atomic orbitals? (c) deoxyribose sugar and thymine
s, px , p y ,d 2 2 (d) deoxyribose sugar and uracil.
(a) s, px , py , dyz (b)
x -y 91. Which of the following are arranged in an increasing order
(c) s, p x , p y ,d (d) s, py , pz , dxy of their bond strengths?
z2 (a) O2– < O2 < O2+ <O22– (b) O22– < O2– < O2 <O2+
79. Polymer formation from monomers starts by (c) O2– < O22– < O2 <O2+ (d) O2+ < O2 < O2– <O22–
(a) condensation reaction between monomers 92. If an endothermic reaction is non-spontaneous at freezing
(b) coordinate reaction between monomers point of water and becomes feasible at its boiling point,
(c) conversion of monomer to monomer ions by protons then
(d) hydrolysis of monomers. (a) H is –ve, S is +ve
80. The type of isomerism present in nitropentamine chromium (b) H and S both are +ve
(III) chloride is
(c) H and S both are –ve
(a) optical (b) linkage
(c) ionization (d) polymerisation. (d) H is +ve, S is -ve
81. Arrangement of (CH3)3 – C –, (CH3)2 – CH –, CH3 – CH2 – 93. A heat engine abosrbs heat Q1 at temperature T1 and heat
when attached to benzyl or an unsaturated group in Q2 at temperature T2 . Work done by the engine is J
increasing order of inductive effect is (Q1 + Q2). This data
(a) (CH3)3 –C – < (CH3)2 – CH – < CH3 – CH2 (a) violates 1st law of thermodynamics
(b) CH3 –CH2– < (CH3)2– CH – < (CH3)3 –C – (b) violates 1st law of themodynamics if Q1 is –ve
(c) (CH3)2 – CH– <(CH3)3 –C –< CH3, –CH2 (c) violates 1st law of thermodynamics of Q2 is –ve
(d) (CH3)3 – C– < CH3 –CH2 –(CH3)2 –CH – (d) does not violate 1st law of themodynamics.
S-6 The Pattern Target AIEEE
94. Most common oxidation states of Ce (cerium) are 104. In a compound C, H and N atoms are present in 9 : 1 : 3.5 by
(a) +2, +3 (b) +2, +4 weight. Molecular weight of compound is 108. Molecular
(c) +3, +4 (d) +3, +5. formula of compound is
95. Arrange Ce+3, La+3, Pm+3 and Yb+3 in increasing order of (a) C2H6N2 (b) C3H4N
their ionic radii. (c) C6H8N2 (d) C9H12N3.
(a) Yb+3 < Pm+3 < Ce+3 < La+3 105. The functional group, which is found in amino acid is
(b) Ce+3 < Yb+3 < Pm+3 < La+3 (a) – COOH group (b) – NH2 group
(c) Yb+3 < Pm+3 < La+3 < Ce+3 (c) – CH3 group (d) both (a) and (b).
(d) Pm+3 < La+3 < Ce+3 < Yb+3. 106. Conductivity (unit Siemen’s S) is directly proportional to
96. KO2 (potassium super oxide) is used in oxygen cylinders in area of the vessel and the concentration of the solution in it
space and submarines because it and is inversely proportional to the length of the vessel
(a) absorbs CO2 and increases O2 content then the unit of the constant of proportionality is
(b) eliminates moisture (a) Sm mol–1 (b) Sm2 mol–1
(c) absorbs CO2 –2
(c) S m mol2 (d) S2m2 mol–2.
(d) produces ozone. 107. In a hydrogen atom, if energy of an electron in ground state
97. A similarity between optical and geometrical isomerism is is 13.6. ev, then that in the 2nd excited state is
that
(a) 1.51 eV (b) 3.4 eV
(a) each forms equal number of isomers for a given
(c) 6.04 eV (d) 13.6 eV.
compound
108. Which of the following statements is true?
(b) if in a compound one is present then so is the other
(a) HF is less polar than HBr
(c) both are included in stereoisomerism
(b) absolutely pure water does not contain any ions
(d) they have no similarity.
(c) chemical bond formation take place when forces of
98. Which of the following does not show geometrical
isomerism? attraction overcome the forces of repulsion
(a) 1,2-dichloro-1-pentene (d) in covalency transference of electron takes place.
(b) 1,3-dichloro-2-pentene 109. Which of the following compounds has wrong IUPAC
(c) 1,1-dichloro-1-pentene name?
(d) 1,4-dichloro-2-pentene (a) CH3–CH2–CH2 –COO–CH2CH3 ethyl butanoate
99. In case of nitrogen, NCl3 is possible but not NCl5 while in (b) CH3 CH CH 2 CHO 3-methyl-butanal
case of phosphorous, PCl3 as well as PCl5 are possible. It is |
due to CH
(a) availability of vacant d orbitals in P but not in N (c) 2-methyl-3-butanol
CH3 CH C H CH
(b) lower electronegativity of P than N | |
3
(c) lower tendency of H-bond formation in P than N OH CH3
(d) occurrence of P in solid while N in gaseous state at
room temperature.
O
100. For an ideal gas, number of moles per litre in terms of its ||
pressure P, gas constant R and temperature T is (d) CH3 CH C CH 2 CH3
(a) PT/R (b) PRT |
(c) P/RT (d) RT/P. CH3
101. The formation of gas at the surface of tungsten due to 2-methyl-3-pentanone
adsorption is the reaction of order
(a) 0 (b) 1 Cl2 alc.KOH
110. CH3CH2COOH A B. What is B?
(c) 2 (d) insufficient data. red P
102. The metallic sodium disolves in liquid ammonia to form a (a) CH3CH2COCl (b) CH3CH2CHO
deep blue coloured solution. The deep blue colour is due to (c) CH2=CHCOOH (d) ClCH2CH2COOH.
formation of: 111. Aluminium is extracted by the electrolysis of
(a) solvated electron, e(NH3 ) x (a) bauxite
(b) alumina
(b) solvated atomic sodium, Na(NH3)y
(c) alumina mixed with molten cryolite
(c) (Na+ + Na–)
(d) molten cryolite.
(d) NaNH2 + H2
103. How do we differentiate between Fe3+ and Cr 3+ in group 112. The metal extracted by leaching with a cyanide is
III? (a) Mg (b) Ag
(a) by taking excess of NH4OH solution (c) Cu (d) Na.
(b) by increasing NH4+ ion concentration 113. Value of gas constant R is
(c) by decreasing OH- ion concentration (a) 0.082 litre atm (b) 0.987 cal mol–1 K–1
–1
(c) 8.3 J mol K –1 (d) 83 erg mol–1 K–1.
(d) both (b) and (c).
AIEEE-2002 Solved Paper S-7
114. Freezing point of an aqueous solution is (–0.186)°C. Elevation 125. Kinetic theory of gases proves
of boiling point of the same solution is Kb = 0.512°C, (a) only Boyle’s law (b) only Charles’ law
Kf = 1.86°C, find the increase in boiling point. (c) only Avogadro’s law (d) all of these.
(a) 0.186°C (b) 0.0512°C 126. A metal M readily forms its sulphate MSO4 which is water-
(c) 0.092°C (d) 0.2372°C. soluble. It forms its oxide MO which becomes inert on
115. EMF of a cell in terms of reduction potential of its left and heating. It forms an insoluble hyroxide M(OH)2 which is
right electrodes is soluble in NaOH solution. Then M is
(a) E = Eleft – Eright (b) E = Eleft + Eright (a) Mg (b) Ba
(c) E = Eright – Eleft (d) E = –(Eright + Eleft). (c) Ca (d) Be.
116. Uncertainty in position of a minute particle of mass 25 g in 127. Following types of compounds (as I, II)
space is 10–5 m. What is the uncertainty in its velocity
(in ms–1)? (h = 6.6 10-34 Js) CH3 CH CHCH3 CH3CHOH
(a) 2.1 10 –34 (b) 0.5 10–34 |
(c) 2.1 10 –28 (d) 0.5 10–23. CH 2CH3
117. Which of these will not react with acetylene? are studied in terms of isomerism in:
(a) NaOH (b) ammonical AgNO3 (a) chain isomerism (b) position isomerism
(c) Na (d) HCl. (c) conformers (d) stereoisomerism
118. Change in volume of the system does not alter which of the 128. What is the product when acetylene reacts with
following equilibria? hypochlorous acid?
(a) N2 (g) + O2 (g) 2NO (g) (a) CH3COCl (b) ClCH2CHO
(b) PCl5 (g) PCl3 (g) + Cl2 (g) (c) Cl2CHCHO (d) ClCHCOOH.
(c) N2 (g) + 3H2 (g) 2NH3 (g) 129. On vigorous oxidation by permanganate solution.
(d) SO2Cl2 (g) SO2 (g) + Cl2 (g). (CH3)2C = CH - CH2 - CHO gives
119. For the reactions,
2C + O2 2 CO2 ; H = -393 J CH OH
| |
2Zn + O2 2ZnO ; H = -412 J (a) CH 3 C CH CH 2CH 3
(a) carbon can oxidise Zn |
(b) oxidation of carbon is not feasible CH
(c) oxidation of Zn is not feasible
CH3
(d) Zn can oxidise carbon. (b)
120. Which of the following ions has the maximum magnetic COOH + CH3CH2COOH
moment? CH3
(a) Mn +2 (b) Fe+2 (c) Ti+2 (d) Cr+2. CH3
121. In which of the following species is the underlined carbon (c) CH – OH + CH2CH2CH2OH
having sp3 hybridisation? CH3
(a) CH3 COOH (b) CH3 CH2 OH
(c) CH3 COCH3 (d) CH2 = CH –CH3 CH3
(d) . C = O + CH2CH2CHO
122. Racemic mixture is formed by mixing two CH3
(a) isomeric compounds
(b) chiral compounds OCOCH3
(c) meso compounds COOH
(d) enantiomers with chiral carbon. 130. The compound is used as
123. The differential rate law for the reaction
H2 + I2 2HI is (a) antiseptic (b) antibiotic
d[ H 2 ] d[ I 2 ] d[ HI ] (c) analgesic (d) pesticide.
(a) 131. What will be the emf for the given cell
dt dt dt
Pt | H2 (P1) | H+ (aq) | | H2 (P2) | Pt
d[H 2 ] d[I 2 ] 1 d[Hl ] RT P
(b)
dt dt 2 dt (a)
RT P
log e 1 (b) loge 1
f P2 2f P2
1 d[H 2 ] 1 d[I 2 ] d[Hl]
(c) RT P
2 dt 2 dt dt (c) log e 2 (c) none of these.
f P1
d[ H 2 ] d[ I 2 ] d[ HI ] 132. When primary amine reacts with chloroform in ethanoic KOH
(d) 2 2
dt dt dt then the product is
124. Number of sigma bonds in P4O10 is (a) an isocyanide (b) an aldehyde
(a) 6 (b) 7 (c) 17 (d) 16. (c) a cyanide (d) an alcohol.
S-8 The Pattern Target AIEEE
133. Which of the following reaction is possible at anode? 140. Hybridisation of the underline atom changes in:
(a) 2 Cr3+ + 7H2O Cr2O72– + 14H+ (a) AlH3 changes to AlH 4
(b) F2 2F – (b) H 2 O changes to H3O+
(c) (1/2) O2 + 2H+ H2O (c) NH 3 changes to NH 4
(d) none of these. (d) in all cases
134. The reaction: 141. When the sample of copper with zinc impurity is to be purified
by electrolysis, the appropriate electrodes are
H O
2
(CH3)3C – Br (CH3)3 – C –OH cathode anode
(a) elimination reaction (b) substitution reaction (a) pure zinc pure copper
(c) free radical reaction (d) displacement reaction. (b) impure sample pure copper
135. If half-life of a substance is 5 yrs, then the total amount of (c) impure zinc impure sample
substance left after 15 years, when initial amount is 64 grams (d) pure copper impure sample.
is 142. The most stable ion is
(a) 16 grams (b) 2 grams (a) [Fe(OH)3]3- (b) [Fe(Cl)6]3-
(c) [Fe(CN)6] 3- (d) [Fe(H2O)6]3+.
(c) 32 grams (d) 8 grams.
136. Picric acid is: 143. -particle is emitted in rdioactivity by
(a) conversion of proton to neutron
COOH (b) from outermost orbit
(c) conversion of neutron to proton
(a) (d) -particle is not emitted.
NO2 144. In mixture A and B components show -ve deviation as
(a) Vmix > 0
COOH (b) Hmix < 0
(c) A – B interaction is weaker than A – A and B – B
(b) interaction
OH (d) A – B interaction is stronger than A – A and B – B
interaction.
OH 145. The heat required to raise the temperature of body by 1 K is
NO2 NO2 called
(a) specific heat (b) thermal capacity
(c) (c) water equivalent (d) none of these.
146. Na and Mg crystallize in BCC and FCC type crystals
NO2 respectively, then the number of atoms of Na and Mg present
in the unit cell of their respective crystal is
COOH (a) 4 and 2 (b) 9 and 14
(c) 14 and 9 (d) 2 and 4.
(d) 147. Number of atoms in 558.5 gram Fe (at. wt. of Fe = 55.85 g
mol–1) is
NH2
(a) twice that in 60 g carbon
137. Oxidation number of Cl in CaOCl2 (bleaching power) is: (b) 6.023 1022
(a) zero, since it contains Cl2 (c) half that in 8 g He
(b) – 1, since it contains Cl – (d) 558.5 6.023 1023
(c) + 1, since it contains ClO– 148. When KMnO4 acts as an oxidising agent and ultimately
(d) + 1 and – 1 since it contains ClO– and Cl– forms [MnO4]–2, MnO2, Mn2O3, Mn+2 then the number of
138. With increase of temperature, which of these changes? electrons transferred in each case respectively is
(a) molality (a) 4, 3, 1, 5 (b) 1, 5, 3, 7
(b) weight fraction of solute (c) 1, 3, 4, 5 (d) 3, 5, 7, 1.
(c) molarity 149. Which of the following is a redox reaction?
(d) mole fraction. (a) NaCl + KNO3 NaNO3 + KCl
139. The integrated rate equation is (b) CaC2O4 + 2HCl CaCl2 + H2C2O4
Rt = log C0 - logCt.
(c) Mg(OH)2 + 2NH4Cl MgCl2 + 2NH4OH
The straight line graph is obtained by plotting
(d) Zn + 2AgCN 2Ag + Zn(CN)2.
1 150. For the reaction
(a) time vs logCt (b) vs C
time t CO (g) + (1/2) O2 (g) = CO2 (g), Kp / Kc is
1 1 (a) RT (b) (RT)–1
(c) time vs Ct (d) vs (c) (RT)–1/2 (d) (RT)1/2
time Ct
AIEEE-2002 Solved Paper S-9
Paper - II
Time : 1¼ hours Max. Marks : 225
42. The sides of a triangle are 3x + 4y, 4x + 3y and 5x + 5y where 52. The number of solution of tanx + secx = 2cosx in [0, 2 ) is
x, y > 0 then the triangle is (a) 2 (b) 3
(a) right angled (b) obtuse angled (c) 0 (d) 1
(c) equilateral (d) none of these 53. Which one is not periodic
43. Locus of mid point of the portion between the axes of x (a) | sin3x | + sin2x (b) cos x + cos2x
cos + y sin = p whre p is constant is (c) cos4x + tan2x (d) cos2x + sinx
4 1p 2 p 3p ..... n p
(a) x2 + y2 = (b) x2 + y2 = 4p2 54. lim is
p2 n np 1
1 1 4 1 1
(c) 1 1 2 (d) 2 2
(a) p 1
(b)
1 p
x 2 y2 p 2 x y p2
44. If the pair of lines ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 1 1 1
(c) (d)
intersect on the y-axis then p p 1 p 2
(a) 2fgh = bg2+ch2 (b) bg2 ch2
log x n [ x ]
(c) abc = 2fgh (d) none of these 55. lim ,n N, ([x] denotes greatest integer less
45. The pair of lines represented by x 0 [x ]
3ax2 + 5xy + (a2 – 2)y2 = 0 than or equal to x)
(a) has value –1 (b) has value 0
are perpendicular to each other for
(c) has vlaue 1 (d) does not exist
(a) two values of a
(b) a f ( x) 1
56. If f(1) = 1, f1(1) = 2, then lim is
(c) for one value of a x 1 x 1
(d) for no values of a (a) 2 (b) 4
46. If the chord y = mx + 1 of the circle x2+y2=1 subtends an (c) 1 (d) 1/2
angle of measure 45° at the major segment of the circle then 57. f is defined in [-5, 5] as
value of m is f(x) = x if x is rational
= –x if x is irrational. Then
(a) 2 2 (b) –2 2 (a) f(x) is continuous at every x, except x = 0
(b) f(x) is discontinuous at every x, except x = 0
(c) –1 2 (d) none of these (c) f(x) is continuous everywhere
47. The centres of a set of circles, each of radius 3, lie on the (d) f(x) is discontinuous everywhere
circle x2+y2=25. The locus of any point in the set is 58. f(x) and g(x) are two differentiable functions on [0, 2] such
(a) 4 x2 + y2 64 (b) x2 + y2 25 that f ( x) g ( x) 0, f (1) 2g (1) 4 f(2) =3g(2) = 9 then
(c) x2 + y2 25 (d) 3 x2 + y2 9 f(x)–g(x) at x = 3/2 is
48. The centre of the circle passing through (0, 0) and (1, 0) and (a) 0 (b) 2
touching the circle x2 + y2 = 9 is (c) 10 (d) 5
59. If f ( x y) f ( x ).f ( y) x.y and f (5) 2, f (0) 3 , then
1 1 1
(a) , (b) , 2 f (5) is
2 2 2
(a) 0 (b) 1
3 1 1 3 (c) 6 (d) 2
(c) , (d) , 60. The maximum distance from origin of a point on the curve
2 2 2 2
at
49. The equation of a circle with origin as a centre and passing x = a sin t–b sin
through equilateral triangle whose median is of length 3a is b
(a) x2 + y2 = 9a2 (b) x2 + y2 = 16a2 at
2 2
(c) x + y = 4a 2 (d) x2 + y2 = a2 y = a cos t – b cos , both a, b > 0 is
b
50. Two common tangents to the circle x2 + y2 = 2a2 and parabola (a) a – b (b) a + b
y2 = 8ax are
(c) a 2 b 2 (d) a 2 b2
(a) x ( y 2a ) (b) y ( x 2a )
61. If 2a + 3b + 6c = 0, (a, b, c R) then the quadratic equation
(c) x (y a) (d) y (x a ) ax2 + bx + c = 0 has
51. In a triangle with sides a, b, c, r 1 > r2 > r3 (which are the ex- (a) at least one root in [0, 1]
radii) then (b) at least one root in [2, 3]
(a) a > b > c (b) a < b < c (c) at least one root in [4, 5]
(c) a > b and b < c (d) a < b and b > c (d) none of these
S-12 The Pattern Target AIEEE
62. If y = f(x) makes +ve intercept of 2 and 0 unit on x and y axes
and encloses an area of 3/4 square unit with the axes then 69. a 3 i 5 j and b 6 i 3 j are two vectors and c is a
2
xf (x)dx is vector such that c a b then | a | :| b |:| c |
0
(a) 34 : 45 : 39 (b) 34 : 45 : 39
(a) 3/2 (b) 1
(c) 5/4 (d) -3/4 (c) 34 : 39 : 45 (d) 39 : 35 : 34
63. The area bounded by the curves y = lnx, y = ln |x|,y=| ln x |
and y = | ln |x| | is 70. If a b b c c a then a b c =
(a) 4sq. units (b) 6 sq. units (a) abc (b) –1
(c) 10 sq. units (d) none of these
(c) 0 (d) 2
64. If | a | 4, | b | 2 and the angle between a and b is / 71. A and B are events such that P(A B)=3/4, P(A B)=1/4,
P( A ) =2/3 then P ( A B) is
6 then ( a b ) 2 is equal to (a) 5/12 (b) 3/8
(a) 48 (b) 16 (c) 5/8 (d) 1/4
72. A dice is tossed 5 times. Getting an odd number is considered
(c) a (d) none of these
a success. Then the variance of distribution of success is
(a) 8/3 (b) 3/8
65. if a , b , c are vectors such that [ a b c ] = 4 then
(c) 4/5 (d) 5/4
[a b b c c a ] 73. The d.r. of normal to the plane through (1, 0, 0), (0, 1, 0)
which makes an angle /4 with plane x + y = 3 are
(a) 16 (b) 64
(c) 4 (d) 8 (a) 1, 2 ,1 (b) 1, 1, 2
A N S W ER KEY
P A P ER 1 - P HYS IC S & C H EM IS T R Y P A P ER II - M A THS
1 (a ) 31 (b ) 61 (b ) 91 (b ) 121 (b ) 1 (a ) 31 (a ) 61 (a )
2 (b ) 32 (a ) 62 (d ) 92 (b ) 122 (d ) 2 (a ) 32 (d ) 62 (d )
3 (b ) 33 (c ) 63 (c) 93 (a ) 123 (d ) 3 (b ) 33 (c ) 63 (a )
4 (b ) 34 (b ) 64 (d ) 94 (c ) 124 (d ) 4 (a ) 34 (d ) 64 (b )
5 (c ) 35 (a ) 65 (a) 95 (a ) 125 (d ) 5 (b ) 35 (b ) 65 (a )
6 (c ) 36 (d ) 66 (b ) 96 (b ) 126 (d ) 6 (d ) 36 (a ) 66 (a )
7 (a ) 37 (c ) 67 (a) 97 (c ) 127 (a ) 7 (d ) 37 (c ) 67 (a )
8 (a ) 38 (b ) 68 (b ) 98 (c ) 128 (c ) 8 (a ) 38 (d ) 68 (a )
9 (b ) 39 (a ) 69 (c) 99 (a ) 129 (b ) 9 (b ) 39 (c ) 69 (b )
10 (c ) 40 (c ) 70 (d ) 100 (c ) 130 (c ) 10 (a ) 40 (c ) 70 (c )
11 (c ) 41 (a ) 71 (a) 101 (a ) 131 (d ) 11 (c ) 41 (b ) 71 (a )
12 (b ) 42 (b ) 72 (a) 102 (a ) 132 (a ) 12 (a ) 42 (b ) 72 (d )
13 (b ) 43 (a ) 73 (c) 103 (d ) 133 (a ) 13 (b ) 43 (d ) 73 (b )
14 (a ) 44 (c ) 74 (a) 104 (c ) 134 (b ) 14 (a ) 44 (a ) 74 (a )
15 (a ) 45 (a ) 75 (c) 105 (d ) 135 (d ) 15 (a ) 45 (a ) 75 (b )
16 (c ) 46 (d ) 76 (c ) 106 (b ) 136 (d ) 16 (b ) 46 (c )
17 (c ) 47 (b ) 77 (c ) 107 (a ) 137 (a ) 17 (b ) 47 (a )
18 (b ) 48 (b ) 78 (b ) 108 (c ) 138 (c ) 18 (a ) 48 (b )
19 (b ) 49 (b ) 79 (a ) 109 (c ) 139 (a ) 19 (b ) 49 (c )
20 (b ) 50 (d ) 80 (b ) 110 (c ) 140 (d ) 20 (c ) 50 (b )
21 (c ) 51 (b ) 81 (b ) 111 (c ) 141 (d ) 21 (b ) 51 (a )
22 (b ) 52 (c) 82 (b ) 112 (b ) 142 (b ) 22 (c ) 52 (b )
23 (b ) 53 (b ) 83 (a ) 113 (c ) 143 (c ) 23 (b ) 53 (b )
24 (c ) 54 (d ) 84 (a ) 114 (b ) 144 (b ) 24 (c ) 54 (a )
25 (a ) 55 (a) 85 (a ) 115 (c ) 145 (b ) 25 (b ) 55 (d )
26 (c ) 56 (d ) 86 (a ) 116 (c ) 146 (d ) 26 (b ) 56 (a )
27 (a ) 57 (b ) 87 (d ) 117 (a ) 147 (a ) 27 (a ) 57 (b )
28 (c ) 58 (c) 88 (a ) 118 (a ) 148 (c ) 28 (a ) 58 (d )
29 (a ) 59 (b ) 89 (a ) 119 (d ) 149 (d ) 29 (c ) 59 (c )
30 (d ) 60 (a) 90 (b ) 120 (a ) 150 (c ) 30 (a ) 60 (b )
1 1 1
MR 2 2mR 2 MR 2 Hence, m v 2 = mAghA or vA = 2gh A ;
2
1
2 2 A A
M 1 Similarly, vB = 2gh or vA = vB
M 4m
39. (a) Let the initial velocity of the body be v. Hence the final
18. (b) The condition to avoid skidding, v = rg velocity = v/2
2
= 0.6 150 10 = 30 m/s. v
Applying v2 = u2 – 2as = v2 – 2.a.3
2
19. (b) v 2gh 2 10 20 20m / s.
a = v2/8
x2 x2
In IInd case when the body comes to rest, final velocity
20. (b) W Fdx Kxdx
v
x1 x1 = 0, initial velocity =
2
x2
x2 K v
2
v2
K [x 22 x1 2 ] Again, (0)2 = 2 s; or s = 1cm
2 2 2 8
x1
So the extra penetration will be 1 cm.
800
[(0.15)2 –(0.05)2] = 8 J. 40. (c) When gravitational force becomes zero so centripetal
2 force on satellite becomes zero so satellite will escape
21. (c) Conserving Linear Momentem its round orbit and becomes stationary.
2Mvc = 2Mv – Mv vc = v/2. 41. (a) The molecular kinetic energy increases, and so
22. (b) It will compress due to the force of attraction between termperature increases.
two adjacent coils carrying current in the same 42. (b)
direction.
43. (a) Because thermal energy decreases, therefore mass
23. (b)
should increase.
24. (c) Semiconductors are insulators at low temperature
44. (c) Maximum in insulators and overlaping in metals
25. (a) 26. (c)
27. (a) Neutrons can’t be deflected by a magnetic field. 45. (a)
46. (d) E = (PE)final – (PE)initial
( 0 )1 (W0 ) 2 4.5
28. (c) hc/ = W0 ; 2 : 1.
0
( GMm GMm GMm
0 )2 ( W0 )1 2. 3
3R R 6R
29. (a) Covalent bond formation is best explained by orbital
theory which uses wave phenomena. 47. (b) Spring constant becomes n times for each piece.
30. (d) 31. (b)
n
T 2 m/k
32. (a) Amount left = N0/2 = N0/8 (Here n=15/5=3)
T T1 nK
33. (c) Use R t R0 or T2 T/ n .
273 T2 K
48. (b) The flux for both the charges exactly cancels the effect
1 1 of each other.
34. (b) E CV 2 nCV 2
2 2
35. (a) Black body also emits radiation whereas nothing V2 (15) 2 (15) 2
49. (b) W ; 150 R 6
escapes a black hole. R net R 2
AIEEE-2002 Solved Paper S-15
50. (d) Resolving power (1 / ) . Hence, 73. (c) -rays are the beam of fast moving electrons.
74. (a) Both have the dimension M1L2T–2.
(R.P)1 2 5 75. (c) 76. (c)
.
(R.P) 2 1 4 77. (c) Alum furnishes Al 3+ ions which bring about
coagulation of negatively charged clay poarticles,
51. (b) T = 2 l eff / 8 ; l eff decreases when the child stands bacteria etc.
up. 78. (b) A square planer complex is formed by hybridisation of
52. (c) Man in the lift is in a non-inertial frame so we have to
s, px, py and d 2 d 2 atomic orbitals
take into account the pseudo acceleration. x y
53. (b) From Faradays law of electrolysis, m it. 79. (a) Polymerisation starts either by condensation or
273 47 T addition reactions between monomers
54. (d) v rms T/m ; or T =20 80. (b) The nitro group can attach to metal through nitrogen
32 2 as (–NO2) or through oxygen as nitrito (–ONO).
K. 81. (b) –CH3 group has +I effect, as number of –CH3 group
55. (a) T = 2 m/Bq. increases, the inductive effect increases.
56. (d) 82. (b) Bond between C of organic molecule and metal atom.
4 140 83. (a) A buffer is a solution of weak acid and its salt with
57. (b) I1N1 = I2N2 I2 2A strong base and vice versa. HCl is strong acid and
280
NaCl is its salt with strong base. pH is less than 7 due
58. (c) Absolute zero temperature is practically not reachable. to HCl
59. (b)
84. (a) (HSO4)– can accept and donate a proton
60. (a) Resultant of F2 and F3 is of magnitude F1.
(HSO4)– + H+ H2SO4
P sin (HSO4)– – H+ SO42–.
61. (b) Use tan
Q P cos 85. (a) Mg(OH)2 [Mg2+] + 2[OH–]
x 2x
P sin Ksp = [Mg] [OH]2 = [x][2x]2 = x.4x2 = 4x3.
tan 90º
Q P cos 86. (a) K = (mol L–1)1–n sec–1, n = 0, 1.
87. (d) XeF2 sp3 d 3 lone pairs
Q + P cos = 0 P cos = –Q.
XeF4 sp3d 2 2 lone pairs
R P2 Q2 2 PQ cos R XeF6 sp3d 3 1 lone pair
88. (a) In NH3 and BF4 the hybridisation is sp3 and the bond
P 2 Q 2 2Q 2 or R P 2 Q 2 12
angle is almost 109º 28'
144 = (P + Q)(P – Q) or P – Q = 144/18 = 8. 89. (a) Order is the sum of the power of the concentrations
P = 13 N and Q = 5 N. terms in rate law expression.
62. (d) Use u2 = 2as. a is same for both cases. 90. (b)
s1 = u2/2a ; s2=16u2/2a = 16s1 s1:s2 = 1: 16. 91. (b) According to bond order values the given order is the
63. (c) for resulting mixture should be in between
1 1
7/5 and 5/3. answer. Bond order values are +1, 1 , +2 and 2 ,
64. (d) Apply the condition for equilibrium of each charge. 2 2
higher bond order means stronger bond.
65. (a) 10
4 0 R 1.1 10 92. (b) At low temperature the S is – ve which makes G
positive ( G = H – T S). But at higher temperature
m1 m 2 1 m1 m 2 S is +ve which makes the G negative (condition for
66. (b) a g ;
m1 m 2 8 m1 m 2 spontaneity)
m1 : m2 = 9: 7. 93. (a) Some mechanical energy is always converted (lost) to
67. (a) Energy radiated R2T4. other forms of energy.
68. (b) Apply Newton’s second law. 94. (c) Common oxidation states of Ce(Cerium) are
F – Tab = ma; Tab – Tbc = ma + 3 and + 4
Tbc = 7.8 N. 95. (a) According to their positions in the periods, these values
69. (c) T – 60g = 60a; T = 3000 N; are in the order:
a= 4 ms-2. Yb+3 < Pm+3 < Ce+3 < La+3
70. (d) Zero, line of motion through the point P. At. Nos. 70 61 58 57
71. (a) This is due to lanthanide contraction.
96. (b) 2KO2 + 2 H2O 2 KOH + H2O2 + O2 eliminates
72. (a) v esc 2gR , where R is radius of the planet. moisture.
Hence escape velocity is independent of m. 97. (c) Optical and geometrical isomers are stereoisomers
S-16 The Pattern Target AIEEE
Cl h
98. (c) C CH CH 2 CH 2 CH 3 does not show 116. (c) x. p ;
Cl 4
geometrical isomerism due to presence of two similar 34
6.62 10 28 1
Cl atoms on the same C-atom. V 2.1 10 ms
5
99. (a) 7N2 = 1s2 2s2 3p3; 2 2 6 2 3 4 3.14 0.025 10
15P = 1s 2s 2p 3s 3p
In phosphorous the 3 d- orbitals are available. 117. (a) Acetylene reacts with the other three as:
100. (c) PV = nRT (number of moles = n/V) CH2
Na +HCl
CH CNa CH CH
n/V = P/RT.. liq. NH3
CHCl
101. (a) It is zero order reaction
CH2 CH3
+HCl
102. (a) Mg(OH)2 Mg 2OH ;
s 2s CHCl CHCl2
Ksp = (s) (2s)2 = 4s3 [AgNO3+NH4OH]
CH CH AgC CAg + NH4NO3
103. (d) NH4+ ions are increased to suppress release of OH- white ppt.
ions, hence solubility product of Fe(OH)3 is attained. 118. (a) In this reaction the ratio of number of moles of reactants
Colour of precipitate is different. to products is same i.e. 2 : 2, hence change in volume
104. (c) According to molecular weight given. will not alter the number of moles.
105. (d) Amino acids contain – NH2 and – COOH groups 119. (d) H negative shows that the reaction is spontaneous.
Higher value for Zn shows that the reaction is more
km 2 mol feasible.
106. (b) S K sm 2 mol 1
m m 3 120. (a) Mn2+ has the maximum number of unpaired electrons
(5) and therefore has maximum moment.
107. (a) 2nd excited state will be the 3rd energy level. 121. (b) In molecules (a), (c) and (d), the carbon atom has a
13.6 13.6 multiple bond, only (b) has sp3 hybridization.
En 2
eV or E eV 1.51 eV. 122. (d) Racemic mixture is formed when enantiomers are mixed
n 9
in equimolar proportion
108. (c) 123. (d)
109. (c) The correct name is 3 - methylbut 2 ol
124. (d)
Cl2 O
110. (c) CH3CH2COOH CH3CHClCOOH ||
red P
P
alc.KOH
CH 2 CHCOOH
HCl Acrylic acid O O
111. (c) Alumina is mixed with cryolite which acts as an O
O P P O
||
electrolyte.
||
112. (b) Silver ore forms a soluble complex with NaCN from O
O P
which silver is precipitated using scrap zinc.
|| O
Zn O
Ag 2S 2 NaCN Na[Ag (CN ) 2 ]
Na 2 [ Zn ( CN ) 4 ] Ag 125. (d)
126. (d) Beryllium shows anomalous properties due to its small
sod. argentocyanide
size.
(soluble)
127. (a) Ecell = Eright (cathode) - Eleft (anode).
113. (c) 8.314 JK–1 mol–1
CHOH
WB 128. (c) CH CH HOCl ||
114. (b) Tb Kb 1000 ;
M B WA CHCl
WB CH (OH) 2 CHO
Tf Kf 1000 ; HOCl
|
H 2O
|
M B WA CHCl 2 CHCl 2
dichloroac etaldehyde
Tb Kb Tb 0.512
0.0512 C . 129. (b) Aldehydic group gets oxidised to carboxylic group.
Tf Kf 0.186 1.86
Double bond breaks and carbon gets oxidised to
115. (c) Ecell = Reduction potential of cathode (right) carboxylic group.
– reduction potential of anode (left) 130. (c) The given compound is aspirin which is antipyretic
= Eright – Eleft. and analgesic.
AIEEE-2002 Solved Paper S-17
2 1 cos 2 2 1
5. (b) sin ; Period cos
2 2 cos
tan–1 =x
6. (d) l = ARp–1 log l = log A + (p – 1) log R 1
1 . cos
m = ARq–1 log m = log A + (q – 1) log R cos
n = ARr–1 log n = log A + (r – 1) log R
1 cos
Now, tan–1 x
2 cos
log l p 1 log A (p 1) log R p 1
1 cos 2 cos
log m q 1 log A (q 1) log R q 1 tan x = OR cot x =
= =0 2 cos 1 cos
log n r 1 log A (r 1) log R r 1
1 cos 1 (1 2 sin 2 / 2)
2 sin x = 2
1 cos 2x 1 (1 2 sin x) 1 cos 1 2 cos /2 1
7. (d) lim lim ;
x 0 2x x 0 2x [Considering a with perpendicular = (1– cos ) and
[1 Form]
2
Lim x 2 5x 3 1 x
x x x 3
Hence, limit e
B C
(-1, -1) (3, 5)
2
9. (b) Total student = 100; for 70 stds. 75 70 = 5250 Lim 2 4x Lim 8x Lim
8
x x 3x 3 x 2x 3 e x
7200 – 5250 = 1950 e e
2
e4
[Using L' Hospital Rule]
1950
Average of girls = 65
30 x
15. (a) f (x)=sin–1 log 3 exists
3
10. (a) cot–1 ( cos ) – tan–1 ( cos )=x
x 1 x
1 if 1 log 1 3 31
tan–1 – tan –1 ( )=x 3 3 3
cos cos
1 x 9 or x [1, 9]
AIEEE-2002 Solved Paper S-19
1 1 2 x sin x dx
S ........ ......(2) =0+4 ;
2 8 16 0 1 cos 2 x
Subtracting (2) from (1)
( x) sin ( x)
I= 4 dx
1 1 1 1 0 1 cos ( 2
x)
S ........
2 4 8 16
( x) sin x
1 1/ 4 1 I=4 dx
or S S 1 0 1 cos2 x
2 1 1/ 2 2
sin x dx x sin x dx
P 2S 2 4 2
4
4
0 1 cos x 1 cos 2 x
17. (b) ar = 2
sin x
a ar ar 2 ar 3 ar 4 ar 5 ar 6 ar 7 ar 8 2 4 dx
= a9 r36 = (ar4)9 = 29 = 512
0 1 cos 2 x
10
Put cos x = t and solve it.
18. (a) I | sin x | dx 10 | sin x | dx 10 sin x dx 22. (c) Apply L H Rule, we have,
0 0 0 xf ( 2) 2f ( x ) 0
lim
[ | sin x| is periodic with period and sin x > 0 x 2 x 2 0
if 0 < x < ]
lim f (2) 2f (x ) f (2) 2f (2)
/2 x 2
/2
I 20 sin x dx 20 cos x 0 20 = 4 – 2 4 = –4.
0 23. (b) Let | z | = | | = r
/4 z = rei = rei where + = .
n 2
19. (b) In In 2 tan x (1 tan x )dx z= rei( – ) = rei .e–i = –re–i = – . = re–i ]
0
24. (c) Given | z – 4 | < | z – 2 | Let z = x + iy
/4 /4 | (x – 4) + iy) | < | (x – 2) + iy |
n 2 tan n 1 x
tan x sec x dx (x – 4)2 + y2 < (x – 2)2 + y2
n 1
0 0 x2 – 8x + 16 < x2 – 4x + 4 12 < 4x
x >3 Re(z) > 3
1 0 1
n 1 n 1 25. (b) Let the circle be |z – z0| = r. Then according to given
conditions |z0 – z1| = r + a and |z0 – z2|=r+b. Eliminating
1 r, we get |z0 – z1| –|z0 – z2| = a – b.
In + In+2 = lim n [In+ In+2]
n 1 x Locus of centre z0 is |z – z1| –|z – z2| = a – b, which
represents a hyperbola.
1 n n
lim n. 1 26. (b) Let a = first term of G.P. and r = common ratio of G.P.;
x n 1 n 1 1
n 1 Then G.P. is a, ar, ar2
n
a
2 1 2 Given S = 20 = 20
1 r
20. (c) [x 2 ]dx [x 2 ]dx [x 2 ]dx
0 0 1 a = 20(1 – r) ... (i)
Also a2 + a2r2 + a2r4 + ... to =100
1 2
0dx 1dx 2 1 a2
=100 a2 = 100(1 – r)(1 + r)... (ii)
0 1 1 r2
S-20 The Pattern Target AIEEE
From (i), a2 = 400(1 – r)2;
1 1 1 1 2
From (ii), we get 100(1 – r)(1 + r) = 400(1 – r)2 1 1 1 1 1 ........
2! n 3! n n
1 + r = 4 – 4r 5r = 3 r = 3/5.
27. (a) 13 – 23 + 33 – 43 + ........+ 93 1 1 1 1
= 13 + 23 + 33 +.......+ 93 – 2(23 + 43 + 63 + 83) 1 ........
1! 2! 3! (9999)!
2
9 10
2.2 3 13 2 3 33 4 3 1
1 1
....... e 3
2 1! 2!
2
39. (c) tr + 2 = 2nCr + 1 xr + 1; t3r = 2nC3r – 1 x3r – 1
4 5 Given 2nCr + 1 = 2nC3r – 1 ;
(45) 2 16. = 2025 – 1600 = 425
2 2nC 2n
2n – (r + 1) = C3r – 1
28. (a) Let and y, are the roots of the equations 2n – r – 1 = 3r – 1 2n = 4r
x + ax + b = 0 and x2 + bx + a = 0.
2
n
1 B
38. (d) (1 0.0001)10000 1 , n = 10000
n M (x1, y1)
1 n(n 1) 1 n (n 1)(n 2) 1
1 n. ........ O A
X
n 2! n 2 3! n3
AIEEE-2002 Solved Paper S-21
2
(d) period of cos2x = at
2 = a2 b2 2ab cos t ;
b
& period of sinx = 2
1p 2 p .... n p at
when. cos t 1
54. (a) We have lim
p 1
; a2 b2 2ab b
n n
=a+b
n 1 1
rp xp 1 1
lim x p dx ax 3 bx 2
p p 1 p 1 61. (a) Let f (x) = cx f (0) = 0 and f (1)
r 1n n
n
0 0 3 2
55. (d) Since lim [ x ] does not exist, hence the required limit a b 2a 3b 6c
x 0 = c .=0
3 2 6
does not exist.
Also f (x) is continuous and differentiable in [0, 1] and
f ( x) 1 0 [0, 1[. So by Rolle’s theorem, f (x) = 0.
56. (a) lim form using L’ Hospital’s rule
x 1 x 1 0 i.e ax2 + bx + c = 0 has at least one root in [0, 1].
2
1 3
f (x) 62. (d) We have f ( x )dx ; Now,,
4
2 f (x ) f (1) 2 0
lim = = =2.
x 1 1/ 2 x f (1) 1
2 2 2
57. (b) Let a is a rational number other than 0, in xf (x )dx x f ( x )dx f ( x )dx
[–5, 5], then 0 0 0
25 16 9 2( a b b c c a) 0
Clearly the bounded area is as shown in the following
figure.
y (a b b c c a) 25
y= – n (–x) y= – n x
|a b b c c a | 25
x
–1 O 1 68. (a) Since a , c, b form a right handed system,
y= n (–x) y= n x
î ĵ k̂
1 c b a 0 1 0 zî xk̂
Required area 4 ( nx)dx x y z
0
3 | a |:| b |:| c | 34 : 45 : 39 .
=4 2 4 3.
2 70. (c) Let a b c r . Then
Now, ( a b )2 ( a b )2 a 2 b2 ; a (a b c ) a r 0 a b a c a r
a b c a a r a r 0
(a b )2 48 16 4 (a b )2 16
Similarly b r 0 & c r 0
Above three conditions will be satisfied for non-zero
65. (a) We have, [ a b b x c c a]
vectors if and only if r 0
3
= ( a b ). (b x c) (c a) 71. (a) P (A B) = P (A) + P (B) – P (A B); =1
4
1
– P( A ) + P(B) –
= ( a b ). (m a ) c (m c ) a ) 4
2 2
1=1– + P(B) P(B) = ;
3 3
(where m b c )
2 1 5
{( a b ). c }.{( a ( b c )} [ a b c ]2 42 16 Now, P( A B ) = P(B) – P ( A B)= – =
3 4 12
S-24 The Pattern Target AIEEE
72. (d) The event follows binomial distribution with P + Q cos = 0 ......(3)
n = 5, p = 3/6 = 1/2. From (2) and (3),
q = 1 – p = 1/2.; Variance = npq = 5/4. Q2 – P2 = 144 (Q – P) (Q + P) = 144
73. (b) Equation of plane through (1, 0, 0) is
144
a (x – 1) + by + cz = 0 ...(i) Q P 8
18
(i) passes through (0, 1, 0).
From (1), On solving, we get Q = 13, P = 5
–a + b = 0 b= a
75. (b) TQW = 180 –
a a RQW = 2
Also, cos 45° =
2(2a 2 c2 ) RQT = 180 –
P
2a = 2a 2 c 2 2a2 = c2
T
c= 2a O
Q
So d.r of normal are a, a 2 a i.e. 1, 1, 2 . 90–2
R
74. (a) Given P + Q = 18 ......(1)
2 2
P + Q + 2PQ cos = 144 ......(2) w
Applying Lami's theorem at Q.
Q sin
tan 90º T R W
P Q cos
sin 2 sin(180 ) sin(180 )
R = W and T = 2W cos
R=12
Q
P
AIEEE - 2003
Paper - I
Time : 2½ hours Max. Marks : 450
(a) less than v 33. A wire suspended vertically from one of its ends is stretched
(b) greater than v by attaching a weight of 200N to the lower end. The weight
stretches the wire by 1 mm. Then the elastic energy stored
(c) v in the direction of the largest force BC in the wire is
(d) v , remainng unchanged (a) 0.2 J (b) 10 J (c) 20 J (d) 0.1 J
26. If the electric flux entering and leaving an enclosed surface 34. The escape velocity for a body projected vertically upwards
from the surface of earth is 11 km/s. If the body is projected
respectively is 1 and 2, the electric charge inside the
surface will be at an angle of 45 o with the vertical, the escape velocity will
( 2 be
(a) 1) o (b) ( 1 2)/ o
(c) ( 2 1) / o (d) ( 1 2) o
(a) 11 2 km / s (b) 22 km/s
27. A horizontal force of 10 N is necessary to just hold a block
stationary against a wall. The coefficient of friction between 11
(c) 11 km/s (d) km / s
the block and the wall is 0.2. The weight of the block is 2
35. A mass M is suspended from a spring of negligible mass.
The spring is pulled a little and then released so that the
mass executes SHM of time period T. If the mass is increased
10N
5T m
by m, the time period becomes . Then the ratio of is
3 M
(a) 20 N (b) 50 N (c) 100 N (d) 2 N
28. A marble block of mass 2 kg lying on ice when given a 3 25 16 5
(a) (b) (c) (d)
velocity of 6 m/s is stopped by friction in 10 s. Then the 5 9 9 3
coefficient of friction is 36. “Heat cannot by itself flow from a body at lower temperature
(a) 0.02 (b) 0.03 (c) 0.04 (d) 0.06 to a body at higher temperature” is a statement or
29. Consider the following two statements : consequence of
A. Linear momentum of a system of particles is zero (a) second law of thermodynamics
B. Kinetic energy of a system of particles is zero.
(b) conservation of momentum
Then
(c) conservation of mass
(a) A does not imply B and B does not imply A
(b) A implies B but B does not imply A (d) first law of thermodynamics
(c) A does not imply B but B implies A 37. Two particles A and B of equal masses are suspended from
(d) A implies B and B implies A two massless springs of spring of spring constant k1 and k2
30. Two coils are placed close to each other. The mitual , respectively. If the maximum velocities, during oscillation,
inductance of the pair of coils depends upon are equal, the ratio of amplitude of A and B is
(a) the rates at which currents are changing in the two
coils k1 k2 k2 k1
(a) k2 (b) (c) k1 (d)
(b) relative position and orientation of the two coils k1 k2
(c) the materials of the wires of the coils
38. The length of a simple pendulum executing simple harmonic
(d) the currents in the two coils
31. A block of mass M is pulled along a horizontal frictionless motion is increased by 21%. The percentage increase in the
surface by a rope of mass m. If a force P is applied at the free time period of the pendulum of increased length is
end of the rope, the force exerted by the rope on the block is (a) 11% (b) 21% (c) 42% (d) 10%
39. The displacement y of a wave travelling in the x -direction is
Pm Pm PM
(a) (b) (c) P (d) given by
M m M m M m
32. A light spring balance hangs from the hook of the other 4
y 10 sin 600 t 2 x metres
light spring balance and a block of mass M kg hangs from 3
the former one. Then the true statement about the scale
reading is where x is expressed in metres and t in seconds. The speed
(a) Both the scales read M kg each of the wave - motion, in ms-1 , is
(b) The scale of the lower one reads M kg and of the upper (a) 300 (b) 600 (c) 1200 (d) 200
one zero 40. When the current changes from +2 A to -2A in 0.05 second,
(c) The reading of the two scales can be anything but the an e.m.f. of 8 V is induced in a coil. The coefficient of self -
sum of the reading will be M kg induction of the coil is
(d) Both the scales read M/2 kg each (a) 0.2 H (b) 0.4 H (c) 0.8 H (d) 0.1 H
S-28 The Pattern Target AIEEE
44. A radioactive sample at any instant has its disintegration 51. A thin spherical conducting shell of radius R has a charge q.
rate 5000 disintegrations per minute. After 5 minutes, the Another charge Q is placed at the centre of the shell. The
rate is 1250 disintegrations per minute. Then, the decay R
constant (per minute) is electrostatic potential at a point P a distance from the
2
(a) 0.4 ln 2 (b) 0.2 ln 2 (c) 0.1 ln 2 (d) 0.8 ln 2 centre of the shell is
45. A nucleus with Z= 92 emits the following in a sequence:
2Q 2Q 2q
, , , , , , , , , , , , (a) (b)
4 oR 4 oR 4 oR
Then Z of the resulting nucleus is
(a) 76 (b) 78 (c) 82 (d) 74 2Q q (q Q ) 2
46. Two identical photocathodes receive light of frequencies f1 (c) 4
(d)
oR 4 oR 4 oR
and f2. If the velocites of the photo electrons (of mass m )
coming out are respectively v1 and v2, then 52. The work done in placing a charge of 8 10 18
coulomb on
2h a condenser of capacity 100 micro-farad is
(a) v12 v 22 (f1 f 2 )
m (a) 16 10 32
joule (b) 3.1 10 26
joule
1/ 2 10 32
2h (c) 4 10 joule (d) 32 10 joule
(b) v1 v 2 ( f1 f 2 )
m 53. The co-ordinates of a moving particle at any time ‘t’are given
1/ 2 (a) 2 2 (b) 3t 2 2 2
2h 3t
(d) v1 v 2 (f1 f 2 )
m
(c) t2 2 2 (d) 2 2
47. Which of the following cannot be emitted by radioactive
substances during their decay ? 54. During an adiabatic process, the pressure of a gas is found
(a) Protons (b) Neutrinoes to be proportional to the cube of its absolute temperature.
(c) Helium nuclei (d) Electrons The ratio C P / C V for the gas is
48. A 3 volt battery with negligible internal resistance is
connected in a circuit as shown in the figure. The current I, 4 5 3
(a) (b) 2 (c) (d)
in the circuit will be 3 3 2
AIEEE-2003 Solved Paper S-29
55. Which of the following parameters does not characterize 63. The wavelengths involved in the spectrum of deuterium
the thermodynamic state of matter? 2
(a) Temperature (b) Pressure ( D) are slightly different from that of hydrogen spectrum,
1
(c) Work (d) Volume because
56. A Carnot engine takes 3 10 6 cal. of heat from a reservoir (a) the size of the two nuclei are different
(b) the nuclear forces are different in the two cases
at 627 o C , and gives it to a sink at 27 o C . The work done (c) the masses of the two nuclei are different
by the engine is (d) the atraction between the electron and the nucleus is
differernt in the two cases
(a) 4.2 10 6 J (b) 8.4 10 6 J 64. In the middle of the depletion layer of a reverse- biased
(c) 6 (d) zero p-n junction, the
16.8 10 J
(a) electric field is zero
3 (b) potential is maximum
57. A spring of spring constant 5 10 N / m is stretched
initially by 5cm from the unstretched position. Then the (c) electric field is maximum
work required to stretch it further by another 5 cm is (d) potential is zero
(a) 12.50 N-m (b) 18.75 N-m 65. If the binding energy of the electron in a hydrogen
atom is 13.6eV, the energy required to remove the electron
(c) 25.00 N-m (d) 6.25 N-m
58. A metal wire of linear mass density of 9.8 g/m is stretched from the first excited state of Li is
with a tension of 10 kg-wt between two rigid supports 1 (a) 30.6 eV (b) 13.6 eV
metre apart. The wire passes at its middle point between the (c) 3.4 eV (d) 122.4 eV
poles of a permanent magnet, and it vibrates in resonance 66. A body is moved along a straight line by a machine
when carrying an alternating current of frequency n. The delivering a constant power. The distance moved by the
frequency n of the alternating source is body in time ‘t’ is proportional to
(a) 50 Hz (b) 100 Hz (a) t3 / 4 (b) t3 / 2
(c) 200Hz (d) 25Hz
59. A tuning fork of known frequency 256 Hz makes 5 beats per
(c) t1/ 4 (d) t1/ 2
second with the vibrating string of a piano. The beat 67. A rocket with a lift-off mass 3.5 10 4 kg is blasted upwards
frequency decreases to 2 beats per second when the tension with an initial acceleration of 10m/s2. Then the initial thrust
in the piano string is slightly increased. The frequency of of the blast is
the piano string before increasing the tension was (a) 3.5 10 5 N (b) 7.0 10 5 N
(a) 256 + 2 Hz (b) 256 – 2 Hz
(c) 256 – 5 Hz (d) 256 + 5 Hz (c) 14.0 10 5 N (d) 1.75 10 5 N
60. A body executes simple harmonic motion. The potential 68. To demonstrate the phenomenon of interference, we require
energy (P.E), the kinetic energy (K.E) and total energy (T.E) two sources which emit radiation
are measured as a function of displacement x. Which of the (a) of nearly the same frequency
following statements is true ? (b) of the same frequency
(a) K.E. is maximum when x = 0 (c) of different wavelengths
(b) T.E is zero when x = 0 (d) of the same frequency and having a definite phase
relationship
(c) K.E is maximum when x is maximum
69. Three charges –q1 , +q2 and –q3 are place as shown in the
(d) P.E is maximum when x = 0 figure. The x - component of the force on -q1 is proportional
61. In the nuclear fusion reaction to
2 3 4 Y
1 H 1H 2 He n -q3
given that the repulsive potential energy between the two
nuclei is ~ 7.7 10 14 J , the temperature at which the gases
a
must be heated to initiate the reaction is nearly b
[Boltzmann’s Constant k 1.38 10 23
J/K ] -q1 +q2 X
(a) 10 K 7 (b) 10 K 5
(c) 10 K 3
(d) 10 K 9
q2 q3 q2 q3
62. Which of the following atoms has the lowest ionization (a) 2 2
cos (b) sin
a2
2
b a b
potential ?
q2 q3 q2 q3
14 133 40 16 (c) cos (d) sin
(a) N (b) Cs (c) Ar (d) O b2 a2 b2 a2
7 55 18 8
S-30 The Pattern Target AIEEE
70. A 220 volt, 1000 watt bulb is connected across a 110 volt 79. The correct order of increasing basic nature for the bases
mains supply . The power consumed will be NH3, CH3NH2 and (CH3)2 NH is
(a) 750 watt (b) 500 watt (a) (CH3)2NH < NH3 < CH3NH2
(c) 250 watt (d) 1000 watt (b) NH3 < CH3NH2 < (CH3)2NH
71. The image formed by an objective of a compound (c) CH3NH2 < (CH3)2NH < NH3
microscope is (d) CH3NH2 < NH3 < (CH3)2NH
(a) virtual and diminished (b) real and diminished 80. Bottles containing C6H5I and C6H5CH2I lost their original
(c) real and enlarged (d) virtual and enlarged labels. They were labelled A and B for testing. A and B were
72. The earth radiates in the infra-red region of the spectrum. separately taken in test tubes and boiled with NaOH solution.
The spectrum is correctly given by The end solution in each tube was made acidic with dilute
(a) Rayleigh Jeans law HNO3 and then some AgNO3 solution was added. Substance
(b) Planck’s law of radiation B gave a yellow precipitate. Which one of the following
(c) Stefan’s law of radiation statements is true for this experiment ?
(d) Wien’s law (a) A and C6H5CH2I
73. To get three images of a single object, one should have two (b) B and C6H5I
plane mirrors at an angle of (c) Addition of HNO3 was unnecessary
(d) A was C6H5I
(a) 60 o (b) 90 o (c) 120 o (d) 30 o
81. The internal energy change when a system goes from state
74. According to Newton’s law of cooling, the rate of cooling
A to B is 40 kJ/mole. If the system goes from A to B by a
of a body is proportional to ( )n , where is the reversible path and returns to state A by an irreversible path
difference of the temperature of the body and the what would be the net change in internal energy ?
surroundings, and n is equal to (a) > 40 kJ (b) < 40 kJ (c) Zero (d) 40 kJ
(a) two (b) three (c) four (d) one 82. If at 298 K the bond energies of C — H, C — C, C = C and H
75. The length of a given cylindrical wire is increased by 100%. — H bonds are respectively 414, 347, 615 and 435 kJ mol–1,
Due to the consequent decrease in diameter the change in the value of enthalpy change for the reaction
the resistance of the wire will be H2C CH 2 (g) H 2 (g) H 3C — CH 3 (g)
(a) 200% (b) 100% (c) 50% (d) 300%
at 298 K will be
(a) – 250 kJ (b) + 125 kJ
Section - 2 (c) – 125 kJ (d) + 250 kJ
234
83. The radionucleide 90 Th undergoes two successive -
decays followed by one -decay. The atomic number and
76. Which of the following could act as a propellant for rockets? the mass number respectively of the resulting radionucleide
(a) Liquid oxygen + liquid argon are
(b) Liquid hydrogen + liquid oxygen (a) 94 and 230 (b) 90 and 230
(c) Liquid nitrogen + liquid oxygen (c) 92 and 230 (d) 92 and 234
(d) Liquid hydrogen + liquid nitrogen 84. The half-life of a radioactive isotope is three hours. If the
77. The reaction of chloroform with alcoholic KOH and p- initial mass of the isotope were 256 g, the mass of it remaining
toluidine forms undecayed after 18 hours would be
(a) 8.0 g (b) 12.0 g (c) 16.0 g (d) 4.0 g
(a) H3C N2Cl
85. If liquids A and B form an ideal solution
(b) H3C NHCHCl2 (a) the entropy of mixing is zero
(c) H3C NC (b) the free energy of mixing is zero
(c) the free energy as well as the entropy of mixing are
(d) H3C CN each zero
78. Nylon threads are made of (d) the enthalpy of mixing is zero
(a) polyester polymer 86. The radius of La3+ (Atomic number of La = 57) is 1.06Å.
(b) polyamide polymer Which one of the following given values will be closest to
(c) polyethylene polymer the radius of Lu3+ (Atomic number of Lu = 71) ?
(d) polyvinyl polymer (a) 1.40 Å (b) 1.06 Å (c) 0.85 Å (d) 1.60 Å
AIEEE-2003 Solved Paper S-31
87. Ammonia forms the complex ion [Cu(NH3)4]2+ with copper 96. Concentrated hydrochloric acid when kept in open air
ions in alkaline solutions but not in acidic solutions. What sometimes produces a cloud of white fumes. The explanation
is the reason for it ? for it is that
(a) In acidic solutions protons coordinate with ammonia (a) oxygen in air reacts with the emitted HCl gas to form a
molecules forming NH 4 ions and NH3 molecules are cloud of chlorine gas
(b) strong affinity of HCl gas for moisture in air results in
not available
forming of droplets of liquid solution which appears
(b) In alkaline solutions insoluble Cu(OH)2 is precipitated
like a cloudy smoke.
which is soluble in excess of any alkali
(c) Copper hydroxide is an amphoteric substance (c) due to strong affinity for water, concentrated
hydrochloric acid pulls moisture of air towards itself.
(d) In acidic solutions hydration protects copper ions
This moisture forms droplets of water and hence the
88. One mole of the complex compound Co(NH3)5Cl3, gives 3
cloud.
moles of ions on dissolution in water. One mole of the same
complex reacts with two moles of AgNO3 solution to yield (d) concentrated hydrochloric acid emits strongly smelling
two moles of AgCl (s). The structure of the complex is HCl gas all the time.
(a) [Co(NH3)3Cl3]. 2 NH3 (b) [Co(NH3)4Cl2] Cl . NH3 97. An ether is more volatile than an alcohol having the same
(c) [Co(NH3)4Cl] Cl2. NH3 (d) [Co(NH3)5Cl] Cl2 molecular formula. This is due to
89. In the coordination compound, K4[Ni(CN)4], the oxidation (a) alcohols having resonance structures
state of nickel is (b) inter-molecular hydrogen bonding in ethers
(a) 0 (b) +1 (c) +2 (d) –1 (c) inter-molecular hydrogen bonding in alcohols
90. In curing cement plasters water is sprinkled from time to (d) dipolar character of ethers
time. This helps in 98. Graphite is a soft solid lubricant extremely difficult to melt.
(a) developing interlocking needle-like crystals of hydrated The reason for this anomalous behaviour is that graphite
silicates (a) is an allotropic form of diamond
(b) hydrating sand and gravel mixed with cement (b) has molecules of variable molecular masses like
(c) converting sand into silicic acid polymers
(d) keeping it cool (c) has carbon atoms arranged in large plates of rings of
91. Which one of the following statements is not true? strongly bound carbon atoms with weak interplate
(a) pH + pOH = 14 for all aqueous solutions bonds
(b) The pH of 1 × 10–8 M HCl is 8 (d) is a non-crystalline substance
(c) 96,500 coulombs of electricity when passed through a 99. According to the Periodic Law of elements, the variation in
CuSO4 solution deposits 1 gram equivalent of copper properties of elements is related to their
at the cathode (a) nuclear masses
(d) The conjugate base of H 2 PO 4 is HPO 24 (b) atomic numbers
92. On mixing a certain alkane with chlorine and irradiating it (c) nuclear neutron-proton number ratios
with ultraviolet light, it forms only one monochloroalkane. (d) atomic masses
This alkane could be 100. Which one of the following statements is correct ?
(a) pentane (b) isopentane (a) From a mixed precipitate of AgCl and AgI, ammonia
(c) neopentane (d) propane solution dissolves only AgCl
93. Butene-1 may be converted to butane by reaction with (b) Ferric ions give a deep green precipitate on adding
(a) Sn – HCl (b) Zn – Hg potassium ferrocyanide solution
(c) Pd/H2 (d) Zn – HCl (c) On boiling a solution having K+, Ca2+ and HCO 3 ions
94. What may be expected to happen when phosphine gas is
mixed with chlorine gas ? we get a precipitate of K2Ca(CO3)2
(a) PCl3 and HCl are formed and the mixture warms up (d) Manganese salts give a violet borax bead test in the
reducing flame
(b) PCl5 and HCl are formed and the mixture cools down
(c) PH3 . Cl2 is formed with warming up 101. Glass is a
(d) The mixture only cools down (a) super-cooled liquid
95. The number of d-electrons retained in Fe2+ (b) gel
(At. no. of Fe = 26) ion is (c) polymeric mixture
(a) 4 (b) 5 (c) 6 (d) 3 (d) micro-crystalline solid
S-32 The Pattern Target AIEEE
102. The orbital angular momentum for an electron revolving in 110. For the reaction equilibrium
h N2O4 (g) 2 NO2 (g)
an orbit is given by l (l 1) . . This momentum for an s- the concentrations of N2O4 and NO2 at equilibrium are 4.8 ×
2
electron will be given by 10–2 and 1.2 × 10–2 mol L–1 respectively. The value of Kc for
the reaction is
h (a) 3 × 10–1 mol L–1 (b) 3 × 10–3 mol L–1
(a) zero (b)
2 3
(c) 3 × 10 mol L –1 (d) 3.3 × 102 mol L–1
111. Consider the reaction equilibrium
h 1 h
(c) 2. (d) . 2SO 2 (g) O 2 (g) 2SO 3 (g); H 198 kJ
2 2 2
103. How many unit cells are present in a cube-shaped ideal On the basis of Le Chatelier’s principle, the condition
crystal of NaCl of mass 1.00 g ? favourable for the forward reaction is
[Atomic masses : Na = 23, Cl = 35.5] (a) increasing temperature as well as pressure
(a) 5.14 × 1021 unit cells (b) 1.28 × 1021 unit cells (b) lowering the temperature and increasing the pressure
21
(c) 1.71 × 10 unit cells (d) 2.57 × 1021 unit cells (c) any value of temperature and pressure
104. In the anion HCOO– the two carbon-oxygen bonds are found (d) lowering of temperature as well as pressure
to be of equal length. what is the reason for it ? 112. Which one of the following is an amphoteric oxide ?
(a) The C = O bond is weaker than the C — O bond (a) Na2O (b) SO2 (c) B2O3 (d) ZnO
(b) The anion HCOO– has two resonating structures 113. A red solid is insoluble in water. However it becomes soluble
(c) The anion is obtained by removal of a proton from the if some KI is added to water. Heating the red solid in a test
acid molecule tube results in liberation of some violet coloured fumes and
droplets of a metal appear on the cooler parts of the test
(d) Electronic orbitals of carbon atom are hybridised
tube. The red solid is
105. Which one of the following characteristics is not correct for
physical adsorption ? (a) HgI2 (b) HgO
(a) Adsorption increases with increase in temperature (c) Pb3O4 (d) (NH4)2Cr2O7
(b) Adsorption is spontaneous 114. Standard reduction electrode potentials of three metals A, B
& C are respectively + 0.5 V, – 3.0 V & –1.2 V. The reducing
(c) Both enthalpy and entropy of adsorption are negative
powers of these metals are
(d) Adsorption on solids is reversible
(a) A > B > C (b) C > B > A
106. For a cell reaction involving a two-electron change, the
(c) A > C > B (d) B > C > A
standard e.m.f. of the cell is found to be 0.295 V at 25ºC. The
equilibrium constant of the reaction at 25ºC will be 115. Which one of the following substances has the highest
proton affinity ?
(a) 29.5 × 10–2 (b) 10
(a) H2S (b) NH3 (c) PH3 (d) H2O
(c) 1 × 1010 (d) 1 × 10–10
116. In a 0.2 molal aqueous solution of a weak acid HX the degree
107. In an irreversible process taking place at constant T and P
of ionization is 0.3. Taking kf for water as 1.85, the freezing
and in which only pressure-volume work is being done, the
point of the solution will be nearest to
change in Gibbs free energy (dG) and change in entropy
(dS), satisfy the criteria (a) – 0.360º C (b) – 0.260º C
(a) (dS)V, E > 0, (dG)T, P < 0 (c) + 0.480º C (d) – 0.480º C
(b) (dS)V, E = 0, (dG)T, P = 0 117. When during electrolysis of a solution of AgNO3 9650
(c) (dS)V, E = 0, (dG)T, P > 0 coulombs of charge pass through the electroplating bath,
the mass of silver deposited on the cathode will be
(d) (dS)V, E < 0, (dG)T, P < 0
(a) 10.8 g (b) 21.6 g (c) 108 g (d) 1.08 g
108. The solubility in water of a sparingly soluble salt AB2 is 1.0
× 10–5 mol L–1. Its solubility product number will be 118. For the redox reaction :
(a) 4 × 10–10 (b) 1 × 10–15 Zn (s) Cu 2 (0.1M ) Zn 2 (1 M) Cu(s)
(c) 1 × 10 –10 (d) 4 × 10–15
º
109. What volume of hydrogen gas, at 273 K and 1 atm. pressure taking place in a cell, E cell is 1.10 volt. Ecell for the cell will
will be consumed in obtaining 21.6 g of elemental boron
RT
(atomic mass = 10.8) from the reduction of boron trichloride be 2.303 0.0591
by hydrogen ? F
(a) 67.2 L (b) 44.8 L (a) 1.80 volt (b) 1.07 volt
(c) 22.4 L (d) 89.6 L (c) 0.82 volt (d) 2.14 volt
AIEEE-2003 Solved Paper S-33
135. 25ml of a solution of barium hydroxide on titration with a 0.1 142. Several blocks of magnesium are fixed to the bottom of a
molar solution of hydrochloric acid gave a litre value of ship to
35ml. The molarity of barium hydroxide solution was (a) make the ship lighter
(a) 0.14 (b) 0.28 (b) prevent action of water and salt
(c) 0.35 (d) 0.07 (c) prevent puncturing by under-sea rocks
136. The correct relationship between free energy change in a (d) keep away the sharks
reaction and the corresponding equilibrium constant Kc is 143. Which one of the following pairs of molecules will have
(a) – G = RT ln Kc (b) Gº = RT ln Kc permanent dipole moments for both members ?
(c) – Gº = RT ln Kc (d) G = RT ln Kc (a) NO2 and CO2 (b) NO2 and O3
137. The rate law for a reaction between the substances A and B (c) SiF4 and CO2 (d) SiF4 and NO2
is given by 144. Which one of the following groupings represents a
Rate = k [A]n [B]m collection of isoelectronic species ?
On doubling the concentration of A and halving the (At. nos. : Cs : 55, Br : 35)
concentration of B, the ratio of the new rate to the earlier (a) N3–, F–, Na+ (b) Be, Al3+, Cl–
rate of the reaction will be as (c) Ca2+, Cs+, Br (d) Na+, Ca2+, Mg2+
(a) (m + n) (b) (n – m) 145. Which one of the following processes will produce hard
water ?
1
(c) 2(n – m) (d) (a) Saturation of water with MgCO3
2 (m n)
(b) Saturation of water with CaSO4
138. Ethyl isocyanide on hydrolysis in acidic medium generates
(c) Addition of Na2SO4 to water
(a) propanoic acid and ammonium salt (d) Saturation of water with CaCO3
(b) ethanoic acid and ammonium salt 146. Which one of the following compounds has the smallest
(c) methylamine salt and ethanoic acid bond angle in its molecule ?
(d) ethylamine salt and methanoic acid (a) OH2 (b) SH2
139. The enthalpy change for a reaction does not depend upon (c) NH3 (d) SO2
(a) use of different reactants for the same product 147. The pair of species having identical shapes for molecules of
(b) the nature of intermediate reaction steps both species is
(c) the differences in initial or final temperatures of (a) XeF2, CO2 (b) BF3, PCl3
involved substances (c) PF5, IF5 (d) CF4, SF4
(d) the physical states of reactants and products 148. The atomic numbers of vanadium (V), Chromium (Cr),
140. A pressure cooker reduces cooking time for food because manganese (Mn) and iron (Fe) are respectively 23, 24, 25,
and 26. Which one of these may be expected to have the
(a) boiling point of water involved in cooking is increased
highest second ionization enthalpy ?
(b) the higher pressure inside the cooker crushes the food
(a) Cr (b) Mn
material
(c) Fe (d) V
(c) cooking involves chemical changes helped by a rise in
149. In Bohr series of lines of hydrogen spectrum, the third line
temperature
from the red end corresponds to which one of the following
(d) heat is more evenly distributed in the cooking space inter-orbit jumps of the electron for Bohr orbits in an atom
141. For the reaction system : of hydrogen
2 NO(g ) O 2 (g) 2 NO 2 (g) volume is suddenly reduced (a) 5 2 (b) 4 1
to half its value by increasing the pressure on it. If the reaction (c) 2 5 (d) 3 2
is of first order with respect to O2 and second order with 150. The de Broglie wavelength of a tennis ball of mass 60 g
respect to NO, the rate of reaction will moving with a velocity of 10 metres per second is
(a) diminish to one-eighth of its initial value approximately
(b) increase to eight times of its initial value Planck’s constant, h = 6.63 × 10–34 Js
(c) increase to four times of its initial value (a) 10–31 metres (b) 10–16 metres
(d) diminish to one-fourth of its initial value (c) 10–25 metres (d) 10–33 metres
AIEEE-2003 Solved Paper S-35
Paper - II
Time : 1¼ hours Max. Marks : 225
(a) 9 (b) 1
d esin x
1. Let F(x) , x 0. (c) 5 (d) 7
dx x
8. If f ( y ) e y , g ( y ) y; y 0 and
4
3 sin x 3
If e dx F(k) F(1) then one of the possible t
x F(t ) f ( t y)g( y) dy, then
1
values of k, is 0
(a) 64 (b) 15
t
(c) 16 (d) 63 (a) F (t ) te (b) F (t ) 1 te t (1 t )
2. The median of a set of 9 distinct observations is 20.5. If
each of the largest 4 observations of the set is increased by
(c) F (t ) et (1 t ) (d) F (t ) tet .
2, then the median of the new set
(a) remains the same as that of the original set 9. The function f ( x ) log x x 2 1 , is
(b) is increased by 2
(a) neither an even nor an odd function
(c) is decreased by 2
(b) an even function
(d) is two times the original median.
(c) an odd function
1 24 34 ...n 4 1 23 33 ...n 3 (d) a periodic function.
3. lim lim
n n4 n n5 10. If the sum of th e roots of the quadratic equation
13. If the pair of straight lines x 2 2 pxy y 2 0 and 19. If f ( x) x n , then the value of
x2 2qxy y 2 0 be such that each pair bisects the angle f ' (1) f ' ' (1) f ' ' ' (1) ( 1) n f n (1)
f (1) .......... is
1! 2! 3! n!
between the other pair, then
(a) pq = -1 (b) p = q (a) 1 (b) 2n
(c) p = -q (d) pq = 1. (c) (d) 0.
2n 1
14. Locus of centroid of the triangle whose vertices are
20. Let u iˆ ˆj , v iˆ ˆj and w iˆ 2 ˆj 3kˆ . If n̂ is a unit
(a cos t , a sin t ), (b sin t , b cos t ) and (1, 0), where t is a
parameter, is vector such that u.nˆ 0 and v .nˆ 0 , then w. nˆ is equal
(a) (3x 1) 2 (3 y ) 2 a 2 b2 to
(a) 3 (b) 0
(b (3x 1) 2 2 2 2 (c) 1 (d) 2.
(3 y ) a b
21. A particle acted on by constant forces 4iˆ ˆj 3kˆ and
(c) (3x 1) 2 (3 y ) 2 a2 b2
3iˆ ˆj kˆ to the point 5iˆ 4 ˆj kˆ . The total work
(d) (3x 1) 2 (3 y ) 2 a 2 b2 .
done by the forces is
log( 3 x ) log( 3 x ) (a) 50 units (b) 20 units
15. If lim k , the value of k is
x 0 x (c) 30 units (d) 40 units.
2s (a) a2 2b (b) a2 3b
(c) (d) 2s(f r)
1 1 35. The solution of the differential equation
f r
1 dy
29. Two stones are projected from the top of a cliff h metres (1 y 2 ) ( x e tan y
) 0 , is
dx
high , with the same speed u, so as to hit the ground at the
same spot. If one of the stones is projected at an angle 1 1
(a) xe 2 tan y
e tan y
k
to the horizontal then tan equals
1
2 2u (b) ( x 2) ke 2 tan y
(a) u (b)
gh gh 1 1
(c) 2 xe tan y
e 2 tan y
k
(c) u (d) u 1
2g 2h (d) xe tan y
tan 1
y k
h g
36. Let f(x) be a function satisfying f '(x) = f(x) with f(0)=1 and
30. If 1, , 2 are the cube roots of unity, th en 2
g(x) be a function that satisfies f(x) + g(x) = x . Then the
n 2n
1 1
n 2n value of the integral f ( x ) g ( x ) dx, is
1 is equal to 0
2n n
1
e2 5 e2 5
(a) e (b) e
2 2 2 2 2
(a) (b) 0
(c) 1 (d) e2 3 e2 3
(c) e (d) e .
31. The sum of the radii of inscribed and circumscribed circles 2 2 2 2
for an n sided regular polygon of side a , is 37. The lines 2 x 3 y 5 and 3x 4 y 7 are diameters of a
a circle having area as 154 sq.units.Then the equation of the
(a) cot (b) a cot
4 2n n circle is
a
(a) x2 y2 2x 2 y 62
(c) cot (d) a cot .
2 2n 2n
(b) x2 y2 2x 2 y 62
32. If x1 , x 2 , x3 and y1, y 2 , y3 are both in G.P. with the same
(c) x2 y2 2x 2 y 47
common ratio, then the points ( x1, y1 ), ( x 2 , y 2 ) and
(d) x2 y2 2x 2 y 47 .
( x3 , y3 )
S-38 The Pattern Target AIEEE
1 2 C A 3b
(c)
2
(d) 47. If in a ABC a cos c cos 2 , then the sides
3 3 2 2 2
41. If x is positive, the first negative term in the expansion of a, b and c
27 5
(1 x ) is (a) satisfy a b c (b) are in A.P
(a) 6th term (b) 7th term (c) are in G..P (d) are in H.P.
(c) 5th term (d) 8th term.
48. a , b , c are 3 vectors, such that a b c 0 ,
42. The number of integral terms in the expansion of
( 3 8
5 )256 is a 1 b 2, c 3, , then a.b b .c c .a is equal to
(a) 35 (b) 32 (a) 1 (b) 0
(c) 33 (d) 34. (c) –7 (d) 7
43. If nCr denotes the number of combination of n things taken 1
49. The value of the integral I x(1 x ) n dx is
r at a time, then the expression 0
n n n
Cr 1 Cr 1 2 Cr equals 1 1 1
(a) (b)
n 1 n 2 n 1 n 2 n 1
(a) Cr 1 (b) Cr
1 1 1
(c) n 2
Cr (d) n 1
Cr . (c) (d) .
1 n 2 n 1 n 2
44. Two particles start simultaneously from the same point
and move along two straight lines, one with uniform x2
sec 2 tdt
velocity u and the other from rest with uniform acceleration 0
50. The value of lim is
x 0 x sin x
f . Let be the angle between their directions of motion.
The relative velocity of the second particle w.r.t. the first is (a) 0 (b) 3
least after a time (c) 2 (d) 1
AIEEE-2003 Solved Paper S-39
f ( a )g ( x ) f ( a ) g ( a ) f ( x ) f (a ) 1 1
lim 4 x x
x a g( x ) f (x ) 59. If f ( x ) xe ,x 0
then the value of k is 0 ,x 0
(a) 0 (b) 4
then f(x) is
(c) 2 (d) 1
(a) discontinuous every where
(b) continuous as well as differentialble for all x
x
1 tan [1 sin x ] (c) continuous for all x but not differentiable at x = 0
2
54. lim is (d) neither differentiable nor continuous at x = 0
x
x
2 1 tan [ 2 x ]3 60. Domain of definition of the function
2
3
f ( x) log10 ( x 3 x) , is
4 x2
1
(a) (b) (a) ( 1,0) (1,2) ( 2, ) (b) ( a , 2)
8
(c) ( 1,0) (a ,2) (d) (1,2) (2, ) .
1
(c) 0 (d)
32 61. If f : R R satisfies f ( x y) f ( x) f ( y ) , for all x,
55. If the equation of the locus of a point equidistant n
from the point (a1, b1 ) and (a2, b2 ) is (a1 b2 ) x y R and f(1) = 7, then f (r ) is
r 1
64. In an experiment with 15 observations on x, the following 71. Two system of rectangular axes have the same origin. If a
results were available: plane cuts them at distances a, b, c and a' , b' , c ' from the
2 origin then
x 2830, x 170
One observation that was 20 was found to be wrong and 1 1 1 1 1 1
(a) 0
was replaced by the correct value 30. The corrected a2 b2 c2 a'2 b' 2 c' 2
variance is
(a) 8.33 (b) 78.00 1 1 1 1 1 1
(b) 0
(c) 188.66 (d) 177.33 a2 b2 c2 a'2 b' 2 c '2
65. A student is to answer 10 out of 13 questions in an
1 1 1 1 1 1
examination such that he must choose at least 4 from the (c) 0
a2 b2 c2 a'2 b' 2 c'2
first five questions. The number of choices available to him
is 1 1 1 1 1 1
(a) 346 (b) 140 (d) 2 2 2 2 2
0.
a b c a' b' c'2
(c) 196 (d) 280
x
a b 1 i
66. If A and A 2
, then 72. If 1 then
1 i
b a
(a) x 2n 1 , where n is any positive integer
(a) 2ab, a2 b2
(b) x 4n , where n is any positive integer
(b) 2 2 (c)
a b , ab x 2n , where n is any positive integer
(d) x 4n 1 , where n is any positive integer..
(c) a 2 b2 , 2ab
73. A function f from the set of natural numbers to integers
(d) 2 2 2 2. defined by
a b , a b
67. The number of ways in which 6 men and 5 women can dine
n 1
at a round table if no two women are to sit together is given , when n is odd
2 is
by f ( n)
n
(a) 7 × 5 (b) 6 × 5 , when n is even
2
(c) 30 (d) 5 × 4
68. Consider points A, B, C and D with position (a) neither one -one nor onto
vectors 7iˆ 4 ˆj 7kˆ, iˆ 6 ˆj 10kˆ , iˆ 3 ˆj 4kˆ and (b) one-one but not onto
(c) onto but not one-one
5iˆ ˆj 5kˆ respectively. Then ABCD is a
(d) one-one and onto both.
(a) parallelogram but not a rhombus 74. Let f(x) be a polynomial function of second degree. If f(1) =
(b) square f(-1) and a, b, c are in A. P , then f '(a),f (b),f '(c) are in
(c) rhombus
(a) Arithmetic -Geometric Progression
(d) rectangle.
(b) A.P
69. If u , v and w are three non- coplanar vectors, then
(c) G..P
(u v w).(u v ) (v w) equals (d) H.P.
(a) 75. The sum of the series
3u .v w (b) 0
(c) u .v w (d) u.w v 1 1 1
.......... .. up to is equal to
1. 2 2.3 3.4
70. The trigonometric equation sin 1 1
x 2 sin a
4
has a solution for (a) log e
e
1 1 1
(a) a (b) a (b) 2 log e 2
2 2 2
(c) log e 2 1
1
(c) all real values of a (d) a
2 (d) log e 2
AIEEE-2003 Solved Paper S-41
A N S W ER K EY
P A P ER I - P H YS IC S & C H EM IS T R Y P A P ER II - M A T H S
1 (b ) 31 (d ) 61 (d ) 91 (b ) 1 21 (c ) 1 (a ) 31 (d ) 61 (a )
2 (a ) 32 (a ) 62 (b ) 92 (c ) 1 22 (a ) 2 (a ) 32 (b ) 62 (c )
3 (b ) 33 (d ) 63 (c ) 93 (c ) 1 23 (b ) 3 (a ) 33 (a ) 63 (a )
4 (a ) 34 (c ) 64 (c ) 94 (b ) 12 4 (a ) 4 (b ) 34 (d ) 64 (b )
5 (d ) 35 (c ) 65 (b ) 95 (c ) 12 5 (a ) 5 (b ) 35 (c ) 65 (c )
6 (a ) 36 (a ) 66 (b ) 96 (a ) 12 6 (a ) 6 (d ) 36 (d ) 66 (c )
7 (a ) 37 (c ) 67 (a ) 97 (c ) 12 7 (d ) 7 (d ) 37 (d ) 67 (a )
8 (d ) 38 (d ) 68 (d ) 98 (c ) 12 8 (a ) 8 (c ) 38 (d ) 68 (c )
9 (c ) 39 (b ) 69 (b ) 99 (b ) 12 9 (b ) 9 (c ) 39 (a ) 69 (c )
10 (b ) 40 (d ) 70 (c ) 100 (a ) 13 0 (c ) 10 (d ) 40 (b ) 70 (b )
11 (a ) 41 (c ) 71 (c ) 101 (a ) 13 1 (d ) 11 (d ) 41 (d ) 71 (a )
12 (c ) 42 (a ) 72 (d ) 102 (a ) 13 2 (a ) 12 (a ) 42 (c ) 72 (b )
13 (c ) 43 (d ) 73 (b ) 103 (d ) 13 3 (d ) 13 (a ) 43 (c ) 73 (d )
14 (d ) 44 (a ) 74 (d ) 104 (b ) 13 4 (b ) 14 (c ) 44 (a ) 74 (b )
15 (c ) 45 (b ) 75 (d ) 105 (a ) 1 35 (d ) 15 (d ) 45 (a ) 75 (a )
16 (a ) 46 (a ) 76 (b ) 106 (d ) 1 36 (c ) 16 (b ) 46 (d )
17 (a ) 47 (a ) 77 (c ) 107 (a ) 1 37 (c ) 17 (c ) 47 (b )
18 (a ) 48 (b ) 78 (b ) 108 (d ) 1 38 (d ) 18 (b ) 48 (c )
19 (c ) 49 (b ) 79 (b ) 109 (a ) 1 39 (b ) 19 (d ) 49 (d )
20 (b ) 50 (c ) 80 (d ) 110 (b ) 1 40 (a ) 20 (a ) 50 (d )
21 (c ) 51 (c ) 81 (c ) 111 (b ) 1 41 (b ) 21 (d ) 51 (d )
22 (d ) 52 (d ) 82 (c ) 112 (d ) 1 42 (b ) 22 (d ) 52 (b )
23 (d ) 53 (b ) 83 (b ) 113 (a ) 1 43 (b ) 23 (d ) 53 (b )
24 (b ) 54 (d ) 84 (d ) 114 (d ) 1 44 (a ) 24 (d ) 54 (d )
25 (d ) 55 (c ) 85 (d ) 115 (b ) 1 45 (b ) 25 (a ) 55 (b )
26 (a ) 56 (b ) 86 (c ) 116 (d ) 146 (b ) 26 (d ) 56 (c )
27 (d ) 57 (b ) 87 (a ) 117 (a ) 14 7 (a ) 27 (a ) 57 (c )
28 (d ) 58 (a ) 88 (d ) 118 (b ) 14 8 (a ) 28 (a ) 58 (d )
29 (c ) 59 (c ) 89 (a ) 1 19 (b ) 14 9 (a ) 29 (a ) 59 (c )
30 (b ) 60 (a ) 90 (a ) 120 (b ) 15 0 (d ) 30 (b ) 60 (a )
T1 T2 1 1
T2 T–1
[C]2 [C] = LT
2 T1 2
0 0
1 9
14. (d) m R 2 or M t R2 S1 + S2 = 9 6S1 = 9 S1 1.5
2 6
S2 = 1.5 × 5 = 7.5
1 1
For disc X, x (m)(R ) 2 = ( r 2 t ).(R ) 2 Note: Maximum distance will be travelled by smaller
2 2 bodies due to the greater acceleration caused by the
same gravitational force
1
for disc y, y [ (4R ) 2 .t / 4][4R ]2 20. (b)
2 21. (c) Energy = Work done by force (F)
x 1 1 2500 m
y 64 x m.(50) 2 (F)(6) F
3
y ( 4) 2 2 6
1
T1
2
R1
3
For v = 100 km/hr .m(100) 2 (F)(S)
15. (c) T2 R3 2
T2 R2
1 2500m
m (100) 2 S
3/ 2 3/ 2 2 2 6
T1 R1 1
T2 R2 4 100 100 6 2
S 24 m
2500 2
T2 22. (d) From, the question if the horizontal distance is none
( 4) 3 / 2 8
T1 other than the horizontal range on the level of the roof
of building
T2 8 T1 8 5 40 = 40 hours
V
10 m/sec
1
16. (a) Angular momentum
Angular frequency
x
K.E. 10m 10m
Kinetic energy L
w
L1 K.E1 w2 L
4 L2 u 2 sin 2 (10) 2 sin(2 30)
L2 w1 KE 2 4 Range =
g g
17. (a) Decrea sin g
10 10 3
RMIVUXGE = 8.66
2 10
R Radio waves ; M Micro waves 23. (d)
I Infra red rays ; V Visible rays 24. (b) [momentum]=[M][L][T–1]=[MLT–1]
1 2
U Ultraviolet rays ; X x rays E [M ][ LT ] ML2 T 1
(Planck’s constant) = =
u T 1
G rays ; C Cosmic rays 25. (d) According to triangle law of forces. The resultant force
rays has least wavelength is zero.
18. (a) Applying the principle of conservation of linear In presence of zero external force, there is no change in
momentum velocity
26. (a) According to Gauss’s Law
4u
(4) (u) = (v) (238) v
238 (E . dA) q0 / 0 q 0( 2 1)
19. (c) Distance between the surface of the spherical bodies
= 12R – R – 2R = 9R [sin ce E. dA]
Force Mass
f= N
Accelerati on Mass
Distance Accelerati on 27. (d) 10N
a1 M 1 S1 1
S2 5S1
a2 SM 5 S2 5 mg
AIEEE-2003 Solved Paper S-43
f = mg N=W . 10 = W
1 k1
0.2 × 10 = W W= 2N K or
2 k2
6
28. (d) a g [using v = u + at] VA max VB max
10
6 6 k1 k2 A1 k2
= 0.06 (A1 ) (A 2 )
10 g 10 10 m m A2 k1
29. (c)
30. (b) T 2 1
38. (d) ; log T log(2 ) log
g 2 g
31. (d) Since the displacement for both block and rope is same
so, the acceleration must be same for both
1 1
log T log(2 ) log( ) log(g)
2 2
M P
Differentiating
T 1 T 1
TT 0 0 100 100
P T 2 T 2
T = Ma....(i) 1
21 10.5 10%
2
P
p = (m + M)a a Note: In this method, the % error obtained is an
m M approximate value on the higher side. Exact value is
PM less than the obtained one.
T M .a
m M 39. (b) y 10 4
sin( 600 t 2 x
)
32. (a) 3
Comparing it with standard equation
1
33. (d) Elastic energy = F x y = A sin(vt-kx); v = 600m/s
2
F = 200 N, x = 1 mm = 10–3 m dI 2 ( 2)
40. (d) e L. 8 ( L) L = 0.1 H
dt 0.05
1
E 200 1 10 3 0.1 J Q
2 41. (c) q
34. (c) Escape velocity of a body is independent of the angle 2
of projection. Hence, changing the angle of projection 42. (a) 43. (d)
is not going to effect the magnitude of escape velocity.
1 N. 1 5000
44. (a) K n K n
M f N. 5 1250
35. (c) T 2 .....(i)
K
1 2
n ( 4) n 2 0.4 n 2
5T M m 5 5
2 ....(ii)
3 K 45. (b) No. of particles emitted = 8
No. of – particles emitted = 4
5 M m No. of + particles emitted = 2
Dividing equation (ii) by equation (i),
3 M z = 92 – 2 × 8 + 4 – 2 = 78
squaring both the sides 46. (a)
47. (a)
25 M m m m 25 16
1 1 3 3
9 M M M 9 9
36. (a) External amount of work must be done in order to flow 3
48. (b) 1.5 A 3
heat from lower temperature to higher temperature. This 2
is according to second law of thermodynamics 3V
37. (c) 2
Vmax A m I k 6
2
2 k k 3
m m 3V 3V
S-44 The Pattern Target AIEEE
49. (b) 57. (b) Required work done
50. (c) x = 4(cos t + sin t) 1 1
K (x 22 x12 = 5 10 3 [10 2 5 2 ] 10 4
2 2
= 4[sin( t )] sin t ]
2
1
5 75 103 10 4
= 18.75
2
t t t t
= 4 2 sin 2 cos 2 1 T
2 2 58. (a) n ; 1m
2
T = 10 Kg wt. = 10 × 10 = 100 N
= 9.8 g/m = 9.8 × 10–3 kg/m
= 8 sin . cos t n = 50 hz
4 4
59. (c) 60. (a)
61. (d) 62. (b)
8
. cos t 4 2 cos t 63. (c) 64. (c)
2 4 4
65. (b)
Comparing it with standard equation 66. (b) Power = F. V
2
1 q2 (8 10 18 ) 2 ds (c.t 1 / 2 ) dt S C. t 3 / 2
52. (d) Work done = = 3
2 c 2 100 10 8
= 32 × 10–32 J c.t 3 / 2
S s t3/ 2
dx dy 3/ 2
53. (b) Vx 3 t 2 , Vy 3 t2 67. (a) Thrust = Mass × Acceleration
dt dt
= 3.5 × 104 × 10 = 3.5 × 105 N
v V V 3t 2 2 2 68. (d)
x2 y2
69. (b) The force body diagram
3
54. (d) 3 P1 T1 –q3
P T
P2 T2
a F2
cp (–q1)
3 +q2
Comparing it with standard eq.
cv 2 F1
a
55. (c)
(627 273) (273 27) 1 q1q 3 1 q1q 3
56. (b) F1 . 2 ; F2 .
627 273 4 o a 4 o b2
900 300 600 2 q1 q 3 q1
= Fx F1 sin F2 sin
900 900 3 4 o a2 b2
work = ( ) × Heat
q3 q2
2 Fx sin
= 3 10 6 4.2 J = 8.4 × 106 J a 2
b2
3
AIEEE-2003 Solved Paper S-45
83. (b) Th X Th Th O O
|| ||
84. (d) t½ = 3hrs. Initial mass (C0) = 256 g 104. (b) H C O H C O
C0 256 256 105. (a) As adsorption is an exothermic process.
Cn 4g.
2 n
( 2) 6 64 Rise in temperature will decrease adsorption.
S-46 The Pattern Target AIEEE
106. (d) The equilibrium constant is related to the standard emf 0.059
of cell by the expression 1.10 log [0.1] 1.10 0.0295 1.07 V
2
n 2
log K E º cell 0.295 119. (b) In equation K = Ae E a / RT ; A = Frequency factor
0.059 0.059
K = velocity constant, R = gas constant and Ea
590 = energy of activation
log K 10 or K = 1 × 1010
59 120. (b) f-block elements show a regular decrease in atomic size
due to lanthanide/actinide contraction.
107. (a) For spontaneous reaction, dS > 0 and G and dG
should be negative i.e. < 0. O CH 3
108. (d) [A] = 1.0 × 10–5, [B] = [1.0 × 10–5] ,
1 2| | 3| 4
121. (c) C H3 C C H C H 3 ; 3- methyl-2-butanone
Ksp = [2.B]2 [A] = [2 × 10–5]2 [1.0 × 10–5]
= 4 × 10–15 122. (a) LiAlH4 can reduce COOH group and not the double
bond.
109. (a) No. of moles of boron = 21.6 2 for BCl3 LAH
10.8 CH 2 CH COOH
1mole of Boron = 3 mole of Cl CH 2
CH CH 2 OH
2 mole of Boron = 6 mole of Cl 123. (b) According to kinetic theory the gas molecules travel in
H2 Cl 2 2HCl a stright line path but show haphazard motion due to
collisions.
3 moles of Hydrogen is required
124. (a) CnH2nO2 is general formula for carboxylic acid
= 3 × 22.4 = 67.2 Litre
125. (a) A chiral object or structure has four different groups
[ NO 2 ]2 [1.2 10 2 ]2 attached to the carbocation.
110. (b) KC = 3 10 3 mol/L
[ N2O4 ] [4.8 10 2 ] 126. (a) Cr2 O 27 + OH– 2Cr O 24 + H+
111. (b) Due to exothermicity of reaction low or optimum The above equilibrium shifts to L.H.S. on addition of
temperature will be required. Since 3 moles are changing acid.
to 2 moles. 127. (d) It is because mercury exists as liquid at room
High pressure will be required. temperature.
112. (d) Na2 O (basic), SO2 and B2O3 (acidic) and ZnO is 128. (a) Gypsum is CaSO4.2H2O
amphoteric H
129. (b) (C H O ) n nH O nC H O
113. (a) HgI 2 KI K 2 HgI 4 D Glu cos e
(inso lub le) (so lub le )
1
On heating HgI2 decomposes as 130. (c) AgNO 3 Ag NO 2 O2
2
HgI2 Hg + I2
114. (d) A B C 131. (d) step 1
CH 3 CH 2 OH H CH 3CH 2 O H
+0.5C –3.0V –1.2V |
The higher the negative value of reduction potential, H
the more is the reducing power. Hence B > C > A. Protonated alcohol
. . 132. (a) The solubility is governed by H solution i.e.
115. (b) Among the given compounds, the N H 3 is most basic.
Hence has highest proton affinity H solution H lattice H Hydration
116. (d) Tf = Kf × m × i ; Due to increase in size the magnitude of hydration
Tf = 1.85 × 0.2 × 1.3 = – 0.480º C energy decreases and hence the solubility.
133. (d) The rain water after thunderstorm contains dissolved
( HX H X , i = 1.3) acid and therefore the pH is less than rain water without
1 0.3 0. 3 0.3
thunderstorm.
9650 1 134. (b) Hydrogen bonding
117. (a) No. of moles of silver = moles 135. (d) 25 × N = 0.1 × 35 ; N = 0.14 M = 0.07 M
96500 10
Ba(OH)2 is diacid base
1 136. (c) Gº = – RT lnKc or – Gº = RT lnKc
Mass of silver deposited = 108 = 10.8 g
10 137 (c) Rate1 = k [A]n [B]m; Rate2 = k [2A]n [½B]m
2 Rate k[ A] n [½ B] m
º 0.059 [Cu ]
118. (b) E cell E cell log Rate k[ A]n [ B] m
n 2
[ Zn ]
= [2]n [½]m = 2n.2–m = 2n–m
AIEEE-2003 Solved Paper S-47
4 4 4 4
CH 3CH 2 N C H 2O
H
CH 3CH 2 NH 2 3. (a) 1 2 3 n
138. (d) lim ........
n n n n n
HCOOH
Therefore it gives only one mono chloroalkane. 1 1 23 n3
lim ............
139. (b) Enthalpy change for a reaction does not depend upon n n n4 n 4
n4
the nature of intermediate reaction steps. 1 1
x5 1
140. (a) On increasing pressure, the temperature is also (x ) 4 dx 0
5 5
increased. Thus in pressure cooker due to increase in 0 0
pressure the b.p. of water increases. 4. (b) Equation of the normal at to a parabola y2 = 4bx at point
141. (b) r = k [O2][NO]2. when the volume is reduced to 1/2, The bt12 , 2bt1 is
conc. will double
New rate = k [2O2][2 NO]2 = 8 k [O2][NO]2 y = – t1x 2bt1 bt13
The new rate increases by eight times. 2
As given, it also passes through bt 2 , 2bt 2 than
142. (b) Magnesium provides cathodic protection and prevent
rusting or corrosion. 2bt 2 – t1 bt 22 2 bt1 bt13
143. (b) Both NO2 and O3 have angular shape and hence will
have net dipole moment. 2t 2 – 2t1 – t1 t 22 – t12 = –t1(t2 + t1) (t2 – t1)
144. (a) N3–, F– and Na+ contain 10 electrons each. 2 = – t1(t2 +t1)
145. (b) Permanent hardness of water is due to chlorides and 2
sulphates of calcium and magnesium. t2 + t1 = –
t1
146. (b) In H2S, due to low electronegativity of sulphur the L.P.
- L. P. repulsion is more than B. P. - B. P. repulsion and 2
t2 = – t 1 –
hence the bond angle is 92º. t1
147. (a) Both XeF2 and CO2 have a linear structure. 5. (b) r1 r2 C1C 2 for intersection
148. (a) Electronic configuration of Cr is
r 3 5 r 8 ..........(1)
3d 4s and r1 r2 C1C 2 , r 3 5 r 2...............(2)
From (1) and (2), 2 < r < 8.
6. (d) y2 4a(x h),
So due to half filled orbital I.P. is high of Cr.
149. (a) The lines falling in the visible region comprise Balmer 2yy1 4a yy1 2a
Differentiating,
series. Hence the third line would be n =2, n = 5 i.e.
5 2. y12 yy2 0
Degree = 1, order = 2.
h 6.6 10 34 x2 y2
150. (d) 3
10 33 m 7. (d) 1
mv 60 10 10 144 81 25
144 81 81 15 5
PAPER II - MATHS a ,b , e 1
25 25 144 12 4
4
d esin x 3 sin x 3 Foci = ( 3 , 0), focus of ellipse
1. (a) Let F(x) or e dx
dx x x 3
1 = (3, 0) e
4
4 2
3x 3
9
3
esin x dx b 2 16 1 7
1 x 16
Let x 3 t, 3x 2dx dt t
8. (c) F(t ) f (t y)g( y)dy
when x 1, t 1& x 4, t 64
0
64
esin t t t
F(t) dt f (t) 164 F(64) F(1)
t e t y ydy e t e y ydy
1
0 0
K 64
t y y t t
th e ye e 0 e t ye y e y 0
9 1
2. (a) n = 9 then median term 5 th term. Last t 1 et
2 et t e t e t 0 1 et
four observations are increased by 2. The median is et
5th observation which is remaining unchanged.
et (1 t )
there will be no change in median.
S-48 The Pattern Target AIEEE
CA r to OB , slope of CA cot 1 1
7
1 1 1
4 p(X 1) 8
C1 8.
2 2 8 5 32
Equation of CA, 2 2
AIEEE-2003 Solved Paper S-49
d P.V of B P.V of A 4i 2 j 2k b b
(a b) f (a b x)dx xf (a b x)dx
W F .d 28 4 8 40 unit
a a
22. (d) A b b
(a b) f (a b x)dx xf (x )dx
a a
3i + 4 k 5i – 2j + 4k b
2I (a b) f (x )dx
a
B D C b b
(a b ) (a b)
(3 5)i (0 2) j (4 4)k I f ( x )dx ; I f (a b x )dx
P.V of AD 2 2
a a
2
28. (a) Portion OA , AB correspond to motions with
4i j 4k or AD 16 16 1 33 acceleration ‘f’ and retardation `r` respectively.
Area of OAB S and OB =t. Let OL = t1, LB= t2and
23. (d) y=–x+1 (0,3) AL= v,
(–1,2) y=x–1 1 1 2S
S OB.AL t.v ; v
(2,1) 2 2 t
(3 x ) ( x 1) dx (3 x) ( x 1) dx v v 2s
Also, f , t1 and
0 1 t1 f tf
0 1 2
(2 2 x )dx 2dx (4 2 x )dx v v 2s 2s 2s
r ,t2 ;t t1 t 2
1 0 1 t2 r tr tf tr
2 0 1 2
2x x 2x 4x x2 1 1 2s 1 1
1 0 1 t t 2s
0 ( 2 1) (2 0) (8 4) (4 1) f r t f r
1 2 4 3 4 sq. units
S-50 The Pattern Target AIEEE
2h
29. (a) R u (u cos ) t dx 1
g (1 y 2 ) x e tan y
dy
1
1 2h dx x e tan y
t ........(1)
cos g dy (1 y 2 ) (1 y 2 )
1
1 2 dy
tan 1 y
Now, h gt
( u sin ) t (1 y 2 )
2 I.F e e
1
Substituting ‘t’from (1), 1 e tan y tan 1 y dy
x (e tan y ) e
u sin 2h 1 2h 1 y2
h g 1
cos g 2 g cos 2 tan 1 y e 2 tan y
x (e ) C
2
1 1
2h 2 2xe tan y e 2 tan y k
h u tan h sec
g
2h f ( x)
h u tan h tan 2 h 36. (d) Given f ( x ) f (x) 1
g f (x)
a a 1 1
31. (d) tan ; sin x 2 e x dx e 2 x dx
n 2r n 2R
0 0
a a
r R cot cos ec r R . cot 1 1 1 2x 1
2 n n 2 2n x 2ex 2 xe x ex e
0 0 2 0
32. (b) Taking co-ordinates as
x y e2 1 e2 3
, ; ( x, y ) & ( xr , yr ) . e 2e e 1 e
r r 2 2 2 2
Above coordinates satisfy the relation y mx
37. (d) r 2 154 r 7
Therefore lies on the straight line. For centre on solving equation
33. (a) | z | | z || | | z || | 1 2x 3y 5& 3x 4y 7 or x 1, y 1
Arg( z ) arg(z ) arg( ) arg(z ) arg centre = (1, –1 )
Equation of circle, ( x 1) 2 ( y 1) 2 72
z 1
2 x2 y2 2x 2y 47
34. (d) 2
z az b 0 ; z1 z 2 a & z1z 2 b 3x 1 1 x 1 2x
38. (d) P ( A) , P(B) , P (C )
0, z1 , z 2 form an equilateral 3 4 2
These are mutually exclusive
0 2 z12 z 2 2 0.z1 z1.z 2 z 2 .0
3x 1 1 x 1 2x
(for an equilateral triangle, 0 1, 0 1 and 0 1
3 4 2
2 2 2
z1 z2 z3 z1z 2 z2z3 z 3 z1 )
1 3x 2, 3 x 1 and 1 2 x 1
z12 z22 z1z 2 (z1 z 2 ) 2 3z1z 2 1 2 1 1
x 3 x 1, and x
a 2
3b 3 3 2 2
1 dy 1 3x 1 x 1 2x
35. (c) (1 y 2 ) ( x e tan y
) 0 Also 0 1
dx 3 4 2
AIEEE-2003 Solved Paper S-51
1 13
0 13 3x 12 1 3x 13 x 45. (a)
3 3 3
h
4
1 1 1 2 1 13
max , 3, , x min ,1, , h
3 2 3 3 2 3
4
1 1 1 1
x x , 40 m
3 2 3 2
5 2 3
39. (a) n( S ) C2 ; n( E ) C1 , tan 1
5
2 2
n( E ) C1 C1 2 or
p( E )
n( S ) 5 5
C2
h h
1 3a 2 2 3
40. (b) 3 &2 tan tan 40 160
a2 5a 3 a2 5a 3 tan or 5 h h
1 tan . tan 1 .
40 160
1 (1 3a ) 2 2
2
9 (a 2 5a 3) 2 a 2
5a 3 h 2 200h 6400 0, h 40 or 160 metre
possible height 40 metre
(1 3a ) 2
9 or 9a 2 6a 1
(a 2 5a 3) A
46. (d)
2 2
9a 45a 27 or 39a 26 or a
3
30º
256 r r 1 8 16
Terms will be integral if & both are +ve Area of ABD 4
2 8 2 3 3 3 3
integer. As 0 r 256 16 32
Area of ABC 2
r 0,8,16,24,........256 , total 33 values. 3 3 3 3
256 r 2 C A 3b
For above values of r, is also an integer.. 47. (b) If a cos c cos 2
2 2 2 2
n n
43. (c) Cr 1 Cr 1 2 nCr a[cos C 1] c[cos A 1] 3b
n n n n (a c) (a cos C c cos B) 3b
Cr 1 Cr Cr Cr 1
n 1 n 1 n 2 a c b 3b or a c 2 b or a , b, c are in A.P.
Cr Cr 1 Cr 1
44. (a) After t; velocity = f t
48. (c) a b c 0 (a b c ).(a b c) 0
2 2 2 2
VBA f t ( u) f t u 2f ut cos 2 2
a b c 2(a.b b.c c.a ) 0
For max. and min.,
d u cos 1 4 9
2
(VBA ) 2f 2 t 2fu cos 0 or t a.b b.c c.a 7
dt f 2
S-52 The Pattern Target AIEEE
1 1 f (a )g ( x ) g (a )f ( x )
n n 53. (b) lim 4
49. (d) I x(1 x ) dx (1 x )(1 1 x ) dx x a g ( x ) f (x )
0 0 (By L’ Hospital rule)
1 k g (x ) k f ( x ) k 4.
1 lim 4
xn 1 xn 2 g (x ) f (x )
(1 x ) x n dx x a
n 1 n 2
0 0
x
tan .(1 sin x)
1 1 4 2
54. (d) lim
n 1 n 2 x ( 2 x) 3
2
2
d x Let x y; y 0
sec 2 tdt
dx 0 sec 2 x 2 .2 x 2
50. (d) lim lim
x 0 d x 0 sin x x cos x
( x sin x)
dx y
tan .(1 cos y)
2
(by L’ Hospital rule) lim
y 0 ( 2 y) 3
2 2
2 sec x 2 1 y y
lim 1 tan 2 sin 2
x 0 sin x 1 1 2 2
cos x lim
x y 0 y3
( 8). .8
8
y
51. (d) tan 2
1 2 . sin y / 2 1
lim
O y 0 32 y y/2 32
2
A C
55. (b) ( x a1 ) 2 ( y b1 ) 2 (x a 2 ) 2 ( y b 2 ) 2
(a 1 a 2 )x (b1 b 2 ) y
Centre of sphere = (-1, 1, 2) 1
(a 2 2 b 2 2 a12 b12 ) 0
Radius of sphere 1 1 4 19 5 2
1
Perpendicular distance from centre to the plane c (a 2 2 b 2 2 a 12 b12 )
2
1 2 4 7 12
OC d 4.
1 4 4 3 a a2 1 a3
56. (c) b b2 1 b3 0
AC 2 AO 2 OC 2 5 2 4 2 9 AC 3 2 3
52. (b) Vector perpendicular to the face OAB c c 1 c
î ˆj k̂ a a2 1 a a2 a3
OA OB 1 2 1 5î ĵ 3k̂ b b2 1 b b2 b3 0
2 1 3 2 2 3
c c 1 c c c
Vector perpendicular to the face ABC
î ˆj k̂ 1 a a2
AB AC 1 1 2 î 5ˆj 3k̂ (1 abc) 1 b b2 0
2 1 1 1 c c2
f (1) 7; m 7, f ( x ) 7x
57. (c) x2 3 x 2 0 | x |2 3 | x | 2 0
n n 7 n (n 1)
(x 2) ( x 1) 0 f (r ) 7 r
r 1 1 2
x 1, 2 or x 1, 2
1 dy 1
No.of solution 4 62. (c) y x or 1
x dx x2
Section - 1 5. A projectile can have the same range ‘R’ for two angles of
projection. If ‘T1’ and ‘T2’ to be time of flights in the two
cases, then the product of the two time of flights is directly
proportional to.
1. Which one of the following represents the correct 1 1
dimensions of the coefficient of viscosity? (a) R (b) (c) (d) R 2
R R2
(a) ML–1T –1 (b) MLT –1 6. Which of the following statements is FALSE for a particle
moving in a circle with a constant angular speed ?
(c) ML–1T –2 (d) ML–2 T –2
(a) The acceleration vector points to the centre of the circle
2. A particle moves in a straight line with retardation
(b) The acceleration vector is tangent to the circle
proportional to its displacement. Its loss of kinetic energy
(c) The velocity vector is tangent to the circle
for any displacement x is proportional to
(d) The velocity and acceleration vectors are perpendicular
(a) x (b) e x (c) x2 (d) loge x to each other.
3. A ball is released from the top of a tower of height h meters. 7. An automobile travelling with a speed of 60 km/h, can brake
It takes T seconds to reach the ground. What is the position to stop within a distance of 20m. If the car is going twice as
fast i.e., 120 km/h, the stopping distance will be
T
of the ball at second (a) 60 m (b) 40 m (c) 20 m (d) 80 m
3
8. A machine gun fires a bullet of mass 40 g with a velocity
8h 1200 ms–1. The man holding it can exert a maximum force of
(a) meters from the ground 144 N on the gun. How many bullets can he fire per second
9
at the most?
7h (a) Two (b) Four (c) One (d) Three
(b) meters from the ground
9 9. Two masses m1 = 5g and
m 2 = 4.8kg tied to a string
h
(c) meters from the ground are hanging over a light
9
frictionless pulley. What is the
17h acceleration of the masses
(d) meters from the ground
18 when left free to move ? (g =
4. If A × B = B× A, then the angle between A and B is 9.8 m/s2)
(a) 5 m/s 2 (b) 9.8 m/s 2
(a) (b) (c) (d)
2 3 4 (c) 0.2 m/s 2 (d) 4.8 m/s 2
S-56 The Pattern Target AIEEE
10. A uniform chain of length 2 m is kept on a table such that a 18. A satellite of mass m revolves around the earth of radius R
length of 60 cm hangs freely from the edge of the table. The at a height x from its surface. If g is the acceleration due to
total mass of the chain is 4 kg. What is the work done in gravity on the surface of the earth, the orbital speed of the
pulling the entire chain on the table ? satellite is
(a) 12 J (b) 3.6 J (c) 7.2 J (d) 1200 J gR 2 gR
11. A block rests on a rough inclined plane making an angle of (a) (b)
R +x R-x
30° with the horizontal. The coefficient of static friction
between the block and the plane is 0.8. If the frictional force 1/2
on the block is 10 N, the mass of the block (in kg) is (take gR 2
(c) gx (d)
2) R+x
g = 10 m/s
(a) 1.6 (b) 4.0 (c) 2.0 (d) 2.5 19. The time period of an earth satellite in circular orbit is
independent of
12. A force F (5i 3 j 2k)N is applied over a particle which (a) both the mass and radius of the orbit
displaces it from its origin to the point r = (2i - j)m. The (b) radius of its orbit
(c) the mass of the satellite
work done on the particle in joules is (d) neither the mass of the satellite nor the radius of its
(a) +10 (b) +7 (c) –7 (d) +13 orbit.
13. A body of mass ‘m’, accelerates uniformly from rest to ‘v1’ 20. If ‘g’ is the acceleration due to gravity on the earth’s surface,
in time ‘t1’. The instantaneous power delivered to the body the gain in the potential energy of an object of mass ‘m’
as a function of time ‘t’ is raised from the surface of the earth to a height equal to the
mv12 t radius ‘R” of the earth is
mv1t 2 mv1t mv12 t
(a) (b) (c) t (d) 1 1
t1 t12 1 t1 (a) mgR (b) mgR (c) 2 mgR (d) mgR
4 2
14. A Particle is acted upon by a force of constant magnitude
which is always perpendicular to the velocity of the particle, 21. Suppose the gravitational force varies inversely as the nth
the motion of the particles takes place in a plane. It follows power of distance. Then the time period of a planet in circular
that orbit of radius ‘R’ around the sun will be proportional to
(a) its kinetic energy is constant n-1
(b) its acceleration is constant (a) R n (b) R 2
(c) its velocity is constant
n+1 n-2
(d) it moves in a straight line
15. A solid sphere is rotating in free space. If the radius of the (c) R 2 (d) R 2
sphere is increased keeping mass same which one of the 22. A wire fixed at the upper end stretches by length by
following will not be affected ? applying a force F. The work done in stretching is
(a) Angular velocity
F F
(b) Angular momentum (a) 2F (b) F (c) (d)
(c) Moment of inertia 2 2
(d) Rotational kinetic energy 23. Spherical balls of radius ‘R’ are falling in a viscous fluid of
16. A ball is thrown from a point with a speed ' v 0 ' at an viscosity ‘ ’ with a velocity ‘v’. The retarding viscous force
acting on the spherical ball is
elevation angle of . From the same point and at the same
(a) inversely proportional to both radius ‘R’ and velocity
' v0 ' ‘v’
instant, a person starts running with a constant speed (b) directly proportional to both radius ‘R’ and velocity
2
to catch the ball. Will the person be able to catch the ball? If ‘v’
yes, what should be the angle of projection ? (c) directly proportional to ‘R’ but inversely proportional
(a) No (b) Yes, 30° to ‘v’
(c) Yes, 60° (d) Yes, 45° (d) inversely proportional to ‘R’ but directly proportional
17. One solid sphere A and another hollow sphere B are of same to velocity ‘v’
mass and same outer radii. Their moment of inertia about 24. If two soap bubbles of different radii are connected by a
their diameters are respectively I A and IB Such that tube.
(a) air flows from the smaller bubble to the bigger
(where d A and d B are their densities). (b) air flows from bigger bubble to the smaller bubble till
the sizes are interchanged
(a) IA < IB (b) IA > IB
(c) air flows from the bigger bubble to the smaller bubble
IA d A till the sizes become equal
(c) IA = IB (d) = (d) there is no flow of air.
IB d B
AIEEE-2004 Solved Paper S-57
25. The bob of a simple pendulum executes simple harmonic 32. If the temperature of the sun were to increase from T to 2T
motion in water with a period t, while the period of oscillation and its radius from R to 2R, then the ratio of the radiant
of the bob is t 0 in air. Nglecting frictional force of water and energy received on earth to what it was previously will be
(a) 32 (b) 16 (c) 4 (d) 64
given that the density of the bob is (4/3)×1000kg/m3 . 33. Which of the following statements is correct for any
What relationship between t and t 0 is true thermodynamic system ?
(a) The change in entropy can never be zero
(a) t = 2t 0 (b) t = t 0 /2
(b) Internal energy and entropy and state functions
(c) t = t 0 (d) t = 4t 0 (c) The internal energy changes in all processes
26. A particle at the end of a spring executes S.H.M with a (d) The work done in an adiabatic process is always zero.
period t1. while the corresponding period for another spring 34. Two thermally insulated vessels 1 and 2 are filled with air at
is t2. If the period of oscillation with the two springs in
temperatures (T1 , T2 ), volume (V1, V2 ) and pressure
series is T then
(a) T –1 = t1–1 + t 2–1 (b) T 2 = t12 + t 22 (P1 , P2 ) respectively. If the valve joining the two vessels is
opened, the temperature inside the vessel at equilibrium will be
(c) T = t1 + t 2 T –2 = t1–2 + t 2–2
(d)
(a) T1T2 (P1V1 + P2 V2 )/(P1V1T1 + P2 V2 T2 )
27. The total energy of a particle, executing simple harmonic
motion is (b) (T1 + T2 )/2
(a) independent of x (b) x 2
(c) T1 + T2
(c) x (d) x1/2 (d) T1T2 (P1V1 + P2 V2 )/(P1V1T1 + P2 V2 T2 )
where x is the displacement from the mean position
28. The displacement y of a particle in a medium can be 35. A radiation of energy E falls normally on a perfectly reflecting
expressed as, surface. The momentum transferred to the surface is
(a) Ec (b) 2E/c (c) E/ c (d) E / c2
y = 10 –6 sin 100t + 20x + m where t is in second and x
4 36. The temperature of the two x 4x
m m 2 1 1
(c) 2 2 (d) 2 2 (a) (b) (c) 1 (d)
0 ( 0 ) 3 2 3
30. In forced oscillation of a particle the amplitude is maximum 37. A light ray is incident
for a frequency 1 of the force while the energy is maximum perpendicularly to one
face of a 90° prism and
for a frequency 2 of the force; then is totally internally 45°
45°
(a) 1 2 when damping is small and 1 2 when reflected at the glass-
damping is large air interface. If the angle
(b) (c) of reflection is 45°, we 45°
1 2 1 2
conclude that the
(d) 1 2 refractive index n
31. One mole of ideal monatomic gas ( 5 / 3) is mixed with 1
(a) n (b) n 2
one mole of diatomic gas ( 7 / 5) . What is for the 2
mixture? Denotes the ratio of specific heat at constant
pressure, to that at constant volume 1
(c) n (d) n 2
(a) 35/23 (b) 23/15 (c) 3/2 (d) 4/3 2
S-58 The Pattern Target AIEEE
38. A plano convex lens of refractive index 1.5 and radius of 46. The total current supplied to the circuit by the battery is
curvature 30 cm. Is silvered at the curved surface. Now this
lens has been used to form the image of an object. At what 2
6V
distance from this lens an object be placed in order to have a 6 3
real image of size of the object
(a) 60 cm (b) 30 cm (c) 20 cm (d) 80 cm 1.5
39. The angle of incidence at which reflected light is totally
polarized for reflection from air to glass (refractive index n), (a) 4 A (b) 2 A
is (c) 1 A (d) 6 A
(a) tan 1 (1/ n) (b) sin 1 (1/ n) 47. The resistance of the series combination of two resistances
is S. when they are joined in parallel the total resistance is P.
(c) sin 1 (n) (d) tan 1 (n) If S n P then the Minimum possible value of n is
40. The maximum number of possible interference maxima for (a) 2 (b) 3
slit-separation equal to twice the wavelength in Young’s (c) 4 (d) 1
double-slit experiment is 48. An electric current is passed through a circuit containing
(a) three (b) five (c) infinite (d) zero two wires of the same material, connected in parallel. If the
41. An electromagnetic wave of frequency = 3.0 MHz 4 2
lengths and radii arein the ratio of and , then the ratio
passes from vacuum into a dielectric medium with 3 3
permittivity 4.0 . Then of the current passing through the wires will be
(a) wavelength is halved and frequency remains (a) 8/9 (b) 1/3
unchanged (c) 3 (d) 2
(b) wavelength is doubled and frequency becomes half 49. In a meter bridge experiment null point is obtained at 20 cm.
(c) wavelength is doubled and the frequency remains from one end of the wire when resistance X is balanced
unchanged against another resistance Y. If X < Y, then where will be the
(d) wavelength and frequency both remain unchanged. new position of the null point from the same end, if one
42. Two spherical conductors B and C having equal radii and decides to balance a resistance of 4 X against Y
carrying equal charges on them repel each other with a force (a) 40 cm (b) 80 cm
F when kept apart at some distance. A third spherical (c) 50 cm (d) 70 cm
conductor having same radius as that B but uncharged is 50. The termistors are usually made of
brought in contact with B, then brought in contact with C (a) metal oxides with high temperature coefficient of
and finally removed away from both. The new force of resistivity
repulsion between B and C is (b) metals with high temperature coefficient of resistivity
(c) metals with low temperature coefficient of resistivity
(a) F/8 (b) 3 F/4 (c) F/4 (d) 3 F/8
(d) semiconducting materials having low temperature
43. A charge particle ‘q’ is shot towards another charged particle
coefficient of resistivity
‘Q’ which is fixed, with a speed ‘v’. It approaches ‘Q’ upto
51. Time taken by a 836 W heater to heat one litre of water from
a closest distance r and then returns. If q were given a
10°C to 40°C is
speed of ‘2v’ the closest distances of approach would be
(a) 150 s (b) 100 s
(a) r/2 (b) 2 r (c) r (d) r/4
(c) 50 s (d) 200 s
44. Four charges equal to -Q are placed at the four corners of a
52. The thermo emf of a thermocouple varies with the temperature
square and a charge q is at its centre. If the system is in
equilibrium the value of q is of the hot junction as E a b 2 in volts where the
ratio a/b is 700°C. If the cold junction is kept at 0°C, then the
Q Q
(a) – (1 + 2 2) (b) (1 + 2 2) neutral temperature is
2 4 (a) 1400°C
Q Q (b) 350°C
(c) – (1 + 2 2) (d) (1 + 2 2) (c) 700°C
4 2
(d) No neutral temperature is possible for this termocouple.
45. Alternating current can not be measured by D.C. ammeter
because 53. The electrochemical equivalent of a metal is 3.35×10 –7 kg
(a) Average value of current for complete cycle is zero per Coulomb. The mass of the metal liberated at the cathode
(b) A.C. Changes direction when a 3A current is passed for 2 seconds will be
(c) A.C. can not pass through D.C. Ammeter (a) 6.6×1057kg (b) 9.9×10–7 kg
(d) D.C. Ammeter will get damaged. –7
(c) 19.8×10 kg (d) 1.1×10–7 kg
AIEEE-2004 Solved Paper S-59
54. A current i ampere flows along an infinitely long straight 61. A coil having n turns and resistance R is connected with
thin walled tube, then the magnetic induction at any point a galvanometer of resistance 4R . This combination is
inside the tube is moved in time t seconds from a magnetic field W1 we be r
to W2 weber. The induced current in the circuit is
0 2i
(a) . Tesla (b) zero (W2 W1) n(W2 W1 )
4 r (a) (b)
Rnt 5 Rt
2i
(c) infinite (d) Tesla
r (W2 W1) n(W2 W1 )
(c) (d)
55. A long wire carries a steady current. It is bent into a circle of 5 Rnt Rt
one turn and the magnetic field at the centre of the coil is B. 62. In a uniform magnetic field of induction B a wire in the form
It is then bent into a circular loop of n turns. The magnetic of a semicircle of radius r rotates about the diameter of the
field at the centre of the coil will be circle with an angular frequency . The axis of rotation is
(a) 2n B (b) n2 B perpendicular to the field. If the total resistance of the circuit
(c) n B (d) 2 n2 B is R, the mean power generated per period of rotation is
56. The magnetic field due to a current carrying circular loop of
radius 3 cm at a point on the axis at a distance of 4 cm from the (B r ) (B r 2 )
(a) (b)
centre is 54 T. What will be its value at the centre of loop? 2R 8R
(a) 125 T (b) 150 T
B r2 2 2
(c) 250 T (d) 75 T (c) (d) (B r )
57. Two long conductors, separated by a distance d carry current 2R 8R
I1 and I2 in the same direction. They exert a force F on each 63. In a LCR circuit capacitance is changed from C to 2 C. For
other. Now the current in one of them is increased to two the resonant frequency to remain unchanged, the
times and its direction is reversed. The distance is also inductance should be changed from L to
increased to 3d. The new value of the force between them is (a) L/2 (b) 2 L (c) 4 L (d) L/4
64. A metal conductor of length 1 m rotates vertically about one
2F F of its ends at angular velocity 5 radians per second. If the
(a) (b)
3 3 horizontal component of earth’s magnetic field is 0.2×10–4T,
then the e.m.f. developed between the two ends of the
F conductor is
(c) –2 F (d)
3 (a) 5 mV (b) 50 V (c) 5 V (d) 50mV
58. The length of a magnet is large compared to its width and 65. According to Einstein’s photoelectric equation, the plot of
breadth. The time period of its oscillation in a vibration the kinetic energy of the emitted photo electrons from a
magnetometer is 2s. The magnet is cut along its length into metal Vs the frequency, of the incident radiation gives as
three equal parts and these parts are then placed on each straight the whose slope
other with their like poles together. The time period of this (a) depends both on the intensity of the radiation and the
combination will be metal used
(b) depends on the intensity of the radiation
2 (c) depends on the nature of the metal used
(a) 2 3s (b) s
3 (d) is the same for the all metals and independent of the
intensity of the radiation
2
(c) 2 s (d) s 66. The work function of a substance is 4.0 eV. The longest
3 wavelength of light that can cause photoelectron emission
59. The materials suitable for making electromagnets should from this substance is approximately.
have (a) 310 nm (b) 400 nm (c) 540 nm (d) 220 nm
(a) high retentivity and low coercivity 67. A charged oil drop is suspended in a uniform field of 3×104
(b) low retentivity and low coercivity v/m so that it neither falls nor rises. The charge on the drop
(c) high retentivity and high coercivity will be (Take the mass of the charge = 9.9×10–15 kg and g =
(d) low retentivity and high coercivity 10 m/s2)
60. In an LCR series a.c. circuit, the voltage across each of the (a) 1.6×10–18 C (b) 3.2×10–18 C
components, L, C and R is 50V. The voltage across the LC (c) 3.3×10–18 C (d) 4.8×10–18 C
combination will be 68. A nucleus disintegrated into two nuclear parts which have
their velocities in the ratio of 2 : 1. The ratio of their nuclear
(a) 100 V (b) 50 2 V sizes will be
(c) 50 V (d) 0 V (zero) (a) 3½ : 1 (b) 1:21/3 (c) 21/3:1 (d) 1:3½
S-60 The Pattern Target AIEEE
2
77. Consider the ground state of Cr atom (X = 24). The number
69. The binding energy per nucleon of deuteron 1 H and of electrons with the azimuthal quantum numbers, = 1 and
2 are, respectively
4
helium nucleus 2 He is 1.1 MeV and 7 MeV respectively.. (a) 16 and 4 (b) 12 and 5
(c) 12 and 4 (d) 16 and 5
If two deuteron nuclei react to form a single helium nucleus, 78. Which one of the following ions has the highest value of
then the energy released is ionic radius ?
(a) 2.36 MeV (b) 26.9 MeV (a) O2– (b) B3+
(c) 13.9 MeV (d) 19.2 MeV (c) Li + (d) F–
70. An -particle of energy 5 MeV is scattered through 180º by a 79. The wavelength of the radiation emitted, when in a hydrogen
fixed uranium nucleus. The distance of closest approach is of atom electron falls from infinity to stationary state 1, would
the order of be (Rydberg constant = 1.097×107 m–1)
(a) 10 12 (b) 10 10 (a) 406 nm (b) 192 nm
cm cm
(c) 91 nm (d) 9.1×10–8 nm
(c) 1A (d) 10 15 cm
80. The correct order of bond angles (smallest first) in H2S,
71. When npn transistor is used as an amplifier NH3, BF3 and SiH4 is
(a) electrons move from collector to base (a) H2S < NH3 < SiH4 < BF3
(b) holes move from emitter to base (b) NH3 < H2S < SiH4 < BF3
(c) electrons move from base to collector (c) H2S < SiH4 < NH3 < BF3
(d) holes move from base to emitter (d) H2S < NH3 < BF3 < SiH4
72. For a transistor amplifier in common emitter configuration 81. Which one of the following sets of ions represents the
for load impedance of 1k (h fe = 50 and h oe = 25) the collection of isoelectronic species?
current gain is (Atomic nos. : F = 9, Cl = 17, Na = 11, Mg = 12, Al = 13, K =
(a) – 24.8 (b) – 15.7 (c) – 5.2 (d) – 48.78 19, Ca = 20, Sc = 21)
73. A piece of copper and another of germanium are cooled (a) K+, Cl–, Mg2+, Sc3+ (b) Na+, Ca2+, Sc3+, F–
+ 2+ 3+
(c) K , Ca , Sc , Cl – (d) Na+, Mg2+, Al3+, Cl–
from room temperature to 77K, the resistance of
(a) copper increases and germanium decreases 82. Among Al2O3, SiO2, P2O3 and SO2 the correct order of acid
(b) each of them decreases strength is
(c) each of them increases (a) Al2O3 < SiO2< SO2 < P2O3
(d) copper decreases and germanium increases (b) SiO2< SO2 < Al2O3 < P2O3
74. The manifestation of band structure in solids is due to (c) SO2< P2O3 < SiO2 < Al2O3
(a) Bohr’s correspondence principle (d) Al2O3 < SiO2< P2O3 < SO2
(b) Pauli’s exclusion principle 83. The bond order in NO is 2.5 while that in NO+ is 3. Which of
(c) Heisenberg’s uncertainty principle the following statements is true for these two species ?
(d) Boltzmann’s law (a) Bond length in NO+ is equal to that in NO
75. When p-n junction diode is forward biased then (b) Bond length in NO is greater than in NO+
(a) both the depletion region and barrier height are reduced (c) Bond length in NO+ is greater than in NO
(b) the depletion region is widened and barrier height is (d) Bond length is unpredictable
reduced 84. The formation of the oxide ion O (2g) requires first an
(c) the depletion region is reduced and barrier height is
increased exothermic and then an endothermic step as shown below
(d) Both the depletion region and barrier height are O (g ) e O (g ) Hº 142 KJmol 1
increased
O (g ) e O (2g ) H º 844 KJmol 1
Section - 2
This is because
(a) O– ion will tend to resist the addition of another
electron
76. Which of the following sets of quantum numbers is correct (b) Oxygen has high electron affinity
for an electron in 4f orbital ? (c) Oxygen is more elecronegative
(a) n = 4, = 3, m = + 1, s = + ½ (d) O– ion has comparatively larger size than oxygen atom
(b) n = 4, = 4, m = – 4, s = – ½ 85. The states of hybridization of boron and oxygen atoms in
(c) n = 4, = 3, m = + 4, s = + ½ boric acid (H3BO3) are respectively
(d) n = 3, = 2, m = – 2, s = + ½ (a) sp3 and sp2 (b) sp2 and sp 3
2
(c) sp and sp 2 (d) sp3 and sp3
AIEEE-2004 Solved Paper S-61
86. Which one of the following has the regular tetrahedral 96. For which of the following parameters the structural isomers
structure ? C2H5OH and CH3OCH3 would be expected to have the same
values?
(a) BF4 (b) SF4
(Assume ideal behaviour)
(c) XeF4 (d) [ Ni(CN) 4 ]2 (a) Boiling points
(b) Vapour pressure at the same temperature
(Atomic nos. : B = 5, S = 16, Ni =28, Xe = 54) (c) Heat of vaporization
87. Of the following outer electronic configurations of atoms, (d) Gaseous densities at the same temperature and pressure
the highest oxidation state is achieved by which one of 97. Which of the following liquid pairs shows a positive
them ? deviation from Raoult’s law ?
(a) (n – 1)d3 ns2 (b) (n – 1)d5 ns1 (a) Water - nitric acid
8
(c) (n – 1)d ns 2 (d) (n – 1)d5 ns2 (b) Benzene - methanol
88. As the temperature is raised from 20ºC to 40ºC, the average (c) Water - hydrochloric acid
kinetic energy of neon atoms changes by a factor of which (d) Acetone - chloroform
of the following ? 98. Which one of the following statements is FALSE?
(a) 313 (b) (313 / 293) (a) The correct order of osmotic pressure for 0.01 M
293 aqueous solution of each compound is
(c) 1 (d) 2 BaCl 2 KCl CH 3COOH sucrose
2
(b) The osmotic pressure ( ) of a solution is given by the
89. The maximum number of 90º angles between bond pair-bond
equation = MRT, where M is the molarity of the
pair of electrons is observed in
solution
(a) dsp2 hybridization (b) sp3d hybridization
3 (c) Raoult’s law states that the vapour pressure of a
(c) dsp hybridization (d) sp3d2 hybridization
component over a solution is proportional to its mole
90. Which one of the following aqueous solutions will exihibit
fraction
highest boiling point ?
(d) Two sucrose solutions of same molality prepared in
(a) 0.015 M urea (b) 0.01 M KNO3
different solvents will have the same freezing point
(c) 0.01 M Na2SO4 (d) 0.015 M glucose depression
91. Which among the following factors is the most important in 99. What type of crystal defect is indicated in the diagram
making fluorine the strongest oxidizing halogen ? below?
(a) Hydration enthalpy
(b) Ionization enthalpy Na Cl Na Cl Na Cl
(c) Electron affinity Cl– Cl– Na+ Na+
+
Na Cl – Cl Na Cl–
– +
(d) Bond dissociation energy
92. In van der Waals equation of state of the gas law, the constant Cl Na Cl Na+
– + – Na+
‘b’ is a measure of (a) Interstitial defect
(a) volume occupied by the molecules (b) Schottky defect
(b) intermolecular attraction (c) Frenkel defect
(c) intermolecular repulsions (d) Frenkel and Schottky defects
(d) intermolecular collisions per unit volume 100. An ideal gas expands in volume from 1×10–3 to 1 × 10–2 m3 at
300 K against a constant pressure of 1×105 Nm–2. The work
93. The conjugate base of H 2 PO 4 is done is
(a) H3PO4 (b) P2O5 (a) 270 kJ (b) – 900 kJ
(c) – 900 kJ (d) 900 kJ
(c) PO 34 (d) HPO 24 101. In a hydrogen-oxygen fuel cell, combustion hydrogen occurs
94. 6.02 × 1020 molecules of urea are present in 100 ml of its to
solution. The concentration of urea solution is (a) produce high purity water
(a) 0.02 M (b) 0.01 M (b) create potential difference between two electrodes
(c) 0.001 M (d) 0.1 M (c) generte heat
(Avogadro constant, NA = 6.02 × 1023 mol–1) (d) remove adsorbed oxygen from elctron surfaces
95. To neutralise completely 20 mL of 0.1 M aqueous solution 102. In a first order reaction, the concentration of the reactant,
of phosphorous acid (H3PO3), the value of 0.1 M aqueous decreases from 0.8 M to 0.4 M is 15 minutes. The time taken
KOH solution required is for the concentration to change from 0.1 M to 0.025 M is
(a) 40 mL (b) 20 mL (a) 7.5 minutes (b) 15 minutes
(c) 10 mL (d) 60 mL (c) 30 minutes (d) 60 minutes
S-62 The Pattern Target AIEEE
103. What is the equilibrium expression for the reaction 110. The enthalpies of combustion of carbon and carbon
P4(s) 5O 2(g ) P4O10(s) ? monoxide are – 393.5 and – 283 kJ mol–1 respectively. The
enthalpy of formation of carbon monoxide per mole is
(a) Kc [O 2 ]5 (a) – 676.5 kJ (b) 676.5 kJ
(c) 110.5 kJ (d) – 110.5 kJ
(b) Kc [P4 O10 ] / 5[P4 ][O 2 ]
111. The limiting molar conductivities º for NaCl, KBr and KCl
(c) Kc [P4 O10 ] /[P4 ][O 2 ]5 are 126, 152 and 150 S cm2 mol–1 respectively. The º for
NaBr is
(d) Kc 1 /[O 2 ]5 (a) 278 S cm2 mol–1 (b) 176 S cm2 mol–1
104. For the reaction, CO ( g ) COCl 2 (g ) the (c) 128 S cm2 mol–1 (d) 302 S cm2 mol–1
Cl 2 ( g )
112. In a cell that utilises the reaction
Kp
Zn (s) 2H (aq) Zn 2 (aq) H 2(g ) addition of H2SO4 to
K c is equal to
cathode compartment, will
(a) (b) RT (c) 1 (d) 1.0 (a) increase the E and shift equilibrium to the right
RT RT
105. The equilibrium constant for the reaction (b) lower the E and shift equilibrium to the right
(c) lower the E and shift equlibrium to the left
N 2( g ) O 2 ( g ) 2NO 2(g ) at temperature T is 4×10–4.
(d) increase the E and shift equilibrium to the left
The value of Kc for the reaction 113. Which one of the following statement regarding helium is
1 1 incorrect ?
NO 2(g) N 2( g ) O 2(g ) at the same temperature is
2 2 (a) It is used to produce an d sustain powerful
superconducting magnets
(a) 4×10–4 (b) 50 (c) 2.5×102 (d) 0.02
106. The rate equation for the reaction 2A + B C is found to be (b) It is used as a cryogenic agent for carrying out
: rate = k[A][B]. The correct statement in relation to this experiments at low temperatures
reaction is that the (c) It is used to fill gas balloons instead of hydrogen
(a) rate of formation of C is twice the rate of disappearance because it is lighter and non-inflammable
of A (d) It is used in gas-cooled nuclear reactors
(b) t1 / 2 is a constant 114. Identify the correct statement regarding enzymes
(a) Enzymes are specific biological catalysts that cannot
(c) unit of k must be s–1
be poisoned
(d) value of k is independent of the initial concentrations
(b) Enzymes are normally heterogeneous catalysts that are
of A and B
very specific in their action
107. Consider the following Eº values
(c) Enzymes are specific biological catalysts that can
Eº 0.77 V ; E º 0.14V normally function at very high temperatures (T~1000K)
Fe3 / Fe 2 Sn 2 / Sn
(d) Enzymes are specific biological catalysts that possess
Under standard conditions the potential for the reaction well-defined active sites
Sn (s ) 2Fe 3 (aq) 2Fe 2 (aq) Sn 2 (aq) is 115. One mole of magnesium nitride on the reaction with an
excess of water gives :
(a) 0.91 V (b) 1.40 V (c) 1.68 V (d) 0.63 V
(a) two moles of ammonia (b) one mole of nitric acid
108. The molar solubility (in mol L–1) of a sparingly soluble salt
(c) one mole of ammonia (d) two moles of nitric acid
MX4 is ‘s’. The corresponding solubility product is Ksp. ‘s’
is given in term of Ksp by the relation : 116. Which one of the following ores is best concentrated by
froth-flotation method ?
(a) s (256 K sp )1 / 5 (b) s (128 K sp )1 / 4 (a) Galena (b) Cassiterite
(c) Magnetite (d) Malachite
(c) s (K sp / 128)1 / 4 (d) s (K sp / 256)1 / 5 117. Beryllium and aluminium exhibit many properties which are
109. The standard e.m.f. of a cell involving one electron change similar. But, the two elements differ in
is found to be 0.591 V at 25ºC. The equilibrium constant of (a) forming covalent halides
the reaction is (F = 96,500 C mol–1; R = 8.314 JK–1 mol–1) (b) forming polymeric hydrides
(a) 1.0 × 1010 (b) 1.0 × 105 (c) exhibiting maximum covalency in compounds
(c) 1.0 × 10 1 (d) 1.0 × 1030 (d) exhibiting amphoteric nature in their oxides
AIEEE-2004 Solved Paper S-63
118. Aluminium chloride exists as dimer, Al2Cl6 in solid state as (a) Cyanocobalamin is B12 and contains cobalt
well as in solution of non-polar solvents such as benzene. (b) Haemoglobin is the red pigment of blood and contains
When dissolved in water, it gives irons
(c) Chlorophylls are green pigments in plants and contain
(a) [Al(OH) 6 ]3 3HCl calcium
(d) Carboxypeptidase - A is an exzyme and contains zinc.
(b) [Al(H 2 O) 6 ]3 3Cl 126. Cerium (Z = 58) is an important member of the lanthanoids.
Which of the following statements about cerium is incorrect?
(c) Al 3 3Cl
(a) The +4 oxidation state of cerium is not known in
(d) Al 2 O3 6HCl solutions
119. The soldiers of Napolean army while at Alps during freezing (b) The +3 oxidation state of cerium is more stable than the
+4 oxidation state
winter suffered a serious problem as regards to the tin
(c) The common oxidation states of cerium are +3 and +4
buttons of their uniforms. White metallic tin buttons got
(d) Cerium (IV) acts as an oxidizing agent
converted to grey power. This transformation is related to
127. Which one of the following has largest number of isomers ?
(a) a change in the partial pressure of oxygen in the air
(b) a change in the crystalline structure of tin (a) [Ir(PR 3 ) 2 H(CO)]2 (b) [Co( NH 3 )5 Cl]2
(c) an interaction with nitrogen of the air at very low
(c) [Ru ( NH 3 ) 4 Cl 2 ] (d) [Co(en ) 2 Cl 2 ]
temperature
(d) an interaction with water vapour contained in the humid (R = alkyl group, en = ethylenediamine)
air 128. The correct order of magnetic moments (spin only values in
B.M.) anong is
120. The E º values for Cr, Mn, Fe and Co are – 0.41, +
M3 / M 2 (a) [Fe(CN ) 6 ]4 [MnCl 4 ]2 [CoCl 4 ]2
1.57, + 0.77 and + 1.97V respectively. For which one of these
metals the change in oxidation state from +2 to +3 is easiest? (b) [MnCl 4 ]2 [Fe(CN ) 6 ]4 [CoCl 4 ]2
(a) Fe (b) Mn (c) Cr (d) Co
(c) [MnCl 4 ]2 [CoCl 4 ]2 [Fe(CN ) 6 ]4
121. Excess of KI reacts with CuSO4 solution and then Na2S2O3
solution is added to it. Which of the statements is incorrect (d) [Fe(CN ) 6 ]4 [CoCl 4 ]2 [MnCl 4 ]2
for this reaction ?
(Atomic nos. : Mn = 25, Fe = 26, Co = 27)
(a) Na2S2O3 is oxidised (b) CuI2 is formed 129. Consider the following nuclear reactions :
(c) Cu2I2 is formed (d) Evolved I2 is reduced
238 x
122. Among the properties (a) reducing (b) oxidising (c) 92 M y N 2 42 He ; xy N A
B L 2
complexing, the set of properties shown by CN– ion towards The number of neutrons in the element L is
metal species is (a) 140 (b) 144
(a) c, a (b) b, c (c) a, b (d) a, b, c (c) 142 (d) 146
123. The coordination number of a central metal atom in a complex 130. The half-life of a radioisotope is four hours. If the initial
is determined by mass of the isotope was 200 g, the mass remaining after 24
(a) the number of ligands around a metal ion bonded by hours undecayed is
sigma and pi-bonds both (a) 3.125 g (b) 2.084 g
(b) the number of ligands around a metal ion bonded by (c) 1.042 g (d) 4.167 g
pi-bonds 131. The compound formed in the positive test for nitrogen with
the Lassaigne solution of an organic compound is
(c) the number of ligands around a metal ion bonded by
(a) Fe4[Fe(CN)6]3 (b) Na3[Fe(CN)6]
sigma bonds
(c) Fe(CN)3 (d) Na4[Fe(CN)5NOS]
(d) the number of only anionic ligands bonded to the metal 132. The ammonia evolved from the treatment of 0.30 g of an
ion. organic compound for the estimation of nitrogen was passed
124. Which one of the following complexes is an outer orbital in 100 mL of 0.1 M sulphuric acid. The excess of acid required
complex ? 20 mL of 0.5 M sodium hydroxide solution for complete
(a) [Co(NH3)6]3+ (b) [Mn(CN)6]4– neutralization. The organic compound is
(c) [Fe(CN)6]4– (d) [Ni(NH3)6]2+ (a) urea (b) benzamide
(Atomic nos. : Mn = 25; Fe = 26; Co = 27, Ni = 28) (c) acetamide (d) thiourea
125. Coordination compounds have great importance in biological 133. Which one of the following has the minimum boiling point ?
systems. In this context which of the following statements (a) 1 - Butene (b) 1 - Butyne
is incorrect ? (c) n- Butane (d) isobutane
S-64 The Pattern Target AIEEE
Paper - II
Time : 1¼ hours Max. Marks : 225
1. Let R = {(1, 3), (4, 2), (2, 4),(2,3), (3,1)} be a relation on the 10. If (1 p) is a root of quadratic equation
set A = {1, 2,3, 4}. The relation R is x 2 px (1 p) 0 then its root are
(a) reflexive (b) transitive (a) –1, 2 (b) –1, 1 (c) 0, –1 (d) 0, 1
(c) not symmetric (d) a function
11. Let S(K ) 1 3 5... (2K 1) 3 K 2 . Then which of
2. The range of the function f ( x ) 7 x Px 3 is
the following is true
(a) {1, 2, 3, 4, 5} (b) {1, 2, 3, 4, 5, 6} (a) Principle of mathematical induction can be used to
(c) {1, 2, 3, 4,} (d) {1, 2, 3,} prove the formula
3. Let z and w be complex numbers such that z i w 0 and (b) S(K ) S(K 1)
arg zw = Then arg z equals
(c) S(K ) / S(K 1)
5 3
(a) (b) (c) (d) (d) S(1) is correct
4 2 4 4
12. How many ways are there to arrange the letters in the word
1
x y 2 2 GARDEN with vowels in alphabetical order
4. If z x i y and z3 p iq , then p (p q ) is
q (a) 480 (b) 240 (c) 360 (d) 120
equal to 13. The number of ways of distributing 8 identical balls in 3
(a) –2 (b) –1 (c) 2 (d) 1 distinct boxes so that none of the boxes is empty is
5. If | z 2 1 | | z |2 1, then z lies on (a) 8
C3 (b) 21 (c) 38 (d) 5
(a) an ellipse (b) the imaginary axis
(c) a circle (d) the real axis 14. If one root of the equation x 2 px 12 0 is 4, while the
(a) (b) (c) (d) 0 x2 2cxy 7 y 2 0 is four times their product c has the
2
4 value
f (a ) (a) –2 (b) –1 (c) 2 (d) 1
ex
41. If f ( x ) , I1 xg{x(1 x )}dx 48. If one of the lines given by 6x 2
xy 4cy 2
0 is 3x + 4y
1 ex f ( a)
= 0, then c equals
(a) –3 (b) –1 (c) 3 (d) 1
f (a )
49. If a circle passes through the point (a, b) and cuts the circle
and I 2 g{x (1 x )}dx,
f ( a) x2 y2 4 orthogonally, then the locus of its centre is
I2
(a) 2ax 2by (a 2 b2 4) 0
then the value of I is
1 (b) 2ax 2by (a 2 b2 4) 0
(a) 1 (b) –3 (c) –1 (d) 2
42. The area of the region bounded by the curves (c) 2ax 2by (a 2 b2 4) 0
y | x 2 |, x 1, x 3 and the x-axis is (d) 2ax 2by (a 2 b2 4) 0
(a) 4 (b) 2 (c) 3 (d) 1
50. A variable circle passes through the fixed point A(p, q) and
43. The – differential equation for the family of circle
touches x-axis . The locus of the other end of the diameter
x2 y2 2ay 0, where a is an arbitrary constant is through A is
(a) (x 2 y2 )y 2xy (b) 2( x 2 y2 )y xy (a) (y q)2 4px (b) ( x q) 2 4py
51. If the lines 2x 3y 1 0 and 3x y 4 0 lie along 59. The intersection of the spheres
diameter of a circle of circumference 10 , then the equation x2 y2 z2 7 x 2 y z 13 and
of the circle is
x2 y2 z 2 3x 3 y 4z 8 is the sa me as the
(a) x2 y2 2x 2 y 23 0
intersection of one of the sphere and the plane
(b) x2 y 2 2 x 2 y 23 0 2x y z 1 x 2y z 1
(a) (b)
(c) 2 2
x y 2x 2 y 23 0 (c) x y 2z 1 (d) x y z 1
(d) 2 2
x y 2x 2 y 23 0 60. Let a, b and c be three non-zero vectors such that no two
52. Intercept on the line y = x by the circle x 2 2
y 2x 0 is of these are collinear. If the vector a 2b is collinear with
AB. Equation of the circle on AB as a diameter is
c and b 3c is collinear with a ( being some non-zero
(a) x2 y2 x y 0 (b) x2 y2 x y 0
scalar) then a 2b 6c equals
(c) 2 2 (d) 2 2
x y x y 0 x y x y 0
(a) 0 (b) b (c) c (d) a
53. If a 0 and the line 2bx 3cy 4d 0 passes through
61. A particles is acted upon by constant forces 4î ˆj 3k̂
the points of intersection of the parabolas
y2 4ax and x 2 4ay, then and 3î ĵ k̂ which displace it from a point î 2ˆj 3k̂ to
(a) d2 (3b 2c) 2 0 (b) d2 (3b 2c) 2 0 the point 5î 4 ĵ k̂ . The work done in standard units by
the forces is given by
(c) d 2 (2b 3c) 2 0 (d) d 2 (2b 3c) 2 0
(a) 15 (b) 30 (c) 25 (d) 40
54. The eccentricity of an ellipse, with its centre at the origin, is
62. If a , b, c are non-coplanar vectors and is a real number,,
1
. If one of the directrices is x = 4 , then the equation of
2 then the vectors a 2 b 3c, b 4c and (2 1)c are
the ellipse is: non coplanar for
(a) 4x 2 3y 2 1 (b) 3x 2 4y 2 12 (a) no value of
(b) all except one value of
(c) 4x 2 3y 2 12 (d) 3x 2 4 y 2 1 (c) all except two value of
55. A line makes the same angle , with each of the x and z axis. (d) all values of
If the angle , which it makes with y-axis, is such that
63. Let u , v, w be such that | u | 1, | v | 2, | w | 3. If the
sin 2 3 sin 2 , then cos2 equals projection v along u is equal to that of w along u and
2 1 3 2 v , w are perpendicular to each other then
(a) (b) (c) (d)
5 5 5 3 | u v w | equals
56. Distance between two parallel planes
2x + y + 2z = 8 and 4x + 2y +4z + 5 = 0 is (a) 14 (b) 7 (c) 14 (d) 2
9 5 7 3 64. Let a, b an d c be non-zero vectors such that
(a) (b) (c) (d)
2 2 2 2
57. A line with direction cosines proportional to 2, 1, 2 meets 1
(a b) c | b || c | a. If is the acute angle between the
each of the lines x y a z and x a 2 y 2z . The 3
co-ordinates of each of the points of intersection are given vectors b and c , then sin equals
by
(a) (2a ,3a ,3a ), (2a , a , a ) (b) (3a ,2a ,3a ), (a , a , a ) 2 2 2 2 1
(a) (b) (c) (d)
3 3 3 3
(c) (3a ,2a ,3a ), (a , a ,2a ) (d) (3a ,3a ,3a ), (a , a , a )
58. If the straight lines 65. Consider the following statements :
t (A) Mode can be computed from histogram
x 1 s, y 3 s, z 1 s and x , y 1 t, z 2 t, (B) Median is not independent of change of scale
2
with parameters s and t respectively, are co-planar, then (C) Variance is independent of change of origin and scale.
equals. Which of these is / are correct ?
1 (a) (A), (B) and (C) (b) only (B)
(a) 0 (b) –1 (c) (d) –2 (c) only (A) and (B) (d) only (A)
2
AIEEE-2004 Solved Paper S-69
(b) 2 3 N and 2 3 N 1
(b) ( 3 1) m / s
4
1 1
(c) 2 2 N and 2 2 N 1
2 2 (c) m/s
4
(d) 2 2 N and 2 2 N 1
(d) m/s
8
71. In a right angle ABC A 90 and sides a, b, c are
respectively, 5 cm, 4 cm and 3 cm. If a force F has moments 75. If t 1 and t 2 are the times of flight of two particles having
0, 9 and 16 in N cm. units respectively about vertices A, B the same initial velocity u and range R on the horizontal ,
and C, then magnitude of F is then t 12 t 22 is equal to
(a) 9 (b) 4 (c) 5 (d) 3 (a) 1
72. Three forces P, Q and R acting along IA, IB and IC, where (b) 4u 2 / g 2
I is the incentre of a ABC, are in equilibrium. Then
(c) u 2 / 2g
P : Q : R is
(d) u2 / g
S-70 The Pattern Target AIEEE
A N S W ER K EY
P H Y S IC S C H EM IS TR Y M A TH EM A TIC S
1 (a) 26 ( b) 51 (a) 76 (a) 101 ( b) 126 (a) 1 (c) 26 ( b) 51 (d)
2 (c) 27 (a) 52 ( d) 77 ( b) 102 (c) 127 ( d) 2 ( d) 27 ( b) 52 (d)
3 (a) 28 ( b) 53 (c) 78 (a) 103 ( d) 128 (c) 3 (c) 28 ( b) 53 (d)
4 (c) 29 ( b) 54 ( b) 79 (c) 104 (c) 129 ( b) 4 (a) 29 (c) 54 (b)
5 (a) 30 (c) 55 ( b) 80 (a) 105 ( b) 130 (a) 5 ( b) 30 (c) 55 (c)
6 (b) 31 ( b) 56 (c) 81 (c) 106 ( d) 131 (a) 6 ( b) 31 (a) 56 (c)
7 (d) 32 ( d) 57 (a) 82 ( d) 107 (a) 132 (a) 7 (a) 32 ( b) 57 (b)
8 (d) 33 ( b) 58 ( b) 83 ( b) 108 ( d) 133 ( d) 8 ( d) 33 ( d) 58 (d)
9 (c) 34 (a) 59 ( b) 84 (a) 109 (a) 134 (a) 9 ( b) 34 ( d) 59 (a)
10 (b) 35 ( b) 60 ( d) 85 ( b) 110 ( d) 135 (a) 10 (c) 35 ( b) 60 (c)
11 (c) 36 ( d) 61 ( b) 86 (a) 111 (c) 136 ( b) 11 ( b) 36 ( b) 61 (d)
12 (b) 37 ( b) 62 ( b) 87 ( d) 112 (a) 137 (c) 12 (c) 37 (a) 62 (c)
13 (b) 38 (c) 63 (a) 88 (a) 113 (c) 138 (a) 13 ( b) 38 ( d) 63 (c)
14 (a) 39 ( d) 64 ( b) 89 ( d) 114 ( d) 139 ( d) 14 ( d) 39 ( b) 64 (a)
15 (b) 40 ( b) 65 ( d) 90 (c) 115 (a) 140 ( d) 15 (c) 40 ( b) 65 (c)
16 (c) 41 (a) 66 (a) 91 ( d) 116 (a) 141 (c) 16 ( b) 41 ( d) 66 (c)
17 (a) 42 ( d) 67 (c) 92 (a) 117 (c) 142 ( b) 17 ( d) 42 ( d) 67 (c)
18 (d) 43 ( d) 68 ( b) 93 ( d) 118 ( b) 143 (c) 18 ( d) 43 (c) 68 (b)
19 (c) 44 ( b) 69 (a) 94 ( b) 119 ( b) 144 (a) 19 ( b) 44 ( b) 69 (a)
20 (b) 45 (a) 70 (a) 95 (a) 120 (c) 145 ( d) 20 ( b) 45 ( d) 70 (c)
21 (c) 46 (a) 71 ( d) 96 ( d) 121 (c) 146 ( b) 21 ( d) 46 (a) 71 (c)
22 (d) 47 (c) 72 ( d) 97 ( b) 122 (a) 147 (a) 22 (a) 47 (c) 72 (d)
23 (b) 48 ( b) 73 ( d) 98 ( d) 123 (c) 148 (c) 23 (c) 48 (a) 73 (c)
24 (a) 49 (c) 74 ( b) 99 ( b) 124 ( d) 149 ( b) 24 ( d) 49 ( b) 74 (a)
25 (a) 50 (a) 75 (a) 100 (c) 125 (c) 150 (a) 25 (a) 50 ( d) 75 (b)
SECTION I – PHYSICS & CHEMISTRY now fort = T/3 second vertical distance moved is given
by
1. (a) From stokes law 2
1 T 1 gT 2 h
F h' g h1
F 6 rv 2 3 2 g 9
6 rv
h 8h
MLT 2 position of ball from ground h
9 9
[ML 1T 1 ]
[L][LT 1 ] 4. (c) A B B A 0 A B A B 0
dv A B 0
2. (c) a x x
dt Angle between them is 0, , or 2
dx dx from the given options.
Also v or dt
dt dv
5. (a) The angle for which the ranges is same are
v2 x complementary.
vdv
x v dv xdx let one angle is , other is 90 -
dx
v1 0 2u sin 2u cos
T1 , T2
2 g
x
v 22 v12
2 4u 2 u sin cos
2
T1T2 =2R
1 1 x g
m v22 v12 m Hence it is proportional to R.
2 2 2
6. (b) Only option (b) is false since acceleration vector is
k x 2 always radial (i.e . torwards the center) for uniform
circular motion.
1 2
3. (a) h = gT 5 50
2 7. (d) Speed v1 60 m/s m/s
18 3
AIEEE-2004 Solved Paper S-71
5 100 15. (b) Angular momentum will remain the same since external
d1 20m, v '1 120 m/s torque is zero.
18 3 16. (c) Yes, the person can catch the ball when horizontal
Let deceleration be a velocity is equal to the horizontal component of ball’s
0 v '12 2a d1 ... (1) velocity, the motion of ball will be only in vertical
direction w.r.t person for that,
or v '12 2ad, vo
v cos or 60
(2v)2 2ad 2 ... (2) 2
(2) divided by (1) gives, 2 1
17. (a) IB MR IA MR 2 ; IA IB
d2 2
4 d 2 4 20 80m
d1 mv 2 GmM GM
18. (d) also g
8. (d) Let n be the number of bullets man can fire in one (R x) (R x) 2
R2
second.
change in momentum per second n mv = F mv 2 GM R2
[ m= mass of bullet, v = velocity] m
( F is the force)
(R x) R2 (R x) 2
F
144 1000 mv 2 R2
n =3 mg
mv
40 1200 (R x) (R x) 2
(5 4.8) 9.8
9. (c) Acceleration m / s 2 = 0.2 m/s2 gR 2 gR 2
1/ 2
(5 4.8) v2 v
R x R x
4
10. (b) m is mass of over hanging chain (0.6)kg
2 mv 2 GmM
Let at the surface PE = 0 19. (c)
C.M.of hanging part = 0.3 m below the table R x (R x) 2
4 x = height of satellite from earth surface
Ui m gx 0.6 10 0.30 m = mass of satellite
2
U m 'gx 3.6J = Workdone in putting the entire GM GM
v or v
chain on the table (R x) R x
fs 2 (R x) 2 (R x)
N T
V GM
11. (c) R x
mg
which is independent of mass of satellite
GmM
Mg sin fs ( for body to beat rest) 20. (b) At earth surface P.C of (m+m) is –
R
m 10 sin 30 10 GmM
m 5 10 m 2.0kg at a distance P.E of (m+m) is
2R
12. (b) W F.x (5iˆ 3jˆ 2k).(2i
ˆ ˆ ˆj) GmM GmM GmM
U ; U
10 3 7 joules 2R R 2R
13. (b) Let acceleration of body be a GM GM
Now g; gR
v1 2 R
R
v1 0 at1 a
t1 1
U mgR
v1t 2
v at v n 2 2 (n 1)
t1 21. (c) F KR MR KR
Pinst = F.v (ma).v (n 1)
or K 'R 2
2
mv1 v1t
v (n 1)
m 1 t 2 (n 1)
t1 t1
t1 R 2
T T R 2
14. (a) Work done by such force is always zero since force is 22. (d) Work done by constant force in displacing the object
acting in a direction perpendicular to velocity by a distance
from work energy theorem k 0 = change in potential energy
k remains constant.
S-72 The Pattern Target AIEEE
1 2 100
stress × strain ×volume 100
2 2
1 F l F 2 2
A L 20
2 A 2 2 20
23. (b) From Stoke‘s law 2 100
viscous force F 6 rv v 5m / s
20 2
hence F is directly proportional to radius & velocity 29. (b) Equation of displacement is given by
24. (a) Let pressure outside be Po
x A sin( t )
2T
P1 ( in smaller bubble ) P0 Fo
r Fo
where A = =
m ( 2 2
0)
P2 ( in bigger bubble ) P0
2T
( R > r) m ( 2 2 2
0)
R here damping effect is considered to be zero
P1 P2 1
hence air moves from smaller bubble to bigger bubble. x 2 2
m( 0)
t 2 t 2 30. (c) Since energy ( Amplitude)2, the maximum for both
25. (a) geff ; 0 of them occurs at the same frequency
g
1 2
t 2 5 7
g/4 31. (b) 1 3 2 5
t 2t 0
Cp1 5 C p1 5 5
Now C p1 R
×1000 Vg C v1 3 C p1 Cv1 2 2
3
Now C v1 Cp1 R R
4 2
1000 Vg
3 7 , 5
similarly Cp2 R Cv2 R
2 2
4 1000 Now let change in temperature be dT
Net force 1 1000 Vg Vg
3 3 change in internal energy of 1st gas is n1Cv1dT
1000 g U 2 n 2C v2 dT
g eff Let molar capacity at constant volume of system is
4 4
3 1000
3 Cv
m m change in internal energy (n1 n 2 )Cv dT
26. (b) t1 2 , t2 2
k1 k2 n1Cv dT n 2C v2dT (n1 n 2 )C v dT
k1k 2 n1Cv1 n 2 Cv2
when springs are in series then, k eff Cv
kl k 2 n1 n 2
m(k l k 2 ) n1Cv1 n 2 C v2
T 2 Cp Cv R R
k1k 2 n1 n 2
n1 (C v1 R) n 2 (C v2 R) n1Cp1 n 2Cp2
m m t 22 t12
T 2 2 n1 n 2 n1 n 2
k2 k1 (2 ) 2 (2 ) 2
Putting n1 n2 1 and other values
T 2 t12 t 22 3
27. (a) At any instant the total energy is Cv 2R & Cp 3R hence
2
1 2 1 1 2
kx mv 2 kA 0 = constant 32. (d) E AT 4 ; A R2 E R 2 T4
2 2 2
A0 = amplitude E2 R 22 T24
hence U is m independent of x.
28. (b) From eqn E1 R12 T14
100 put R 2 2R, R1 R
AIEEE-2004 Solved Paper S-73
E 2 (2R)2 (2T) 4
T2 2T, T1 T 64
E1 R 2 T4
33. (b) Internal energy and entropy are state function, they 38. (c)
do not depend upon path but on the state.
P1V1 P2 V2
34. (a) n1 , n2 For the image to be real and of same size as object, final
T1 T2
image should be formed at the position of object itself.
P(V1 V2 ) Let x be the distance of object from plane surface.
also (n1 n 2 ) Apparent distance from surface = x
T
This should be centre of curve
P P1V1 P2 V2 1
x 30 1.5x 30 x 20
T T1 T2 (V1 V2 )
39. (d) The angle of incidence for total polarization is given
P (P1V2 T2 P2 V2 T1 ) by tan n tan 1 n
T (T1T2 )(V1 V2 ) ... (1)
40. (b) For constructive interference d sin n
internal energy of gas before connection
n
Pv given d 2 sin
U nCv T Cv 2
R
n 0,1, 1, 2, 2 hence five maxima are possible
Internal energy of gases before the connection
C v P1V1 PV 1 1 1 1 c
Cv 2 2 2 41. (a) v .
R R 0 0 0 0 .4 2 0 0 2
Cv P c v
And after connection (V1 V2 ) remains constant, hence 0
since v
R 2 2
C v P1V1 C v2 P2 V2 CvP(V1 V2 ) wavelength is halved and frequency remains
R R R unchanged
C
P1V1 P2 V2 B A C
or P ... (2) 42. (d) F
V1 V2 r r 2
x
x is distance between the spheres. After first operation
T1T2 (P1V1 P2 V2 )
T from eqn (1) & (2)) charge on B is halved i.e /2 and charge on third sphere
P1V1T2 P2 V2 T1
E becomes . Now it is touched to C, charge then equally
35. (b) P of photon 2
C distributes them selves to make potential same, hence
2E 1 3
Change in momentum charge on C becomes .
C 2 2 4
= momentum transferred to the surface
(the photon will reflect with same magnitude of 3
momentum in opposite direction) ' '
c B 4 2 3 2
36. (d) The thermal resistance is given by Fnew
x 2
x2 8 x2
x 4x x 2x 3x
3
kA 2kA kA kA kA or Fnew F
8
dQ T (T2 T1 )kA 1 A(T2 T1 )k
1 k q 1 kq r
dt 3x 3x 3 x 43. (d) mv 2 m(2v)2 r'
kA 2 r 2 r' 4
44. (b) Net field along,
1
f
3 kQ 2 kQ kq
AB =0
37. (b) The incident angle is 45° incident angle > critical angle, a 2
( 2a ) 2 2
a
i ic
2
sin i sin ic or sin 45 sin i c
–Q –Q
1
sin i c
n
1 1 1
sin 45 or n 2 –Q –Q
n 2 n
S-74 The Pattern Target AIEEE
4 r1 2 n. 0i
2 1 B' ... (2)
R1 1
; R2 3 ; r2 2 r'
A 2
r1 r22 2 3
n 0 i.n
V V From (1) and (2), B ' n 2B
i1 , i2 2 R
R1 R2
0i a2
i1 V R2 56. (c) B
i2 V R1 2(x 2 a 2 )3 / 2
R1
R2 a2 (x 2 a 2 )3/ 2
0i 0i
2 B'
R2 2 ( 2) r12 2 r1 2a 2(x 2 a 2 )3 / 2 a2
R1 1 ( 1) r22 r2
1
B.(x 2 a 2 )3/ 2
B'
3 4 1 a3
4 9 3
R1 54(53 )
1 Put x 4& a 3 B' 250 T
49. (c) R2 Where 2 = 100 1 3 3 3
2
x 20 0 I1 0 2I1I2 0 I1I 2 2
in the first case 57. (a) F I2 –
2 r 2 .3r 2 r 3
y 80
4x 2
in the second case 50 F
y 20 3
50. (a) Metaloxides with high temperature coefficient of
I I
resistivity 58. (b) T 2 2
m B mB
51. (a) Q mCp T
AIEEE-2004 Solved Paper S-75
–H+
1 1 1 H 2 PO 4 HPO 4
79. (c) R 93. (d)
n12 n 22 Acid congugate base
1 1 1 6.02 10 20 1000
1.097 10 7 1.097 10 7 94. (b) M 0.01M
1 6.02 10 23 100
95. (a) N1V1 N 2 V2 (H 3PO 3 is dibasic M = 2N)
91.15 10 9 m 91nm
80. (a) The order of bond angles 20 0.2 0.1 V
V = 40 ml
BF3 SiH 4 NH 3 H 2S
120º 109º 28 107 º 92.5º
96. (d) Gaseous densities of ethanol and dimethyl ether would
be same at same temperature and pressure. The heat of
81. (c) , 20 Ca 2 , 21 Sc 3 ,17 Cl each contains 18 vaporisation, V.P. and bpts will differ due to H - bonding
19 K
electrons in ethanol.
97. (b) A mixture of benzene and methanol show positive
82. (d) SO 2 P2 O 3 SiO 2 Al 2 O 3 deviation from Raoult’s law
Acidic Weak acidic Amphoteric
98. (d) Tf K f m i . Since Kf has different values for
83. (b) The lower the bond order the greater the bond length
and vice verse different solvents, hence even if the m is the same Tf
84. (a) O– ion exerts a force of repulsion on the incoming will be different
electron. The energy is required to overcome it. 99. (b) When equal number of cations and anions are missing
85. (b) In H 3BO 3 hybridisation of B is sp2 and O is sp3. from their regular lattice positions, we have sehottky
defect
86. (a) XeF4 (Sp 3d 2 square planar), 100. (c) w P V 10 5 (1 10 2
1 10 3 ) 900J
2 (dsp2 square planar ),
[ Ni(CN) 4 ] 101. (b) In H 2 O 2 full cell, the combustion of H2 occurs to
create potential difference between the two electrodes
BF4 (sp3 tetrahedral), SF4 (sp3d see saw shaped)
87. (d) (n–1)d5ns2 attains the maximum O.S. of + 7 102. (c) The t1 / 2 is 15 minutes. To fall the concentration from
0.1 to 0.025 we need two half lives i.e., 30 minutes.
3
K 313 103. (d) For P4 (s) 5O 2 (g) P4O10 (s)
88. (a) K.E of neon at 40 C 2 313
K.E of neon at 20 C 3 293 1
K 293 Kc . The solids have concentration unity
2 (O 2 ) 5
104. (c) Kp K c (RT ) n ; n 1 2 1
Kp 1
M
89. (d)
Kc RT
[ NO]2 4
dsp2 hybridisation sp3d or dsp3 hybristion 105. (b) Kc 4 10
Number of 90° angle Number of 90° angle [ N 2 ][O 2 ]
between bonds = 4 between bonds = 6 [ N 2 ]1 / 2 [O 2 ]1 / 2 1 1
Kc 50
[ NO] Kc 4 10 4
106. (d) The velocity constant depends on temperature only. It
is independent of concentration of reactants.
107. (a) Fe 3 e Fe 2 G 1 F 0.77
Sn 2 2e Sn (s) G 2 F( 0.14)
for
sp3d2 hybridisation
Number of 90° angle Sn (s) 2Fe 3 (aq) 2Fe 2 (aq) Sn 2 (aq)
between bonds = 12 G ( 2F ( 0.14) 2( 1 F 0.77) 1.82 F
90. (c) Tb K b m i . Standard potentel for the given reaction
The value of i for Na 2SO 4 3, 1.82F
0.91V
for KNO3 = 2, for urea = 1 and for glucose = 1 2 F
91. (d) The fluorine has low dissociation energy of F - F bond M4 4X
108. (d) MX4
and reaction of atomic fluorine is exothermic in nature S 4S
92. (a) In van der walls equation ‘b’ is for volume correction Ksp = [S] [4S]4 = 256 S5
AIEEE-2004 Solved Paper S-77
K sp
1/ 5 [Fe(CN) 6 ]4, [Mn (CN ) 6 ]4 ,
S d 2sp 3 d 2sp 3
256
[CO( NH 3 ]3 , [ Ni( NH 3 ) 6 ]2
0.059
109. (a) Ecell = log K c d 2sp 3 sp 3d 2
n
Hence [ Ni( NH 3 ) 6 ]2 is outer orbital complex
0.591 0.059 log K c or 10 log K c Kc 1 1010
110. (d) 125. (c) Chlorophyll contains Mg and not Ca
126. (a) The +4 oxidation state of cerium is also known in
(i) C O2 1
CO 2 H 393.5 KJmol solution
127. (d) isomers
1 1
(ii) CO O2 CO 2 H 283.0 KJmol
2 [Ru ( NH 3 ) 4 Cl 2 ] , [Co ( NH 3 )5 Cl 2 ],
(i) - (ii), cis and trans none
1 1 [Ir(PR 3 ) 2 H(CO)] 2
, [Co(en ) 2 Cl 2 ]
C O2 CO H 110.5 KJmol
2 cis and trans cis and trans and
optical isomers
111. (c) º NaCl Na Cl ....(i) 128. (c) Number of Unparied electrons
KBr K Br ....(ii) [Fe(CN ) 6 ]4 , [CoCl]2 , [MnCl 4 ]2
zero three five
KCl K Cl ....(iii) The greater the number of unparied electrons, the
(i) + (ii) - (iii) higher the value of magnetic moment
238 230
NaBr Na Br 129. (b) 92 M 2 42 He
88 N
230 230
126 152 150 128 S cm 2 mol 1
88 N 86 L
2
The number of neutrons in element L = 230 – 86 = 144
112. (a) Zn (s) 2H (aq ) Zn 2 (aq) H 2 (g ) n
1
0.059 [Zn ][H 2 ] 2 130. (a) Nt N0 where n is number of half life periods.
Ecell E cell log 2
2 [ H ]2
24
Addition of H2SO4 will increase [H+]and Ecell will also n 6
increase and the equilibrium will shift towards RHS 2
113. (c) Helium is heavier than hydrogen although it is non- 6
1
inflammable Nt 200 3.125g
2
114. (d) Enzymes are very specific biological catalysts
possessing well - defined active sites 131. (a) Prussian blue Fe 4 [Fe(CN) 6 ]3 is formed in lassagne
115. (a) Mg 3 N 2 6H 2 O 3Mg(OH ) 2 2 NH 3 test for nitrogen
116. (a) Galena is PbS and purified by froth floatation method 132. (a) H 2SO 4 is dibasic.
117. (c) The valancy of Beryllium is +2 while that of aluminium 0.1 MH 2SO 4 0.2NH 2SO 4
is +3
Meq of H 2SO 4 taken = 100 0.2 20
118. (b) AlCl 6 12H 2 O 2[Al(H 2 O) 6 ]3 6Cl
Meq of H 2SO 4 neutralised by NaOH 20 0.5 10
119. (b) Grey tin white tin
Meq of H 2SO 4 neutralised by NH3 12 10 10
Grey tin is brittle and crumbles down to powder in very
cold climate 1.4 Meq of acid neutrialis ed by NH 3
% of N 2
120. (c) The given values show that Cr has maximum oxidation wt. of organic compound
potental, therefore its oxidation will be easiest. (Change
the sign to get the oxidation values) 1.4 10
46.6
0.3
121. (c) 4KI1 2CuSO 4 I 02 Cu 2 I 2 2K 2SO 4
% of nitrogen in urea 14 2 100
2 2.5
46.6
60
I02 2Na 2 S2 O 3 Na 2 S4 O 6 2NaI 1
133. (d) Bpt. follows the order
122. (a) CN ion acts good complexing as well as reducing Alkynes > Alkenes > Alkanes (Straight chain) > Alkanes
agent. branched chain) of comparable mol. wts
123. (c) The coordination number of central metal atom in a
complex is equal to number of monovalent ligands,
twice the number of bidentate ligands and so on, 134. (a) 1 3 IUPAC name – 3, 3-Dimethyl -1
2
around the metal ion bonded by coordinate bonds
124. (d) Hybridisation HO
cyclohexanol
S-78 The Pattern Target AIEEE
O O CH 3
sp 3 || sp 3 sp 3 || |
135. (a) H3 C C CH 3 ; C H 3 C OH ; CH 3 C OH
sp 2 sp 2 |
Acetone Acetic acid CH 3
2 methyl 2 propanol
O
sp3 sp sp 3 || 145. (d) It is Clemmensen’s reduction
C H3 C N ; C H3 C NH 2 O
sp 2 ||
Acetonitrile Acetamide CH 3 C CH 2 CH 3
136. (b) The compounds containing two similar assymmetric
C-atoms have plane of symmetry and exist in Meso
Zn Hg
CH 3 CH 2 CH 2 CH 3
Conc. HCl
from plane of symmetry CH2 OH
CHO
50% NaOH
146. (b)
Meso 2, 3 dichlorobutane + –
COONa
137. (c) Cl– is the best leaving group among the given option.
+
138. (a) Only 2- cylcopropyl butane has a chiral centre.
COOH COOH 147. (a) 3-methyl pentanol-3 will be dehydrated most readily
since it produces tertiary carbonium ion as intermediate.
139. (d) 148. (c) 1-chloropentane is not chiral while others are chiral in
< nature
NO2 Cl Cl
| |
COOH COOH C C C C C; C C C C C,
1 chloropenta ne *
NO2 2 chloropent ane
< < Cl C C Cl
| | | |
C C C C C C C C C C
NO2 * *
1 chloro -2-methyl pentane 3 chloro-2-methyl pentane
140. (d) Lone pair of electrons present on the nitrogen of benzyl 149. (b) Insulin is a biochemically active peptide harmone
amine is not involved in resonance. secreted by pancreas.
141. (c) RNA contains cytosine and uracil as pyrimidine bases 150. (a) Photo chemical smog is caused by oxides of sulpher
while DNA has cytosine and thymine. and nitrogen.
Both have the same purine bases i.e., Guanine and
ademine. PAPER II - MATHS
142. (b) DDT is prepared by heating chlorbenzene and chloral
with concentrated sulphuric acid 1. (c) (2,3) R but (3, 2) R
R is not symmetric
CCl3CHO + 2 H Cl 2. (d) 7 x is defined if
Px 3
7 x 0, x 3 0 and 7 x x 3
3 x 5 and x I
Cl
x 3, 4, 5
H2SO4
CCl3CH 7 3 4
–H2O f (3) P3 3 P0 1
7 4 3
Cl f ( 4) P4 3 P1 3
7 5 2
f (5) P5 3 P2 2
1,1,1-trichloro–2,2 bis
(p-chlorophyenyl) ethane Hence range = {1, 2, 3}
or 3. (c) arg zw =
DDT
arg z arg w ...(1)
143. (c) There is no reaction hence the resultant mixture contains
CH3 COOC2H5 + NaCl. z iw 0 z iw
O
|| (i)CH 3 MgI z iw arg z arg w
144. (a) CH 3 C Br 2
(ii) Saturated NH 4 Cl
AIEEE-2004 Solved Paper S-79
1 0 0
arg z arg z (from (1))
2 0 1 0 I
3 0 0 1
arg z
4
1 1 1
1
(a) 7. (a) A 2 1 3
z3 p iq
1 1 1
z p 3 (iq) 3 3p(iq)(p iq) Cofactors of various entries are
x iy p3 3pq 2 i (3p 2 q q 3 ) 4, 5, 1; 2, 0, 2; 2, 5, 3
x | A | 1 4 ( 1) 5 1 1 10
x p 3 3pq 2 p 2 3q 2
p 4 5 1
y Cofactor Matrix C 2 0 2
y q 3 3p 2 q q 2 3p 2
q 2 5 3
x y 4 2 2
2p 2 2q 2 T
p q Adj A C 5 0 5
x y 1 2 3
(p 2 q2) 2
p q 4 2 2
2 2 1 AdjA 1
5. (b) | z 1| | z | 1 A 5 0 5
|A| 10
1 2 3
| z 2 1 |2 (zz 1) 2
Comparing , we get = 5
(z 2 1)(z 2 1) (zz 1) 2 8. (d) Let r be the common ratio, then
1 3 5.... (2k 1) 2k 1
1 1 1
3 k2 2k 1 3 (k 1) 2 Add, 2 t n (n )
n n
....
n
nSn
C0 C1 Cn
S(k) S(k 1) tn n
12. (c) Total number of arrangements of letters in the word
GARDEN = 6 ! = 720 there are two vowels A and E, in sn 2
half of the arrangements A preceeds E and other half A 1
follows E. 18. (d) Tm = a+ (m – 1) d = ... (1)
n
1
So, vowels in alphabetical order in 720 360 1
2 Tn a (n 1)d ... (2)
13. (b) Required number of ways m
= coefficient of x 2 in ( x x 2 ...x 6 ) 3 1 1 1
(1) - (2) (m n )d d
[ each box can receive minimum 1 and maximum 6 n m mn
balls ] 1
8 2 2 5 3
From (1) a a d 0
= coeff of x in x (1 x x .... x ) mn
19. (b) If n is odd, the required sum is
3
6
1 x 12 + 2.22 + 32 2.4 2 ...... 2.(n 1) 2 n 2
coeff of x 5 in
1 x
(n 1)(n 1 1) 2
n 2 [ (n 1) is even ]
= coeff of x 5 in (1 x ) 3 2
= coeff of x 5 in (1 3 4
C 2 x 2 ...) n 1 n 2 (n 1)
C1x 1 n2
7
2 2
C5 21
1 1 1
14. (d) 4 is a root of x 2 px 12 0 20. (b) e 1 .......
1! 2! 3!
16 4p 12 0 p 7 1 1 1 1
e 1 .......
Now, the equation x 2 1! 2! 3!
px q 0
has equal roots. 1 1 1
e e 21 ....
2 2! 4!
p 49
p2 4q 0 q
4 4 1 1 1 e e 1
15. (c) The middle term in the expansion of ...... 1
2! 4! 6! 2
4 4 2 2 2
(1 x) C2 ( x) 6 x
The middle term in the expansion of e 2 1 2e (e 1) 2
2e 2e
(1 x ) 6 6 C3 ( x) 3 20 3 x 3
21. (d) 3
According to the given equation
3 3
6 2 20 3 cos 0
10 2 2 2 2
16. (b) Coeff. of xn in ( 1+x) (1 – x)n 21
sin sin
= coeff of xn in 65
n
(1 x )(1 n C1x n C 2 x 2 .... ( 1) n C n x n ) 21
2 sin cos ... (1)
n 1n 2 2 65
( 1) n C n ( 1) n Cn 1 ( 1) n ( 1) n 1.n
27
n
( 1) (1 n) cos cos
65
1 1 1 1 27
17. (d) Sn = n n n
.... n 2 cos cos ... (2)
C0 C1 C2 Cn 2 2 65
Square and add (1) and (2)
0 1 2 n
tn n n n
.... n (21) 2 (27) 2 1170
C0 C1 C2 Cn 4 cos 2
2 2 65 65
(65)
n n 1 n 2 0
tn .... 9 3
n
Cn n
Cn 1
n
Cn 2
n
C0 cos 2 cos
2 130 2 130
AIEEE-2004 Solved Paper S-81
(a 4 b 4 ) cos 2 sin 2 y
22. (a) u2 a2 b2 2 tan 60° y 3x ....................(1)
2 2 4 4 x
a b (cos sin )
y x 40
Now (a 4 b 4 ) cos 2 sin 2 tan 30 y ...............(2)
x 40 3
2 2 4 4
a b (cos sin )
x 40
(a 4 b 4 ) cos 2 sin 2 From (1) and (2), 3x x 20m
3
a 2 b 2 (1 2 cos 2 sin 2 ) 25. (a) f ( x ) is onto
4 4 2 2 2 2 2 2 S = range of f (x)
(a b 2a b ) cos sin a b
sin 2 2 Now f(x) = sin x 3 cos x 1
(a 2 b2 ) 2. a 2b 2
4
2 sin x 1
0 sin 2 2 1 3
sin 2 2 (a 2 b2 ) 2
0 (a 2 b2 ) 2 1 sin x 1
4 4 3
sin 2 2
a 2b 2 (a 2 b 2 ) 2 a 2b 2 1 2 sin x 1 3
4 3
1
(a 2 b 2 )2 . a 2b 2 f ( x) [ 1, 3] S
4
from (1) 26. (b) Y
Minimum value of u2
-x x
a 2 b 2 2 a 2 b 2 (a b) 2
Maximum value of u2
1
a2 b2 2 (a 2 b 2 ). a 2 b2
4
2
a2 b2
(a 2 b 2 ) 2 2 (a 2 b2 ) x2 X
2 x1
Max value - Min value x=2
2 2 2 2 From the figure if
2 (a b ) (a b ) (a b)
23. (c) Let a sin , b cos f ( x1 ) f ( x 2 ), then x1 2 x and x 2 2 x
and c f (2 x ) f (2 x)
1 sin cos
Clearly a and b < 1 but sin 1 (x 3)
c > 1 as sin > 0 and cos > 0 27. (b) f (x) is defined
c is the greatest side and greatest angle is C 9 x2
a2 b2 c2 if 1 x 3 1 2 x 4 ...(i)
cos C
2ab and 9 x 2 0 3 x 3 ...(ii)
Taking common solution of (i) and (ii),
sin 2 cos 2 1 sin cos 1
we get 2 x 3
2 sin cos 2
Domain = [2, 3)
C 120
2x
T Lim a b
28. (b) x 1 e2
x x2
y Lim a b
ex 1 1 2x = e2
24. (d) x x2
30° 60°
R Lim 2b
P 40m Q x x 2a 2
From the figure x
2a 0 2, b R a 1, b R
S-82 The Pattern Target AIEEE
The equation of the normal at is
1 tan x
29. (c) f (x) is continuous in 0, y a sin tan ( x a a cos )
4x 2
y cos a sin cos x sin a sin
Lim
f f (x) Lim f (x) a sin cos
4 x x
4 4 x sin y cos a sin
y ( x a ) tan
Lim f h
h 0 4 which always passes through (a, 0)
34. (d) Let us define a function
1 tan h
Lim 4 ax 3 bx 2
h 0 ,h 0 f (x) cx
3 2
4 h
4 Being polynomial, it is continuous and differentiable,
also,
1 tanh
1 a b
Lim 1 tanh f (0) 0 and f (1) c
h 0 4h 3 2
Lim 2 tanh 2 1 2a 3b 6c
h 0 . f (1) 0 (given )
1 tanh 4 h 4 2 6
30. (c) y .... f (0) f (1)
x ey e x ey x .
Taking log on both sides, f(x) satisfies all conditions of Rolle’s Mean value
theorem f’(x) = 0 has a root in (0, 1)
1 dy
log x = y + x 1 i.e. ax 2 bx c 0 has at lease one root in (0, 1)
x dx
n r
dy 1 1 x 1 n
1 Lim
e
dx x x 35. (b) n
n
r 1
dy dy 9
31. (a) y2 18x 2 y 18 [Using definite integrals as limit of sum]
dx dx y
1
Given
dy
2
9
2 y
9 e x dx e 1
dx y 2 0
9 sin x sin(x )
Putting in y 2 18 x x dx dx
8 36. (b)
sin(x ) sin(x )
9 9
Required point is , sin(x ) cos cos( x ) sin
8 2 dx
sin(x )
32. (b) f (x) 6(x 1). Inegrating, we get
{cos sin cot( x )}dx
f (x) 3x 2 6x c
Slope at (2, 1) = f (2) c 3 (cos )x (sin ) log sin(x ) C
[ slope of tangent at (2,1) is 3] A = cos , B = sin
f ( x) 3x 2 6 x 3 3(x 1) 2 dx dx
Inegrating again, we get 37. (a) cos x sin x
2 cos x
f ( x ) (x 1)3 D 4
The curve passes through (2, 1) 1
3 sec x dx
1 (2 1) D D 0 2 4
f (x) = ( x – 1)3
1 x
dx dy log tan C
33. (d) a sin and a cos 2 4 2 8
d d
dy 1 x 3
cot . log tan C
dx 2 2 8
The slope of the normal at = tan
AIEEE-2004 Solved Paper S-83
3 3 f (a )
2 2
38. (d) |1 x | dx |x 1| dx = (1 x )g{x(1 x )}dx
2 2 f ( a)
x 2 1 if x 1 I 2 I1 2I1 I 2
Now | x 2
1| 1 x if 2
1 x 1 42. (d) The required area is shown by shaded region
2 Y
x 1 if x 1
Integral is (y = 2 – x) (y = x – 2)
1 1 3
( x 2 1)dx (1 x 2 )dx (x 2 1)dx
2 1 1
1 1 3
x3 x3 x3
x x x X
3 3 3 3
2 1 1
1 8 2 27 1 Required Area
1 2 2 3 1
3 3 3 3 3 3 3 3
A | x 2 |dx 2 ( x 2)dx x2
2 2 4 2 28 2 2x 1
= 6 2
1 2 2
3 3 3 3 3
43. (c) x 2 y 2 2ay 0 ...........(1)
2
(sin x cos x)2 Differentiate,
39. (b) I dx x yy
1 sin 2x dy dy
0 2x 2 y 2a 0 a
dx dx y
We know [(sin x cos x)2 1 sin 2x] , so
2 x yy
Put in (1), x y2 2 y 0
y
2 2 2
(sin x cos x)
I dx (sin x cosx )dx (x 2 y 2 ) y' 2xy 2y 2 y' 0
(sin x cos x)
0 0
(x 2 y 2 ) y' 2xy
sin x cos x 0 if 0 x
2 44. (b) ydx (x x 2 y)dy 0
dx x dx x
x2 x2,
or I cos x sin x 2 2 dy y dy y
0
It is Bernoullis form. Divide by x2
xf (sin x )dx ( x )f (sin x)dx 2 dx 1 1
40. (b) Let I = x x 1.
0 0 dy y
1 2 dx dt
2 put x t, x
2I f (sin x )dx dy dy
.2 f (sin x )dx
2 0 dt 1 dt 1
We get, t 1 t 1
dy y dy y
2
It is linear in t.
I f (sin x )dx A
1
0 dy
Integrating factor = e y e log y
y 1
x x
e e 1
41. (d) f (x) f ( x) Solution is t ( y 1 ) ( y 1 )dy c
x x x
1 e 1 e e 1
f ( x) f ( x ) 1 x 1 1 1
. log y c log y D
f (a ) x y xy
Now I1 xg{x (1 x )}dx 45. (d) Let the vertex C be (h, k), then the
f ( a) 2 2 h 3 1 k
centroid of ABC is ,
3 3
S-84 The Pattern Target AIEEE
4 3 h 2 p 2 2hp 4kq 0
(1) passes through (4, 3) , 1
a b locus of (h, k) is
4b 3a ab .................(3) x2 p 2 2 xp 4 yq 0
Eliminating b from (2) and (3), we get
2
( x p) 4qy
a2 4 0 a 2 b 3 or 1
51. (d) Two diameters are along
Equations of straight lines are
2x 3y 1 0 and 3x y 4 0
x y x y solving we get centre (1, –1)
1 or 1
2 3 2 1 circumference = 2 r = 10
47. (c) Let the lines be y = m1x and y = m2x then r 5.
2c 1
m1 m 2 and m1m 2 Required circle is, ( x 1) 2 ( y 1) 2 52
7 7
Given m1 m 2 4 m1m 2 x 2 y 2 2x 2y 23 0
52. (d) Solving y = x and the circle
2c 4
c 2 x2 y2 2x 0, we get
7 7
48. (a) 3x 4 y 0 is one of the lines of the pair x 0, y 0 and x 1, y 1
Extremities of diameter of the required circle are
6x 2 xy 4cy 2 0 (0, 0) and (1, 1). Hence, the equation of circle is
3 ( x 0)(x 1) ( y 0)( y 1) 0
Put y x,
4
2
x 2 y2 x y 0
2 3 2 3 53. (d) Solving equations of parabolas
we get 6 x x 4c x 0
4 4
y 2 4ax and x 2 4ay
3 9c we get (0, 0) and ( 4a, 4a)
6 0 c 3
4 4 Substituting in the given equation of line
49. (b) Let the variable circle is 2bx 3cy 4d 0,
x 2 y 2 2gx 2fy c 0 ... (1) we get d = 0 and 2b +3c = 0
It passes through (a, b) d2 (2b 3c) 2 0
2 2 ... (2)
a b 2ga 2fb c 0 1 a
2 2 54. (b) e . Directrix , x 4
(1) cuts x y 4 orthogonally 2 e
2(g 0 f 0) c 4 c 4 1
a 4 2
2 2 2
from (2) a b 2ga 2fb 4 0
Locus of centre (- g,- f) is 1
b 2 1 3
4
a2 b2 2ax 2by 4 0 Equation of ellipse is
or 2ax 2by a 2 b 2 4 x 2 y2
50. (d) Let the variable circle be 1 3x 2 4 y 2 12
4 3
x2 y2 2gx 2fy c 0 ... (1) 55. (c) The direction cosines of the line are
cos , cos , cos
p 2 q 2 2gp 2fq c 0 ... (2)
Circle (1) touches x-axis, cos 2 cos 2 cos 2 1
g2 c 0 c g 2 . From (2) 2 cos 2 sin 2 3 sin 2 (given)
AIEEE-2004 Solved Paper S-85
5 a 3b 3c t c sa
or 2 x y 2 z 0 ... (2) a 2b 6c tc sa b 3c
2
Distance between (1) and (2) tc (b 3c) b 3c ( t 6)c
5 c, where t 6
8
2 21 7
= 61. (d) Resultant of forces F 7 î 2ˆj 4k̂
2 2 12 22 2 9 2
Disptacement d 4î 2 ĵ 2k̂
57. (b) Let a point on the line x y a z is
Work done F.d 28 4 8 40
( , a , ) and a point on the line x a 2y 2z
62. (c) Vectors a 2 b 3c, b 4c, and (2 1)c are
is a , , , then direction ratio of the line 1 2 3
2 2
coplanar if 0 4 0
joining these points are a, a , 0 0 2 1
2 2
If it respresents the required line, then 1
(2 1) 0 0 or
a 2
a 2 2 Forces are noncoplanar for all ,
2 1 2 1
3a, 2a except 0,
on solving we get 2
The required points of intersection are (3a, 3a-
v. u v. u
2a 2a 63. (c) Projection of v along u
|u | 2
a,3a) and 2a a , ,
2 2
w. u w.u
or (3a ,2a ,3a ) and (a , a , a ) projection of w along u
|u | 2
58. (d) The given lines are
v.u w.u
y 3 z 1 Given ....(1)
x 1 s ... (1) 2 2
Also , v.w 0 ....(2)
z 2
and 2x y 1 t ... (2) Now | u v w | 2
1
The lines are coplanar, if = | u | 2 | v |2 | w |2 2u.v 2v.w 2u.w
= 1 + 4 + 9 + 0 [ From (1) and (2)] = 14
0 ( 1) 1 3 2 ( 1)
1 0 |u v w| 14
1
1 1 1
2 64. (a) Given (a b ) c | b || c | a
3
4 3 4 3
Probability 1 1
5 4 5 4 P
I
4 1 1 3 7 R
Q
5 4 5 4 20
B C
68. (b) P(E) = P ( 2 or 3 or 5 or 7)
0.23 0.12 0.20 0.07 0.62 P Q R
A B C
P(F) P(1 or 2 or 3) Sin (90 ) sin 90 sin 90
2 2 2
0.15 0.23 0.12 0.50
A B C
P( E F) P(2 or 3) 0.23 0.12 0.35 P:Q:R cos
: cos : cos
2 2 2
P(EUF) P(E ) P(F) P(E F) 73. (c) Time taken by the particle in complete journey
0.62 0.50 0.35 0.77 12 5
T 4 hr.
69. (a) mean = np 4 and variance = npq = 2 4 5
1
p q and n 8
2
5 km
6 2
8 1 1 28
P(2 success ) C2
2 2 28 A 12km B
Y must be and
2
C(0,4)
F 2u sin
2u sin 2 2u cos
t1 and t 2
g g g
(y = mx)
4u 2
t 12 t 22
A B(3,0) g2
AIEEE - 2005
Paper - I
Time : 2½ hours Max. Marks : 450
(Note: Q1 to 10 & 76 to 85 carry 1½ marks, 11 to 65 & 86 to 140 carry 3 marks and 66 to 75 & 141 to 150 carry 4½ marks)
1 2 1 2 1 1
(c) S = ft (d) S = ft (c) s = 1 (d) s =1
4 2 n 2
n2
2. A particle is moving eastwards with a velocity of 5 ms–1. In 6. A parachutist after bailing out falls 50 m without friction. When
10 seconds the velocity changes to 5 ms–1 northwards. The parachute opens, it decelerates at 2 m/s2 . He reaches the
average acceleration in this time is ground with a speed of 3 m/s. At what height, did he bail out?
1 2 (a) 182 m (b) 91 m (c) 111 m (d) 293 m
(a) ms towards north
2 7. A bullet fired into a fixed target loses half of its velocity after
penetrating 3 cm. How much further it will penetrate before
1 2 coming to rest assuming that it faces constant resistance to
(b) ms towards north - east
2 motion ?
(a) 2.0 cm (b) 3.0 cm (c) 1.0 cm (d) 1.5 cm
1 2
(c) ms towards north - west 8. An annular ring with inner and outer radii R1 and R2 is
2 rolling without slipping with a uniform angular speed. The
(d) zero ratio of the forces experienced by the two particles situated
3. Out of the following pair , which one does NOT have identical
dimensions is on the inner and outer parts of the ring , F1 is
(a) impulse and momentum F2
(b) angular momentum and Planck’s constant
2
(c) work and torque R1 R2 R1
(a) (b) (c) (d) 1
(d) moment of inertia and moment of a force R2 R1 R2
S-88 The Pattern Target AIEEE
9. A projectile can have the same range ‘R’ for two angles of
projection. If ‘t1’ and ‘t2’ be the times of flights in the two
cases, then the product of the two time of flights is
proportional to
1 1 a
(a) (b) R2 (c) R (d)
R2 R
10. The upper half of an inclined plane with inclination is (a) g cosec (b) g / tan
perfectly smooth while the lower half is rough. A body starting
(c) g tan (d) g
from rest at the top will again come to rest at the bottom if the
15. A spherical ball of mass 20 kg is stationary at the top of a hill
coefficient of friction for the lower half is given by
of height 100 m. It rolls down a smooth surface to the ground,
(a) 2 cos (b) 2 sin then climbs up another hill of height 30 m and finally rolls
(c) tan (d) 2 tan down to a horizontal base at a height of 20 m above the
11. A mass ‘m’ moves with a velocity ‘v’ and collides inelastically ground. The velocity attained by the ball is
with another identical mass . After collision the lst mass (a) 20 m/s (b) 40 m/s
v (c) 10 30 m/s (d) 10 m/s
moves with velocity in a direction perpendicular to the
3 16. A body A of mass M while falling vertically downwards
nd
initial direction of motion. Find the speed of the 2 mass 1
under gravity breaks into two parts; a body B of mass M
after collision. 3
2
v and a body C of mass M. The centre of mass of bodies
3 3
Aafter
collision B and C taken together shifts compared to that of body A
m m towards
A before (a) does not shift
collision (b) depends on height of breaking
(c) body B
v 2
(a) 3v (b) v (c) (d) v (d) body C
3 3
17. The moment of inertia of a uniform semicircular disc of mass
12. A particle of mass 0.3 kg subject to a force F = – kx with k = M and radius r about a line perpendicular to the plane of the
15 N/m . What will be its initial acceleration if it is released disc through the centre is
from a point 20 cm away from the origin ?
2 1 1
(a) 15 m/s2 (b) 3 m/s2 (c) 10 m/s2 (d) 5 m/s2 (a) Mr 2 (b) Mr (c) Mr 2 (d) Mr 2
5 4 2
13. The block of mass M moving on the frictionless horizontal
surface collides with the spring of spring constant K and 18. The change in the value of ‘g’ at a height ‘h’ above the
compresses it by length L. The maximum momentum of the surface of the earth is the same as at a depth ‘d’ below the
block after collision is surface of earth. When both ‘d’ and ‘h’ are much smaller
than the radius of earth, then which one of the following is
correct ?
M
3h h
(a) d = (b) d = (c) d = h (d) d =2 h
2 2
19. A particle of mass 10 g is kept on the surface of a uniform
KL 2 sphere of mass 100 kg and radius 10 cm. Find the work to be
(a) (b) MK L done against the gravitational force between them to take
2M
the particle far away from the sphere (you may take G =
ML2 11
(c) (d) zero 6.67× 10 Nm 2 / kg 2 )
K
10 10
14. A block is kept on a frictionless inclined surface with angle (a) 3.33 × 10 J (b) 13.34 × 10 J
of inclination ‘ ’ . The incline is given an acceleration ‘a’ to 10 9
keep the block stationary. Then a is equal to (c) 6.67 × 10 J (d) 6.67 × 10 J
AIEEE-2005 Solved Paper S-89
20. A ‘T’ shaped object with dimensions shown in the figure, is 27. The figure shows a system of two concentric spheres of
radii r 1 and r 2 are kept at temperatures T 1 and T 2 ,
lying on a smooth floor. A force ‘ F ’ is applied at the point
respectively. The radial rate of flow of heat in a substance
P parallel to AB, such that the object has only the translational between the two concentric spheres is proportional to
motion without rotation. Find the location of P with respect
to C.
l r1
T1
A B
r2 T2
P
2l
F r (r2 r1 )
(a) In 2 (b)
r1 (r1 r2 )
r1 r2
C (c) ( r2 r1 ) (d)
(r2 r1 )
3 2 4
(a) (b) (c) (d) 28. A system goes from A to B via two processes I and II as
2 3 3
shown in figure. If U1 and U 2 are the changes in
21. Which of the following is incorrect regarding the first law
internal energies in the processes I and II respectively, then
of thermodynamics?
(a) It is a restatement of the principle of conservation of
p II
energy
(b) It is applicable to any cyclic process A B
(c) It introduces the concept of the entropy I
(d) It introduces the concept of the internal energy
22. Consider a car moving on a straight road with a speed of 100 v
m/s . The distance at which car can be stopped is [ k 0.5 ] (a) relation between U1 and U 2 can not be determined
(a) 1000 m (b) 800 m (c) 400 m (d) 100 m
23. A body of mass m is accelerated uniformly from rest to a (b) U1 = U 2
speed v in a time T . The instantaneous power delivered to (c) U 2 < U1
the body as a function of time is given by
(d) U 2 > U1
mv 2 2 mv 2
(a) .t (b) .t 29. The temperature-entropy diagram of a reversible engine
T2 T2 cycle is given in the figure. Its efficiency is
1 mv 2 2 1 mv 2 T
(c) .t (d) .t
2 T2 2 T2
24. Average density of the earth 2T0
(a) is a complex function of g T0
(b) does not depend on g
(c) is inversely proportional to g
S
(d) is directly proportional to g S0 2S0
25. If ‘S’ is stress and ‘Y’ is young’s modulus of material of a
wire, the energy stored in the wire per unit volume is 1 2
1 1
2
(a) (b) (c) (d)
S S 2Y 4 2 3 3
(a) (b) 2S 2 Y (c) (d)
2Y 2Y S2 27
30. If radius of the 13 Al nucleus is estimated to be 3.6 fermi
26. A 20 cm long capillary tube is dipped in water. The water
rises up to 8 cm. If the entire arrangement is put in a freely then the radius of 125 nucleus be nearly
52 Te
falling elevator the length of water column in the capillary
(a) 8 fermi (b) 6 fermi
tube will be
(c) 5 fermi (d) 4 fermi
(a) 10 cm (b) 8 cm (c) 20 cm (d) 4 cm
S-90 The Pattern Target AIEEE
38. If the kinetic energy of a free electron doubles, it’s deBroglie
7
66
31. Starting with a sample of pure Cu ,
of it decays into wavelength changes by the factor
8
Zn in 15 minutes. The corresponding half life is 1 1
(a) 2 (b) (c) 2 (d)
(a) 15 minutes (b) 10 minutes 2 2
(c)
1
minutes (d) 5 minutes 39. A nuclear transformation is denoted by X (n, ) 73 Li . Which
7
2 of the following is the nucleus of element X ?
32. A photocell is illuminated by a small bright source placed 1 10 12 11 9
(a) 5 Be (b) C6 (c) 4 Be (d) 5B
1
m away. When the same source of light is placed m away,, 40. In a full wave rectifier circuit operating from 50 Hz mains
2 frequency, the fundamental frequency in the ripple would
the number of electrons emitted by photocathode would be
(a) increase by a factor of 4 (a) 25 Hz (b) 50 Hz (c) 70.7 Hz (d) 100 Hz
(b) decrease by a factor of 4 41. Two simple harmonic motions are represented by the
(c) increase by a factor of 2
(d) decrease by a factor of 2 equations y1 = 0.1 sin 100 t and y 2 = 0.1 cos t .
3
33. The electrical conductivity of a semiconductor increases
when electromagnetic radiation of wavelength shorter than The phase difference of the velocity of particle 1 with respect
2480 nm is incident on it. The band gap in (eV) for the to the velocity of particle 2 is
semiconductor is
(a) 2.5 eV (b) 1.1 eV (c) 0.7 eV (d) 0.5 eV (a) (b) (c) (d)
3 6 6 3
34. The intensity of gamma radiation from a given source is I.
42. A fish looking up through the water sees the outside world
I contained in a circular horizon. If the refractive index of
On passing through 36 mm of lead, it is reduced to . The
8 4
water is and the fish is 12 cm below the surface, the radius
I 3
thickness of lead which will reduce the intensity to will
2 of this circle in cm is
be 36
(a) 9 mm (b) 6mm (c) 12mm (d) 18mm (a) (b) 36 7 (c) 4 5 (d) 36 5
7
35. A gaseous mixture consists of 16 g of helium and 16 g of
43. Two point white dots are 1 mm apart on a black paper. They
Cp are viewed by eye of pupil diameter 3 mm. Approximately,
oxygen. The ratio of the mixture is
Cv what is the maximum distance at which these dots can be
resolved by the eye? [Take wavelength of light = 500 nm]
(a) 1.62 (b) 1.59 (c) 1.54 (d) 1.4 44. A thin glass (refractive index 1.5) lens has optical power of
36. In a common base amplifier, the phase difference between – 5 D in air. Its optical power in a liquid medium with refractive
the input signal voltage and output voltage is index 1.6 will be
(a) – 1D (b) 1 D (c) – 25 D (d) 25 D
(a) (b) (c) (d) 0
4 2 45. The function sin 2 ( t ) represents
37. The diagram shows the energy levels for an electron in a (a) a periodic, but not simple harmonic motion with a period
certain atom. Which transition shown represents the
emission of a photon with the most energy?
46. A Young’s double slit experiment uses a monochromatic 54. Two thin wire rings each having a radius R are placed at a
source. The shape of the interference fringes formed on a distance d apart with their axes coinciding. The charges on
screen is the two rings are +q and -q. The potential difference between
(a) circle (b) hyperbola the centres of the two rings is
(c) parabola (d) straight line
47. The bob of a simple pendulum is a spherical hollow ball q 1 1
(a)
filled with water. A plugged hole near the bottom of the 2 0 R R 2
d2
oscillating bob gets suddenly unplugged. During
observation, till water is coming out, the time period of
qR
oscillation would (b)
4 0 d2
(a) first decrease and then increase to the original value
(b) first increase and then decrease to the original value
(c) increase towards a saturation value q 1 1
(c)
(d) remain unchanged 4 0 R R2 d2
48. If I 0 is the intensity of the principal maximum in the single (4) zero
slit diffraction pattern, then what will be its intensity when 55. A parallel plate capacitor is made by stacking n equally
the slit width is doubled? spaced plates connected alternatively. If the capacitance
I0 between any two adjacent plates is ‘C’ then the resultant
(a) 4 I 0 (b) 2 I 0 (c) (d) I0 capacitance is
2
(a) (n + 1) C (b) (n – 1) C
49. When two tuning forks (fork 1 and fork 2) are sounded
(c) nC (d) C
simultaneously, 4 beats per second are heard. Now, some
tape is attached on the prong of the fork 2. When the tuning 56. A charged ball B hangs from a silk thread S, which makes an
forks are sounded again, 6 beats per second are heard. If the angle with a large charged conducting sheet P, as shown
frequency of fork 1 is 200 Hz, then what was the original in the figure. The surface charge density of the sheet is
frequency of fork 2? proportional to
(a) 202 Hz (b) 200 Hz (c) 204 Hz (d) 196 Hz
d2x
50. If a simple harmonic motion is represented by x 0, P
dt 2
its time period is S
2 2
(a) (b) (c) 2 (d) 2
B
51. An observer moves towards a stationary source of sound,
(a) cot (b) cos (c) tan (d) sin
with a velocity one-fifth of the velocity of sound. What is
the percentage increase in the apparent frequency ? 57. One conducting U tube can slide inside another as shown
(a) 0.5% (b) zero (c) 20 % (d) 5 % in figure, maintaining electrical contacts between the tubes.
The magnetic field B is perpendicular to the plane of the
52. Two point charges + 8q and – 2q are located at x = 0 and x =
figure . If each tube moves towards the other at a constant
L respectively. The location of a point on the x axis at which
speed v, then the emf induced in the circuit in terms of B, l
the net electric field due to these two point charges is zero is
and v where l is the width of each tube, will be
L
(a) (b) 2 L (c) 4 L (d) 8 L X X
4 X A X
BX
53. When an unpolarized light of intensity I 0 is incident on a X v X v X
polarizing sheet, the intensity of the light which does not X X X
X
get transmitted is X C
X X
1 1
(a) I0 (b) I0 (c) I0 (d) zero
4 2 (a) – Blv (b) Blv (c) 2 Blv (d) zero
S-92 The Pattern Target AIEEE
58. A fully charged capacitor has a capacitance ‘C’. It is 64. Two voltameters, one of copper and another of silver, are
discharged through a small coil of resistance wire embedded joined in parallel. When a total charge q flows through the
in a thermally insulated block of specific heat capacity ‘s’ voltameters, equal amount of metals are deposited. If the
and mass ‘m’. If the temperature of the block is raised by electrochemical equivalents of copper and silver are z 1and
‘ T ’, the potential difference ‘V’ across the capacitance is z2 respectively the charge which flows through the silver
voltameter is
mC T 2mC T
(a) (b) q q
s s (a) (b)
z z
1 2 1 1
2ms T ms T z1 z2
(c) (d)
C C
z2 z
59. Two thin, long, parallel wires, separated by a distance ‘d’ (c) q (d) q 1
z1 z2
carry a current of ‘i’ A in the same direction. They will
65. Two concentric coils each of radius equal to 2 cm are placed
2
(a) repel each other with a force of 0i /(2 d) at right angles to each other. 3 ampere and 4 ampere are the
2
currents flowing in each coil respectively. The magnetic
(b) attract each other with a force of 0i /(2 d) induction in Weber/m2 at the centre of the coils will be
2
(c) repel each other with a force of 0i /(2 d 2 ) 0 4 10 7
Wb / A.m
2
(d) attract each other with a force of 0i /(2 d 2 ) (a) 10 5
(b) 12 10 5
60. A heater coil is cut into two equal parts and only one part is (c) 7 10
5
(d) 5 10 5
now used in the heater. The heat generated will now be
66. In a potentiometer experiment the balancing with a cell is at
(a) four times (b) doubled length 240 cm. On shunting the cell with a resistance of
(c) halved (d) one fourth
2 , the balancing length becomes 120 cm. The internal
61. In the circuit , the galvanometer G shows zero deflection. If resistance of the cell is
the batteries A and B have negligible internal resistance, the
(a) 0.5 (b) 1 (c) 2 (d) 4
value of the resistor R will be -
67. A charged particle of mass m and charge q travels on a
500
G circular path of radius r that is perpendicular to a magnetic
field B. The time taken by the particle to complete one
2V revolution is
12V R
B A 2 q 2B 2 mq
(a) (b)
m B
(a) 100 (b) 200 (c) 1000 (d) 500 2 m 2 qB
(c) (d)
62. A moving coil galvanometer has 150 equal divisions. Its qB m
current sensitivity is 10-divisions per milliampere and voltage 68. A magnetic needle is kept in a non-uniform magnetic field. It
sensitivity is 2 divisions per millivolt. In order that each
experiences
division reads 1 volt, the resistance in ohms needed to be
(a) neither a force nor a torque
connected in series with the coil will be -
(b) a torque but not a force
(a) 105 (b) 103 (c) 9995 (d) 99995
(c) a force but not a torque
63. Two sources of equal emf are connected to an external
(d) a force and a torque
resistance R. The internal resistance of the two sources are
69. The resistance of hot tungsten filament is about 10 times
R1 and R 2 (R 2 R1 ) . If the potential difference across the cold resistance. What will be the resistance of 100 W
the source having internal resistance R 2 is zero, then and 200 V lamp when not in use ?
(a) 20 (b) 40 (c) 200 (d) 400
(a) R= R 2 R1
70. The self inductance of the motor of an electric fan is 10 H. In
(b) R= R 2 (R1 R 2 ) /(R 2 R1 ) order to impart maximum power at 50 Hz, it should be
(c) R= R 1R 2 /( R 2 R 1 ) connected to a capacitance of
(a) 8 F (b) 4 F (c) 2 F (d) 1 F
(d) R= R 1R 2 /( R 1 R 2 )
AIEEE-2005 Solved Paper S-93
71. An energy source will supply a constant current into the 82. For a spontaneous reaction the G, equilibrium constant
load if its internal resistance is
(a) Very large as compared to the load resistance (K) and E oCell will be respectively
(b) equal to the resistance of the load (a) –ve, >1, –ve (b) –ve, <1, –ve
(c) non-zero but less than the resistance of the load (c) +ve, >1, –ve (d) –ve, >1, +ve
(d) zero 83. Which of the following is a polyamide?
72. The phase difference between the alternating current and (a) Bakelite (b) Terylene
(c) Nylon-66 (d) Teflon
emf is . Which of the following cannot be the constituent 84. Which one of the following types of drugs reduces fever ?
2
of the circuit? (a) Tranquiliser (b) Antibiotic
(a) R, L (b) C alone (c) Antipyretic (d) Analgesic
(c) L alone (d) L, C 85. Due to the presence of an unpaired electron, free radicals
73. A circuit has a resistance of 12 ohm and an impedance of 15 are:
ohm. The power factor of the circuit will be – (a) Cations (b) Anions
(a) 0.4 (b) 0.8 (c) 0.125 (d) 1.25 (c) Chemically inactive (d) Chemically reactive
74. A uniform electric field and a uniform magnetic field are 86. Lattice energy of an ionic compound depends upon
acting along the same direction in a certain region. If an (a) Charge on the ion and size of the ion
electron is projected along the direction of the fields with a (b) Packing of ions only
certain velocity then (c) Size of the ion only
(a) its velocity will increase (d) Charge on the ion only
(b) Its velocity will decrease 87. The highest electrical conductivity of the following aqueous
(c) it will turn towards left of direction of motion solutions is of
(d) it will turn towards right of direction of motion (a) 0.1 M difluoroacetic acid
75. A coil of inductance 300 mH and resistance 2 is connected (b) 0.1 M fluoroacetic acid
to a source of voltage 2 V. The current reaches half of its (c) 0.1 M chloroacetic acid
steady state value in
(d) 0.1 M acetic acid
(a) 0.1 s (b) 0.05 s (c) 0.3 s (d) 0.15 s
88. Aluminium oxide may be electrolysed at 1000°C to furnish
aluminium metal (At. Mass = 27 amu; 1 Faraday = 96,500
Section - 2 Coulombs). The cathode reaction is– Al
3
3e Al
To prepare 5.12 kg of aluminium metal by this method would
require
(a) 5.49 × 101 C of electricity
76. Which of the following oxides is amphoteric in character?
(b) 5.49 × 104 C of electricity
(a) SnO 2 (b) SiO 2 (c) CO 2 (d) CaO
(c) 1.83 × 107 C of electricity
77. Which of the following species is diamagnetic in nature? (d) 5.49 × 107 C of electricity
(a) H2 (b) H 2 (c) (d) 89. Consider an endothermic reaction X Y with the
H2 He 2
activation energies E b and E f for the backward and
78. If is the degree of dissociation of Na 2SO 4 , the Vant
forward reactions, respectively. In general
Hoff’s factor (i) used for calculating the molecular mass is
(a) 1 – 2 (b) 1 + 2 (c) 1 – (d) 1 + (a) there is no definite relation between E b and E f
(b) Eb = Ef
79. The oxidation state Cr in [Cr ( NH 3 ) 4 Cl 2 ] is
(c) Eb > Ef
(a) 0 (b) + 1 (c) + 2 (d) + 3 (d) Eb < Ef
80. Hydrogen bomb is based on the principle of
90. Consider the reaction : N 2 3H 2 2 NH 3 carried out at
(a) artificial radioactivity (b) nuclear fusion
(c) natural radioactivity (d) nuclear fission constant temperature and pressure. If H and U are the
81. An ionic compound has a unit cell consisting of A ions at the enthalpy and internal energy changes for the reaction, which
corners of a cube and B ions on the centres of the faces of the of the following expressions is true ?
cube. The empirical formula for this compound would be
(a) H U (b) H U
(a) A 3B (b) AB3 (c) A 2B (d) AB
(c) H U (d) H 0
S-94 The Pattern Target AIEEE
91. Which one of the following statements is NOT true about (a) Whether Kp is greater than, less than or equal to Kc
the effect of an increase in temperature on the distribution depends upon the total gas pressure
of molecular speeds in a gas? (b) KP = Kc
(a) The area under the distribution curve remains the same (c) KP is less than Kc
as under the lower temperature (d) KP is greater than Kc
(b) The distribution becomes broader 98. Hydrogen ion concentration in mol/L in a solution of
(c) The fraction of the molecules with the most probable pH = 5.4 will be :
speed increases (a) 3.98 × 10–6 (b) 3.68 × 10–6
(d) The most probable speed increases (c) 3.88 × 106 (d) 3.98 × 108
92. The volume of a colloidal particle, VC as compared to the 99. A reaction involving two different reactants can never be
volume of a solute particle in a true solution VS, could be (a) bimolecular reaction (b) second order reaction
VC ~ 3 VC ~ 3 (c) first order reaction (d) unimolecular reaction
(a) 10 (b) 10
VS VS 100. If we consider that 1/6, in place of 1/12, mass of carbon atom
is taken to be the relative atomic mass unit, the mass of one
VC ~ 23 VC ~ mole of the substance will
(c) 10 (d) 1
VS VS (a) be a function of the molecular mass of the substance
(b) remain unchanged
93. The solubility product of a salt having general formula MX2,
(c) increase two fold
in water is : 4 × 10–12. The concentration of M2+ ions in the
aqueous solution of the salt is (d) decrease twice
(a) 4.0 × 10–10 M (b) 1.6 × 10–4 M 101. In a multi-electron atom, which of the following orbitals
described by the three quantum members will have the same
(c) 1.0 × 10–4 M (d) 2.0 × 10–6 M
energy in the absence of magnetic and electric fields ?
94. Benzene and toluene form nearly ideal solution. At 20°C,
A. n = 1, l = 0, m = 0 B. n = 2, l = 0, m = 0
the vapour pressure of benzene is 75 torr and that of toluene
is 22 torr. The partial vapour pressure of benzene at 20°C for C. n = 2, l = 1, m = 1 D n = 3, l = 2, m = 1
a solution containing 78 g of benzene and 46 g of toluene in E. n = 3, l = 2, m = 0
torr is (a) (D) and (E) (b) (C) and (D)
(a) 53.5 (b) 37.5 (c) 25 (d) 50 (c) (B) and (C) (d) (A) and (B)
95. The exothermic formation of CIF3 is represented by the 102. During the process of electrolytic refining of copper, some
metals present as impurity settle as ‘anode mud’. These are
equation :
(a) Fe and Ni (b) Ag and Au
CI 2(g ) 3F2(g ) 2ClF3 (g ) (c) Pb and Zn (d) Sn and Ag
mixture?
(a) 2.70 M (b) 1.344 M (c) 1.50 M (d) 1.20 M
97. For the reaction 2.0
1.5 10 3 1 (K 1 ) 2.0 10 3
2 NO 2(g) 2NO (g ) O 2( g ) , T
The reaction must be
6
(K c 1.8 10 at 184 C) (a) highly spontaneous at ordinary temperature
(R = 0.0831 kJ/ (mol. K)) (b) one with negligible enthalpy change
When K p and K c are compared at 184°C, it is found that (c) endothermic
(d) exothermic
AIEEE-2005 Solved Paper S-95
105. The disperse phase in colloidal iron (III) hydroxide and 116. In which of the following arrangements, the order is NOT
colloidal gold is positively and negatively charged, according to the property indicated against it?
respectively. Which of the following statements is NOT (a) Li < Na < K < Rb :
correct? Increasing metallic radius
(a) Coagulation in both sols can be brought about by (b) I < Br < F < Cl :
electrophoresis Increasing electron gain enthalpy
(b) Mixing the sols has no effect (with negative sign)
(c) Sodium sulphate solution causes coagulation in both (c) B < C < N < O
sols Increasing first ionization enthalpy
(d) Magnesium chloride solution coagulates, the gold sol (d) Al3+ < Mg2+ < Na+ < F–
more readily than the iron (III) hydroxide sol Increasing ionic size
106. Based on lattice energy and other considerations which 117. In silicon dioxide
one of the following alkali metal chlorides is expected to
(a) there are double bonds between silicon and oxygen
have the highest melting point ?
atoms
(a) RbCl (b) KCl (c) NaCl (d) LiCl
(b) silicon atom is bonded to two oxygen atoms
107. Heating mixture of Cu2O and Cu2S will give
(c) each silicon atom is surrounded by two oxygen atoms
(a) Cu2SO3 (b) CuO + CuS and each oxygen atom is bonded to two silicon atoms
(c) Cu + SO3 (d) Cu + SO2 (d) each silicon atom is surrounded by four oxygen atoms
108. The molecular shapes of SF4, CF4 and XeF4 are and each oxygen atom is bonded to two silicon atoms.
(a) different with 1, 0 and 2 lone pairs of electrons on the 118. Of the following sets which one does NOT contain
central atom, respectively isoelectronic species?
(b) different with 0, 1 and 2 lone pairs of electrons on the
central atom, respectively (a) BO 33 , CO 32 , NO 3 (b) SO 32 , CO 32 , NO 3
(c) the same with 1, 1 and 1 lone pair of electrons on the 2
central atoms, respectively (c) CN , N 2 , C 2 (d) PO 34 , SO 24 , ClO 4
(d) the same with 2, 0 and 1 lone pairs of electrons on the 119. The lanthanide contraction is responsible for the fact that
central atom, respectively (a) Zr and Zn have the same oxidation state
109. The number and type of bonds between two carbon atoms (b) Zr and Hf have about the same radius
in calcium carbide are (c) Zr and Nb have similar oxidation state
(a) Two sigma, two pi (b) Two sigma, one pi (d) Zr and Y have about the same radius
(c) One sigma, two pi (d) One sigma, one pi 120. The IUPAC name of the coordination compound
110. The oxidation state of chromium in the final product formed K3[Fe(CN)6] is
by the reaction between Kl and acidified potassium (a) Tripotassium hexacyanoiron (II)
dichromate solution is and acidified potassium dichromate (b) Potassium hexacyanoiron (II)
solution is (c) Potassium hexacyanoferrate (III)
(a) + 3 (b) + 2 (c) + 6 (d) + 4 (d) Potassium hexacyanoferrate (II)
111. The number of hydrogen atom(s) attached to phosphorus 121. Which of the following compounds shows optical isomerism?
atom in hypophosphorous acid is
(a) three (b) one (c) two (d) zero (a) [Co(CN) 6 ]3 (b) [Cr(C 2 O 4 )3 ]3
O
(b) C'1 and C'5 respectively of the sugar molecule
||
||
R–C + Nu R–C +X
(c) C'2 and C'5 respectively of the sugar molecule
X Nu
(d) C'5 and C'2 respectively of the sugar molecule
is fastest when X is
128. Reaction of one molecule of HBr with one molecule of
(a) OCOR (b) OC2H5 (c) NH2 (d) Cl
1, 3-butadiene at 40°C gives predominantly
139. Elimination of bromine from 2-bromobutane results in the
(a) 1-bromo-2-butene under kinetically controlled
formation of –
conditions
(a) Predominantly 2-butyne
(b) 3-bromobutene under thermodynamically controlled
conditions (b) Predominantly 1-butene
(c) 1-bromo-2-buteneunder (c) Predominantly 2-butene
thermodynamically controlled conditions (d) equimolar mixture of 1 and 2-butene
(d) 3-bromobutene under kinetically controlled conditions 140. Equimolar solutions in the same solvent have
129. Among the following acids which has the lowest pKa value? (a) Different boiling and different freezing points
(b) Same boiling and same freezing points
(a) CH 3CH 2 COOH (b) (CH 3 ) 2 CH COOH
(c) Same freezing point but different freezing
(c) HCOOH (d) CH3COOH (d) Same boiling point but different freezing points
130. The decreasing order of nucleophilicity among the 141. Which of the following statements in relation to the
nucleophiles hydrogen atom is correct ?
(a) CH 3C O (b) (c) CN (a) 3s, 3p and 3d orbitals all have the same energy
CH 3 O
|| (b) 3s and 3p orbitals are of lower energy than 3d orbital
O (c) 3p orbital is lower in energy than 3d orbital
(d) 3s orbital is lower in energy than 3p orbital
O
|| 142. The structure of diborane (B2H6) contains
(d) H C
3 S – O – is (a) four 2c-2e bonds and four 3c-2e bonds
||
O (b) two 2c-2e bonds and two 3c-3e bonds
(c) two 2c-2e bonds and four 3c-2e bonds
(a) (c), (b), (a), (d) (b) (b), (c), (a), (d)
(d) four 2c-2e bonds and two 3c-2e bonds
(c) (d), (c), (b), (a) (d) (a), (b), (c), (d)
143. The value of the ‘spin only’ magnetic moment for one of the
131. Which one of the following methods is neither meant for
following configurations is 2.84 BM. The correct one is
the synthesis nor for separation of amines?
(a) d5 (in strong ligand field)
(a) Curtius reaction (b) Wurtz reaction
(b) d3 (in weak as well as in strong fields)
(c) Hofmann method (d) Hinsberg method
(c) d4 (in weak ligand fields)
132. Which of the following is fully fluorinated polymer ?
(d) d4 (in strong ligand fields)
(a) PVC (b) Thiokol (c) Teflon (d) Neoprene
AIEEE-2005 Solved Paper S-97
144. Which of the following factors may be regarded as the main 147. An organic compound having molecular mass 60 is found
cause of lanthanide contraction? to contain C = 20%, H = 6.67% and N = 46.67% while rest is
(a) Greater shielding of 5d electrons by 4f electrons oxygen. On heating it gives NH3 alongwith a solid residue.
(b) Poorer shielding of 5d electrons by 4f electrons The solid residue give violet colour with alkaline copper
(c) Effective shielding of one of 4f electrons by another in sulphate solution. The compound is
the subshell (a) CH3CH2CONH2 (b) (NH2)2CO
(d) Poor shielding of one of 4f electron by another in the (c) CH3CONH2 (d) CH3NCO
subshell 148. If the bond dissociation energies of XY, X2 and Y2 (all
145. Reaction of cyclohexanone with dimethylamine in the diatomic molecules) are in the ratio of 1 : 1 : 0.5 and Hf for
presence of catalytic amount of an acid forms a compound the formation of XY is – 200 kJ mol–1. The bond dissociation
if water during the reaction is continuously removed. The energy of X 2 will be
compound formed is generally known as (a) 400 kJ mol–1 (b) 300 kJ mol–1
(a) an amine (b) an imine (c) 200 kJ mol –1 (d) 100 kJ mol–1
(c) an anemine (d) a Schiff’s base 149. t1/4 can be taken as the time taken for the concentration of
146. p -cresol reacts with chloroform in alkaline medium to give a reactant to drop to 3 / 4 of its initial value. If the rate
the compound A which adds hydrogen cyanide to form, the constant for a first order reaction is K, the t1/4 can be written
compound B. The latter on acidic hydrolysis gives chiral as
carboxylic acid. The structure of the carboxylic acid is (a) 0.75/K (b) 0.69/K
CH 3 CH 3 (c) 0.29/K (d) 0.10/K
CH 2 COOH 150. An amount of solid NH 4 HS is placed in a flask already
(a) (b) containing ammonia gas at a certain temperature and 0.50
CH 2 COOH atm pressure. Ammonium hydrogen sulphide decomposes
OH OH to yield NH3 and H2 S gases in the flask. When the
decomposition reaction reaches equilibrium, the total
CH 3 CH 3 pressure in the flask rises to 0.84 atm? The equilibrium
constant for NH4HS decomposition at this temperature is
CH(OH)COOH
(c) (d) (a) 0.11 (b) 0.17
CH(OH)COOH (c) 0.18 (d) 0.30
OH OH
Paper - II
Time : 1¼ hours Max. Marks : 225
1. If A2 – A + I = 0, then the inverse of A is 4. Area of the greatest rectangle that can be inscribed in the
(a) A + I (b) A (c) A – I (d) I – A x2 y2
ellipse = 1 is
2. If the cube roots of unity are 1, , 2 then the roots of the a2 b2
a
equation ( x – 1) 3 + 8 = 0, are (a) 2ab (b) ab (c) ab (d)
b
(a) –1, –1 + 2 , – 1 – 2 2 (b) –1, – 1, – 1 5. The differential equation representing the family of curves
(c) – 1, 1 – 2 , 1 – 2 2 (d) – 1, 1 + 2 , 1 + 2 2
y 2 2c ( x c ) , where c > 0, is a parameter, is of order and
3. Let R = {(3, 3), (6, 6), (9, 9), (12, 12), (6, 12), (3, 9), degree as follows :
(3, 12), (3, 6)} be a relation on the set (a) order 1, degree 2 (b) order 1, degree 1
(c) order 1, degree 3 (d) order 2, degree 2
A = {3, 6, 9, 12}. The relation is
1 1 2 4 1
(a) reflexive and transitive only 6. Lim sec 2 sec 2 .......... .. sec 2 1 equals
n 2 2 2 2 n
n n n n
(b) reflexive only
1 1
(c) an equivalence relation (a) sec 1 (b) cosec 1
2 2
(d) reflexive and symmetric only
1
(c) tan 1 (d) tan 1
2
S-98 The Pattern Target AIEEE
7. ABC is a triangle. Forces P , Q , R acting along IA, IB, and 15. If the roots of the equation x 2 – bx + c = 0 be two
IC respectively are in equilibrium, where I is the incentre of consecutive integers, then b 2 – 4c equals
ABC. Then P : Q : R is (a) – 2 (b) 3 (c) 2 (d) 1
(a) sin A : sin B : sin C 16. If the letters of the word SACHIN are arranged in all possible
A B C ways and these words are written out as in dictionary, then
(b) sin : sin : sin
2 2 2 the word SACHIN appears at serial number
A B C (a) 601 (b) 600 (c) 603 (d) 602
(c) cos : cos : cos
2 2 2 50
6
56 r
17. The value of C4 C 3 is
(d) cos A : cos B : cos C r 1
8. If in a frequency distribution, the mean and median are 21
and 22 respectively, then its mode is approximately 55 55 56 56
(a) C4 (b) C3 (c) C3 (d) C4
(a) 22.0 (b) 20.5 (c) 25.5 (d) 24.0
1 0 1 0
9. Let P be the point ( 1, 0 ) and Q a point on the locus y 2 8x . 18. If A = and I = , then which one of the
1 1 0 1
The locus of mid point of PQ is
following holds for all n 1, by the principle of mathematical
(a) y 2 – 4x + 2 = 0 (b) y 2 + 4x + 2 = 0 induction
(c) x 2 + 4y + 2 = 0 (d) x 2 – 4y + 2 = 0 (a) A n = nA – (n – 1) I (b) A n = 2 n 1 A – (n – 1) I
10. If C is the mid point of AB and P is any point outside AB,
then (c) A n = nA + (n – 1) I (d) A n = 2 n 1 A + (n – 1) I
11
(a) PA + PB = 2 PC (b) PA + PB = PC 1
19. If the coefficient of x 7 in ax 2 equals the
bx
(c) PA + PB + 2 PC = 0 (d) PA + PB + PC = 0
11. If the coefficients of rth, (r + 1)th, and (r + 2)th terms in the 11
7 1
m coefficient of x in ax , then a and b satisfy
the binomial expansion of (1 y) are in A.P., then m are r
bx 2
satisfy the equation
the relation
2
(a) m 2 – m (4r – 1) + 4 r – 2 = 0 (a) a – b = 1 (b) a + b = 1
2 2
(b) m – m (4r + 1) + 4 r + 2 = 0 a
(c) =1 (d) ab = 1
(c) 2 2
m – m (4r + 1) + 4 r – 2 = 0 b
2 20. Let f : (– 1, 1) B, be a function defined by
(d) m 2 – m (4r – 1) + 4 r + 2 = 0
1 2x
P Q f (x) = tan ,
12. In a triangle PQR, R=
2
. If tan
2
and – tan
2
are 1 x2
then f is both one - one and onto when B is the interval
2
the roots of ax + bx + c = 0, a 0 then
(a) a = b + c (b) c = a + b (a) 0, (b) 0, (c) , (d) ,
(c) b = c (d) b = a + c 2 2 2 2 2 2
13. The system of equations
21. If z1 and z 2 are two non- zero complex numbers such that
x+y +z= –1
x + y+ z = – 1 | z1 z 2 | = | z1 | + | z 2 | , then arg z1 – arg z 2 is equal to
x+ y+ z = –1
has n solution, if is (a) (b) – (c) 0 (d)
2 2
(a) – 2 (b) either – 2 or 1
(c) not – 2 (d) 1 z
14. The value of a for which the sum of the squares of the roots 22. If and | | = 1, then z lies on
1
z i
of the equation x 2 – (a – 2) x – a – 1 = 0 assume the least 3
value is (a) an ellipse (b) a circle
(a) 1 (b) 0 (c) 3 (d) 2 (c) a straight line (d) a parabola
AIEEE-2005 Solved Paper S-99
3 2 x 3 2
(c) x (d) x
(c) it passes through a , a 8 2 8
2
(d) It is at a constant distance from the origin 31. If x = an , y = bn , z = c n where a, b, c are in A.P
25. A function is matched below against an interval where it is n 0 n 0 n 0
supposed to be increasing. Which of the following pairs is and |a | < 1, | b | < 1, | c | < 1 then x, y, z are in
incorrectly matched? (a) G. P. (b) A.P.
Interval Function (c) Arithmetic - Geometric Progression
(a) (– , ) x 3 3x 2 3x 3 (d) H.P.
39. The line parallel to the x- axis and passing through the 2 1
(a) (b) 2 1
intersection of the lines ax + 2by + 3b = 0 and bx – 2ay – 3a 4 4
= 0, where (a, b) (0, 0) is
3 (c) 1 2 (d) 1 2
(a) below the x - axis at a distance of from it 4 4
2
48. The distance between the line 55. An ellipse has OB as semi minor axis, F and F' its focii and
r 2 î 2ˆj 3k̂ (i j 4 k ) and the plane the angle FBF' is a right angle. Then the eccentricity of the
ellipse is
r . ( î 5 ĵ k̂ ) 5 is 1 1
1 1
(a) (b) (c) (d)
10 10 3 10 2 2 4 3
(a) (b) (c) (d)
9 3 3 10 3 56. Let a, b and c be distinct non- negative numbers. If the
49. For any vector a , vectors a î aĵ ck̂ , î k̂ and c î cˆj bk̂ lie in a plane,
then c is
the value of (a î ) 2 (a ĵ) 2 (a k̂ ) 2 is equal to (a) the Geometric Mean of a and b
2 2
(b) the Arithmetic Mean of a and b
2 2
(a) 3a (b) a (c) 2a (d) 4a (c) equal to zero
50. If non zero numbers a, b, c are in H.P., then the straight line (d) the Harmonic Mean of a and b
x y 1 57. If a , b , c are non coplanar vectors and is a real number
0 always passes through a fixed point. That
a b c then
point is
2
[ (a b ) b c ] = [a b c b ] for
1
(a) (–1, 2) (b) (–1, –2) (c) (1, – 2) (d) 1, (a) exactly one value of
2
(b) no value of
51. If a vertex of a triangle is (1, 1) and the mid points of two
sides through this vertex are (–1, 2) and (3, 2) then the (c) exactly three values of
centroid of the triangle is (d) exactly two values of
7 1 7 7 1 7 58. Let a = î k̂ , b = x î ĵ + (1 – x) k̂ and
(a) 1, (b) , (c) 1, (d) ,
3 3 3 3 3 3
c = y î + x ĵ + (1 + x – y) k̂ . Then [ a , b , c ] depends on
2 2 2 2
52. If the circles x y + 2ax + cy + a = 0 and x y – 3ax + (a) only y (b) only x
dy – 1 = 0 intersect in two distinct points P and Q then the (c) both x and y (d) neither x nor y
line 5x + by – a = 0 passes through P and Q for 59. Three houses are available in a locality. Three persons apply
(a) exactly one value of a for the houses. Each applies for one house without
consulting others. The probability that all the three apply
(b) no value of a
for the same house is
(c) infinitely many values of a
2 1 8 7
(d) exactly two values of a (a) (b) (c) (d)
9 9 9 9
53. A circle touches the x- axis and also touches the circle with
centre at (0,3 ) and radius 2. The locus of the centre of the 60. A random variable X has Poisson distribution with mean 2.
circle is Then P (X > 1.5) equals
(a) an ellipse (b) a circle 2 3 3
(a) (b) 0 (c) 1 (d)
(c) a hyperbola (d) a parabola e 2
e 2 e2
54. If a circle passes through the point (a, b) and cuts the circle
1
x2 y2 p 2 orthogonally, then the equation of the locus 61. Let A and B be two events such that P( A B) ,
6
of its centre is
1 1
(a) x2 y 2 – 3ax – 4by + ( a 2 b2 p2 ) = 0 P( A B) an d P(A ) , where A stands for
4 4
complement of event A. Then events A and B are
(b) 2ax + 2by – ( a 2 b2 p2 ) = 0
(a) equally likely and mutually exclusive
(c) x2 y 2 – 2ax – 3by + ( a 2 b2 p2 ) = 0 (b) equally likely but not independent
(c) independent but not equally likely
2
(d) 2ax + 2by – ( a b2 p2 ) = 0 (d) mutually exclusive and independent
S-102 The Pattern Target AIEEE
ANSW ER KEY
PHYSICS CHEMIS TRY MATHEMATICS
1 N 26 c 51 c 76 a 101 a 126 a 1 d 26 a 51 c
2 c 27 d 52 b 77 c 102 b 127 b 2 c 27 c 52 b
3 d 28 b 53 b 78 b 103 b 128 c 3 a 28 a 53 d
4 d 29 d 54 a 79 d 104 d 129 c 4 a 29 b 54 d
5 b 30 b 55 b 80 b 105 b 130 b 5 c 30 c 55 a
6 d 31 d 56 c 81 b 106 c 131 b 6 d 31 d 56 a
7 c 32 a 57 c 82 d 107 d 132 c 7 c 32 b 57 b
8 c 33 d 58 c 83 c 108 a 133 c 8 d 33 c 58 d
9 c 34 c 59 b 84 c 109 c 134 b 9 a 34 b 59 b
10 d 35 a 60 b 85 d 110 a 135 c 10 a 35 b 60 c
11 d 36 d 61 a 86 a 111 c 136 d 11 c 36 a 61 c
12 c 37 b 62 c 87 a 112 a 137 c 12 b 37 d 62 c
13 b 38 d 63 c 88 d 113 c 138 d 13 b 38 c 63 d
14 c 39 a 64 a 89 d 114 b 139 c 14 a 39 a 64 b
15 b 40 d 65 d 90 b 115 d 140 b 15 d 40 b 65 d
16 a 41 b 66 c 91 c 116 c 141 a 16 a 41 d 66 d
17 c 42 a 67 c 92 a 117 d 142 d 17 d 42 d 67 b
18 d 43 b 68 d 93 c 118 b 143 d 18 a 43 d 68 b
19 c 44 b 69 b 94 d 119 b 144 d 19 d 44 d 69 d
20 d 45 c 70 d 95 a 120 c 145 c 20 c 45 a 70 b
21 b, c 46 d 71 d 96 b 121 b 146 c 21 c 46 b 71 d
22 a 47 b 72 a 97 d 122 a 147 b 22 c 47 c 72 c
23 b 48 b 73 b 98 a 123 c 148 N 23 d 48 b 73 b
24 d 49 d 74 b 99 d 124 b 149 c 24 d 49 c 74 a
25 a 50 a 75 a 100 d 125 a 150 a 25 c 50 c 75 b
15 S
S f t1t 2S 15 S
S
f t1t 12 S .... (i)
v (v 2 v 1) = v12 v 22 2 v1v 2 cos 90
1 2
f t1 S ... (ii)
2 = 5 2 52 0 [As | v1 | = | v 2 | = 5 m/s]
Dividing (i) by (ii),
2 = 5 2 m/s
t 1 t f t2
we get t1 = S f
6 2 6 72 v 5 2 1
Avg. acc. = m / s2
Note : None of the given options is correct. t 10 2
S-104 The Pattern Target AIEEE
5 6. (d) v = 2gh
tan 1
5
v= 2 9.8 50 = 14 5 50 m
which means is in the second quadrant.
v
(towards north-west) v2 u 2 32 980
S= = a 2 m / s2
2 2 2 4
3. (d) Moment of Inertia, I = Mr
243 m
Dimensions = [ML2 ] Initially he has fallen 50 m.
3m / s
Total height from where
Moment of force = r F he bailed out = 244 + 50 = 293 m
2
7. (c)
Dimensions = [L ][MLT ] [ML2 T 2
]
3 cm x
4. (d) t ax 2 bx ; Diff. with respect to time (t) u
u v 0
d d dx dx
t a (x 2 ) b a.2 x b.v. 2
dt dt dt dt
1 = 2axv + bv = v (2ax + b) (v = velocity) 2
u
1 u2 2. a . 3
2ax + b = . 2
v
Again differentiating, 3u 2 u2
or = 2. a. 3 a=
4 8
dx 1 dv dx
2a 0 v
dt v 2 dt dt u
2
u2 u2
0 = 2. a. x or – = 2. ×x
2 4 8
dv
=f= 2av3 x = 1 cm
dt
8. (c)
a2
5. (b) g sin g cos R2
R1 V2 R2
n
g si d a1 V1 R1
45 45
smooth rough
1 1
d= (g sin ) t1 , d = (g sin g cos ) t 2 v12 2
R12 2
2 2 a1 R1
R1 R1
2d 2d
t1 , t2 v 22 2
g sin g sin g cos a2 R2
R2
According to question, t 2 nt1 Taking particle masses equal
F1 ma1 a1 R1
2d 2d
n = F2 ma 2 a2 R2
g sin g sin g cos
2R
, applicable here, is kinetic friction as the block moves 9. (c) t1t 2 (It is a formula)
over the inclined plane. g
t 1t 2 R
1 1
n= cos 45 sin 45 10. (d) Acceleration of block while sliding down upper half =
1 k 2
g sin ;
1 retardation of block while sliding down lower half = –
n2
1 k
(g sin g cos )
For the block to come to rest at the bottom, acceleration
1 1 in I half = retardation in II half.
or 1 k = 2
or k 1
n n2 g sin (g sin g cos ) 2 tan
AIEEE-2005 Solved Paper S-105
16. (a) Does not shift as no external force acts. The CoM of
11. (d) v
V the system continues its original path. It is only the
v
V
3 internal forces which comes into play while breaking.
17. (c) Let I be the moment of inertia of the uniform semicircular
disc
u1 V u2 0 2mr 2 Mr 2
m m 2I I
2 2
In x direction : mv + 0 = m (0) + m(V2 ) x 2h d
18. (d) gh g1 ; gd g1
V R R
In y direction : 0 + 0 = m m(V2 ) y is
3 Equating g h and g d , we get d = 2h
V GMm
(V2 ) y and (V2 ) x V 19. (c) W
3 R
2 6.67 10 11 100 10
V W = 6.67 × 10–10 J
V2 V2 0.1 1000
3
20. (d)
V2 4 2V A B
V2 = V2 = V =
3 3 3
12. (c) m = 0.3, F = m.a = – 15 x
P
2(0, ) 2
15 150
a=– x x 50 x F
0.3 3
a = – 50 × 0.2 = 10 m / s 2
13. (b) C
a 1
100 2 210 s s = 1000 m
a 2
23. (b) u = 0; v = u + aT; v = aT
g g sin
2
Instantaneous power = F × v = m. a. at = m. a . t
g sin a cos a = g tan
v2
15. (b) Instantaneous power = m t
T2
100 4
30 20 G R3
Gm G v 3
24. (d) g = g=
mgh 1 2 2 R2
mv 2 mgh R R
2
4
Using conservation of energy, g= G. R where average density
3
1 2
m (10 × 100) m v 10 20 3g
2 =
4 GR
1 2
or v 800 or v = 1600 = 40 m/s ‘ ’ is directly proportional to (g).
2
S-106 The Pattern Target AIEEE
1 1 3
25. (a) Energy stored per unit volume, E = stress strain Q1 T0S0 T0S0 T0S0
2 2 2
Stress Stress Q2 T0S0 and Q 3 0
We know that, Y or Strain
Strain Y W Q1 Q 2 Q2 T0S0 1
=1 1
1 Stress 1 S2 Q1 Q1 Q1 3 3
E= Stress = . T0S0
2 Y 2 Y 2
26. (c) Water fills the tube entirely in gravity less condition. 1/ 3 1/ 3
R1 A1 27 3
20 cm. 30. (b)
27. (d) R2 A2 125 5
T T
R2 3.6 6 fermi
dr 3
T1
7
r1
r 31. (d) days of Cu decays into Zn in 15 minutes.
8
3
T2 7 1 1
Cu undecayed = N = 1 – =
r2 8 8 2
No. of half lifes = 3
Consider a shell of thickness (dr) and of radii (r). t 15
If the temperature of inner and outer surfaces of this n= or 3 =
shell be T, (T – dT) T T
dQ 15
T = half life period = = 5 minutes
= rate of flow of heat though it 3
dt
2
kA[(T dT ) T] kAdT I I1 r2 1
= = 32. (a) I ;
dr dr 2 I2 r1 4
r
dT I2 4 times I1
= 4 kr 2 ( A 4 r2)
dr When intensity is 4 times, no. of photoelectrons emitted
would increase by 4 times.
(dQ / dT) dr
dt = hc
4 k r2 33. (d) E
Integrating between the proper limits
T2 r2 6.63 10 34 3 108 1
dQ 1 dr = eV = 0.5 eV
dt 2480 10 9
1.6 10 19
dt 4 k 2
T1 r1
r
34. (c) I I0.e d ,
dQ 1 1 Applying logarithm on both sides,
(T2 T1 )
dt. 4 k r1 r2
I
d log
dQ 4 k r1 r2 (T1 T2 ) dQ r1 r2 I0
=
dt r2 r1 dt r2 r1 I /8
28. (b) Change in internal energy do not depend upon the 36 log ..........(i)
I
path followed by the process.
29. (d) I/2
T d log
...........(ii)
I
Dividing (i) by (ii),
2T0
Q1
Q3 1 1
log 3 log
T0 36 8 2 36
Q2 3 or d = = 12 mm
d 1 1 3
S log log
S0 2S0 2 2
AIEEE-2005 Solved Paper S-107
1 1
energy level. Clearly, is maximum for the
n1 2
n 22 fm 1.5 1
Dividing (i) by (ii), =–8
fa 1.5
third transition, i.e. 2 1. 1
1.6
h h 1
38. (d) 1 1
P 2. m. (K. E) K. E Pa 5 fa
fa fa 5
If K.E is doubled, becomes 1 8
fm = 8 fa 8
2 5 5
A 1 7 4 1.6
39. (a) zX 0n 3 Li 2 He 5 1D
Pm =
On comparision, A = 7 + 4 – 1 = 10, z = 3 + 2 – 0 = 5 fm 8
It is boran 10 45. (c) Clearly sin2 wt is a periodic function as sin wt is periodic
5B
with period / w
1
40. (d) Input frequency, f = 50 Hz T
50
T 1
For full wave rectifier, T1 f1 = 100 Hz
2 100
0 /w 2 /w 3 /w
dy1
41. (b) v1 0.1 100 cos 100 t
dt 3 d2y
For SHM y
dy 2 dt 2
v2 0.1 sin t 0.1 cos t
dt 2 dy
2 w sin wt cos wt w sin 2 wt
2 3 dt
Phase diff. = 1 2 = = =–
3 2 6 6 d2y
2 2 cos 2 t which is not proportional to –y
1 3 3 3 R dt 2
42. (a) sin c or tan c .Hence it is not in SHM.
4 16 9 7 12
46. (d) The shape of interference fringes formed on a screen in
R case of a monochromatic source is a straight line.
Remember for double hole expt.a hyperbola is generated.
47. (b) Centre of mass of combination of liquid and hollow
c c 12 cm
portion (at position ), first goes down ( to ) and
when total water is drained out, centre of mass regain
its original position (to ),
36 T 2
R= cm g
7
S-108 The Pattern Target AIEEE
56. (c) P
d2x 2
50. (a) = x = x
dt 2 T cos
T
2 2
or T = F Eq q
0K
T sin
v
v
v v0 5 6
51. (c) n' n n n mg
v v 5
R R
2 2
1 F 0 i1 i 2 0i
d =
2 d 2 d
(attractive as current is in the same direction)
At (1) using, potential (V1 ) Vself Vdue to ( 2)
AIEEE-2005 Solved Paper S-109
10 1
12 – 2 = (500 ) i i
500 50
(2)
12 1
Again, I =
500 R 50
500 + R = 600 R = 100
62. (c) Resistance of Galvanometer, 0i 2 4
0 i1 0 3 0
B1 = ; B2 2( 2 ) 4
Current sensitivity 10 2 (2 ) 4
G= G= 5
Voltage sensitivity 2
B= B12 B 22 = 0
.5
150 4
Here i g = Full scale deflection current = = 15 mA
10 7
B = 10 5 10 2
V = voltage to be measured = 150 volts
(such that each division reads 1 volt) B = 5 10 5 Wb / n 2
150 1 2 240 120
R= 3
5 9995 66. (c) r = R 2=2
15 10 2 120
63. (c) 67. (c) Equating magnetic force to centripetal force,
R1 R2
I
mv 2
qvB
r
2 r 2 m
Time to complete one revolution, T =
v qB
68. (d) A magnetic needle kept in non uniform magnetic field
R experience a force and torque due to unequal forces
2 acting on poles.
I=
R R1 R 2 V2
69. (b) P Vi
Potential difference across second cell = V iR 2 0 R
2 V2 200 200
= .R2 0 R hot 400 ;
R R1 R 2 P 100
R R1 R 2 2R 2 0 400
R cold 40
R R1 R 2 = 0 R = R 2 R1 10
70. (d) For maximum power, X L X C , which yields
1
64. (a) m = zq z
q 1 1
C
2 2
z1 q2 (2 n ) L 4 50 50 10
............ (i)
z2 q1 C = 0.1 10 5
F 1 F
S-110 The Pattern Target AIEEE
85. (d) Free radicals are electron deficient species, unstable
E
71. (d) I = , Internal resistance (r) is and very reactive.
R r
Product of charges
E 86. (a) Lattic energy =
zero, I = = constant. inter ionic distance
R 87. (a) Difluoro acetic acid being strongest will furnish
maximum number of ions showing highest electrical
72. (a) Phase difference for R–L circuit lies between 0, conductivity.
2
88. (d) 1 mole of e– = 1F = 96500 C
R 12 4 27g of Al is deposited by 3 × 96500 C
73. (b) Power factor = cos 0. 8 5120 g of Al will be deposited by
Z 15 5
74. (b) Due to electric field, it experiences force and accelerates 3 96500 5120
= 5.49 10 7 C
i.e. its velocity decreases. 27
Rt 89. (d) For endothermic reaction Ef = H + Eb. Hence Ef < Eb.
75. (a) i i0 1 e L
90. (b) H U nRT for N 2 3H 2 2 NH 3
n=2–4=–2
Rt Rt H U 2RT U H
i0 L ) L 1
i 0 (1 e e 91. (c) With the increase in temperature the most probable
2 2 velocity increase but the fraction of such molecules
Rt decreases.
Taking log both the sides, log 1 log 2
L Vc
92. (a) The relation 103 is true
L 300 10 3 Vs
t= log 2 0.69
R 2 93. (c) MX 2 M 2X
t = 0.1 sec.
If S is the solubility of MX2 then Ksp = 4S3;
4 × 10–12 = 4s3; s = 1 × 10–4
CHEMISTRY
[M ] 1 10 4
76. (a) SnO2 anphoteric, CaO basic, CO2 and SiO2 acidic. 94. (d) Vapour pressure of benzene is 75 torr and toluene 22
77. (c) H2 is diamagnetic as it contains all paired electrons
78
H2 2
b, H2 1
b, H2 2
b,
*1
a ; He 2 2
b,
*1
a Torr. Moles of Benzene = 1 ; Moles of Toluene =
78
78. (b) Na 2 SO 4 2 Na SO 4 – 46
1 2 0.5
92
1 2 Total moles 1 + 0.5 = 1.5
Vant. Hoff’s factor i 1 2
1 0 1
Benzene = x B p B 75 50
79. (d) Oxidation state of Cr in [Cr ( NH 3 ) 4 Cl 2 ] . Let it be x 1.5
1 × x + 4 × 0 + 2 × (–1) = 1 95. (a) Lechatellier’s principle. Adding F2 will favour the
Therefore x = 3. formation of ClF3 (g)
80. (b) Hydrogen bomb is based on the principle of nuclear 480 1.5 520 1.2
fusion. 96. (b) M 1.344 M
480 520
1 97. (d) Kp = Kc(RT) n; In the given equation n = 3 – 2 = 1
81. (b) Number of A ions in the unit cell. = 8 1
8 Kp K c (RT )1 It shows Kp > Kc
1
Number of B ions in the unit cell = 6 3 1
2 98. (a) pH = –log [H+] = log
Hence empirical formula of the compound = AB3 [H ]
82. (d) For spontaneous reaction G should be negative.
1
Equilibrium constant should be more than one 5. 4 log On solving [H+] = 3.98 × 10–6
( G = – 2.303 RT log Kc, If Kc = 1 then G = 0; If Kc < 1 [H ]
99. (d) Reaction involving two different reactants can never
then G = +ve). Again G nFE 0cell . E 0cell must be
be unimolecular.
+ve to have G –ve. 101. (a) The energy of an orbital is given by (n + l) in (d) and
83. (c) Nylon 66 is polyamide (see polymers). (c). (n + l) value is (3 + 2) = 5 hence they will have same
84. (c) Antipyretics are used to reduce fever. energy.
AIEEE-2005 Solved Paper S-111
102. (b) During the process of electrolytic refining Ag and Au 113. (c) The order of thermal stability HF > HCl > HBr > HI.
are obtained as anode mud. 114. (b) Al (OH) Cl2.
103. (b) 426.2 (i) 115. (d) Hg 2 Cl 2 2 NH 4 OH
HCl
91.0 (ii) NH 2 Hg Hg Cl NH 4 Cl 2 H 2 O
AcONa
116. (c) B < C < N < O
NaCl 126.5 (iii) Increasing first ionization enthalpy (wrong order)
Correct order : C < B < O < N
AcOH (i) (ii) (iii) [426.2 91.0 126.5] 117. (d) In SiO2 (quartz), each of O-atom is shared between two
390.7 SiO44– tetrahedra.
104. (d) The graphs show that reaction is exothermic. 118. (b) SO 32 , CO 32 , NO3 are not iso electornic.
H 119. (b) Lanthanide contraction is responsible for Zr and Hf to
log k I
RT have about same radius. These are known as twin
For exothermic reaction H 0 or elements.
120. (c) K 3[Fe(CN ) 6 ] is potassium hexacyano ferrate (III).
1
log K Vs would be negative straight line with
T 121. (b) C2O4 C2O4
positive slope.
105. (b) Coagulation will occur due to neutralization of opposite
electric charges of sol. particles. Cr C2O4 C2O4 Cr
106. (c) LiCl has partly covalent character. Other halides are
ionic in nature. Lattice energy decreases with increase C2O4
C2O4
of ionic radius of cation, anion being the same. Larger
is the lattice energy, the higher will be m. pt. NaCl will Non-superimposable mirror images, hence optical
have highest lattice energy. isomers.
122. (a) (a) Cr 3+, No. of unpaired electrons, 3, 3d3
107. (d) 2Cu 2 O Cu 2S 6Cu SO 2 self reduction. (b) Mn3+, Unpaired electrons = 4, 3d4
108. (a) SF4 – Configuration of excited S atom : (c) Fe3+, Unpaired electrons = 5, 3d5
3s2 3p
3 1
3d
(d) Co3+, Unpaired electrons = 4, 3d4
3
sp d CH3
|
shape-square pyramidal, one lone pair 123. (c) CH 3 C H CH 2 CH 3
CF4 – Configuration of excited C-atom : CH3
2s1 2p
3 |
Br2
3 CH 3 C CH 2 CH 3
sp sun light |
Br
shape-tetrahedral; no lone pair 2 bromo 2 methyl butane
XeF6 – configuration of excited Xe atom : Ease of replacement of H-atom 3° > 2° > 1°.
4 2
5s2 5p 5d
3 2
124. (b) 12 Mg 24 11 Na
23 1
1H .
sp d
hybridization 125. (a) Secondary alcohols are oxidised to ketones, at which
shape-square planar, 2 lone pairs stage reaction usually stops because further oxisation
requires breaking of C–C bond. Secondary alcohols
C
can be oxidised to ketone either by (i) chromic acid
109. (c) Ca ; Two carbon atoms will have one two (H2CrO4), prepared by adding CrO3 or Na2Cr2O7 to
C aqueous sulphuric acid, or by a complex of chromium
bonds. trioxide and pyridine (Collins reagent) or by (iii) CrO3
in aqueous acetone (Jones reagent); or by (iv) PDC/
110. (a) Cr2 O 72 6I 14H 2Cr 3 3I 2 7 H 2O
PCC in CH2Cl2.
+3 oxidation state of Cr.
OH
H |
| CH 3 C H CH CH CH 3
111. (c) Hypophosphorous acid H O P O
| O
H ||
Two H-atoms are attached to P atom. CH 3 C CH CH CH 3
Pyridinium chloro-chromate (PCC) is specific for the
112. (a) OH H O2 conversion.
126. (a) Due to steric hindrance tertiary alkyl halide do not react
Conjugate base of OH– is O 2 .
by SN2 mechanism they react by SN1 mechanism.
127. (b) In DNA and RNA heterocyclic base and phosphate ester
are at C1' and C5' respectively of the sugar molecule.
128. (c) CH 2 CH CH CH 2 HBr
Br Br
S-112 The Pattern Target AIEEE
129. (c) pKa = –log Ka; HCOOH is the strongest acid and hence O
it has the highest Ka or lowest pKa value. ||
130. (a) In moving down a group, the basicity and 138. (d) R C X ; when X is Cl the C–X bond is more polar
nucleophilicity are inversely related, i.e. nucleophilicity and ionic which leaves the compound more reactive
increases while basicity decreases. In going from left for nucleophilic substitution reaction.
to right across a period, the basicity and nucleophilicity
Br
are directly related. Both of the characteristics |
decrease as the electronegativity of the atom bearing 139. (c) CH 3 C H CH 2 CH 3
lone pair of electrons increases. If the nucleophilic
Alc. KOH
centre of two or more species is same, nucleophilicity CH 3 CH CH CH 3 HBr
parallels basicity, i.e. more basic the species, stronger The formation of 2-butene is in accordance to
is its nucleophilicity. Saytzeff’s rule. The more substituted alkene is formed.
hence based on the above fact, the correct order of 140. (b) Equimolar solutions of normal solutes in the same
nucleophilicity will be solvent will have the same B. pts and same F. pts.
O 141. (a) In one electron species, such as H-atom, the energy of
orbital depends only on the principal quantum number,
– – –
H3C S— O– > CN > CH3O > CH3C– O n. Hence answer (d)
142. (d)
O O 143. (d) In diborane structure B2H6 there are two 2C-2e bonds
(d) (c) (b) (a) and two 3C–2e bonds (see structure of diborane).
131. (b) Wurtz reaction is for the preparation of hydrocarbons 144. (d) In lanthanides, there is poorer shielding of 5d electrons
from alkyl halide by 4f electrons resulting in greater attraction of the
nucleus over 5d electrons and contraction of the atomic
RX 2 Na XR R – R 2 NaX radii.
132. (c) Teflon is polymer of CF2 = CF2. OH
CH3 CH3 145. (c) O + HN(CH3)2 N (CH3)2
| |
133. (c) CH 3 CH C H CH 3 . Since it contains only two
types of H-atoms hence it will give only two mono –H2O
chlorinated compounds viz. N (CH3)2
CH 3 CH 3 enamine
| |
Cl.CH 2 C H C H CH 3 and CH3 CH3
1 chloro 2,3 dimethyl butane
CHCl3 + NaOH HCN
CH 3 CH 3 146. (c) Reimer Tiemman reaction C=O
| |
CH 3 C C H CH 3 OH OH H
|
Cl CH3 CH3
2 chloro 2,3 dimethyl butane
OH HOH
134. (b) Corly House alkane, synthesis C CH(OH).COOH
CN
R ' X LiR 2 Cu R ' R RCu LiX OH H OH
135. (c) Water adds directly to the more reactive alkene in Cyanohydrin
presence of a strongly acidic catalyst forming alcoholis. 147. (b) Element % Relative no Simple Ratio
Addition occurs according to Markonikov’s rule. C 20 % 20/12 = 1.66 1.66 / 1.66 = 1
CH 3 CH CH 2 H 2O H 6.67% 6.67 / 1=6.67 6.67 / 1.66 = 4
N 46.67% 46.67/14=3.33 3.33 / 1.66 = 2.0
H 2SO 4 O 26.64% 26.64/16=1.66 1.66 / 1.66 = 1.0
CH 3 CH CH 3
| The compound is CH4N2O
OH 60
2 alcohol Empirical weight = 60; Mol. wt. = 60; n 1
60
CH O
|
3
||
CH 3 C CH 2 H 2O Molecular formula = CH4N2O; NH 2 C NH 2
Biuret with alkaline CuSO4 gives violet colour. Test for Area of greatest rectangle is equal to 2ab
–CONH– group. When sin 2 1 .
148. (N) X 2 Y2 2 XY , H = 2(–200). Let x be the bond y2 2c ( x c)
5. (c) ........ (i)
dissociation energy of X2. Then
2yy' 2c.1 or yy ' c ........ (ii)
H 400 x x y y 2 x y
y2
2 yy ' ( x yy ' )
x 0.5x – 2x 0.5x
[On putting value of c from (ii) in (i)]
400 On simplifying, we get
or x 800 kJ mol 1
0.5 ( y 2xy' ) 2 4 yy'3 ........ (iii)
(In the question paper, this option was not mentioned. So the Hence equation (iii) is of order 1 and degree 3.
answer has been marked ‘N’)
1 1 2 4 3 9
t1 / 4
2 .303
log
1 2.303
log
4 sec 2 sec 2 sec 2
149. (c) 2 2 2 2 2
n2
K 3/ 4 K 3 6. (d) lim n n n n n
n 1
.... sec 2 1
2.303 2.303 n
(log 4 log 3) (2 log 2 log 3)
K K is equal to
2.303 0.29 r r2 2
(2 0.301 0.4771) 1 r 2 r
K K lim sec 2 . sec = lim
n n2 n2 n n n n2
150. (a) NH 4 HS(s ) NH 3 (g ) H2S (g ) Given limit is equal to value of integral
start 0.5 atm 0 atm 1
At equib . 0.5 x atm x atm.
x sec 2 x 2 dx
Then 0.5 + x + x = 2x + 0.5 = 0.84 (given)
0
x 0.17 atm.
1 1
p NH3 0.5 0.17 0.67 atm ; p H 2S 0.17 atm 1 1
or 2 x sec x 2 dx sec 2 tdt [put x 2 t]
2 2
2 0 0
K = p NH3 p H 2S 0.67 0.17 atm 0.1139 0.11
1 1
= (tan t )10 tan 1 .
PAPER II : MATHEMATICS 2 2
7. (c) Using Lami’s Theorem
1. (d) Given A 2 A I 0
A B C
A 1A 2 A 1A A 1.I A 1. 0 P : Q : R = cos : cos : cos .
2 2 2
1
(Multiplying A on both sides)
A
1 1
A 1 A 0 or A 1 A.
P
2. (c) ( x 1)3 8 0 ( x 1) ( 2) (1)1 / 3
2
x – 1 = – 2 or 2 or 2 Q
I
R
or x = – 1 or 1 – 2 or 1 – 2 2 . B C
3. (a) Reflexive and transitive only.
e.g. (3, 3), (6, 6), (9, 9), (12, 12) [Reflexive] 8. (d) Mode + 2Mean = 3 Median
(3, 6), (6, 12), (3, 12) [Transitive]. Mode = 3 × 22 – 2 × 21 = 66 – 42 = 24.
4. (a) Area of rectangle ABCD 2a cos 9. (a) P = (1, 0)
2
10. (a) PA AC CP = 0 PB BC CP = 0 [ 2 2] = 0
P Q = 50
C4 55
C3 54
C3 53
C3 52
C3
tan tan
2 2 P Q 51
C3 50
C3
tan 1
P Q 2 2
1 tan tan
2 2 n
Cr n
Cr n 1
Cr
We know 1
b
( 50 C 4 50
C3 )
a 1 b a c
c a a a 51 52 53 54 55
1 C3 C3 C3 C3 C3
a
– b = a – c or c = a + b. = (51 C 4 51
C3 ) 52
C3 53
C3 54
C3 55
C3
13. (b) x y z 1 = 55 55 56
C4 C3 C4
x + y+ z = – 1; x + y + z = –1
18. (a) By the principle of mathematical induction (a) is true.
1 1
19. (d) Tr 1 in the expansion
1 1 ( 2
1) 1( 1) 1(1 )
=
11 r
1 1 1 1
ax 2 11
C r (ax 2 )11 r
bx bx
= ( 1)( 1) 1( 1) 1( 1)
2 = 11 C r (a )11 r (b) r
( x) 22 2r r
( 1)[ 1 1] = 0
22 – 3r = 7 r=5
2
( 1)[ 2] = 0
7 11
coefficient of x C 5 ( a ) 6 ( b) 5
...(1)
AIEEE-2005 Solved Paper S-115
f (x) = ( x 1)
2
| f ' ( x ) | = lim
f (x h) f (x) (h) 2
lim
Hence, degree = 2 h 0 h h 0 h
S-116 The Pattern Target AIEEE
1 1
1 1
bn
2 3
y=
1 b
b= 1 = I3 2 x dx, I 4 2 x dx
n 0 y
0 0
1 1 0 x 1, x 2 x3
z= cn c= 1
1 c z 1 1
n 0
2 3
a, b, c are in A.P. Or 2b = a + c 2 x dx 2 x dx I1 I2
0 0
1 1 1
2 1 =1 1 0
y x y
36. (a) Required area (OAB) = ln(x e)dx
1 e
2 1 1
x, y, z are in H.P..
y x z 1
1
= x ln(x e) xdx = 1.
x e 0
2ab
32. (b) 2r + 2R = c +
(a b c ) 37. (d) y2 4x and x2 = 4y are symmetric about line y = x
area bounded between y2 = 4x and y = x is
( a b ) 2 c( a b )
= =a+b
(a b c) 4
8
(2 x x) dx
3
(Since c 2 a2 b2 ) 0
1 1 y 16 16
33. (c) cos x cos As2 and A s1 A s3
2 3 3
A s1 : A s 2 : A s3 : : 1 : 1 : 1
xy y2
cos 1 (1 x 2 ) 1
2 4 xdy
38. (c) = y (log y – log x + 1)
dx
xy 4 y2 4x 2 x 2 y2 dy y y
cos 1 log 1
2 dx x x
Put y = vx
4 y 2 4x 2 x 2 y2 dy
v
xdv xdv
v+ = v (log v + 1)
dx dx dx
= 4 cos 2 x 2 y2 4xy cos
xdv dv dx
= v log v
4x 2 y2 4xy cos 4 sin 2 dx v log v x
AIEEE-2005 Solved Paper S-117
2a 2 3a 2 f
2
1
4
sin
2
2 1
4
2
y 2b 3b =0
b b
x2 y2
2 2 2 2 44. (d) Tangent to the hyperbola = 1 is
y
2b 2a 3b 3a a2 b2
b b
y = mx a 2m2 b2
3(a 2 b2 ) 3 Given that y = x+ is the tangent of hyperbola
y= 2 2
2(b a ) 2
and a 2 m 2 b2 2
m=
So it is 3/2 units below x-axis.
a2 2
b2 2
dv
40. (b) = 50 and 4 r 2 dr = 50
dt dt Locus is a 2 x 2 y 2 b 2 which is hyperbola.
45. (a) Angle between line and normal to plane is
dr 50 1
= (here r = 10 + 5)
dt 4 (15) 2 18 2 2 2
cos
2 3 5
(log x 1) 2 Where is angle between line and plane
41. (d) dx
(1 (log x ) 2 ) 2
2 1 5
sin .
1 2 log x 3 5 3 3
= 2 2 2
dx
(1 (log x) ) (1 (log x) ) 46. (b) Angle between the lines 2x 3y z
et 2t e t
and 6x y 4z is 90
= dt put log x = t
1 t2 (1 t 2 ) 2 Since a1a 2 b1b 2 c1c 2 0 .
47. (c) Plane 2ax 3ay 4az 6 0 passes through the mid
dx = e t dt
point of the centre of spheres
1 2t x2 y2 z2 6x 8y 2z 13 and
et 2 2 2
dt
1 t (1 t )
x 2 y 2 z 2 10x 4 y 2z 8 respectively
et x center of spheres are (– 3, 4, 1) and (5, – 2, 1). Mid point
= c = c
1 t 2
1 (log x ) 2 of centre is (1, 1, 1).
Satisfying this in the equation of plane, we get
2a 3a 4a 6 0 a 2
S-118 The Pattern Target AIEEE
48. (b) Distance between the line s1 s 2 0 5ax (c d) y a 1 0
r 2i 2 j 3k (i j 4k) Given that 5x + by – a = 0 passes through P and Q
(a i ) 2 y2 z2 ( , )
Similarly,
(a j) 2 x2 z 2 and (a k ) 2 x2 y2
Circle touch externally
(a i ) 2 (a j) 2 (a k ) 2
2
( 3) 2 2
2 (x 2 y2 z2 ) 2a 2|
2
( 3) 2 2
4 4
2 1 1
50. (c) a, b, c are in H.P. 0 2
a a c 10( 1 / 2)
x y 1 1
0 Locus is x 2 10 y
a a c 2
x y 1 Which is parabola.
1 2 1 54. (d) Let the centre be ( , )
x 1, y 2 It cut the circle x2 + y2 = p2 orthogonally
51. (c) Vertex of tringle is (1, 1) and midpoint of sides through 2( ) 0 2( ) 0 c1 p 2
this vertex is (– 1, 2) and (3, 2).
c1 p 2
A (1, 1) Let equation of circle is
x 2 y2 2 x 2 y p2 0
It passes through
( 1, 2) (3, 2)
(a , b ) a2 b2 2 x 2 y p2 0
Locus
B C 2ax 2by (a 2 b2 p2 ) 0
vertex B and C come out to be (– 3, 3) and (5, 3) 55. (a) FBF ' 90
2 2
1 3 5 1 3 5 7 a 2e 2 b2 a 2e2 b 2 (2ae) 2
Centroid is , 1,
3 3 3
s2 x2 y 2 3ax dy 1 0 b2
e2
Equation of radical axis of s1 and s 2 a2
AIEEE-2005 Solved Paper S-119
B (0, b) 1 1 1
61. (c) P(A B) , P(A B) and P(A)
6 4 4
[a b c] a.(b c) F
3F
î ĵ k̂
3
b c = x 1 1 x
2
y x 1 x y F'
a.(b c) = 1 ex e x
x2 x4 x6
66. (d) 1 .............
which does not depend on x and y. 2 2! 4! 6!
59. (b) For a particular house being selected
1 1
Probability = Putting x = we get
3 2
P (all the persons apply for the same house)
e 1
1 1 1 1 .
3 . 2 e
3 3 3 9
k cos 2 x
60. (c) P( x k) e 67. (b) dx cos 2 x d
x 2
k! 1 a
0
P( x 2) 1 P ( x 0) P( x 1)
1 1
68. (b) Perpendicular distance of centre ,0,
3 2 2
1 e e 1 .
1! e2 from (x + 2y – 2 = 4 ).
S-120 The Pattern Target AIEEE
[ x = 0, y = 0, f (0) f 2 ( 0) f 2 ( a )
30
f 2 (a ) 0 f (a) = 0]
60 30
f (2a x ) f (x)
Given a1 0 f(0) = 0
0 x axis Again f (x) has root , f( ) 0
x=5
As is a positive root and 0 is another root of f(x), then
f ( x) 0 between 0 and .
then (i) Discriminant 0
Hence f ( x ) has a positive root smaller than .
AIEEE - 2006
Time : 2 ½ Hours Max. Marks : 450
25. The current I drawn from the 5 volt source will be 29. Two insulating plates are both uniformly charged in such a
way that the potential difference between them is
10 V2 – V1 = 20 V. (i.e., plate 2 is at a higher potential). The
plates are separated by d = 0.1 m and can be treated as
5 10 20 infinitely large. An electron is released from rest on the inner
surface of plate 1. What is its speed when it hits plate 2? (e
= 1.6 × 10–19 C, me = 9.11 × 10–31 kg)
I 10
+–
5 volt Y
(a) 0.33 A (b) 0.5 A
0.1 m
(c) 0.67 A (d) 0.17 A
26. The energy spectrum of -particles [number N(E) as a X
function of -energy E] emitted from a radioactive source is
1 2
(a) N(E)
(a) 2.65 × 106 m/s (b) 7.02 × 1012 m/s
(c) 1.87 × 10 m/s6 (d) 32 × 10–19 m/s
E
E0 30. In an AC generator, a coil with N turns, all of the same area
A and total resistance R, rotates with frequency in a
magnetic field B. The maximum value of emf generated in
the coil is
(b) N(E) (a) N.A.B.R. (b) N.A.B.
(c) N.A.B.R. (d) N.A.B.
E 31. A solid which is not transparent to visible light and whose
E0
conductivity increases with temperature is formed by
(a) Ionic bonding
(b) Covalent bonding
(c) Vander Waals bonding
(c) N(E)
(d) Metallic bonding
32. The refractive index of a glass is 1.520 for red light and 1.525
E
E0 for blue light. Let D1 and D2 be angles of minimum deviation
for red and blue light respectively in a prism of this glass.
Then,
(a) D1 < D2
(d) N(E) (b) D1 = D2
(c) D1 can be less than or greater than D2 depending upon
E the angle of prism
E0 (d) D1 > D2
27. In a series resonant LCR circuit, the voltage across R is 100 33. If the ratio of the concentration of electrons to that of holes
volts and R = 1 k with C = 2 F. The resonant frequency 7 7
is 200 rad/s. At resonance the voltage across L is in a semiconductor is and the ratio of currents is ,
5 4
(a) 2.5 × 10–2 V (b) 40 V
then what is the ratio of their drift velocities?
(c) 250 V (d) 4 × 10–3 V
28. The resistance of a bulb filmanet is 100 at a temperature of 5 4
100°C. If its temperature coefficient of resistance be 0.005 (a) (b)
8 5
per °C, its resistance will become 200 at a temperature of
(a) 300°C (b) 400°C 5 4
(c) (d)
(c) 500°C (d) 200°C 4 7
S-124 The Pattern Target AIEEE
P 2R P R (S1 S2 ) –10 V
(a) Q S1 S2
(b) Q S1S 2
+5 V
P R (S1 S2 ) P R
(c) (d) R
Q 2S1S 2 Q S1 S2
R
35. The flux linked with a coil at any instant 't' is given by (d)
4
(b) I
D1 D2
12V
3 2 O
+10 V
O
(a) R
+5 V
(b) (d) I
–10 V
R
–5 V O
AIEEE-2006 Solved Paper S-125
51. Two rigid boxes containing different ideal gases are placed Section - 2
on a table. Box A contains one mole of nitrogen at temperature
T0 , while Box contains one mole of helium at
7
temperature T0 . The boxes are then put into thermal 56. How many molesof magnesium phosphate, Mg3(PO4)2 will
3
contain 0.25 mole of oxygen atoms?
contact with each other, and heat flows between them until (a) 1.25 × 10–2 (b) 2.5 × 10–2
the gases reach a common final temperature (ignore the heat
(c) 0.02 (d) 3.125 × 10–2
capacity of boxes). Then, the final temperature of the gases,
Tf in terms of T0 is 57. Total volume of atoms present in a face-centred cubic unit
cell of a metal is (r is atomic radius)
3 7
(a) Tf T0 (b) Tf T0 12 3 16 3
7 3 (a) r (b) r
3 3
3 5
(c) Tf T0 (d) Tf T0 20 3 24 3
2 2 (c) r (d) r
3 3
52. The work of 146 kJ is performed in order to compress one
kilo mole of gas adiabatically and in this process the 58. According to Bohr's theory, the angular momentum of an
temperature of the gas increases by 7°C. The gas is electron in 5th orbit is
(R = 8.3 J mol–1 K–1) (a) 10 h / (b) 2.5 h /
(a) diatomic
(c) 25 h / (d) 1.0 h /
(b) triatomic
(c) a mixture of monoatomic and diatomic 59. A reaction was found to be second order with respect to the
concentration of carbon monoxide. If the concentration of
(d) monoatomic
carbon monoxide is doubled, with everything else kept the
53. If the lattice constant of this semiconductor is decreased, same, the rate of reaction will
then which of the following is correct? (a) increase by a factor of 4
(b) double
conduction
Ec (c) remain unchanged
band width
band gap (d) triple
Eg
60. Which of the following molecules/ions does not contain
valence
Ev unpaired electrons?
band width
(a) N2 (b) O2
(a) All Ec, Eg, Ev increase
(b) Ec and Ev increase, but Eg decreases (c) O 22 (d) B2
(c) Ec and Ev decrease, but Eg increases
61. TheIUPAC name for the complex [Co(NO2)(NH3)5]Cl2 is :
(d) All Ec, Eg, Ev decrease
(a) pentaammine nitrito-N-cobalt(II) chloride
54. The rms value of the electric field of the light coming from
(b) pentaammine nitrito-N-cobalt(III) chloride
the Sun is 720 N/C. The average total energy density of the
electromagnetic wave is (c) nitrito-N-pentaamminecobalt(III) chloride
(a) 4.58 × 10–6 J/m3 (b) 6.37 × 10–9 J/m3 (d) nitrito-N-pentaamminecobalt(II) chloride
(c) 81.35 × 10–12 J/m3 (d) 3.3 × 10–3 J/m3 62. Which of the following chemical reactions depict the
oxidizing beahviour of H2SO4?
55. If the terminal speed of a sphere of gold (density = 19.5 kg/
m3) is 0.2 m/s in a viscous liquid (density = 1.5 kg/m3), find (a) NaCl H 2 SO 4 NaHSO 4 HCl
the terminal speed of a sphere of silver (density = 10.5 kg/ (b) 2PCl 5 H 2SO 4 2 POCl3 2 HCl SO 2 Cl 2
m3) of the same size in the same liquid
(c) 2 HI H 2 SO 4 I2 SO 2 2H 2 O
(a) 0.4 m/s (b) 0.133 m/s
(d) Ca (OH ) 2 H 2 SO 4 CaSO 4 2H 2 O
(c) 0.1 m/s (d) 0.2 m/s
AIEEE-2006 Solved Paper S-127
63. The IUPAC name of the compound shown below is 70. The standard enthalpy of formation ( fHº) at 298 K for
methane, CH4 (g) is –74.8 kJ mol –1 . The additional
Cl information required to determine the average energy for
C – H bond formation would be
(a) the first four ionization energies of carbon and electron
Br gain enthalpy of hydrogen
(a) 3-bromo-1-chlorocyclohexene (b) the dissociation energy of hydrogen molecule, H2
(b) 1-bromo-3-chlorocyclohexene (c) the dissociation energy of H2 and enthalpy of
(c) 2-bromo-6-chlorocyclohex-1-ene sublimation of carbon
(d) 6-bromo-2-chlorocyclohexene (d) latent heat of vapourization of methane
64. HBr reacts with CH2 = CH – OCH3 under anhydrous 71. Phosphorus pentachloride dissociates as follows, in a closed
conditions at room temperature to give reaction vessel,
(a) BrCH2 – CH2 – OCH3 (b) H3C – CHBr – OCH3 PCl5(g) PCl3(g) + Cl2(g)
(c) CH3CHO and CH3Br (d) BrCH2CHO and CH3OH If total pressure at equilibrium of the reaction mixture is P
65. The increasing order of the rate of HCN addition to and degree of dissociation of PCl5 is x, the partial pressure
compound A – D is of PCl3 will be
(1) HCHO (2) CH3COCH3
x x
(3) PhCOCH3 (4) PhCOPh (a) P (b) P
x 1 1 x
(a) D < C < B < A (b) C < D < B < A
(c) A < B < C < D (d) D < B < C < A x 2x
(c) P (d) P
66. In Langmuir's model of adsorption of a gas on a solid surface x 1 1 x
(a) the mass of gas striking a given area of surface is
o o
proportional to the pressure of the gas
72. The molar conductivities NaOAc and HCl at infinite
(b) the mass of gas striking a given area of surface is
dilution in water at 25ºC are 91.0 and 426.2 S cm2/mol
independent of the pressure of the gas
o
(c) the rate of dissociation of adsorbed molecules from respectively. To calculate HOAc , the additional value
the surface does not depend on the surface covered required is
(d) the adsorption at a single site on the surface may
o o
involve multiple molecules at the same time (a) NaOH (b) NaCl
67. An ideal gas is allowed to expand both reversibly and
o o
irreversibly in an isolated system. If T i is the initial (c) H2O (d) KCl
temperature and Tf is the final temperature, which of the
following statements is correct? 73. Rate of a reaction can be expressed by Arrhenius equation
(a) (Tf)rev = (Tf)irrev as :
(b) Tf = Ti for both reversible and irreversible processes k = A e–E/RT
(c) (Tf)irrev > (Tf)rev In this equation, E represents
(a) the total energy of the reacting molecules at a
(d) Tf > Ti for reversible process but Tf = Ti for irreversible
temperature, T
process
(b) the fraction of molecules with energy greater than the
68. Uncertainty in the position of an electron (mass = 9.1 × 10–31
activation energy of the reaction
kg) moving with a velocity 300 ms–1, accurate upto 0.001%
(c) the energy above which all the colliding molecules will
will be
react
(h = 6.63 × 10–34 Js)
(d) the energy below which all the colliding molecules will
(a) 1.92 × 10–2 m (b) 3.84 × 10–2 m react
(c) 19.2 × 10–2 m (d) 5.76 × 10–2 m 74. A metal, M forms chlorides in its +2 and +4 oxidation states.
69. Among the following mixtures, dipole-dipole as the major Which of the following statements about these chlorides is
interaction, is present in correct?
(a) KCl and water (a) MCl2 is more ionic than MCl4
(b) benzene and carbon tetrachloride (b) MCl2 is more easily hydrolysed than MCl4
(c) benzene and ethanol (c) MCl2 is more volatile than MCl4
(d) acetonitrile and acetone (d) MCl2 is more soluble in anhydrous ethanol than MCl 4
S-128 The Pattern Target AIEEE
75. In which of the following molecules/ions are all the bonds 85. The pyrimidine bases present in DNA are
not equal? (a) cytosine and thymine
(a) XeF4 (b) BF4– (b) cytosine and uracil
(c) SF4 (d) SiF4 (c) cytosine and adenine
76. Which of the following statements is true? (d) cytosine and guanine
(a) HClO4 is a weaker acid than HClO3 86. The increasing order of stability of the following free radicals
(b) HNO3 is a stronger acid than HNO2 is
(c) H3PO3 is a stronger acid than H2SO3 • • • •
(a) (C6H5)2 C H < (C6H5)3 C < (CH3)3 C < (CH3)2 C H
(d) In aqueous medium HF is a stronger acid than HCl • • • •
77. Which one of the following sets of ions represents a (b) (CH3)2 C H < (CH3)3 C < (C6H5)3 C < (C6H5)2 C H
collection of isoelectronic species? • • • •
(c) (CH3)2 C H < (CH3)3 C < (C6H5)2 C H < (C6H5)3 C
(a) N3–, O2–, F–, S2– • • • •
(d) (C6H5)3 C < (C6H5)2 C H < (CH3)3 C < (CH3)2 C H
(b) Li+, Na+, Mg2+, Ca2+
(c) K+, Cl–, Ca2+, Sc3+ 87. Fluorobenzene (C6H5F) can be synthesized in the laboratory
(d) Ba2+, Sr2+, K+, Ca2+ (a) by direct fluorination of benzene with F2 gas
78. Nickel (Z = 28) combines with a uninegative monodentate (b) by reacting bromobenzene with NaF solution
ligand X– to form a paramagnetic complex [NiX4]2–. The (c) by heating phenol with HF and KF
number of unpaired electron(s) in the nickel and geometry (d) from aniline by diazotisation followed by heating the
of this complex ion are, respectively : diazonium salt with HBF4
(a) one, square planar (b) two, square planar 88. CH 3 Br Nu CH 3 Nu Br
(c) one, tetrahedral (d) two, tetrahedral The decreasing order of the rate of the above reaction with
79. The increasing order of the first ionization enthalpies of the nucleophiles (Nu–) A to D is
elements B, P, S and F (lowest first) is [Nu– = (A) PhO–, (B) AcO–, (C) HO–, (D) CH3O–]
(a) B < P < S < F (b) B < S < P < F (a) A > B > C > D (b) B > D > C > A
(c) F < S < P < B (d) P < S < B < F
(c) D > C > A > B (d) D > C > B > A
80. What products are expected from the disproportionation
89. Among the following the one that gives positive iodoform
reaction of hypochlorous acid?
test upon reaction with I2 and NaOH is
(a) HCl and Cl2O (b) HCl and HClO3
(c) HClO3 and Cl2O (d) HClO2 and HClO4 CH3
|
81. In Fe(CO)5, the Fe – C bond possesses (a) CH 3 C HCH 2 OH
(a) ionic character
(b) -character only (b) PhCHOHCH3
(c) -character (c) CH3CH2CH(OH)CH2CH3
(d) both and characters (d) C6H5CH2CH2OH
238
90. The correct order of increasing acid strenght of the
82. In the transformation of to 234 , if one emission is
92 U 92 U compounds is
an -particle, what should be the other emission(s)? (1) CH3CO2H (2) MeOCH2CO2H
(a) one – and one (b) one + and one –
–
Me
(c) two (d) two – and one + (3) CF3CO2H (4) CO2H
83. The term anomers of glucose refers to Me
(a) enantiomers of glucose (a) d < a < b < c (b) a < d < c < b
(b) isomers of glucose that differ in configuration at carbon (c) b < d < a < c (d) d < a < c < b
one (C-1) 91. Density of a 2.05 M solution of acetic acid in water is 1.02 g/
(c) isomers of glucose that differ in configurations at mL. The molality of the solution is
carbons one and four (C-1 and C-4) (a) 2.28 mol kg–1 (b) 0.44 mol kg–1
(d) a mixture of (D)-glucose and (L)-glucose (c) 1.14 mol kg–1 (d) 3.28 mol kg–1
84. Phenyl magnesium bromide reacts with methanol to give 92. 18 g of glucose (C6H12O6) is added to 178.2 g of water. The
(a) a mixture of toluene and Mg(OH)Br vapour pressure of water for this aqueous solution at 100ºC
(b) a mixture of phenol and Mg(Me)Br is
(c) a mixture of anisole and Mg(OH)Br (a) 76.00 Torr (b) 752.40 Torr
(d) a mixture of benzene and Mg(OMe)Br (c) 759.00 Torr (d) 7.60 Torr
AIEEE-2006 Solved Paper S-129
93. The enthalpy changes for the following processes are listed 100. Lanthanoid contraction is caused due to
below : (a) the same effective nuclear charge from Ce to Lu
Cl2(g) = 2Cl(g), 242.3 kJ mol–1 (b) the imperfect shielding on outer electrons by 4f
I2(g) = 2I(g), 151.0 kJ mol–1 electrons from the nuclear charge
ICl(g) = I(g) + Cl(g), 211.3 kJ mol–1 (c) the appreciable shielding on outer electrons by 4f
I2(s) = I2(g), 62.76 kJ mol–1 electrons from the nuclear charge
Given that the standard states for iodine and chlorine are (d) the appreciable shielding on outer electrons by 5d
I2(s) and Cl2(g), the standard enthalpy of formation for ICl(g) electrons from the nuclear charge
is : 101. How many EDTA (ethylenediaminetetraacetic acid)
(a) +16.8 kJ mol–1 (b) +244.8 kJ mol–1 molecules are required to make an octahedral complex with
(c) –14.6 kJ mol –1 (d) –16.8 kJ mol–1 a Ca2+ ion?
94. ( H – U) for the formation of carbon monoxide (CO) from (a) One (b) Two
its elements at 298 K is (c) Six (d) Three
(R = 8.314 J K–1 mol–1) 102. The "spin-only" magnetic moment [in units of Bohr
(a) –2477.57 J mol–1 (b) 2477.57 J mol–1 magneton, (µB)] of Ni2+ in aqueous solution would be (At.
(c) –1238.78 J mol –1 (d) 1238.78 J mol–1 No. Ni = 28)
95. The equilibrium constant for the reaction (a) 6 (b) 1.73
(c) 2.84 (d) 4.90
1 103. The decreasing values of bond angles from NH3 (106º) to
SO3(g) SO 2 (g ) O 2 (g)
2 SbH3 (101º) down group-15 of the periodic table is due to
is Kc = 4.9 × 10–2. The value of Kc for the reaction (a) decreasing lp-bp repulsion
2SO2(g) + O2(g) 2SO3(g) (b) decreasing electronegativity
will be (c) increasing bp-bp repulsion
(a) 9.8 × 10–2 (b) 4.9 × 10–2 (d) increasing p-orbital character in sp3
(c) 416 (c) 2.40 × 10–3 104. Following statements regarding the periodic trends of
chemical reactivity of the alkali metals and the halogens are
96. Given the data at 25ºC
given. Which of these statements gives the correct picture?
Ag I AgI e E º 0.152 V (a) Chemical reactivity increases with increase in atomic
Ag Ag e Eº 0.800 V number down the group in both the alkali metals and
halogens
What is the value of log Ksp for AgI?
(b) In alkali metals the reactivity increases but in the
(2.303 RT/F = 0.059 V)
halogens it decreases with increase in atomic number
(a) –37.83 (b) –16.13 (c) –8.12 (d) +8.612 down the group
97. Resistance of a conductivity cell filled with a solution of an (c) The reactivity decreases in the alkali metals but
electrolyte of concentration 0.1 M is 100 . The conductivity increases in the halogens with increase in atomic
of this solution is 1.29 S m–1. Resistance of the same cell number down the group
when filled with 0.2 M of the same solution is 520 . The
(d) In both the alkali metals and the halogens the chemical
molar conductivity of 0.2 M solution of electrolyte will be
reactivity decreases with increase in atomic number
(a) 1.24 × 10–4 S m2 mol–1 (b) 12.4 × 10–4 S m2 mol–1 down the group
(c) 124 × 10–4 S m2 mol–1 (d) 1240 × 10–4 S m2 mol–1 105. The structure of the compound that gives a tribromo
98. The following mechanism has been proposed for the derivative on treatment with bromine water is
reaction of NO with Br 2 to form NOBr :
CH 3 CH3
NO(g) + Br2(g) NOBr2(g)
OH
NOBr 2 (g ) NO (g ) 2 NOBr ( g ) (a) (b)
If the second step is the rate determining step, the order of
the reaction with respect to NO(g) is OH
(a) 3 (b) 2 (c) 1 (d) 0
99. The ionic mobility of alkali metal ions in aqueous solution is CH3 CH2OH
maximum for
(a) Li+ (b) Na+ (c) (d)
(c) K+ (d) Rb+ OH
S-130 The Pattern Target AIEEE
Me Section - 3
106.
N Me
OH
n-Bu Et 111. If the roots of the quadratic equation
The alkene formed as a major product in the above
x2 px q 0 are tan30° and tan15°,
elimination reaction is
respectively, then the value of 2 + q – p is
Me Me
(a) (b) (a) 2 (b) 3
(c) 0 (d) 1
6
(c) Me (d) CH2 = CH2 x
112. The value of integral,
I dx is
107. Increasing order of stability among the three main 9 x x
conformations (i.e. Eclipse, Anti, Gauche) of 2-fluoroethanol 3
is
1 3
(a) Eclipse, Anti, Gauche (a) (b)
2 2
(b) Anti, Gauche, Eclipse
(c) Eclipse, Gauche, Anti (c) 2 (d) 1
(d) Gauche, Eclipse, Anti 113. ABC is a triangle, right angled at A. The resultant of the
108. Reaction of trans 2-phenyl-1-bromocyclopentane on 1 1
forces acting along AB, BC with magnitudes and
reaction with alcoholic KOH produces AB AC
(a) 1-phenylcyclopentene
respectively is the force along AD , where D is the foot of
(b) 3-phenylcyclopentene
the perpdicular from A onto BC. The magnitude of the
(c) 4-phenylcyclopentene
resultant is
(d) 2-phenylcyclopentene
109. The structure of the major product formed in the following AB 2 AC 2 (AB)(AC )
(a) (b)
reaction is 2
(AB) (AC) 2 AB AC
CH2Cl
NaCN 1 1 1
(c) (d)
DMF AB AC AD
I 114. If A and B are square matrices of size n × n such that
117. Suppose a population A has 100 observations 101, 102, 123. At a telephone enquiry system the number of phone cells
............., 200 and another population B has 100 obsevrations regarding relevant enquiry follow Poisson distribution with
151, 152, ................ 250. If VA and VB represent the variances an average of 5 phone calls during 10 minute time intervals.
VA The probability that there is at the most one phone call
of the two populations, respectively then is during a 10-minute time period is
VB
6 5
9 (a) (b)
(a) 1 (b) e 6
4 5
4 2 6 6
(c) (d) (c) (d)
9 3 55 e5
118. The number of values of x in the interval [0, 3 ] satisfying 124. A particle has two velocities of equal magnitude inclined to
each other at an angle . If one of them is halved, the angle
the equation 2 sin 2 x 5 sin x 3 0 is
between the other and the original resultant velocity is
(a) 4 (b) 6
bisected by the new resultant. Then is
(c) 1 (d) 2
(a) 90° (b) 120°
119. If ( a b) c a (b c) where a, b and c are any three
(c) 45° (d) 60°
vectors such that a.b 0 , b. c 0 then a and c are 125. A body falling from rest under gravity passes a certain point
P. It was at a distance of 400 m from P, 4s prior to passing
(a) inclined at an angle of between them through P. If g 10m / s 2 , then the height above the point
3
P from where the body began to fall is
(b) inclined at an angle of between them (a) 720 m (b) 900 m
6
(c) 320 m (d) 680 m
(c) perpendicular
126. The locus of the vertices of the family of parabolas
(d) parallel
a 3x 2 a2x
10
2k 2k y 2a is
120. The value of sin i cos is 3 2
11 11
k 1
105 3
(a) i (b) 1 (a) xy (b) xy
64 4
(c) – 1 (d) – i
35 64
(c) xy (d) xy
121. xf (sin x )dx is equal to 16 105
0
x 2
127. The function f ( x ) has a local minimum at
2 x
(a) f (cos x)dx (b) f (sin x)dx (a) x 2 (b) x 2
0 0
(c) x 0 (d) x 1
/2 /2
128. Let a1 , a 2 , a 3 ............ be terms on A.P. If
(c) f (sin x)dx (d) f (cos x)dx
2
0 0 a1 a 2 ...........a p p2
, p q , then a 6 equals
122. The values of a, for which points A, B, C with position a1 a 2 ........... a q 2 a 21
q
vectors 2iˆ ˆj kˆ , ˆi 3jˆ 5kˆ and aiˆ 3jˆ kˆ respectively
41 7
(a) (b)
are the vertices of a right angled triangle with C are 11 2
2
(a) 2 and 1 (b) – 2 and – 1 2 11
(c) (d)
(c) – 2 and 1 (d) 2 and – 1 7 41
S-132 The Pattern Target AIEEE
129. If the expansion in powers of x of the function 134. In an ellipse, the distance between its foci is 6 and minor
axis is 8. Then its eccentricity is
1
is a 0 a1x a 2 x 2 a 3x 3...... then a n is
(1 ax)(1 bx ) 3 1
(a) (b)
5 2
bn a n a n bn 4 1
(a) (b) (c) (d)
b a b a 5 5
135. At an election, a voter may vote for any number of
an 1
bn 1 bn 1
an 1
(c) (d) candidates, not greater than the number to be elected. There
b a b a are 10 candidates and 4 are of be selected, if a voter votes
for at least one candidate, then the number of ways in
x which he can vote is
130. The set of points where f (x) is differentiable is
1 |x| (a) 5040 (b) 6210
(c) 385 (d) 1110
(a) ( ,0) (0, ) (b) ( , 1) ( 1, )
136. Angle between the tangents to the curve y x 2 5x 6
(c) ( , ) (d) (0, )
at the points (2, 0) and (3, 0) is
(a) (b)
2 2
131. [(x )3 cos 2 (x 3 )]dx is equal to
3 (c) (d)
2 6 4
3x 2 9 x 17
4 4 137. If x is real, the maximum value of is
2
(a) (b) 3x 9x 7
32 32 2
1
(a) (b) 41
4
(c) (d) 1
2 4
17
132. A triangular park is enclosed on two sides by a fence and on (c) 1 (d)
7
the third side by a straight river bank. The two sides having
138. A straight line the point A (3, 4) is such that its intercept
fence are of same length x. The maximum area enclosed by
between the axes is bisected at A. Its equation is
the park is
(a) x y 7 (b) 3x 4y 7 0
3
3 2 x (c) 4x 3y 24 (d) 3x 4y 25
(a) x (b)
2 8
139. For natural numbers m, n if (1 y) m (1 y) n
1 2
(c) x (d) x2
2 = 1 a1y a 2 y 2 ....... and a1 a 2 = 10, then (m, n) is
(a) (20, 45) (b) (35, 20)
1 2 a 0 (c) (45, 35) (d) (35, 45)
133. Let A and B , a, b N . Then
3 4 0 b
140. The two lines x ay b , z cy d ; and x a ' y b' ,
(a) there cannot exist any B such that AB = BA z c' y d' are perpendicular to each other if
(b) there exist more then one but finite number of B s such
(a) aa ' cc' 1 (b) aa ' cc' 1
that AB = BA
(c) there exists exactly one B such that AB = BA (c) a c (d) a c
1 1
a' c' a' c'
(d) there exist infinitely many B s such that AB = BA
AIEEE-2006 Solved Paper S-133
SECTION I : PHYSICS +q
F1
1. (d) A function in a electrical circuit cannot act as a source
or sink of charges and charges therefore cannot be E1
created or destroyed. 5. (c) F2
If the sum of emf's and potential difference across the E2
resistors is not zero then one can continuously gain –q
energy by circulating charge around a closed loop in a The electric field will be different at the location of
particular direction, which is not possible. Thus this force on the two charges. Therefore the two charges
law is based on conservation of charge. will be unequal. This will result in a force as well as
2. (b) Ferromagnetic substance has magnetic domains torque.
whereas paramagnetic substances have magnetic 6. (a) v x
dipoles which get attracted to a magnetic field.
Diamagnetic substances do not have magnetic dipole dx
x
but in the presence of external magnetic field due to dt
their orbital motion these substance are repelled. dx
3. (d) B = 2 A dt
x
dB = 2dA
RB = RA x t
dx
dt
B B A A x
0 0
AB AA
x
2 x
dB 2 2
4d A [ t ]0t
B A A 1
2 0
A B d 2A 2 A d 2A
2
4. (b) The charged particle will move along the lines of electric 2 x t t2 x
field (and magnetic field). Magnetic field will exert no 4
force. The force by electric field will be along the lines 7. (b) Let the velocity and mass of 4 Kg piece be v1 and m1
of uniform electric field. Hence the particle will move in and that of 12 Kg piece be v2 and m2.
Applying conservation of linear momentum
a straight line.
AIEEE-2006 Solved Paper S-135
T
1 1000
200 C V 2 (220) 2
R
0.005 5 P 100
Final temperature = 300°C The power consumed when operated at 110 V is
NOTE : We may use this expression as an approximation
because the difference in the answers is appreciable. (110) 2
100
P 25 W
For accurate results one should use R = R0e T (220) /100 42
1 2eV 2 1.6 10 19 20 42. (c) The risk posed to a human being by any radiation
29. (a) eV mv 2 v exposure depends partly upon the absorbed dose, the
2 m 9.1 10 31 amount of energy absorbed per gram of tissue.
Absorbed dose is expressed in rad. A rad is equal to
2.65 10 6 m / s
100 ergs of energy absorbed by 1 gram of tissue. The
d d ( N.B.A) more modern, internationally adopted unit is the gray
30. (d) e (named after the English medical physicist L. H. Gray);
dt dt one gray equals 100 rad.
AIEEE-2006 Solved Paper S-137
mg x
mean
position
mg – N = m 2x
where x is the distance from mean position T T
For block to leave contact N = 0
2 g
mg m x x 2
44. (c) n W W
D C At equilibrium T = W
W/A W/A
Y Y
/2 /L
B
A L/2
Elongation is the same.
n' 49. (a) Initially, when steady state is achieved,
2 E
2 i=
I nn ' 2 m m( 2 ) R
2 Let E is short circuited at t = 0. Then
m 2
2m 2
3m 2 E
At t = 0, i0 =
R
45. (b) Torque r F (î ˆj) ( Fk̂ ) Let during decay of current at any time the current
(î ˆj) (Fk̂ ) F[ î k̂ ˆj k̂] flowing is L
di
iR 0
dt
F( ĵ î ) F( î ˆj) i t
46. (a) Velocity is max. when K.E. is max. di R di R
dt dt
For min. P.E., i L i L
i0 0
dv
0 x3 x 0 x 1 R
dx i R t
log e t i i0 e L
1 1 1 i0 L
Min. P.E.
4 2 4 R 100 10 3
1 9 E Lt 100 100 10 3 1
K.E. + P.E. = 2 (Given) K.E. 2 i e e
4 4 R 100 e
1 1 9 3 +Q +Q
K.E. mv 2 1 v2 v
2 2 4 2
47. (d) Applying conservation of angular moment 50. (c) r1 r2
I' ' = I
A B
(mR 2 2MR 2 ) ' mR 2 After connection, v1 = v2
m Q Q Q1 Q 2
' K 1 K 2
m 2M r1 r2 r1 r2
48. (a) Case (i) The ratio of electric fields
Q
K 1
E1 r12 E1 Q1 r22
E2 Q E 2 r12 Q 2
T K 2
2
r2
E1 r1 r22 E1 r2 2
W E2 r12 r2 E2 r1 1
S-138 The Pattern Target AIEEE
Since the distance between the spheres is large as Angular momentum of electron
compared to their diameters, the induced effects may 5h 2.5h
be ignored.
51. (c) Heat lost by He = Heat gained by N2 2
59. (a) Rate = K[CO]2, this is rate law equation
n1Cv1 T1 n 2 Cv2 T2 New rate = K[2CO]2 = 4K[CO]2
increase by a factor of 4.
3 7 5
R T0 Tf R Tf T0 60. (c) The distribution of electrons in MOs is as follows :
2 3 2 2 1 * *
7T0 3Tf 5Tf 5T0 N2+(electrons 13) 2 *2 2 *2
2 *
12 3 2 *1
2 *2 2 *2 2 *
12T0 8Tf Tf T0 Tf T0 O2 (electrons 16) 2 *1
8 2
2 *
nR T 1000 8.3 7 O22– (electrons 18) 2 *2 2 *2 2 *
52. (a) W 146000 2 *
1 1
1
58.1 58.1 B2 (electrons 10) 2 *2 2 *2 * *
or 1 1.4 1
146 146
53. (c) A crystal structure is composed of a unit cell, a set of Only O22– does not contain any unpaired electron.
atoms arranged in a particular way; which is periodically 61. (b) [Co(NO)2(NH3)5]Cl2
repeated in three dimensions on a lattice. The spacing pentaammine nitro-N-cobalt (III) chloride
between unit cells in various directions is called its 6 4
lattice parameters or constants. Increasing these lattice 62. (c) 2HI 1 H 2SO 4 I 02 SO 2 2H 2 O
constants will increase or widen the band-gap (Eg), Cl
which means more energy would be required by 1
electrons to reach the conduction dand from the 6 2
63. (a)
valence band. Automatically Ec and Ev decreases. 5
54. (a) Erms = 720 3 Br
4
1 3-bromo-1chlorocyclohexene double bond is senior to
The average total energy density 0 E 20 X and the name is written in alphabetical order.
2
64. (c) CH2=CH–O–CH3 + Hbr CH2=CH.OH + CH3Br
1 2 2 Tautomerisation
0 [ 2 E rms ] 0 E rms
2 CH3.CHO
12 65. (a) Addition of HCN to carbonyl compounds is
8.85 10 (720)2 nucleophilic addition reaction. The order of reactivity
4.58 10 6
J / m3 of carbonyl compounds is
Aldehydes (smaller to higher) Ketones (smaller to
2r 2 (d1 d 2 )g higher), Then
55. (c) VT HCHO > CH3COCH3 > Ph.COCH3 > PhCOPh
9
x
VT2 (10.5 1.5) 9 66. (a) In the physical adsorption P
VT2 0.2 m
0.2 (19.5 1.5) 18 67. (c) In a reversible process the work done is greater than in
irreversible process. Hence the heat absorbed in
VT2 0.1 m / s reversible process would be greater than in the latter
case. So
SECTION II : CHEMISTRY Tf(rev.) < Tf (irr.)
56. (d) 1 Mole of Mg3(PO4)2 contains 8 mole of oxygen atoms 68. (c) Accurate velocity
8 mole of oxygen atoms 1 mole of Mg3(PO4)2 0.001 300
0.25 mole of oxygen atom 0.003 ms 1
100
1
0.25 3.125 10 2 6.62 10 34
8 x
31
57. (b) The face centered cubic unit cell contains 4 atom 4 3.14 0.003 9.1 10
Total volume of atoms 1.92 10 2
m
4 3 16 3
4 r r 69. (d) Acetonitrile CH 3 C N and acetone
3 3
58. (b) Angular momentum of an electron, +
H3C –
nh C = O are polar molecules and have dipole-
mvr +
2 H3C
when n = 5 diple interactions.
AIEEE-2006 Solved Paper S-139
o
Hence NaCl is required.
73. (c) In Arrhenius equation K = A e–E/RT, E is the energy of
activation, which is required by the colliding molecules
to react resulting in the formation of products.
74. (a) Metal atom in the lower oxidation state forms the ionic 84. (d) CH 3 OH C 6 H 5 MgBr CH 3 O.MgBr C 6 H 6
bond and in the higher oxidation state the covalent 85. (a) The pyrimidine bases present in DNA are cytosine and
bond. Hence MCl2 is more ionic than MCl4. thymine.
75. (d) In SF4 the hybridisation is sp3d and the shape of 86. (d) The order of stability of free radicals
molecule is • •
F (C H ) C (C H ) CH ]
6 5 3 6 5 2
F
• •
(CH3 )3 C (CH3 )2 CH
S
The stabilisation of first two is due to resonance and
F last two is due to inductive effect.
F NH2 N2Cl
5 3 •
Na NO 2 HCl HBF4
76. (b) The HNO3 is stronger than HNO 2 . The more the 87. (d) 0 5 º diazotisation
oxidation state of N, the more is the acid character.
77. (c) Isoelectronic species have same number of electrons. N2+BF4 F
K+, Cl–, Ca2+, Sc3+ each contains 18 electrons.
78. (d) [Ni X4]2–, the electronic configuration of Ni 2+ is + BF3 + N2
Cl ( x m 1)( x m 1) 0
| ••
H – C – Cl + NaOH Cl – C – Cl + NaCl + H2O x m 1, m 1
| dichlorocarbene m – 1 > –2 and m 1 4
Cl m 1 and m 3
-elimination or, 1 m 3
SECTION III - MATHEMATICS
2
d i2
117. (a) x (Here deviations are taken from the
111. (b) x 2 px q 0 n
Sum of roots = tan30° + tan15° = – p mean). Since A and B both have 100 consecutive
Products of root = tan30° . tan15° = q integers, therefore both have same standard deviation
and hence the variance.
tan 30 tan15 p
tan 45 1 VA
1 tan 30 . tan15 1 q 1 (As d i2 is same in both the cases)
VB
– p 1 q
y
q p 1
2 q p 3
6
x 118. (a) x
112. (b) I dx O 3
9 x x
3
6
9 x 2 sin 2 x 5 sin x 3 0
I dx (sin x 3)(2 sin x 1) 0
9 x x
3
1
6 sin x and sin x 3
3 2
2I dx 3 I
2 In [0, 3 ] , x has 4 values.
3
113. (d) Magnitude of Resultant 119. (d) (a b ) c a ( b c) , a.b 0 , b. c 0
1
2
1
2
AB 2 AC 2 (a. c ). b ( b. c )a (a. c ). b (a. b ).c
AB AC AB.AC (a. b).c ( b. c)a
BC BC 1 a || c .
=
AB.AC AD.BC AD 10 2k 2k
120. (d) sin i cos
C k 1 11 11
10 2k 10 2k
= sin i cos = 0 i ( 1) i
k 1 11 k 1 11
D
121. (d) I xf (sin x)dx ( x)f (sin x)dx
0 0
A B f (sin x )dx I
114. (b) 2 2
A B (A B)(A B) 0
A2 B 2 A 2 AB BA B 2
2I f (sin x)dx
AB BA
115. (b) Clearly ( x, x ) R x W . So R is relexive. 0
/2 /2
Let ( x, y) R , then ( y, x ) R as x and y have at least
one letter in common. So, R is symmetric. I f (sin x )dx f (sin x )dx = f (cos x)dx
2
But R is not transitive for example 0 0 0
Let x = INDIA, y = BOMBAY ˆi 2ˆj 6kˆ ; CA
and z = JUHU 122. (a) BA (2 a)iˆ 2 ˆj ;
then ( x, y) R and ( y, z ) R but ( x, z ) R . CB (1 a)iˆ 6kˆ
116. (c) Equation x 2 2 mx m2 1 0 CA.CB 0 (2 a )(1 a ) 0
2 a 2, 1
( x m) 1 0
S-142 The Pattern Target AIEEE
e mmr a4 a3
123. (d) P(X r) 4. .2a 1 8 4
r! 4 3 a
4 3
= P(X 1) P(X 0) P(X 1) 4 3
a3 a
6 4. 3
=e 5 5 e 5 3
e5 35 a 35
= 3 a
u 12 4 16
sin
124. (b) tan 2
3 35 105
4 u a
u cos 4a 16 64
2
1 1 x 2 1
sin sin cos sin cos 127. (a) is of the form x 2 and equality holds for
4 2 4 2 4 2 x x
x= 1
3
2 sin sin 3sin 4 sin3 p
4 4 4 [2a1 (p 1)d]
2 p2
1 128. (d)
sin 2 q q2
30 or 120 [2a1 (q 1)d]
4 4 4 2
R2 R1 2a1 (p 1)d p
2a1 (q 1)d q
p 1
a1 d
u 2 p
4 4 q 1 q
/2 a1 d
2
u/2 u
a6 a6 11
For , p 11, q 41
1 2 a 21 a 21 41
125. (a) Using h gt and
2
129. (d) (1 ax) 1 (1 bx ) 1
1
h 400 g ( t 4) 2 (1 ax a 2 x 2 ...)(1 bx b 2 x 2 ...)
2
Coefficient of xn
xn bn abn 1 a 2bn 2 ....... a n 1b a n
h bn 1 a n 1
b a
Q(t) bn 1
an 1
400m an
b a
P(t+4) x
, x 0
Subtracting, we get 400 = 8g + 4gt 130. (c) f (x) 1 x
t = 8 sec x
, x 0
1 1 x
h 10 64 320m
2 x
Desired height = 320 + 400 = 720 m , x 0
(1 x)2
a 3x 2 a 2x f '(x)
126. (a) Given, Parabola : y 2a x
3 2 , x 0
(1 x)2
Vertex : ( , )
f ' ( x ) exist at everywhere.
a2
2 3
2a 3 4a
3
AIEEE-2006 Solved Paper S-143
3x 2 9 x 17
2 137. (b) y
2
131. (c) I [(x )3 cos2 (x 3 )]dx 3x 9x 7
3 3x 2 ( y 1) 9x ( y 1) 7 y 17 0
2 x is real
D 0
Put x t
81(y 1) 2 4 3(y 1)(7y 17) 0
2 2 ( y 1)( y 41) 1 y 41 0
I [t 3 cos2 t)dt 2 cos2 tdt 138. (c) The equation of axes is xy 0
Hence the equation of the line is
2 2
x.4 y.3
12 4x 3y 24
2
2
= (1 cos 2t)dt 0 139. (d) (1 y)m (1 y)n
2
m m
0 [1 C1y C 2 y 2 ......] [1 n
C1y n
C 2 y 2 .....]
1 2 m(m 1) n(n 1)
132. (c) Area = x sin = 1 (n m) mn y 2 .....
2 2 2
x a1 n m 10
x
m2 n2 m n 2 mn
and a 2 10
2
Maximum value of sin is 1 at So, n – m = 10
2
and (m n ) 2 (m n) 20
1 2
A max x at sin 1, m n 80
2 2
m = 35, n = 45
1 2 a 0 x b z d
133. (c) A B 140. (a) Equation of lines y
3 4 0 b a c
a 2b x b' z d'
AB y
3a 4b a' c'
Line are perpendicular aa ' 1 cc' 0
a 0 1 2 a 2a
BA 141. If ( , , ) be the image, then
0 b 3 4 3b 4 b
Hence, AB = BA only when a = b 1 3
2 0
134. (a) 2ae 6 ae 3 2 2
2b 8 b 4 1 2 6 0 2 7 ..... (1)
2 2 2
b a (1 e ) 1 3 4
..... (2)
1 2 0
16 a2 a 2e2
From (1) and (2)
a 2 16 9 25 9 13
a 5 , , 4
5 5
3 3 None of the options matches.
e=
a 5 142. (a) x m .y n (x y) m n
135. (c) 10 10 10 10 = mlnx + nlny = (m + n)ln(x + y)
C1 C2 C2 C2
differentiating both sides.
10 45 120 210 385
dy m n dy m n dy
136. (b) 2x 5 1
dx x y dx x y dx
m1 (2x 5) ( 2, 0) 1, m m n m n n dy
m2 (2x 5)(3, 0) 1 x x y x y y dx
m1m 2 1 my nx my nx dy dy y
i.e. the tangents are perpendicular to each other. x( x y) y( x y) dx dx x
S-144 The Pattern Target AIEEE
AIEEE - 2007
Time : 2 ½ Hours Max. Marks : 360
27. Carbon, silicon and germanium have four valence electrons 34. A sound absorber attenuates the sound level by 20 dB. The
each. At room temperature which one of the following intensity decreases by a factor of
statements is most appropriate ? (a) 100 (b) 1000 (c) 10000 (d) 10
(a) The number of free electrons for conduction is 35. If CP and CV denote the specific heats of nitrogen per unit
significant only in Si and Ge but small in C. mass at constant pressure and constant volume respectively,
(b) The number of free conduction electrons is significant then
in C but small in Si and Ge.
(a) CP – CV = 28R (b) CP – CV = R/28
(c) The number of free conduction electrons is negligibly
(c) CP – CV = R/14 (d) CP – CV = R
small in all the three.
36. A charged particle moves through a magnetic field
(d) The number of free electrons for conduction is
significant in all the three. perpendicular to its direction. Then
28. A charged particle with charge q enters a region of constant, (a) kinetic energy changes but the momentum is constant
(b) the momentum changes but the kinetic energy is constant
uniform and mutually orthogonal fields E and B with a
(c) both momentum and kinetic energy of the particle are
velocity v perpendicular to both E and B , and comes out not constant
without any change in magnitude or direction of v . Then (d) both momentum and kinetic energy of the particle are
(a) v B E / E2 (b) v E B / B2 constant
37. Two identical conducting wires AOB and COD are placed
(c) v B E / B 2 (d) v E B / E 2
at right angles to each other. The wire AOB carries an electric
29. The potential at a point x (measured in m) due to some
current I1 and COD carries a current I2. The magnetic field
charges situated on the x-axis is given by V(x) = 20/(x2 – 4)
on a point lying at a distance d from O, in a direction
volt. The electric field E at x = 4 m is given by
perpendicular to the plane of the wires AOB and COD, will
(a) (10/9) volt/ m and in the +ve x direction be given by
(b) (5/3) volt/ m and in the –ve x direction
1
(c) (5/3) volt/ m and in the +ve x direction
0 I1 I2 2
(d) (10/9) volt/ m and in the –ve x direction (a) 0
(I12 2
I2 ) (b)
2 d 2 d
30. Which of the following transitions in hydrogen atoms emit
photons of highest frequency? 1
(a) n = 1 to n = 2 (b) n = 2 to n = 6 (c) 0
I12 I 22 2 (d) 0
I1 I2
(c) n = 6 to n = 2 (d) n = 2 to n = 1 2 d 2 d
31. A block of mass m is connected to another block of mass M 38. The resistance of a wire is 5 ohm at 50°C and 6 ohm at
by a spring (massless) of spring constant k. The block are 100°C. The resistance of the wire at 0°C will be
kept on a smooth horizontal plane. Initially the blocks are (a) 3 ohm (b) 2 ohm (c) 1 ohm (d) 4 ohm
at rest and the spring is unstretched. Then a constant force F
39. A parallel plate condenser with a dielectric of dielectric
starts acting on the block of mass M to pull it. Find the
constant K between the plates has a capacity C and is charged
force of the block of mass m.
to a potential V volt. The dielectric slab is slowly removed
MF mF mF from between the plates and then reinserted. The net work
(a) (b) (c) (M m)F (d)
(m M) M m (m M) done by the system in this process is
32. Two lenses of power –15 D and +5 D are in contact with
1
each other. The focal length of the combination is (a) zero (b) (K 1) CV 2
(a) + 10 cm (b) – 20 cm (c) – 10 cm (d) + 20 cm 2
33. One end of a thermally insulated rod is kept at a temperature 2
T1 and the other at l2. The rod is composed of two sections (c) CV (K 1) (d) (K 1) CV 2
of length l1 and l2 and thermal conductivities K1 and K2 K
respectively. The temperature at the interface of the two 40. If gE and gM are the accelerations due to gravity on the
section is surfaces of the earth and the moon respectively and if
T1 l1 l2 T2
Millikan’s oil drop experiment could be performed on the
two surfaces, one will find the ratio
K1 K2 electronic charge on the moon
to be
(K 2 l2 T1 K1l1T2 ) electronic charge on the earth
(a) (K1l1T1 K 2l2 T2 ) (b)
(K1l1 K 2l2 ) (K1l1 K 2l2 )
(a) g M / g E (b) 1
(K 2 l1T1 K1l2 T2 ) (K1l2 T1 K 2l1T2 )
(c) (d) (c) 0 (d) g E / g M
(K 2 l1 K1l2 ) (K1l2 K 2 l1 )
S-148 The Pattern Target AIEEE
48. The charge/size ratio of a cation determines its polarizing
power. Which one of the following sequences represents
the increasing order of the polarizing power of the cationic
species, K+, Ca2+, Mg2+, Be2+?
(a) Ca2+ < Mg2+ < Be+ < K+
41. The equivalent conductances of two strong electrolytes at (b) Mg2+ < Be2+ < K+ < Ca2+
infinite dilution in H2O (where ions move freely through a (c) Be2+ < K+ < Ca2+ < Mg2+
solution) at 25°C are given below : (d) K+ < Ca2+ < Mg2+ < Be2+.
49. The density (in g mL–1) of a 3.60 M sulphuric acid solution
CH3COONa 91.0 S cm 2 / equiv. that is 29% H2SO4 (molar mass = 98 g mol–1) by mass will
be
HCl 426.2 S cm 2 / equiv. (a) 1.45 (b) 1.64 (c) 1.88 (d) 1.22
What additional information/ quantity one needs to calculate 50. The first and second dissociation constants of an acid H2 A
A° of an aqueous solution of acetic acid? are 1.0 × 10–5 and 5.0 × 10–10 respectively. The overall
dissociation constant of the acid will be
(a) of chloroacetic acid (ClCH2COOH) (a) 0.2 × 105 (b) 5.0 × 10–5
(c) 5.0 × 1015 (d) 5.0 × 10–15 .
(b) of NaCl 51. A mixtuve of ethyl alcohol and propyl alcohol has a vapour
(c) of CH3COOK pressure of 290 mm at 300 K. The vapour pressure of propyl
alcohol is 200 mm. If the mole fraction of ethyl alcohol is
(d) the limiting equivalent coductance of H ( ). 0.6, its vapour pressure (in mm) at the same temperature
H
will be
42. Which one of the following is the strongest base in aqueous (a) 360 (b) 350 (c) 300 (d) 700
solution ? 52. In conversion of lime-stone to lime,
(a) Methylamine (b) Trimethylamine CaCO3(s) CaO(s) CO2(g)
(c) Aniline (d) Dimethylamine.
43. The compound formed as a result of oxidation of ethyl the values of H and S are + 179.1 kJ mol-1 and 160.2 J/
benzene by KMnO4 is K respectively at 298 K and 1 bar. Assuming that H and
(a) benzyl alcohol (b) benzophenone S do not change with temperature, temperature above
(c) acetophenone (d) benzoic acid. which conversion of limestone to lime will be spontaneous
is
(a) 1118 K (b) 1008 K (c) 1200 K (d) 845 K.
44. The IUPAC name of is 53. The energies of activation for forward and reverse reactions
for A2 + B2 2AB are 180 kJ mol–1 and 200 kJ mol–1
(a) 3-ethyl-4-4-dimethylheptane respectively. The presence of a catalyst lowers the activation
(b) 1, 1-diethyl-2,2-dimethylpentane energy of both (forward and reverse) reactions by 100 kJ
(c) 4, 4-dimethyl-5,5-diethylpentane mol–1. The enthalpy change of the reaction (A2 + B2 ®
(d) 5, 5-diethyl-4,4-dimethylpentane. 2AB) in the presence of a catalyst will be (in kJ mol–1)
(a) 20 (b) 300 (c) 120 (d) 280
45. Which of the following species exhibits the diamagnetic
behaviour ? 54. The cell, Zn | Zn 2 (1 M) || Cu 2 (1 M) | Cu (E cell 1.10 V)
(a) NO (b) O22– (c) O2 + (d) O2. was allowed to be completely discharged at 298 K. The
46. The stability of dihalides of Si, Ge, Sn and Pb increases
[Zn 2 ]
steadily in the sequence relative concentration of Zn 2+ to Cu2+ is
[Cu 2 ]
(a) PbX 2 SnX 2 GeX 2 SiX 2
(a) 9.65 × 104 (b) antilog (24.08)
(b) GeX2 << SiX2 << SnX2 << PbX2 (c) 37.3 (d) 1037.3.
(c) SiX2 << GeX2 << PbX2 << SnX2 55. The pKa of a weak acid (HA) is 4.5. The pOH of an aqueous
(d) SiX2 << GeX2 << SnX2 << PbX2. buffer solution of HA in which 50% of the acid is ionized is
47. Identify the incorrect statement among the following. (a) 7.0 (b) 4.5 (c) 2.5 (d) 9.5
(a) Br2 reacts with hot and strong NaOH solution to give 56. Consider the reaction, 2A + B products. When
NaBr and H2O. concentration of B alone was doubled, the half-life did not
(b) Ozone reacts with SO2 to give SO3. change. When the concentration of A alone was doubled,
the rate increased by two times. The unit of rate constant for
(c) Silicon reacts with NaOH(aq) in the presence of air to
this reaction is
give Na2SiO3 and H2O. (a) s–1 (b) L mol–1 s–1
(d) Cl2 reacts with excess of NH3 to give N2 and HCl. (c) no unit (d) mol L–1 s–1.
AIEEE-2007 Solved Paper S-149
57. Identify the incorrect statement among the following: 65. In which of the following ionization processes, the bond
(a) 4f and 5f orbitals are equally shielded. order has increased and the magnetic behaviour has
(b) d-Block elements show irregular and erratic chemical changed?
properties among themselves.
(a) N 2 N2 (b) C2 C2
(c) La and Lu have partially filled d-orbitals and no other
partially filled orbitals. (c) NO NO (d) O 2 O2 .
(d) The chemistry of various lanthanoids is very similar.
58. Which of the following has a square planar geometry? 66. The actinoids exhibit more number of oxidation states in
general than the lanthanoids. This is because
(a) [PtCl4]2– (b) [CoCl4]2– (c) [FeCl4]2– (d) [NiCl4]2–
(a) the 5f orbitals extend further from the nucleus than the
(At. nos.: Fe = 26, Co = 27, Ni = 28, Pt = 78)
59. Which of the following molecules is expected to rotate the 4f orbitals
(b) the 5f orbitals are more buried than the 4f orbitals
plane of plane-polarised light?
(c) there is a similarity between 4f and 5f orbitals in their
CHO angular part of the wave function
COOH (d) the actinoids are more reactive than the lanthanoids.
(a) H2N (b) HO H 67. Equal masses of methane and oxygen are mixed in an empty
H
container at 25°C. The fraction of the total pressure exerted
H by oxygen is
CH2OH
1 273
(a) 1/2 (b) 2/3 (c) (d) 1/3.
3 298
(c)
68. A 5.25% solution of a substance is isotonic with a 1.5% solution
SH of urea (molar mass = 60 g mol–1) in the same solvent. If the
densities of both the solutions are assumed to be equal to 1.0
H2N NH2 g cm–3, molar mass of the substance will be
(a) 210.0 g mol–1 (b) 90.0 g mol–1
(d) H H –1
(c) 115.0 g mol (d) 105.0 g mol–1.
Ph Ph 69. Assuming that water vapour is an ideal gas, the internal
60. The secondary structure of a protein refers to energy change ( U) when 1 mol of water is vapourised at
(a) fixed configuration of the polypeptide backbone
1 bar pressure and 100°C, (given : molar enthalpy of
(b) helical backbone
vapourisation of water at 1 bar and 373 K = 41 kJ mol –1 and
(c) hydrophobic interactions
R = 8.3 J mol–1 K–1) will be
(d) sequence of amino acids.
(a) 41.00 kJ mol–1 (b) 4.100 kJ mol–1
61. Which of the following reactions will yield 2, 2- –1
(c) 3.7904 kJ mol (d) 37.904 kJ mol–1
dibromopropane?
70. In a saturated solution of the sparingly soluble strong
(a) CH3 – CH = CH2 + HBr
electrolyte AgIO3 (molecular mass = 283) the equilibrium
(b) CH3 – C CH + 2HBr
(c) CH3CH = CHBr + HBr which sets in is AgIO3 (s) Ag (aq) IO3 (aq) . If the
(d) CH CH + 2HBr
62. In the chemical reaction, solubility product constant K sp of AgIO3 at a given
CH3CH2NH2 + CHCl3 + 3KOH (A) + (B) + 3H2O, the temperature is 1.0 × 10–8 , what is the mass of AgIO3
compounds (A) and (B) are respectively contained in 100 ml of its saturated saolution?
(a) C2H5NC and 3KCl (b) C2H5CN and 3KCl (a) 1.0 × 10– 4 g (b) 28.3 × 10–2 g
(c) CH3CH2CONH2 and 3KCl(d) C2H5NC and K2CO3. (c) 2.83 × 10–3 g (d) 1.0 × 10–7 g.
63. The reaction of toluene with Cl2 in presence of FeCl3 gives 71. A radioactive element gets spilled over the floor of a room.
predominantly Its half-life period is 30 days. If the initial velocity is ten
(a) m-chlorobenzene (b) benzoyl chloride times the permissible value, after how many days will it be
(c) benzyl chloride (d) o- and p-chlorotoluene. safe to enter the room?
64. Presence of a nitro group in a benzene ring (a) 100 days (b) 1000 days (c) 300 days (d) 10 days.
(a) deactivates the ring towards electrophilic substitution 72. Which one of the following conformations of cyclohexane
(b) activates the ring towards electrophilic substitution is chiral?
(c) renders the ring basic (a) Boat (b) Twist boat
(d) deactivates the ring towards nucleophilic substitution. (c) Rigid (d) Chair.
S-150 The Pattern Target AIEEE
73. Which of the following is the correct order of decreasing
SN2 reactivity?
(a) R2CH X > R3C X > RCH2 X
(b) RCH X > R3C X > R2CH X
(c) RCH2 X > R2CH X > R3C X
(d) R3C X > R2CH X > RCH2 X. 81. The resultant of two forces Pn and 3n is a force of 7n. If the
(X is a halogen) direction of 3n force were reversed, the resultant would be
74. In the following sequence of reactions, 19 n. The value of P is
P I2 Mg HCHO H 2O (a) 3 n (b) 4 n (c) 5 n (d) 6 n.
CH3 CH 2 OH A B C D
ether 82. Two aeroplanes I and II bomb a target in succession. The
the compound D is probabilities of I and II scoring a hit correctly are 0.3 and
(a) propanal (b) butanal 0.2, respectively. The second plane will bomb only if the
(c) n-butyl alcohol (d) n-propyl alcohol. first misses the target. The probability that the target is hit
75. Which of the following sets of quantum numbers represents by the second plane is
the highest energy of an atom? (a) 0.2 (b) 0.7 (c) 0.06 (d) 0.14.
(a) n = 3, l = 0, m = 0, s = +1/2 1 1 1
(b) n = 3, l = 1, m = 1, s = +1/2 83. If D = 1 1 x 1 for x 0, y 0 , then D is
(c) n = 3, l = 2, m = 1, s = +1/2 1 1 1 y
(d) n = 4, l = 0, m = 0, s = +1/2. (a) divisible by x but not y (b) divisible by y but not x
76. Which of the following hydrogen bonds is the strongest? (c) divisible by neither x nor y(d) divisible by both x and y
(a) O – H - - - F (b) O – H - - - H
(c) F – H - - - F (d) O – H - - - O. x2 y2
84. For the Hyperbola 1 , which of the
77. In the reaction, cos 2 sin 2
following remains constant when varies =?
2A (s) 6HC (aq) 2A 3 (aq) 6C (aq) 3H 2(g)
(a) abscissae of vertices (b) abscissae of foci
(a) 11.2 L H2(g) at STP is produced for every mole HCl(aq) (c) eccentricity (d) directrix.
consumed 85. If a line makes an angle of / 4 with the positive directions
(b) 6 L HCl(aq) is consumed for every 3 L H2(g) produced of each of x- axis and y- axis, then the angle that the line
(c) 33.6 L H2(g) is produced regardless of temperature and makes with the positive direction of the z-axis is
pressure for every mole Al that reacts
(d) 67.2 H2(g) at STP is produced for every mole Al that (a) (b) (c) (d)
4 2 6 3
reacts.
78. Regular use of the following fertilizers increases the acidity 86. A value of c for which conclusion of Mean Value Theorem
of soil? holds for the function f (x) = loge x on the interval [1, 3] is
(a) Ammonium sulphate (b) Potassium nitrate 1
(a) log3 e (b) loge3 (c) 2 log3e (d)
log3e
(c) Urea (d) Superphosphate of lime 2
79. Identify the correct statement regarding a spontaneous 87. The function f (x) = tan –1(sin x + cos x) is an increasing
process: function in
(a) Lowering of energy in the process is the only criterion
for spontaneity. (a) 0, (b) , (c) , (d) ,
2 2 2 4 2 2 4
(b) For a spontaneous process in an isolated system, the
change in entropy is positive.
5 5
(c) Endothermic processes are never spontaneous. 88. Let A = 0 5 . If A 2 25 , then equals
(d) Exothermic processes are always spontaneous. 0 0 5
80. Which of the following nuclear reactions will generate an (a) 1/5 (b) 5 (c) 52 (d) 1
isotope?
1 1 1
(a) - particle emission (b) Neutron praticle emission 89. The sum of series ....... upto infinity is
2! 3! 4!
(c) Positron emission (d) particle emission.
1 1
(a) (b) (c) e–2 (d) e–1
e 2 e 2
AIEEE-2007 Solved Paper S-151
90. If û and v̂ are unit vectors and is the acute angle between x
1 log t
them, then 2 û ×3 v̂ is a unit vector for 99. Let F(x) = f (x) + f ,where f (x) dt, Then F(e)
x 1 t
(a) no value of l
(a) (2, 4) (b) (–2, 0) (c) (–1, 1) (d) (0, 2) (a) 3/2 (b) 2 2 (c) 2 (d) none
94. If (2, 3, 5) is one end of a diameter of the sphere x2 + y2 + z2
– 6x – 12y – 2z + 20 = 0, then the cooordinates of the other
end of the diameter are dx
103. equals
(a) (4, 3, 5) (b) (4, 3, – 3) (c) (4, 9, – 3) (d) (4, –3, 3). cos x 3 sin x
x 5 1 1 1
107. If sin 1 cosec 1
, then the values of x is (a) k (b) k
5 4 2 2 2 2
(a) (b) 0, 117. A tower stands at the centre of a circular park. A and B are
,
4 2 2 two points on the boundary of the park such that AB (= a)
subtends an angle of 60° at the foot of the tower, and the
angle of elevation of the top of the tower from A or B is 30°.
(c) [0, ] (d) ,
2 2 The height of the tower is
111. A body weighing 13 kg is suspended by two strings 5m and (a) a / 3 (b) a 3 (c) 2a / 3 (d) 2a 3
12m long, their other ends being fastened to the extremities 118. The sum of the series
of a rod 13m long. If the rod be so held that the body hangs 20 20 20 20 20
C0 C1 C2 C3 ..... ..... C10 is
immediately below the middle point, then tensions in the
strings are (a) 0 (b) 20
C10
(a) 5 kg and 12 kg (b) 5 kg and 13 kg
(c) 12 kg and 13 kg (d) 5 kg and 5 kg 1 20
(c) 20
C10 (d) C10
112. A pair of fair dice is thrown independently three times. The 2
probability of getting a score of exactly 9 twice is 119. The normal to a curve at P(x, y) meets the x-axis at G. If the
(a) 8/729 (b) 8/243 (c) 1/729 (d) 8/9. distance of G from the origin is twice the abscissa of P, then
113. Consider a family of circles which are passing through the the curve is a
point (– 1, 1) and are tangent to x-axis. If (h, k) are the (a) circle (b) hyperbola
coordinate of the centre of the circles, then the set of values (c) ellipse (d) parabola.
of k is given by the interval 120. If | z + 4 | 3, then the maximum value of | z + 1 | is
(a) 6 (b) 0 (c) 4 (d) 10
AIEEE-2007 Solved Paper S-153
SECTION I – PHYSICS m R2 1
x= R .R = R
M' 2 3
1. (b) Let O be the centre of mass of the disc having radius 3 R
2R. O' is the new C.M.
But x =
R
2R
1
M' m =.R
R
O' x O R
3
There appears misprint in this question. There must be
1 1
R instead of . Then R R
R 3 3
Let m = mass of disc of radius R
M' = mass of disc when the disc of radius R is removed. 1
| |
M = mass of disc of radius 2R 3
2. (b) The acceleration of a solid sphere of mass M, radius R
Now, m ( R 2 ). ,
and moment of inertia I rolling down (without slipping)
M M an inclined plane making an angle with the
where 2
= the mass per unit area
(2R) 4 R2 horizontal is given by
A( 2, 2) a r12
r1 I , where I is total current
r1 2 a2
Q X
(0,0) r2 B (2,0)
40x
or, E = K1 K2
2
(x 4) 2
K1A(T1 T) K 2 A(T T2 )
At x = 4 m , ,
1 2
40 4 160 10 where A is the area of cross-section.
E= 2 2 volt / m.
(4 4) 144 9 or, K1A(T1 T) 2 K 2 A(T T2 ) 1
or, K1T1 2 K1T 2 K 2T 1 K 2T2 1
Positive sign indicates that E is in +ve x-direction.
30. (d) We have to find the frequency of emitted photons. For or, (K 2 1 K1 2 )T K1T1 2 K 2 T2 1
emission of photons the transition must take place from
K1T1 2 K 2 T2 1 K1 2 T1 K 2 1T2
a higher energy level to a lower energy level which are T .
given only in options (c) and (d). K2 1 K1 2 K1 2 K 2 1
1 1 I1
34. (a) We have, L1 10 log
Frequency is given by, h 13.6 I0
n12 n 22
For transition from n = 6 to n = 2, I2
L2 10 log
I0
13.6 1 1 2 13.6
1 2 2
h 6 2 9 h I1 I2
L1 – L2 = 10 log 10 log
For transition from n = 2 to n = 1, I0 I0
13.6 1 1 3 13.6
2 . I1 I2 I1
h 22 1 2 4 h or, L 10 log or, L 10log
I0 I0 I2
1 2
I1 I1
Hence option (d) is the correct answer. or, 20 10log or, 2 log
31. (d) Writing free body-diagrams for m & M, I2 I2
M I1 I1
m or, 102 or, I2
K .
F I2 100
N Intensity decreases by a factor 100.
N 35. (b) We have,
M a
m Molar heat capacity
T T = Molar mass × Specific heat capacity per unit mass
F
Cp = 28 Cp (for nitrogen)
mg and Cv = 28 Cv
Mg
Now, Cp – Cv = R or, 28 Cp – 28 Cv = R
we get T = ma and F – T = Ma
where T is force due to spring R
Cp – Cv = .
F – ma = Ma or, F = Ma + ma 28
S-158 The Pattern Target AIEEE
36. (b) When a charged particle enters a magnetic field at a SECTION II - CHEMISTRY
direction perpendicular to the direction of motion, the
path of the motion is circular. In circular motion the 41. (b) According to Kohlrausch’s law, molar conductivity of
direction of velocity changes at every point (the weak electrolyte acetic acid (CH3COOH) is given as
magnitude remains constant). Therefore, the follows:
momentum will change at every point. But kinetic
CH3COOH CH3COONa HCl NaCl
1
energy will remain constant as it is given by mv2
2
and v2 is the square of the magnitude of velocity which Value of NaCl should also be known for calculating
does not change.
value of CH3COOH .
37. (c) Clearly, the magnetic fields at a point P, equidistant
from AOB and COD will have directions perpendicular 42. (d) Aromatic amines are less basic than aliphatic amines.
to each other, as they are placed normal to each other. Among aliphatic amines the order of basicity is 2° > 1°
A > 3° ( of decreased electron density due to crowding
B2 B in 3° amines)
dimethylamine (2° aliphatic amine) is strongest base
B1 among given choices.
I1 d P 43. (d) When alkyl benzene are oxidised with alkaline KMnO4,
the entire alkyl group is oxidised to –COOH group
C O I2 D regardless of length of side chain.
CH2CH3 COOH
KMnO4
B
Resultant field, B B12 B22 Ethyl benzene Benzoic aicd
0 I1 0 I2 CH3
But B1 and B2
7 6 5 4| 3 2 1
2 d 2 d
44. (a) CH3 CH 2 CH 2 C CH CH 2 CH3
2 1/ 2 | |
B 0
I12 I22 or,, B 0
I12 I22 CH3 CH 2
2 d 2 d |
CH3
38. (d) We know that, Rt = R0 (1 + t )
3 ethyl 4,4 dimethyl heptane
R50 = R0 (1 + 50 ) ... (i)
R100 = R0 (1 + 100 ) ... (ii) 45. (b) Diamagnetic species have no unpaired electrons
From (i), R50 – R0 = 50 R 0 ... (iii) O 22 1s 2 , *1s 2 ,
2
s 2 , 2pz , 2p x2 , 2p y 2 ,
From (ii), R100 – R0 = 100 R 0 ... (iv) *2px2, *2py2
46. (d) When ns2 electrons of outermost shell do not participate
R 50 R 0 1
Dividing (iii) by (iv), we get in bonding it is called inert pair and the effect is called
R100 R 0 2 inert pair effect. The last three elements have a tendecy
Here, R50 = 5 and R100 = 6 to form M2+ ions as well as M4+ ions. Since the inert pair
effect increases from Ge to Pb, the stability of M4+ ions
5 R0 1 decreases and that of M2+ ions increases. Thus, the
or, 6 – R0 = 10 – 2 R0 or, R0 = 4 .
6 R0 2 stability of these ions follow the following order :
39. (a) The potential energy of a charged capacitor is given Ge2+ < Sn2+ < Pb2+
47. (d) Chlorine reacts with excess of ammonia to produce
Q2 ammonium chloride and nitrogen.
by U .
2C 8NH3 + 3Cl2 N2 + NH4Cl
If a dielectric slab is inserted between the plates, the energy 48. (d) Smaller the size and higher the charge more will be
polarising power of cation. So the correct order of
Q2 polarising power is K+ < Ca2+ < Mg2+ < Be2+
is given by , where K is the dielectric constant.
2KC 49. (d) Mass of 3.6 moles of H2SO4 = Moles × Molecular mass
Again, when the dielectric slab is removed slowly its = 3.6 × 98 g = 352.8 g
energy increases to initial potential energy. Thus, work 1000 ml solution has 352.8 g of H2SO4
done is zero. Given that 29 g of H2SO4 is present in = 100 g of solution
40. (b) Electronic charge does not depend on acceleration due 352.8 g of H2SO4 is present in
to gravity as it is a universal constant. So,
electronic charge on earth = electronic charge on moon 100
= 352.8 g of solution = 1216 g of solution
Required ratio = 1. 29
Mass 1216
Density = = = 1.216 g/ml = 1.22 g/ml
Volume 1000
AIEEE-2007 Solved Paper S-159
56. (b) From the given data we can say that order of reaction
50. (d) H2 A H HA with respect to B = 1 because change in concentration
of B does not change half life. Order of reaction with
[H ][HA ]
K1 = 1.0 × 10–5 = respect to A = 1 because rate of reaction doubles when
[H 2 A] concentration of A is doubled keeping concentration of
HA H A A constant.
Order of reaction = 1 + 0 = 1 and units of first order
10 [H ][A ] reaction are L mol–1 sec–1.
K2 5.0 10
[HA ] 57. (a) 4f orbital is nearer to nucleus as compared to 5f orbital
therefore, shielding of 4f is more than 5f.
[H ]2 [A 2 ] 58. (a) Complexes with dsp2 hybridisation are square planar.
K K1 K 2 So [PtCl4]2– is square planar in shape.
[H 2 A]
= (1.0 × 10–5) × (5 × 10–10) = 5 × 10–15 59. (b) The organic compounds which have chiral carbon atom
and do not have plane of symmetry rotate plane polarised
51. (b) Given p0A ? , p0B 200mm , xA = 0.6, light.
xB = 1 – 0.6 = 0.4, P = 290
CHO
P = pA + pB = p0A x A p0B x B |
HO C* H (* is asymmetric carbon)
290 = p0A × 0.6 + 200 × 0.4 p0A = 350 mm |
CH 2 OH
52. (a) G° = H° – T S°
For a spontaneous reaction G° < 0 60. (b) Proteins have two types of secondary structures -helix
H and -plated sheet.
or H° – T S° < 0 T
S 61. (b) The reaction follows Markownikoff rule which states
that when unsymmetrical reagent adds across
179.3 103 unsymmetrical double or triple bond the negative part
T> 1117.9K 1118K
18K
160.2 adds to carbon atom having lesser number of hydrogen
atoms.
53. (a) H R E f E b = 180 – 200 = – 20 kJ/mol
The nearest correct answer given in choices may be CH3 C CH HBr CH3 C CH 2
obtained by neglecting sign. |
54. (d) Ecell = 0; when cell is completely discharged. Br
Zn 2 Br
0.059 |
Ecell = E°cell log HBr
2 Cu 2 CH3 C CH3
|
Br
0.059 Zn 2 2, 2-dibromo-propane
or 0 = 1.1 log 62. (a) This is carbylamine reaction.
2 Cu 2
CH3CH2NH2 + CHCl3 + 3KOH
C2H5NC + 3KCl + 3H2O
Zn 2 2 1.1
63. (d) FeCl3 is Lewis acid. In presence of FeCl3 side chain
log 37.3 hydrogen atoms of toluene are substituted.
2 0.059
Cu
CH3 CH3 CH3
Cl
FeCl3
+ Cl2 +
Zn 2
37.3
10 Toluene o-chloro toluene
Cl
Cu 2 p-chloro toluene
SECTION III - MATHEMATICS 85. (b) Let the angle of line makes with the positive direction of
z-axis is direction cosines of line with the +ve directions
of x-axis, y-axis, and z-axis is l, m, n respectively.
81. (c) Given : Force P = Pn, Q = 3n, resultant R = 7n &
P' = Pn, Q' = (–3)n, R' = 19 l = cos, m = cos , n = cos
4 4
as we know that, l + m2 + n2 = 1
2
We know that R2 = P2 + Q2 + 2PQ cos Hence, angle with positive direction of the z-axis is .
(7)2 = P2 + (3)2 + 2 × P × 3 cos 2
49 = P2 + 9 + 6P cos .....(i) 86. (c) Using Lagrange's Mean Value Theorem
40 = P2 + 6P cos Let f(x) be a function defined on [a, b]
2 f (b) f (a)
and 19 = P2 + (–3)2 + 2P × –3 cos then, f '(c) ....(i)
b a
19 = P2 + 9 – 6P cos c [a, b]
10 = P2 – 6P cos .....(ii) Given f(x) = logex
Adding (i) and (ii), 50 = 2P2 P2 = 25 P = 5n 1
82. (d) Required probability = 0.7 × 0.2 + (0.7) × 0.8 × 0.2 + (0.7)3
2 f'(x) =
x
× (0.8)2 × 0.2 + (0.7)4 (0.8)3 × 0.2 +
1 f (3) f (1)
= 0.7 × 0.2 × [1 + (0.7 × 0.8) + (0.7 × 0.8)2 + (0.7 × 0.8)3 +..] equation (i) become
c 3 1
= 0.14 × [1 + (.56) + (.56)2 + (.56)3 + ..]
1 loge 3 log e 1 log e3 2
1 0.14 7 c
= 0.14 0.318 c 2 2 loge 3
1 0.56 0.47 22
c = 2 log3e
1 1 1 87. (d) Given f(x) = tan –1 (sin x + cos x)
83. (d) Given, D = 1 1 x 1
1 1 1 y 1
f '(x) = .(cos x sin x)
Apply R2 R2 – R1 and R R3 – R1 1 (sin x cos x) 2
1 1 1 1 1
D = 0 x 0 = xy 2. cos x sin x
0 0 y 2 2
25 25 5 2 5 25 2 5 bc
2 tan =
= 0 5 2 25 a(c a)
0 0 25
bc
1 The angle of projection, = tan–1
| A2 | = 25 (25 2
), 25 = 25 (25 2
) | | a(c a)
5
92. (a) Let the number of boys be x and that of girls be y.
x2 x3 52x + 42y = 50(x + y) 52x – 50x = 50y – 42y
89. (d) We know that ex = 1 + x + ........
2! 3! x 4
Put x = – 1 x 4
2x = 8y and
y 1 x y 5
1 1 1
e–1 = 1 1 ........
2! 3! 4! x 4
Required % of boys = 100 = 100 = 80 %
1 1 1 1 x y 5
e–1 = ........
2! 3! 4! 5! 93. (b) Parabola y2 = 8x
90. (b) Given | 2uˆ 3vˆ | 1 and is acute angle between û Y y2 8x
and v̂ , | uˆ | 1, | vˆ | 1
1 (2,0)
6 | uˆ | | vˆ | | sin | 1 6 | sin | 1 sin X' X
6 F
0
Hence, there is exactly one value of for which
x 2
2 û × 3 v̂ is a unit vector..
91. (a) Let B be the top of the wall whose coordinates will be Y'
(a, b). Range (R) = c Point must be on the directrix of parabola
u B (a,b) equation of directrix x + 2 = 0 x = –2
Hence the point is (–2, 0)
b 94. (c) We know that equation of sphere is
A x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0
C
a D where centre is (–u, –v, –w)
c
Given x2 + y2 + z2 – 6x – 12y – 2z + 20 = 0
B lies on the trajectory
centre (3, 6, 1)
1 x2 Coordinates of one end of diameter of the sphere are
y = x tan – g
2 u 2 cos 2 (2, 3, 5). Let the coordinates of the other end of diameter
are ( , , )
1 a2
b = a tan – g
2 u 2 cos 2 2 3 5
3, 6, 1
2 2 2
ga
b = a tan 1
2u 2 cos 2 tan = 4, = 9 and = –3
a Coordinate of other end of diameter are (4, 9, –3)
= a tan 1
2u 2
cos2 .
sin 95. (b) Given a ˆi ˆj kˆ , b i j 2k and
g cos
c xi (x 2) j k
= a tan a
1
u 2 .2 sin cos If c lies in the plane of a and b , then [a b c] = 0
g
1 1 1
a i.e. 1 1 2 0
= a tan 1
u 2 sin 2 x (x 2) 1
g
1[1 – 2(x – 2)] – 1[– 1 – 2x] + 1[x – 2 + x] = 0
a 2
u sin 2 1 – 2x + 4 + 1 + 2x + 2x – 2 = 0
= a tan 1 R 2x = –4 x=–2
R g
AIEEE-2007 Solved Paper S-163
96. (a) Given : The vertices of a right angled triangle A(l, k), B 1 x log t
(1, 1) and C (2, 1) and Area of ABC = 1 square unit 99. (c) Given f(x) = f(x) + f ,wheref(x) = dt
x 1 1 t
Y
A (1, k) 1
F(e) = f(e) + f
e
e log
t 1/ e logt
F(e) = dt dt ....(A)
1 1 t 1 1 t
C (2, 1)
B (1, 1) 1/ e log t
O Now for solving, I = dt
X 1 1 t
We know that, area of right angled triangle 1 1 dz
Put z 2
dt dz dt = – 2
1 1 t t z
= × BC × AB = 1 = (1) | (k – 1)|
2 2 and limit for t = 1 z = 1 and for t = 1/e z=e
(k 1) 2 k = – 1, 3 1
log e (log1 log z).z dz
97. (c) Given : The coordinates of points P, Q, R are (–1, 0), e z dz
I=
(0, 0), (3,3 3) respectively.. 1 1 z 2 1 z 1 z2
1
z
Y R (3, 3 3 ) e log z dz
[ log1 = 0]
M 1 (z 1) z
e log z
dz
1 z(z 1)
2 /3 /3
X' X e
log t
P (-1, 0) Q (0, 0) b b
I= dt [By property f (t)dt f (x)dx ]
t(t 1)
1 a a
Y' Equation (A) be,
y2 y1 3 3 e log t e log t
Slope of QR = F(e) = dt dt
x2 x1 3 1 1 t 1 t(1 t)
e t.log t e (log t)(t 1)
tan 3 log t dt
dt
1 t(1 t) 1 t(1 t)
RQX e log t
3 3 F(e) = dt
1 t
2
RQC Let log t = x
3 3
1
2 dt dx [for limit t = 1, x = 0 and t = e, x = log e = 1]
Slope of the line QM = tan =– 3 t
3 1
Equation of line QM is (y – 0) = –
1 x2 1
3 (x – 0) F(e) = x dx ; F(e) = F(e) =
0 2 2
y=– 3x 3x+y=0 0
98. (a) Equation of bisectors of lines, xy = 0 are y = x 100. (a) f(x) = min {x + 1, | x | + 1}
y f(x) = x + 1 x R
Y
(0, 1)
x
(0, 0)
X' X
Put y = x in the given equation (-1, 0)
my2 + (1 – m2)xy – mx2 = 0
mx2 + (1 – m2)x2 – mx2 = 0 Y'
Hence, f(x) is differentiable everywhere for all x R.
1 – m2 = 0 m= 1
S-164 The Pattern Target AIEEE
y=x
1 2 Y
101. (b) Given, f(x) = (1, 1)
x e 2x 1 y = -x A y2 x
1 2 y2
f (0) = lim y1
x 2x X' X
x 0 e 1 (0, 0) O (1, 0)
(e2x 1) 2x 0
lim form
2x 0
x(e x 1) 0
Y'
using, L'Hospital rule,
1
4e 2x required area = (y 2 y1 )dx
f (0) lim 0
x 0 2(xe 2x 2 e 2x .1) e2x .2 1
1 x3 / 2 x2
4e 2x 0 ( x x)dx
lim form 0 3/ 2 2
0
x 0 4xe 2x 2e 2x 2e2x 0
2 3/ 2 1 1 2 1
required area x x
4e 2x 4.e0 3 0 2 0
lim =1
0 4(xe 2x 2x 0
x e ) 4(0 e ) 2 1 1
x dt 3 2 6
102. (c) , dx 1
2 2 2 sec x 105. (c) Let and are roots of the equation x2 + ax + 1 = 0
t t 1 x x 2
1
+ = – a and =1
1 x
sec t sec–1 x – sec–1 2 = given | | 5
2 2 2
sec–1x – = sec–1x = ( )2 4 5
4 2 2 4
( )2 ( )2 4
3 3
sec–1x = x = sec x=– 2
4 4
a2 4 5 a2 – 4 < 5
103. (c) I dx a2
–9<0 2
a <9 –3<a<3 a (–3, 3)
cos x 3 sin x 106. (b) Let the series a, ar, ar 2, ..... are in geometric progression.
given, a = ar + ar 2
dx 1 1 4 1
I
1 3 1= r + r2 r2 + r – 1 = 0 r=
2 cos x sin x 2
2 2
1 5
r= (taking +ve value)
1 dx 2
2 5 1
sin cos x cos sin x r=
6 6 2
1 dx x 5
. 107. (d) sin 1 cosec 1
2 5 4 2
sin x
6 x 5
1 1
sin cosec
1 5 2 4
I= . cosec x dx
2 6 1 x 1 4
sin sin
5 2 5
But we know that cosec x dx log | (tan x / 2) | C
1 1
[ sin x cos x / 2]
1 x
I = . log tan C x 4
2 2 2 1 1 ....(1)
sin cos
104. (a) The area enclosed between the curves 5 5
y2 = x and y = | x | 1 4 4
From the figure, area lies between y2 = x and y = x Let cos A cos A
5 5
AIEEE-2007 Solved Paper S-165
5m 4 – 4(–2k + 2) 0 1 + 2k – 2 0 k
X'
/2
X 2
T2 sin O T1 cos 114. (c) Let the direction cosines of line L be l, m, n, then
13 Kgs
2l + 3m + n = 0 ....(i)
Y' and l + 3m + 2n = 0 ....(ii)
S-166 The Pattern Target AIEEE
C
On solving equation (i) and (ii), we get
In
l m n l m n
h
6 3 1 4 6 3 3 3 3 tan 30
a h
l m n l2 m2 n2
OBA = AOB
Now l2 + m2 + n 2 = 1 = OAB = 60°
3 3 3 3 2
( 3) 2
3 2
30° a
1 h a A
h= 60° O
l m n 1 3 a 3 a a
30°
3 3 3 27
118. (d) We know that, B
3 1 1 1 (1 + x)20 = 20C0 + 20C1x + 20C2 x2 + ......
l= ,m ,n
27 3 3 3 20C x10 + ..... 20C x20
10 20
Line L, makes an angle with +ve x-axis Put x = –1, (0) = 20C0 – 20C1 + 20C2 – 20C3 + ......
1 + 20C10 – 20C11 .... + 20C20
l = cos cos =
3 0 = 2[20C0 – 20C1 + 20C2 – 20C3
115. (a) General equation of circles passing through origin and + ..... – 20C9] + 20C10
having their centres on the x-axis is 20C = 2[20C – 20C + 20C – 20C
10 0 1 2 3
x2 + y2 + 2gx = 0 ...(i) + ...... – 20C9 + 20C10]
On differentiating w.r.t x, we get 20C 1 20
0 – 20C1 + 20C2 – 20C3 + .... + 20C10 = C10
2
dy dy
2x + 2y . + 2g = 0 g=– x y dx
dx dx 119. (b,c) Equation of normal at p(x, y) is Y – y = – (X x)
dy
dy Coordinate of G at X axis is (X, 0) (let)
equation (i) be x2 + y2 + 2 x y .x 0
dx dx dy dy
0 – y= – (X x) y X x X=x+y
dy dx dx
dy dy
x2 + y2 – 2x2 – 2x .y = 0 y2 = x2 + 2xy dy
dx dx Co-ordinate of G x y ,0
116. (c) Since, p and q are positive real numbers dx
p2 + q2 = 1 (Given) Given distance of G from origin = twice of the abscissa of p.
dy
Using AM GM x y | 2x |
dx
p q
2 p2 q 2 2pq dy dy
(pq) 2 pq x+y = 2x or x + y = – 2x
2 4 dx dx
1 2pq dy dy
pq 1 + 2pq 4pq y = x or y = – 3x
4 dx dx
ydy = xdx or ydy = – 3xdx
1 2pq or, 2pq 1 On Integrating,
1 1 y2 x2 y2 3x 2
pq or, pq c1 or c2
2 2 2 2 2 2
x2 y2
– = –2c1 or 3x2
+ = 2c2 y2
Now, (p + q)2 = p2 + q2 + 2pq
the curve is a hyperbola and ellipse both
1 120. (a) z lies on or inside the circle with centre (–4, 0) and radius
(p + q)2 1+2× p+q 2 3 units. Y
2
Im.
117. (a) In the AOB, AOB = 60°, and OBA = OAB
(since OA = OB = AB radius of same circle).
AOB is a equilateral triangle. ( 7, 0) ( 4, 0) ( 1, 0) Real
X
Let the height of tower is h m. X'
Section - 1 5. A parabola has the origin as its focus and the line x = 2 as
the directrix. Then the vertex of the parabola is at
(a) (0, 2) (b) (1, 0) (c) (0, 1) (d) (2, 0)
6. The point diametrically opposite to the point
1. AB is a vertical pole with B at the ground level and A at the P(1, 0) on the circle x2 + y2 + 2x + 4y – 3 = 0 is
top. A man finds that the angle of elevation of the point A (a) (3, – 4) (b) (–3, 4) (c) (–3, –4) (d) (3, 4)
from a certain point C on the ground is 60°. He moves away 7. Let f: N Y be a function defined as f(x) = 4x + 3, where
from the pole along the line BC to a point D such that CD =7 Y = {y N : y = 4x + 3 for some x N}.
m. From D the angle of elevation of the point A is 45°. Then Show that f is invertible and its inverse is
the height of the pole is
3y 4 y 3
(a) g(y) (b) g(y) 4
7 3 1 7 3 3 4
(a) m (b) ( 3 1)m
2 3 –1 2 y 3 y–3
(c) g(y) (d) g(y)
4 4
7 3 7 3 1 1
(c) ( 3 –1)m (d) m 8. The conjugate of a complex number is then that
2 2 3 1 i –1
2. It is given that the events A and B are such that complex number is
–1 1 –1 1
1 1 2 (a) (b) (c) (d)
P(A) , P(A | B) and P(B | A) . i –1 i 1 i 1 i –1
4 2 3
9. Let R be the real line. Consider the following subsets of the
Then P(B) is plane R × R:
1 1 2 1 S ={(x, y): y = x + 1 and 0 < x < 2}
(a) (b) (b) (c) T ={(x, y): x – y is an integer},
6 3 3 2
Which one of the following is true?
3. A die is thrown. Let A be the event that the number obtained
(a) Neither S nor T is an equivalence relation on R
is greater than 3. Let B be the event that the number obtained
(b) Both S and T are equivalence relation on R
is less than 5. Then P(A B) is
(c) S is an equivalence relation on R but T is not
3 2 (d) T is an equivalence relation on R but S is not
(a) (b) 0 (c) 1 (d)
5 5 10. The perpendicular bisector of the line segment joining P(1,
4. A focus of an ellipse is at the origin. The directrix is the line 4) and Q(k, 3) has y-intercept –4. Then a possible value of k
is
1 (a) 1 (b) 2 (c) –2 (d) – 4
x = 4 and the eccentricity is . Then the length of the semi-
2 dy x y
major axis is 11. The soluton of the differential equation
dx x
8 2 4 5 satisfying the condition y(1) =1 is
(a) (b) (c) (c) (a) y = ln x + x (b) y = x ln x + x2
3 3 3 3 (x – 1)
(c) y = xe (d) y = x ln x + x
S-168 The Pattern Target AIEEE
12. The mean of the numbers a, b, 8, 5, 10 is 6 and the variance (a) Statement -1 is false, Statement-2 is true
is 6.80. Then which one of the following gives possible (b) Statement -1 is true, Statement-2 is true; Statement -2 is
values of a and b ? a correct explanation for Statement-1
(a) a = 0, b = 7 (b) a = 5, b = 2 (c) Statement -1 is true, Statement-2 is true; Statement -2
(c) a = 1, b = 6 (d) a = 3, b = 4 is not a correct explanation for Statement-1
(d) Statement -1 is true, Statement-2 is false
13. The vector a iˆ 2ˆj kˆ lies in the plane of the vectors
n
b ˆi ˆj and c ˆj kˆ and bisects the angle between b 19. Statement -1 : (r 1) n C r (n 2)2n –1.
r 0
and c . Then which one of the following gives possible
n
values of and ? (r 1) n C r x r
(a) = 2, = 2 (b) = 1, = 2 Statement-2 : (1 x) n nx(1 x) n–1.
r 0
(c) = 2, = 1 (d) = 1, = 1 (a) Statement -1 is false, Statement-2 is true
14. The non-zero vectors a , b and c are related by a 8b (b) Statement -1 is true, Statement-2 is true; Statement -2 is
a correct explanation for Statement-1
and c –7b . Then the angle between a and c is (c) Statement -1 is true, Statement-2 is true; Statement -2
is not a correct explanation for Statement-1
(a) 0 (b) (c) (d) (d) Statement -1 is true, Statement-2 is false
4 2
15. 20. Let p be the statement “x is an irrational number”, q be the
The line passing through the points (5, 1, a) and (3, b, 1)
statement “y is a transcendental number”, and r be the
17 –13 statement “ x is a rational number if f y is a transcendental
crosses the yz-plane at the point 0, , . Then number”.
2 2
(a) a = 2, b = 8 (b) a = 4, b = 6 Statement-1 : r is equivalent to either q or p
(c) a = 6, b = 4 (d) a = 8, b = 2 Statement-2 : r is equivalent to ~(p ~q).
16. If the straight lines (a) Statement -1 is false, Statement-2 is true
(b) Statement -1 is true, Statement-2 is true; Statement -2 is
x –1 y – 2 z – 3 x–2 y–3 z –1 a correct explanation for Statement-1
and intersect at
k 2 3 3 k 2 (c) Statement -1 is true, Statement-2 is true; Statement -2
a point, then the integer k is equal to is not a correct explanation for Statement-1
(a) –5 (b) 5 (d) Statement -1 is true, Statement-2 is false
(c) 2 (d) –2 21. In a shop there are five types of ice-creams available. A
child buys six ice-creams.
DIRECTIONS : Q. No. 17 to 21 are Assertion-Reason type Statement-1 : The number of different ways the child can
questions. Each of these questions contains two buy the six ice-creams is 10C5.
statements: Statement-1(Assertion) and Statement- Statement -2 : The number of different ways the child can
2(Reason). Each of these questions also has four alternative buy the six ice-creams is equal to the number of different
choices, only one of which is the correct answer. You have ways of arranging 6 A’s and 4 B’s in a row.
to select the correct choice. (a) Statement -1 is false, Statement-2 is true
(b) Statement -1 is true, Statement-2 is true; Statement -2 is
17. Statement-1 : For every natural number n 2, a correct explanation for Statement-1
(c) Statement -1 is true, Statement-2 is true; Statement -2
1 1 1 is not a correct explanation for Statement-1
.........n.
1 2 n (d) Statement -1 is true, Statement-2 is false
Statement-2 : For every natural number n 2,
1
(x – 1) sin ,if x 1
n(n 1) n 1. 22. Let f (x) x –1
(a) Statement -1 is false, Statement-2 is true 0 , if x 1
(b) Statement -1 is true, Statement-2 is true; Statement -2 is
Then which one of the following is true?
a correct explanation for Statement-1
(a) f is neither differentiable at x = 0 nor at x =1
(c) Statement -1 is true, Statement-2 is true; Statement -2 (b) f is differentiable at x = 0 and at x =1
is not a correct explanation for Statement-1 (c) f is differentiable at x = 0 but not at x = 1
(d) Statement -1 is true, Statement-2 is false (d) f is differentiable at x = 1 but not at x = 0
18. Let A be a 2 × 2 matrix with real entries. Let I be the 2 × 2 23. The first two terms of a geometric progression add up to 12.
identity matrix. Denote by tr(A), the sum of diagonal entries the sum of the third and the fourth terms is 48. If the terms of
of A. Assume that A2 = I. the geometric progression are alternately positive and
Statement-1 : If A I and A –I, then det(A) = –1 negative, then the first term is
Statement-2 : If A I and A –I, then tr (A) 0. (a) –4 (b) –12 (c) 12 (d) 4
AIEEE-2008 Solved Paper S-169
24. Suppose the cubic x3 – px + q has three distinct real roots 32. How many different words can be formed by jumbling the
where p > 0 and q > 0. Then which one of the following letters in the word MISSISSIPPI in which no two S are
holds? adjacent?
p p (a) 8. 6C4. 7C4 (b) 6.7. 8C4
(a) The cubic has minima at and maxima at – (c) 6. 8. 7C4. (d) 7. 6C4. 8C4
3 3
33. Let a, b, c be any real numbers. Suppose that there are real
p p numbers x, y, z not all zero such that x = cy + bz, y = az + cx,
(b) The cubic has minima at – and maxima at and z = bx + ay. Then a2 + b2 + c2 + 2abc is equal to
3 3
(a) 2 (b) –1 (c) 0 (d) 1
p p 34. Let A be a square matrix all of whose entries are integers.
(c) The cubic has minima at both and –
3 3 Then which one of the following is true?
(a) If det A = ± 1, then A–1 exists but all its entries are not
p p necessarily integers
(d) The cubic has maxima at both and –
3 3 (b) If det A ± 1, then A–1 exists and all its entries are non
25. How many real solutions does the equation integers
x7 + 14x5 + 16x3 + 30x – 560 = 0 have ? (c) If det A = ± 1, then A–1 exists but all its entries are
(a) 7 (b) 1 (c) 3 (d) 5 integers
26. The statement p (q p) is equivalent to (d) If det A = ± 1, then A–1 need not exists
(a) p (p q) (b) p (p q) 35. The quadritic equations
(c) p (p q) (d) p (p q)
x2 – 6x + a = 0 and x2 – cx + 6 = 0
–1 5 2 have one root in common. The other roots of the first and
27. The value of cot cos ec tan –1 is
3 3 second equations are integers in the ratio 4 : 3. Then the
common root is
6 3 4 5 (a) 1 (b) 4 (c) 3 (d) 2
(a) (b) (c) (d)
17 17 17 17
28. The differential equation of the family of circles with fixed
radius 5 units and centre on the line y = 2 is Section - 2
(a) (x – 2) y'2 = 25 –(y – 2)2
(b) (y – 2) y'2 = 25 –(y – 2)2
(c) (y – 2)2 y'2 = 25 –(y – 2)2
(d) (x – 2)2 y'2 = 25 –(y – 2)2
1 1 36. At 80° C, the vapour pressure of pure liquid ‘A’ is 520 mm
sin x cos x Hg and that of pure liquid ‘B’ is 1000 mm Hg. If a mixture
29. Let I dx and J dx. Then which one of
x x solution of ‘A’ and ‘B’ boils at 80° C and 1 atm pressure, the
0 0
amount of ‘A’ in the mixture is (1 atm = 760 mm Hg)
the following is true?
(a) 50 mol percent (b) 34 mol percent
2 2 (c) 48 mol percent (d) 52 mol percent
(a) I and J 2 (b) I and J 2
3 3 1
37. For a reaction A 2B, rate of disappearance of ‘A’ is
2 2 2
(c) I and J 2 (d) I and J 2
3 3 related to the rate of appearance of ‘B’ by the expression
30. The area of the plane region bounded by the curves x + 2y2
d[A] 1 d[B] d[A ] d[B]
= 0 and x + 3y2 = 1 is equal to (a) – (b) –
dt 2 dt dt dt
5 1 2 4
(a) (b) (c) (d) d[A] 1 d[B] d[A] d[B]
3 3 3 3 (c) – (d) – 4
sin xdx dt 4 dt dt dt
31. The value of 2 is
38. The equilibrium constants K p and K p2 for the reactions
sin x – 1
4 X 2Y and Z P + Q, respectively are in the ratio of 1 :
9. If the degree of dissociation of X and Z be equal then the
(a) x log | cos x – | c ratio of total pressures at these equilibria is
4
(a) 1 : 1 (b) 1: 36 (c) 1 : 3 (d) 1 : 9
x – log | sin x – | c 39. Oxidising power of chlorine in aqueous solution can be
(b)
4 determined by the parameters indicated below:
1
x log | sin x – | c 1 diss H eg H
(c) Cl2 (g) 2 Cl(g) Cl– (g)
4 2
(a) K1 K 2 K3 (b) K 2 K3 K1
OH
HO H H (c) K3 .K 32 K12 (d) K3 = K1 K2
(a) S, S (b) R, S 49. Standard entropy of X2, Y2 and X Y3 are 60, 40 and 50 J K–1
(c) R, R (d) S, R mol –1 , respectively. For the reaction,
42. Which of the following factors is of no significance for
roasting sulphide ores to the oxides and not subjecting the 1 3
X2 Y2 XY3 , H –30kJ , to be at equilibrium, the
sulphide ores to carbon reduction directly? 2 2
(a) Metal sulphides are thermodynamically more stable temperature will be
than CS2 (a) 1250 K (b) 500 K (c) 1000 K (d) 750 K
(b) CO2 is thermodynamically more stable than CS2 50. Phenol, when first reacts with concentrated sulphuric acid
(c) CO2 is more volatile than CS2 and then with concentrated nitric acid, gives
(d) Metal sulphides are less stable than the corresponding (a) 2, 4, 6-trinitrobenzene (b) p-nitrophenol
oxides (c) o-nitrophenol (d) nitrobenzene
43. The electrophile, E attacks the benzene ring to generate 51. The hydrocarbon which can react with sodium in liquid
the intermediate -complex. Of the following, which - ammonia is
complex has lowest energy?
(a) CH 3CH 2 CH 2 C CCH 2CH 2 CH3
NO2
NO2 (b) CH 3CH CHCH3
H
+ (c) CH 3CH 2C CH
(a) (b) + E
(d) CH 3CH 2C CCH 2CH3
H E
NO2 52. The treatment of CH3MgX with CH 3C C H produces
H (a) CH4
+ E (b) CH 3C C CH3
(c) (d) + H
E H H
44. Bakelite is obtained from phenol by reacting with | |
(a) HCHO (b) CH3CHO (c) CH 3 C C CH3
(c) CH3 COCH3 (d) (CH2OH)2
(d) CH 3 CH CH 2
45. The organic chloro compound, which shows complete
sterochemical inversion during a SN2 reaction, is 53. The correct decreasing order of priority for the functional
(a) (C2H5)2CHCl (b) CH3Cl groups of organic compounds in the IUPAC system of
(c) (CH3)2 CHCl (d) (CH3)3CCl nomenclature is
46. Toluene is nitrated and the resulting product is reduced (a) – SO3H, – COOH,– CONH2, – CHO
with tin and hydrochloric acid. The product so obtained is (b) – COOH, – SO3H, – CONH2, – CHO
diazotised and then heated with cuprous bromide. The (c) – CHO, – COOH, – SO3H, – CONH2
reaction mixture so formed contains (d) – CONH2, – CHO, – SO3H, – COOH
(a) mixture of o- and p-dibromobenzenes 54. The pKa of a weak acid, HA, is 4.80. The pKb of a weak base,
(b) mixture of o- and p-bromotoluenes BOH, is 4.78. The pH of an aqueous solution of the
(c) mixture of o- and p-bromoanilines corresponding salt, BA, will be
(d) mixture of o- and m-bromotoluenes (a) 9.58 (b) 4.79 (c) 9.22 (d) 7.01
AIEEE-2008 Solved Paper S-171
55. In context with the industrial preparation of hydrogen from Cr|Cr3+ (0.1M)|| Fe2 + (0.01 M)| Fe is
water gas (CO + H2), which of the following is the correct (a) – 0.26 V (b) 0.26 V (c) – 0.339 (d) 0.336 V
statement? 65. Amount of oxalic acid present in a solution can be
(a) CO is oxidised to CO2 with steam in the presence of a determined by its titration with KMnO4 solution in the
catalyst followed by absorption of of CO2 in alkali presence of H2SO4.The titration gives unsatisfactory result
(b) CO is removed by absorption in aqueous Cu 2Cl 2 when carried out in the presence of HCl, because HCl
solution (a) gets oxidised by oxalic acid to chlorine
(c) H2 is removed through occlusion with pd (b) furnishes H+ ions in addition to those from oxalic acid
(d) CO and H2, are fractionally separated using differences (c) Oxidises oxalic acid to carbon dioxide and water
in their densities (d) reduces permanganate to Mn 2+
56. In a compound, atoms of element Y form ccp lattice and 66. Among the following substituted silanes the one which will
those of element X occupy 2/3rd of tetrahedral voids. The give rise to cross linked silicone polymer on hydrolysis is
formula of the compound will be (a) R4Si (b) R2SiCl2 (c) RSiCl3 (d) R3SiCl
(a) X2 Y3 (b) X4 Y3 (c) X2 Y (d) X3 Y4 67. In which of the following complexes of the Co (at. no. 27),
57. Which one of the following pairs of species have the same will the magnitude of o be the highest?
bond order? (a) [Co(C2O4)3]3– (b) [Co(CN)6]3–
(a) CN– and CN+ (b) CN– and NO+ (c) [Co(H2O)6] 3+ (d) [Co(NH3)6]3+
–
(c) O 2 and CN – (d) NO+ and CN+
68. The coordination number and the oxidation state of the
58. The vapour pressure of water at 20° C is 17.5 mm Hg. If 18 g
element ‘E’ in the complex [E (en)2 (C2O4)]NO2 (where (en)
of glucose (C6H12O 6) is added to 178.2 g of water at 20° C,
is ethylene diamane) are, respectively,
the vapour pressure of the resulting solution will be
(a) 6 and 3 (b) 4 and 2 (c) 4 and 3 (d) 6 and 2
(a) 17.325 mm Hg (b) 15.750 mm Hg
69. Identify the wrong statement in the following:
(c) 16.500 mm Hg (d) 17.500 mm Hg
(a) Chlorofluorocarbons are responsible for ozone layer
59. Four species are listed below:
– depletion
i. HCO 3 ii. H3O+ (b) Greenhouse effect is responsible for global warming
iii. HSO4– iv. HSO3F (c) Acid rain is mostly because of oxides of nitrogen and
Which one of the following is the correct sequence of their sulphur
acid strength? (d) Ozone layer does not permit infrared radiation from the
(a) iv < ii < iii < i (b) ii < iii < i < iv sun to reach the earth
(c) iii < i < iv < ii (d) i < iii < ii < iv 70. Which one of the following is the correct statement?
60. The ionization enthalpy of hydrogen atom is 1.312 × 106 J (a) B2H6.2NH3 is known as ‘inorganic benzene’
mol–1. The energy required to excite the electron in the atom (b) Boric acid is a protonic acid
from n = 1 to n = 2 is (c) Beryllium exhibits coordination number of six
(a) 9.84 × 105 J mol–1 (b) 6.56 × 105 J mol–1 (d) Chlorides of both beryllium and aluminium have bridged
5
(c) 7.56 × 10 J mol –1 (d) 8.51 × 105 J mol–1 chloride structures in solid phase
61. Which one of the following constitutes a group of the
isoelectronic species?
Section - 3
(a) C2– –
2 , O2 , CO, NO (b) CN – , N 2 ,O 2– 2–
2 , C2
(c) NO , C2– –
2 , CN , N 2 (d) N 2 , O 2– , NO , CO
62. Gold numbers of protective colloids A, B, C and D are 0.50,
0.01, 0.10 amd 0.005, respectively. the correct order of their Directions: Question No. 71, 72 and 73 are based on the
protective powers is following paragraph.
(a) D < A < C < B (b) C < B < D < A Wave property of electrons implies that they will show
(c) B < D < A < C (d) A < C < B < D diffraction effects. Davisson and Germer demonstrated this by
63. Larger number of oxidation states are exhibited by the diffracting electrons from crystals. The law governing the
actinoids than those by the lanthanoids, the main reason diffraction from a crystal is obtained by requiring that electron
being waves reflected from the planes of atoms in a crystal interfere
(a) 4f orbitals more diffused than the 5f orbitals constructively (see figure).
(b) more energy difference between 5f and 6d than between
4f and 5d orbitals Inco m g
(c) lesser energy difference between 5f and 6d than Electr ing i Outgoin s
ons le c tro n
between 4f and 5d orbitals E
(d) more reactive nature of the actionoids than the
d
lanthanoids
64. Given Eº –0.72 V, Eº = –0.42 V. The
Cr 3 / Cr Fe 2+ / Fe Crystal plane
potential for the cell
S-172 The Pattern Target AIEEE
71. Electrons accelerated by potential V are diffracted from a 76. Shown in the figure below is a meter-bridge set up with null
crystal. If d = 1Å and i = 30°, V should be about deflection in the galvanometer.
(h = 6.6 × 10 – 34 Js, me = 9.1 × 10–31 kg, e = 1.6 × 10 – 19 C)
(a) 2000 V (b) 50 V 55 R
(c) 500 V (d) 1000 V
72. If a strong diffraction peak is observed when electrons are
incident at an angle ‘i’ from the normal to the crystal planes G
with distance ‘d’ between them (see figure), de Broglie
wavelength dB of electrons can be calculated by the 20 cm
relationship ( n is an integer)
(a) d sin i = n dB (b) 2d cos i = n dB
(c) 2d sin i = n dB (d) d cos i = n dB
73. In an experiment, electrons are made to pass through a narrow
slit of width ‘d’ comparable to their de Broglie wavelength. The value of the unknown resistor R is
They are detected on a screen at a distance ‘D’ from the slit (a) 13.75 (b) 220 (c) 110 (d) 55
(see figure). 77. A thin rod of length ‘L’ is lying along the x-axis with its ends
at x = 0 and x = L. Its linear density (mass/length) varies with
n
x
d x as k , where n can be zero or any positive number. If
y=0 L
the position xCM of the centre of mass of the rod is plotted
D
against ‘n’, which of the following graphs best approximates
Which of the following graphs can be expected to represent the dependence of xCM on n?
the number of electrons ‘N’ detected as a function of the
xCM xCM
detector position ‘y’(y = 0 corresponds to the middle of the
slit) L
L L
y y (a) (b) 2
2
n n
O O
81. A body of mass m = 3.513 kg is moving along the x-axis with (a) Statement -1 is false, Statement-2 is true
a speed of 5.00 ms–1. The magnitude of its momentum is (b) Statement -1 is true, Statement-2 is true; Statement -2 is
recorded as a correct explanation for Statement-1
(a) 17.6 kg ms–1 (b) 17.565 kg ms–1 (c) Statement -1 is true, Statement-2 is true; Statement -2
(c) 17.56 kg ms–1 (d) 17.57 kg ms–1 is not a correct explanation for Statement-1
82. An athlete in the olympic games covers a distance of 100 m (d) Statement -1 is true, Statement-2 is false
in 10 s. His kinetic energy can be estimated to be in the 87. A jar is filled with two non-mixing liquids 1 and 2 having
range densities 1 and, 2 respectively. A solid ball, made of a
(a) 200 J - 500 J (b) 2 × 105 J - 3 × 105 J material of density 3 , is dropped in the jar. It comes to
(c) 20, 000 J - 50,000 J (d) 2,000 J - 5, 000 J equilibrium in the position shown in the figure.
83. A parallel plate capacitor with air between the plates has
capacitance of 9 pF. The separation between its plates is
‘d’. The space between the plates is now filled with two
Liquid 1
dielectrics. One of the dielectrics has dielectric constant k1
d
= 3 and thickness while the other one has dielectric
3
2d
constant k2 = 6 and thickness . Capacitance of the
3
capacitor is now Liquid 2
(a) 1.8 pF (b) 45 pF
(c) 40.5 pF (d) 20.25 pF
84. The speed of sound in oxygen (O2) at a certain temperature
is 460 ms–1. The speed of sound in helium (He) at the same
temperature will be (assume both gases to be ideal)
(a) 460 ms–1 (b) 500 ms–1 Which of the following is true for 1, 1and 3?
(c) 650 ms–1 (d) 330 ms–1 (a) 3 < 1 2 (b) 1 > 3 2
85. This question contains Statement-1 and statement-2. Of the (c) 1 < 2 3 (d) 1< 3 2
four choices given after the statements, choose the one 88. A working transistor with its three legs marked P, Q and R is
that best describes the two statements. tested using a multimeter. No conduction is found between
Statement-1: Energy is released when heavy nuclei undergo P and Q. By connecting the common (negative) terminal of
fission or light nuclei undergo fusion and the multimeter to R and the other (positive) terminal to P or
Statement-2 : For heavy nuclei, binding energy per nucleon Q, some resistance is seen on the multimeter. Which of the
increases with increasing Z while for light nuclei it decreases following is true for the transistor?
with increasing Z. (a) It is an npn transistor with R as base
(a) Statement-1 is false, Statement-2 is true (b) It is a pnp transistor with R as collector
(b) Statement-1 is true, Statement-2 is true; Statement-2 is (c) It is a pnp transistor with R as emitter
a correct explanation for Statement-1 (d) It is an npn transistor with R as collector
(c) Statement-1 is true, Statement-2 is true; Statement-2 is
not a correct explanation for Statement-1 DIRECTIONS : Question No. 89 and 90 are based on the
(d) Statement-1 is true, Statement-2 is false following paragraph.
86. This question contains Statement-1 and Statement-2. Of the Consider a block of conducting material of resistivity ‘ ’
four choices given after the statements, choose the one shown in the figure. Current ‘I’ enters at ‘A’ and leaves from ‘D’.
that best describes the two statements. We apply superposition principle to find voltage ‘ V’ developed
Statement-1 : For a mass M kept at the centre of a cube of between ‘B’ and ‘C’. The calculation is done in the following
side ‘a’, the flux of gravitational field passing through its steps:
sides 4 GM. (i) Take current ‘I’ entering from ‘A’ and assume it to spread
and over a hemispherical surface in the block.
Statement-2: If the direction of a field due to a point source (ii) Calculate field E(r) at distance ‘r’ from A by using Ohm’s law
is radial and its dependence on the distance ‘r’ from the E = j, where j is the current per unit area at ‘r’.
1 (iii) From the ‘r’ dependence of E(r), obtain the potential V(r) at
source is given as , its flux through a closed surface r.
r2 (iv) Repeat (i), (ii) and (iii) for current ‘I’ leaving ‘D’ and superpose
depends only on the strength of the source enclosed by the results for ‘A’ and ‘D’.
surface and not on the size or shape of the surface.
S-174 The Pattern Target AIEEE
92. A block of mass 0.50 kg is moving with a speed of 2.00 ms –
1 on a smooth surface. It strikes another mass of 1.00 kg and
I V I
then they move together as a single body. The energy loss
during the collision is
(a) 0.16 J (b) 1.00 J (c) 0.67 J (d) 0.34 J
a b a
93. A capillary tube (A) is dipped in water. Another identical
A B C D tube (B) is dipped in a soap-water solution. Which of the
following shows the relative nature of the liquid columns in
the two tubes?
A B
89. V measured between B and C is
I I I I
(a) – (b) –
a (a b) a (a b) (a)
I I I
(c) – (d)
2 a 2 (a b) 2 (a b)
A B
90. For current entering at A, the electric field at a distance ‘r’
from A is
I I I I (b)
(a) 2 (b) 2 (c) 2 (d) 2
8 r r 2 r 4 r
91. A student measures the focal length of a convex lens by
putting an object pin at a distance ‘u’ from the lens and
measuring the distance ‘v’ of the image pin. The graph A B
between ‘u’ and ‘v’ plotted by the student should look like
v(cm)
(c)
(a)
O u(cm) A B
v(cm) (d)
(b)
94. Suppose an electron is attracted towards the origin by a
O u(cm)
k
force where ‘k’ is a constant and ‘r’ is the distance of the
r
v(cm) electron from the origin. By applying Bohr model to this
system, the radius of the nth orbital of the electron is found
to be ‘rn’ and the kinetic energy of the electron to be ‘T n’.
(c) Then which of the following is true?
O u(cm) 1
(a) Tn , rn n2
2
n
v(cm) (b) Tn independent of n, rn n
1
(c) Tn , rn n
n
(d) 1
O u(cm) (d) Tn , rn n2
n
AIEEE-2008 Solved Paper S-175
95. A wave travelling along the x-axis is described by the (x1 – x2)
equation y(x, t) = 0.005 cos ( x – t). If the wavelength and
the time period of the wave are 0.08 m and 2.0s, respectively,
then and in appropriate units are
(a) = 25.00 , =
0.08 2.0 (c) t
(b) , O
(d) 12.50 ,
2.0
96. Two coaxial solenoids are made by winding thin insulated (d) t
wire over a pipe of cross-sectional area A = 10 cm 2 and O
length = 20 cm. If one of the solenoid has 300 turns and the
other 400 turns, their mutual inductance is 99. An experment is performed to find the refractive index of
( 0 = 4 × 10 –7 Tm A–1) glass using a travelling microscope. In this experiment
(a) 2.4 × 10–5 H (b) 4.8 × 10–4 H distances are measured by
–5
(c) 4.8 × 10 H (d) 2.4 × 10–4 H (a) a vernier scale provided on the microscope
97. In the circuit below, A and B represent two inputs and C (b) a standard laboratory scale
represents the output. (c) a meter scale provided on the microscope
(d) a screw gauge provided on the microscope
A 100. A thin spherical shell of radus R has charge Q spread
uniformly over its surface. Which of the following graphs
C most closely represents the electric field E(r) produced by
the shell in the range 0 r < , where r is the distance from
B the centre of the shell?
E(r)
The circuit represents
(a) NOR gate (b) AND gate
(c) NAND gate (d) OR gate (a)
98. A body is at rest at x = 0. At t = 0, it starts moving in the r
positive x-direction with a constant acceleration. At the same O R
instant another body passes through x = 0 moving in the
positive x-direction with a constant speed. The position of E(r)
the first body is given by x1(t) after time ‘t’; and that of the
second body by x2(t) after the same time interval. Which of
the following graphs correctly describes (x1 – x2) as a (b)
function of time ‘t’? r
O R
(x1 – x2)
E(r)
(a) t (c)
O r
O R
(x1 – x2)
E(r)
(d)
t r
(b) O O R
S-176 The Pattern Target AIEEE
101. A 5V battery with internal resistance 2 and a 104. Two full turns of the circular scale of a screw gauge cover a
2V battery with internal resistance 1 are connected to a distance of 1mm on its main scale. The total number of
10 resistor as shown in the figure. divisions on the circular scale is 50. Further, it is found that
P2 the screw gauge has a zero error of – 0.03 mm. While
measuring the diameter of a thin wire, a student notes the
main scale reading of 3 mm and the number of circular scale
divisions in line with the main scale as 35. The diameter of
the wire is
5V 2V
2 10 (a) 3.32 mm (b) 3.73 mm
1
(c) 3.67 mm (d) 3.38 mm
105. An insulated container of gas has two chambers separated
by an insulating partition. One of the chambers has volume
V1 and contains ideal gas at pressure P1 and temperature
T1. The other chamber has volume V2 and contains ideal
The current in the 10 resistor is gas at pressure P2 and temperature T2. If the partition is
(a) 0.27 A P2 to P1 (b) 0.03 A P1 to P2 removed without doing any work on the gas, the final
(c) 0.03 A P2 to P1 (d) 0.27 A P1 to P2 equilibrium temperature of the gas in the container will be
102. A horizontal overhead powerline is at height of 4m from the
ground and carries a current of 100A from east to west. T1T2 (P1V1 P2 V2 )
(a)
The magnetic field directly below it on the ground is P1V1T2 P2 V2 T1
( 0 = 4 ×10 –7 Tm A–1)
(a) 2.5 × 10–7 T southward (b) 5 × 10–6 T northward P1V1T1 P2 V2 T2
(c) 5 × 10–6 T southward (d) 2.5 × 10–7 T northward (b) P1V1 P2 V2
103. Relative permittivity and permeability of a material r and
P1V1T2 P2 V2 T1
r,
respectively. Which of the following values of these (c) P1V1 P2 V2
quantities are allowed for a diamagnetic material?
(a) r = 0.5, r = 1.5 (b) r = 1.5, r = 0.5 T1T2 (P1V1 P2 V2 )
(d)
(c) r = 0.5, r = 0.5 (d) r = 1.5, r = 1.5 P1V1T1 P2 V2 T2
AIEEE-2008 Solved Paper S-177
ANSWER KEY
1 (b) 16 (a) 31 (c) 46 (b) 61 (c) 76 (b) 91 (c)
2 (b) 17 (b) 32 (d) 47 (a) 62 (d) 77 (a) 92 (c)
3 (c) 18 (d) 33 (d) 48 (d) 63 (c) 78 (b) 93 (c)
4 (a) 19 (b) 34 (c) 49 (d) 64 (b) 79 (c) 94 (b)
5 (b) 20 (N) 35 (d) 50 (c) 65 (d) 80 (d) 95 (a)
6 (c) 21 (a) 36 (a) 51 (c) 66 (c) 81 (a) 96 (d)
7 (d) 22 (c) 37 (c) 52 (a) 67 (b) 82 (d) 97 (d)
8 (c) 23 (b) 38 (b) 53 (b) 68 (a) 83 (c) 98 (b)
9 (d) 24 (a) 39 (c) 54 (d) 69 (d) 84 (N) 99 (a)
10 (d) 25 (b) 40 (d) 55 (a) 70 (d) 85 (d) 100 (a)
11 (d) 26 (d) 41 (c) 56 (b) 71 (b) 86 (b) 101 (c)
12 (d) 27 (a) 42 (d) 57 (b) 72 (b) 87 (d) 102 (c)
13 (d) 28 (c) 43 (c) 58 (a) 73 (d) 88 (N) 103 (b)
14 (d) 29 (b) 44 (a) 59 (d) 74 (c) 89 (a) 104 (d)
15 (c) 30 (d) 45 (b) 60 (a) 75 (a) 90 (c) 105 (a)
5 5 SECTION - I1 CHEMISTRY
2 4
= 1– 3 – –1 3 =2 sq. units.
3 3 36. (a) At 1 atmospheric pressure the boiling point of mixture
sin xdx is 80°C.
31. (c) Let I = 2 At boiling point the vapour pressure of mixture, PT = 1
sin x – atmosphere = 760 mm Hg.
4 Using the relation,
put x – t PT PAo X A PBo X B , we get
4
dx = dt PT 520X A 1000(1 X A )
PT2 n
nQ nP
Now K P2
n2 n
(b )(b ).PT2
or K P2
[b(1 )][b(1 )]
2
K P1 4 .PT1 (1 )2 4PT1
or K P2 2 2 PT2
(1 ) PT2 .
PT1 1 K P1 1
or given
PT2 9 K P2 9
PT1 1
or PT2 36 or 1 : 36 45. (b) SN2 reaction is favoured by small groups on the carbon
39. (c) The energy involved in the conversion of atom attached to halogen.
So, the order of reactivity is
1 CH 3Cl (CH 3 ) 2 CHCl (CH 3 ) 3 CCl
Cl 2 (g) to Cl–1 (aq) is given by
2
(C 2 H 5 ) 2 CHCl
1 (–) (–) (–) SN2 reaction is shown to maximum extent by primary
H diss H Cl2 eg H Cl hyl H Cl
2 halides. The only primary halides given is CH3Cl. So
Substituting various values from given data, we get the correct answer is (b).
1 CH3
–1
H 240 (–349) (–381) kJmol
2 46. (b) Toluene ( ) contains –CH3 group which is o-,
= (120 – 349 – 381) kJ mol–1
= – 610 kJ mol–1 p- directing group so on nitration of toluene the –NO2
i.e., the correct answer is (c). group will occupy o-, p- positions.
40. (d) Since D ( ) glucose and D ( ) glucose CH3
differ in configuration at C – 1 atom so they are anomers. (HNO3 + H2SO4)
Note:- Anomers are those diastereomers that differ in
configuration at C – 1 atom.
i.e., (d) is the correct answer. CH3 CH3
41. (c) The absolute configuration is (R, R) NO2
(using priority rules to get the absolute configuration) (HNO + HSO)
So the correct answer is (c). +
42. (d) The reduction of metal sulphides by carbon reduction
process is not spontaneous because G for such a NO2
process is positive. The reduction of metal oxide by o- p-
carbon reduction process is spontaneous as G for on reduction with sn HCl they will form corresponding
anilines in which –NO2 group changes to –NH2. The
such a process is negative. From this we find that on
CH3 CH3
thermodynamic considerations CO2 is more stable than
CS2 and the metal sulphides are more stable than NH2
corresponding oxides. mixture now contains and . These
In view of above the factor listed in choice (d) is
NH2
incorrect and so is of no significance.
anilines when diazotized and then treated with CuBr
Hence the correct answer is (d). forms o-, p- bromotoluenes.
43. (c) In option (c) the complex formed is with benzene where 47. (a) Completing the sequence of given reactions,
as in other cases it is formed with nitrobenzene with – O3
NO2 group in different position (o-, m-, p-). The complex CH 3 – CH CH CH 3
formed with nitrobenzene in any position of –NO2 O
group is less stable than the complex formed with
CH3 – CH CH – CH3 Zn / H 2O
benzene so the correct answer is (c)
(Note : The most stable complex has lowest energy)
O O
44. (a) Bakelite is formed by the reaction of formaldehyde
‘A’
(HCHO) and phenol so the correct answer is (a). (ozoxide)
AIEEE-2008 Solved Paper S-183
Q
AIEEE-2008 Solved Paper S-185
n = 2d cos i
Also by de-broglie concept
h h h
p 2mK.E 2meV
nh
2d cos i
2meV W=V 1 g
h2
Here n =1 : V Vg ( 1 1)
8med 2 cos 2 i V 1g V 2g kv t2 vt
k
(6.6 10 34)2 76. (b) According to the condition of balancing
31 19 10 2 2
= 50 V
8 9.1 10 1.6 10 (10 ) cos 30 55 R
R = 220
Alternatively 20 80
Using Bragg's equation 2d sin = n 77. (a) When n = 0, x = k where k is a constant. This means
Here n = 1, = 90 – i = 90 – 30 = 60° that the linear mass density is constant. In this
case the centre of mass will be at the midelle of the
2d Sin = .......(i) rod ie at L/2. Therefore (c) is ruled out
n is positive and as its value increases, the rate of
increase of linear mass density with increase in x
increases. This shows that the centre of mass will
shift toward that end of the rod where n = L as the
d value of n increases. Therefore graph (b) is ruled
out.
n
x
The linear mass density k
L
12.27 10
Also, 10 m .......(ii) x
V Here 1
L
From (i) & (ii) With increase in the value of n, the centre of mass shift
12.27 towards the end x = L such that first the shifting is at a
10
2 10 sin 60o 10 10
higher rate with increase in the value of n and then the
V rate decreases with the value of n.
These characteristics are represented by graph (a).
(12.27)2
V 50V L L L n
3 x
72. (b) 2d cos i = n dB x dm x ( dx) k .xdx
L
73. (d) The electron beam will be diffracted and the maxima is 0 0 0
x CM
L L L n
obtained at y = 0. Also the distance between the first x
dm dm k dx
minima on both side will be greater than d. L
0 0 0
2GM p
L
(ve)p Rp Mp Re xn 2
74. (c) k
(ve)e 2GM e Me Rp (n 2)Ln 0 L(n 1)
Re L n 2
k xn 1
(n 1)Ln 0
10M e Re
10
Me Re/ 10 L
For n 0 , x CM ; n 1,
2
(ve ) p 10 (ve )e 10 11 110 km / s
2L 3L
75. (a) The condition for terminal speed (vt) is x CM ;n 2, x CM ;....
Weight = Buoyant force + Viscous force 3 4
S-186 The Pattern Target AIEEE
v v 1
78. (b) For first resonant length (in winter) K.E. 40 (10) 2 2000 J
4 1 4 18 2
For second resonant length If mass is 100 kg then,
3v ' 3v ' 1
' (in summer) K.E. 100 (10) 2 5000 J
4 2 4x 2
v 3v'
=
4 ×18 4× x
v' 83. (c)
x 3 18
v
v'
x 54 cm The given capacitance is equal to two capacitances
v
connected in series where
v' > v because velocity of light is greater in summer as
compared to winter (v T)
x 54cm and
79. (c) We know that F = q v B
F MLT 2
B MT 1C 1
qv C LT 1 The equivalent capacitance Ceq is
1 1 1 d d 2d
1 Ma 2
80. (d) Inn' M(a 2 a2 ) Ceq C1 C2 9 0 A 9 0 A 9 0 A
12 6
n m 9 0 A 9
Ceq 9 pF 40.5pF
2 d 2
A RT
D 84. (N) The speed of sound in a gas is given by v
M
vO2 O2 M He
v He M O2 He
1.4 4
0.3237
B 32 1.67
C
vO2
460
v He 1421 m / s
n m 1
0.3237 0.3237
None of the option is correct.
DB 2a a 85. (d) We know that energy is released when heavy nuclei
Also, DO undergo fission or light nuclei undergo fusion.
2 2 2
Therefore statement (1) is correct.
According to parallel axis theorem
The second statement is false because for heavy nuclei
2 the binding energy per nucleon decreases with
a Ma 2 Ma 2
Imm ' Inn ' M increasing Z and for light nuclei, B.E/nucleon increases
2 6 2
with increasing Z.
Ma 2
3Ma 2 2 86. (b) Gravitational flux through a closed surface is given by
Ma 2
6 3 E dS 4 GM
g
81. (a) Momentum, p = m × v
= (3.513) × (5.00) = 17.565 kg m/s where, M = mass enclosed in the closed surface
= 17.6 (Rounding off to get three significant figures) 1
82. (d) The average speed of the athelete This relationship is valid when | E g | .
r2
100 87. (d) From the figure it is clear that liquid 1 floats on liquid 2.
v 10m / s
10 The lighter liquid floats over heavier liquid. Therefore
1 we can conclude that 1 2
K.E. mv 2
2 Also 3 < 2 otherwise the ball would have sink to the
If mass is 40 kg then, bottom of the jar.
AIEEE-2008 Solved Paper S-187
0 2I 7 2 100
B 10 = 5 × 10–6 T
4 r 4
v/a
t
W N
For the body moving with constant speed
x 2 vt 100A
1 2
x1 x2 at vt
2
at t = 0, x1–x2 = 0
4m
v
For t < ; the slope is negative E
a S
v Ground
For t = ; the slope is zero
a
B
v
For t > ; the slope is positive
a
These characteristics are represented by graph (b). According to right hand palm rule, the magnetic field is
99. (a) To find the refractive index of glass using a travelling directed towards south.
micrscope, a vernier scale is provided on the 103 (b) For a diamegnetic material, the value of µr is less than
microscope one. For any material, the value of r is always greater
100. (a) The electric field inside a thin spherical shell of radius R than 1.
has charge Q spread uniformly over its surface is zero.
104. (d) Least count of screw gauge
Q +
++ + +
=
0.5
mm 0.01mm
+
R 50
+
+ + +
+++
+ +
+ + +
++
thermodynamics U = Q + W = 0
Q Internal energy of the system with partition = Internal
Outside the shell the electric field is E . These k energy of the system without partition.
r2
characteristics are represented by graph (a). n1Cv T1 n 2 Cv T2 (n1 n 2 )Cv T
101. (c) Applying kirchoff's loop law in AB P2P1A we get
n1T1 n 2 T2
2i 5 10 i1 0 .....(i) T
n1 n 2
B i P2 i–i1 C
i1 P1V1 P2 V2
But n1 and n 2
2V
RT1 RT2
10
5V 1
2 P1V1 P2 V2
T1 T2
RT1 RT2
A P1 D T
P1V1 P2 V2
Again applying kirchoff's loop law in P2 CDP1P2 we get RT1 RT2
10 i1 + 2 – i + i1= 0 .....(ii)
Part A : Chemistry (144 marks) : Question No. 1 to 24 are of 4 marks each and Question No. 25 to 30 are of 8 marks each.
Part B : Mathematics (144 marks) : Question No.31 to 32 and 39 to 60 are of 4 marks each and Question No.33 to 38 are of 8 marks each.
Part C : Physics (144 marks) : Question No. 61 to 84 are of 4 marks each and Question No. 85 to 90 are of 8 marks each.
Negative Marking : One fourth (¼) marks will be deducted for indicating incorrect response of each question.
26. Copper crystallises in fcc with a unit cell length of 361 pm. 32. Let A be a 2 × 2 matrix
What is the radius of copper atom? Statement -1 adj (adj A) = A
(a) 127 pm (b) 157 pm (c) 181 pm (d) 108 pm Statement -2 |adj A |= |A|
27. In a fuel cell methanol is used as fuel and oxygen gas is (a) Statement-1 is true, Statement-2 is true.
used as an oxidizer. The reaction is
Statement-2 is not a correct explanation for Statement-1.
CH 3 OH(l ) + 3/2O 2 (g) CO 2 (g) + 2H 2 O(l ) (b) Statement-1 is true, Statement-2 is false.
At 298 K standard Gibb’s energies of formation for CH3OH(l), (c) Statement -1 is false, Statement-2 is true.
H2O(l) and and CO2 (g) are –166.2 –237.2 and –394.4 kJ (d) Statement-1 is true, Statement -2 is true.
mol–1 respectively. If standard enthalpy of combustion of Statement-2 is a correct explanation for Statement-1.
methonal is – 726 kJ mol–1, efficiency of the fuel cell will be
(a) 87% (b) 90% (c) 97% (d) 80% 33. Let f(x) = ( x 1)2 –1, x –1
28. Two liquids X and Y form an ideal solution. At 300 K, vapour Statement -1 The set {x f(x) = f –1(x) = {0, –1}
pressure of the solution containing 1 mol of X and 3 mol of Statement-2 f is a bijection.
Y is 550 mmHg. At the same temperature, if 1 mol of Y is
(a) Statement-1 is true, Statement-2 is true.
further added to this solution, vapour pressure of the
solution increases by 10 mmHg. Vapour pressure ( in mmHg) Statement-2 is not a correct explanation for Statement-1.
of X and Y in their pure states will be, respectively (b) Statement-1 is true, Statement-2 is false.
(a) 300 and 400 (b) 400 and 600 (c) Statement-1 is false, Statement-2 is true.
(c) 500 and 600 (d) 200 and 300 (d) Statement-1 is true, Statement-2 is true.
Statement-2 is not a correct explanation for Statement-1.
29. Given E°Fe3+ / Fe = –0.036V, E°Fe2+ / Fe = –0.439 V
34. Statement-1
The value of standard electrode potential for the change,
n2 – 1
Fe3+(aq) +e –
Fe 2+
(aq) The variance of first n even natural numbers is .
4
will be
(a) 0.385 V (b) 0.770 V (c) –0.270 V (d) –0.072 V
Statement-2
30. The half life period of a first order chemical reaction is 6.93
minutes. The time required for the completion of 99% of the n (n 1)
chemical reaction will be (log 2 = 0.301) The sum of first n natural numbers is and the sum
2
(a) 23.03 minutes (b) 46.06 minutes
(c) 460.6 minutes (d) 230.03 minutes of squares of first n natural numbers is n (n 1)(2n 1) .
6
(a) Statement-1 is true, Statement-2 is true
Section - 2
Statement-2 is not a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is false.
(c) Statement-1 is false, Statement-2 is true.
Directions Question number 31 to 35 are Assertion–Reason (d) Statement-1 is true, Statement-2 is true.
type question. Each of these questions contains two Statement-2 is a correct explanation for Statement-1.
statements
35. Let f (x) = x | x | and g (x) = sin x.
Statement-1 (Assertion ) and Statement-2 (Reason).
Statement-1 gof is differentiable at x = 0 and its derivative
Each of these questions also has four alternative choices,
only one of which is the correct answer. You have to select is continuous at that point.
the correct choice. Statement-2 gof is twice differentiable at x = 0.
(a) Statement-1 is true, Statement-2 is true;
Statement-2 is not a correct explanation for Statement-1.
31. Statement-1 ~ ( p ~ q) is equivalent to p q.
(b) Statement-1 is true, Statement-2 is false.
Statement-2 ~ ( p ~ q) is a tantology (c) Statement-1 is false, Statement-2 is true.
(a) Statement-1 is true, Statement-2 is true; (d) Statement-1 is true, Statement-2 is true;
Statement-2 is not a correct explanation for Statement 1. Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is false. 36. The area of the region bounded by the parabola
(c) Statement-1 is false, Statement-2 is true. (y – 2)2 = x –1, the tangent of the parabola at the point (2, 3)
(d) Statement-1 is true, Statement-2 is true, Statement-2 is and the x-axis is
a correct explanation for statement -1 (a) 6 (b) 9 (c) 12 (d) 3
S-192 The Pattern Target AIEEE
37. Given P(x) = x4 + ax3 + bx2 + cx + d such that x = 0 is the only 45. The differential equation which represents the family of
real root of P' (x) = 0. If P(–1) < P(1), then in the interval
curves y c1ec2 x , where c1, and c2 are arbitrary constants,
[ –1, 1]
(a) P(–1) is not minimum but P(1) is the maximum of P is
(b) P(–1) is the minimum but P(1) is not the maximum of P (a) y" = y'y (b) yy" = y'
(c) yy" = (y')2 (d) y' = y2
(c) Neither P(–1) is the minimum nor P(1) is the maximum
of P 46. Let a, b, c be such that b(a + c) 0 if
(d) P(–1) is the minimum and P(1) is the maximum of P
a a 1 a –1 a 1 b 1 c –1
38. The shortest distance between the line y – x = 1 and the
curve x = y2 is –b b 1 b –1 a –1 b 1 c 1 0,
+
c c –1 c 1 n 2 n 1 n
( 1) a ( 1) b ( 1) c
2 3 3 2 3 3 2
(a) (b) (c) (d)
8 5 4 8 then the value of n is
(a) any even integer (b) any odd integer
x – 2 y –1 z 2
39. Let the line lie in the plane (c) any integer (d) zero
3 –5 2 47. The remainder left out when 82n – (62)2n+1 is divided by 9
x + 3y – z + = 0. Then ( , ) equals is
(a) (–6, 7) (b) (5, –15) (c) (–5, 5) (d) (6, –17) (a) 2 (b) 7 (c) 8 (d) 0
40. From 6 different novels and 3 different dictionaries,4 novels 48. Let y be an implicit function of x defined by
and 1 dictionary are to be selected and arranged in a row on x2x – 2xx cot y – 1= 0. Then y (1) equals
a shelf so that the dictionary is always in the middle. Then
(a) 1 (b) log 2 (c) –log 2 (d) –1
the number of such arrangement is
49. If the roots of the equation bx2 + cx + a = 0 be imaginary,
(a) at least 500 but less than 750
then for all real values of x, the expression
(b) at least 750 but less than 1000 3b2x2 + 6bcx + 2c2 is
(c) at least 1000 (a) less than 4ab (b) greater than – 4ab
(d) less than 500 (c) 1ess than – 4ab (d) greater than 4ab
1 50. The sum to infinite term of the series
41. In a binomial distribution B n, p , if the probability of
4 2 6 10 14 ... is.
1+ + + + +
3 32 33 34
9
at least one success is greater than or equal to , then n is (a) 3 (b) 4 (c) 6 (d) 2
10
51. The projections of a vector on the three coordinate axis are
greater than 6, –3, 2 respectively. The direction cosines of the vector are
1 9 6 3 2 6 3 2
(a) (b) (a) , , (b) , ,
log10 4 log10 3 log10 4 – log10 3 5 5 5 7 7 7
4 1 6 3 2
(c) (d) (c) , , (d) 6, – 3, 2
log10 4 – log10 3 log10 4 – log10 3 7 7 7
42. The lines p(p2 +1)x – y + q = 0 and 52. Let A and B denote the statements
(p2 + 1)2x + (p2 + 1)y + 2q = 0 are perpendicular to a common A cos + cos + cos = 0
line for B sin + sin + sin = 0
(a) exactly one values of p (b) exactly two values of p 3
(c) more than two values of p (d) no value of p If cos ( – ) + cos ( – ) + cos ( – ) = , then
2
43. If A, B and C are three sets such that A B= A C and (a) A is false and B is true (b) both A and B are true
A B A C , then (c) both A and B are false (d) A is true and B is false
(a) A = C (b) B = C (c) A B (d) A = B 53. One ticket is selected at random from 50 tickets numbered
00,01,02,...,49. Then the probability that the sum of the digits
44. For real x, let f (x) = x3 + 5x + 1, then on the selected ticket is 8, given that the product of these
(a) f is onto R but not one-one digits is zero, equals
(b) f is one-one and onto R
1 5 1 1
(c) f is neither one-one nor onto R (a) (b) (c) (d)
7 14 50 14
(d) f is one-one but not onto R
AIEEE-2009 Solved Paper S-193
5 5 5 –v1
(a) ,0 (b) ,0 (c) ,0 (d) (0, 0)
4 2 3
55. If the mean deviation of the numbers 1, 1 + d, 1 + 2d, .... y
1 + 100d from their mean is 255, then d is equal to
(a) 20.0 (b) 10.1 (c) 20.2 (d) 10.0 h
56. 2 2
The ellipse x + 4y = 4 is inscribed in a rectangle aligned
with the coordinate axes, which in turn is inscribed in another
ellipse that passes through the point (4, 0). Then the equation t
of the ellipse is
(a) x2 + 12y2 = 16 (b) 4x2 + 48y2 = 48 v
2 2
(c) 4x + 64y = 48 (d) x2 + 16y2 = 16 +v1
4 O t
57. If z 2 , then the maximum value of |Z| is equal to (b) t1 2t1 4t1
z
–v1
(a) 5 1 (b) 2 (c) 2 2 (d) 3 1
58. If P and Q are the points of intersection of the circles y
2 2 x2 + y2 + 2x + 2y – p2=0
x y 3 x 7 y 2 p 5 0 and h
then there is a circle passing through P, Q and (1, 1) for t
(a) all except one value of p
(b) all except two values of p t
(c) exactly one value of p
(d) all values of p
59. If u, v, w are non-coplanar vectors and p, q are real numbers,
then the equality [3u pv p ] [ pv qu ] [2 qv qu ] 0 (c) O t
t1 2t1
holds for
(a) exactly two values of (p, q)
(b) more than two but not all values of (p, q) y
(c) all values of (p, q)
(d) exactly one value of (p, q) h
(d) O t
Section - 3
y
61. Consider a rubber ball freely falling from a height h = 4.9 m h
onto a horizontal elastic plate. Assume that the duration of
collision is negligible and the collision with the plate is totally
elastic. t
Then the velocity as a function of time and the height as a
function of time will be
S-194 The Pattern Target AIEEE
62. The height at which the acceleration due to gravity becomes Directions Question numbers 65 and 66 are based on
g the following paragraph.
(where g = the acceleration due to gravity on the surface
9
A current loop ABCD is held fixed on the plane of the paper as
of the earth) in terms of R, the radius of the earth, is shown in the figure. The arcs BC (radius = b) and DA (radius = a)
R of the loop are joined by two straight wires AB and CD. A steady
(a) (b) R / 2 (c) 2R (d) 2 R current I is flowing in the loop. Angle made by AB and CD at the
2 origin O is 30°. Another straight thin wire with steady current I1
63. A long metallic bar is carrying heat from one of its ends to flowing out of the plane of the paper is kept at the origin.
the other end under steady–state. The variation of
temperature along the length x of the bar from its hot end B
is best described by which of the following figures? a A
I1 30° I
O
D
b C
(a)
x 65. The magnitude of the magnetic field (B) due to the loop
ABCD at the origin (O) is
o I (b a)
(a)
24 ab
oI b a
(b)
4 ab
(b) oI
(c) [2(b a) (a b)]
4 3
x
(d) zero
66. Due to the presence of the current I1 at the origin
(a) The forces on AD and BC are zero.
(b) The magnitude of the net force on the loop is given by
I1I
o [2(b a ) 3
(a b] .
4
(c) The magnitude of the net force on the loop is given by
(c)
o II1
(b a).
x 24 ab
(d) The forces on AB and DC are zero.
Two moles of helium gas are taken over the cycle ABCDA, as
(d) shown in the P-T diagram.
x 5 A B
2 × 10
67. Assuming the gas to be ideal the work done on the gas in Statement 1 The temperature dependence of resistance is
taking it from A to B is usually given as R = Ro (1 + t). The resistance of a wire
(a) 300 R (b) 400 R (c) 500 R (d) 200 R changes from 100 to 150 when its temperature is
increased from 27°C to 227°C. This implies th at
68. The work done on the gas in taking it from D to A is
= 2.5 × 10–3/°C.
(a) + 414 R (b) – 690 R (c) + 690 R (d) – 414 R
Statement 2 R = Ro (1+ t) is valid only when the change
69. The net work done on the gas in the cycle ABCDA is
in the temperature T is small and
(a) 276 R (b) 1076 R (c) 1904 R (d) zero
R = (R – R0) << Ro.
70. In an experiment the angles are required to be measured (a) Statement-1 is true, Statement-2 is true; Statement-2 is
using an instrument. 29 divisions of the main scale exactly the correct explanation of Statement-1.
coincide with the 30 divisions of the vernier scale. If the (b) Statement-1 is true, Statement-2 is true; Statement-2 is
smallest division of the main scale is half- a degree (= 0.5°), not the correct explanation of Statement-1.
then the least count of the instrument is
(c) Statement-1 is false, Statement-2 is true.
(a) half minute (b) one degree (d) Statement-1 is true, Statement-2 is false.
(c) half degree (d) one minute 75. The transition from the state n = 4 to n = 3 in a hydrogen like
71. A charge Q is placed at each of the opposite corners of a atom results in ultraviolet radiation. Infrared radiation will
square. A charge q is placed at each of the other two corners. be obtained in the transition from
If the net electrical force on Q is zero, then Q/q equals (a) 3 2 (b) 4 2 (c) 5 4 (d) 2 1
1 76. A mixture of light, consisting of wavelength 590 nm and an
(a) –1 (b) 1 (c) (d) 2 2 unknown wavelength, illuminates Young’s double slit and
2
gives rise to two overlapping interference patterns on the
72. One kg of a diatomic gas is at a pressure of 8 × 104N/m2. The screen. The central maximum of both lights coincide. Further,
density of the gas is 4kg/m3. What is the energy of the gas it is observed that the third bright fringe of known light
due to its thermal motion? coincides with the 4th bright fringe of the unknown light.
(a) 5 × 104 J (b) 6 × 104 J From this data, the wavelength of the unknown light is
(c) 7 × 104 J (d) 3 × 104 J (a) 885.0 nm (b) 442.5 nm (c) 776.8 nm (d) 393.4 nm
73. An inductor of inductance L = 400 mH and resistors of 77. A particle has an initial velocity of 3iˆ 4 ˆj and an
resistance R1 = 2 and R2 = 2 are connected to a battery
of emf 12 V as shown in the figure. acceleration of 0.4iˆ 0.3 ˆj . Its speed after 10 s is
2
. It is surrounded by air. A light ray is incident at the
3
Output is
mid-point of one end of the rod as shown in the figure.
(a)
The incident angle for which the light ray grazes along the
(b)
wall of the rod is
1 3 1 2
(a) sin 2 (b) sin
3 (c)
1 1 1 1
(c) sin (d) sin
3 2 (d)
83. Two wires are made of the same material and have the same
86. If x, v and a denote the displacement, the velocity and the
volume. However wire 1 has cross-sectional area A and wire
acceleration of a particle executing simple harmonic motion
2 has cross-sectional area 3A. If the length of wire 1 increases
of time period T, then, which of the following does not change
by x on applying force F, how much force is needed to with time?
stretch wire 2 by the same amount?
(a) aT/x (b) aT + 2 v
(a) 4 F (b) 6 F (c) 9 F (d) F (c) aT/v (d) a2T2 + 4 2v2
AIEEE-2009 Solved Paper S-197
1l 1 l2 2
(a) (b)
6 g 2 g
I
2 2 2
1l 1l
(c) (d) (b)
6 g 3 g
f f I
(a) , (b) ( f, f )
2 2
(c) ( 4 f, 4 f ) (d) ( 2 f, 2 f ) (d)
89. A p-n junction (D) shown in the figure can act as a rectifier.
t
An alternating current source (V) is connected in the circuit.
Q
90. Let P (r ) r be the charge density distribution for a
D R4
R solid sphere of radius R and total charge Q. For a point ‘p’
inside the sphere at distance r1 from the centre of the sphere,
the magnitude of electric field is
V ~ Q Qr12
(a) 2 (b)
4 0 r1 4 R4
0
ANSWER KEY
1 (a) 11 (b) 21 (b) 31 (b) 41 (d) 51 (b) 61 (b) 71 (d) 81 (d)
2 (d) 12 (d) 22 (b) 32 (a) 42 (a) 52 (b) 62 (d) 72 (a) 82 (c)
3 (b) 13 (c) 23 (b) 33 (b) 43 (b) 53 (d) 63 (a) 73 (c) 83 (c)
4 (c) 14 (c) 24 (a) 34 (c) 44 (b) 54 (a) 64 (c) 74 (c) 84 (a)
5 (c) 15 (b) 25 (a) 35 (b) 45 (c) 55 (b) 65 (a) 75 (c) 85 (d)
6 (a) 16 (c) 26 (a) 36 (b) 46 (b) 56 (a) 66 (d) 76 (b) 86 (a)
7 (a) 17 (b) 27 (c) 37 (a) 47 (a) 57 (a) 67 (b) 77 (a) 87 (c)
8 (a) 18 (b) 28 (b) 38 (d) 48 (d) 58 (a) 68 (a) 78 (a) 88 (d)
9 (b) 19 (a) 29 (b) 39 (a) 49 (b) 59 (d ) 69 (a) 79 (b) 89 (b)
10 (a) 20 (b) 30 (b) 40 (c) 50 (a) 60 (c) 70 (d) 80 (a) 90 (b)
SECTION I : CHEMISTRY 8. (a) Lower oxidation state of an element forms more basic
oxide and hydroxide, while the higher oxidation state
1. (a) CH3 will form more acidic oxide/hydroxide. For example,
| 1
2 3 2 3 7
H3C C CH3
| MnO Mn 2 O3 Mn 2 O7
CH3 (basic) acidic
Neopentane
9. (b) According to Heisenberg uncertainty principle.
or 2, 2- Dimethylpropane
h h
2. (d) The products of the concerned reaction react each x.m v x
4 4 m v
other forming back the reactants.
600 0.005
XeF6 3H 2 O XeO3 6HF Here v 0.03
100
OH OH
NaOH 34
6.6 10
3. (b) CO2 So, x
COOH 4 3.14 9.1 10 31 0.03
4. (c) Adsorption is an exothermic process, hence H will = 1.92 × 10–3 meter
always be negative 10. (a) The SCN– ion can coordinate through S or N atom
giving rise to linkage isomerism
5. (c) H3N NH3 NH3 M SCN thiocyanato
3+ 3+ NH3 M NCS isothiocyanato.
Co en Co
11. (b) The delocalised p p bonding between filled p-
en en orbital of F and vacant p-orbital of B leads to shortening
en
of B–F bond length which results in higher bond
dissociation energy of the B–F bond.
3
Enantiomers of cis- Co(en)2 (NH3 )2
6. (a) Na 2 CO 3 2Na CO 32 F
4 4 4
1 10 M 1 10 M 1 10 M B F
F
KSP(BaCO3 ) [Ba 2 ][CO32– ]
9 Vacant Filled
5.1 10
[Ba 2 ] 4
5.1 10 5 M 2p-orbital 2p -orbital
1 10
+ +1/3
h 6.63 10 34 F F F F
7. (a) B=F
+
B–F B–F B F
+1/3
mv 1.67 10 27 1 103 +1/3
+
= 3.97 × 10–10 meter = 0.397 nanometer F F F F
AIEEE-2009 Solved Paper S-199
Y a2 a 1 1 1 2 3
D (a )
(2, 3) 2 2 2 4
D
P
1 3 3 2
(0,2)A It is min when a = and Dmin =
2 4 2 8
x 2 y 1 z 2
39. (a) The line lie in the plane
3 5 2
x + 3y – z + = 0
B Pt (2, 1, – 2) lies on the plane
X
(–4, 0) O i.e. 2 + 3 + 2 + = 0
C(5,0)
or 2 + + 5 = 0 ....(i)
Also normal to plane will be perpendicular to line,
3 × 1 – 5 × 3 + 2 × (– ) = 0
=–6
Then required area = Ar BOA + Ar (OCPD) – Ar From equation (i) then, = 7
( APD) ( ) = (– 6, 7)
40. (c) 4 novels, out of 6 novels and 1 dictionary out of 3 can
1 3 1
= ×4×2+ xdy 2 1 be selected in 6 C4 3
C1 ways
2 0 2
3
Then 4 novels with one dictionary in the middle can be
= 3 ( y 2) 2 1 dy arranged in 4! ways.
0
Total ways of arrangement 6 3
C4 C1 4! = 1080
3 3
( y 2) 1 8 41. (d) We have
= 3 y = 3 3
3 3 3 9 9
0
P (x 1) 1 – P (x = 0)
= 3 + 6 = 9 Sq. units 10 10
S-202 The Pattern Target AIEEE
n 1
0
3
n
9 45.(c) We have y c1e c2 x
1 C0
4 4 10 y c1c2 e c2 x c2 y
n
9 3 y y y ( y )2
1 c2 O
10 4 y y2
n
3 1 y y ( y )2
4 10
Taking log to the base 3/4, on both sides, we get a a 1 a 1 a 1 b 1 c 1
b b 1 b 1 a 1 b 1 c 1 0
3 1 46. (b)
n log3/4 log3/4 c c 1 c 1 ( 1)n 2 a ( 1)n 1b ( 1)n c
4 10
log10 10 1 n 2
n log 3/4 10 a 1 a 1 1 a
3 log10 3 log10 4 a a 1 a 1
log10 n 1
4 b b 1 b 1 b 1 b 1 1 b 0
c c 1 c 1 c 1 c 1 ( 1)n c
1
n
log10 4 log10 3
(Taking transpose of second determineat)
42. (a) If thelines p (p2 + 1) x – y + q = 0
C1 C3
and (p2 + 1)2 x + (p2 + 1) y +2q = 0
are perpendicular to a common line then these lines
n 2
must be parallel to each other, 1 a a 1 a 1
a a 1 a 1
n 2
p ( p 2 1) ( p 2 1) 2 b b 1 b 1 1 ( b) b 1 b 1 0
m1 = m2
1 p2 1 c c 1 c 1 ( 1)n 2 c c 1 c 1
(p2 + 1)
(p+1)=0
p=–1 In second matrix, C2 C3
p can have exactly one value.
43. (b) Let x A and x B x A B a a 1 a 1 a a 1 a 1
n 2
b b 1 b 1 ( 1) b b 1 b 1 0
x A C ( A B A C)
c c 1 c 1 c c 1 c 1
x C
B = C. a a 1 a 1
Let x A and x B x A B 1 ( 1)n 2
b b 1 b 1 0
x A C ( A B A C) c c 1 c 1
x C
In second matrix, C2 – C1, C3 – C1
B C
44. (b) Given that f (x) = x3 + 5x + 1 a 1 1
n 2
f' (x) = 3x2 + 5 > 0, x R 1 ( 1) b 2b 1 2b 1 0
f (x) is strictly increasing on R c 1 1
f (x) is one one
R1 + R3
being a polynomial f (x) is cont. and inc.
a c 0 0
on R with xlim f ( x ) n 2
1 ( 1) b 2b 1 2b 1 0
and xlim f ( x ) c 1 1
[1+ (– 1)n + 2](a + c) (2b + 1+ 2b – 1) = 0
Range of f = ( , ) R
4b (a + c) [1 + (–1)n + 2] = 0
Hence f is onto also.
1 + (–1)n + 2 = 0 as b (a + c) 0
So, f is one one and onto R.
n should be an odd integer.
AIEEE-2009 Solved Paper S-203
– 63 ×
2n 1
C0 (63)2n 2n 1
C1 (63)2 n 1
....... 1 1 1 2 6 10
S ........ ....(2)
3 3 32 33 34
63 × some integral value + 2
Subtracting eqn. (2) from eqn. (1) we get
82n – (62)2n+1when divided by 9 leaves 2 as the
remainder. 2 1 4 4 4
S 1 ........
48. (d) x2x – 2xx cot y – 1 = 0 3 3 32 3 3
34
2 cot y = xx – x – x
2 4 4 4 4
S ........
1 3 3 3 2 3
3 34
2 cot y = u – where u = xx
u
Differentiating both sides with respect to x, we get 4
2 3 4 3
dy 1 du S
– 2cosec y 2
1 3 1 3 2
1
dx u2 dx 3
where u = xx log u = x log x
S =3
1 du 51. (b) Let P (x1, y1, z1) and Q (x2, y2, z2) be the initial and final
1 log x
u dx points of the vector whose projections on the three
coordiante axes are 6, – 3, 2
du
x x (1 log x) then
dx x2 – x1, = 6 ; y2 – y1 = – 3 ; z2 – z1 = 2
We get
So that direction ratios of PQ are 6, – 3, 2
dy 2x x
–2 cosec2 y (1 x ).x (1 log x) Direction cosines of PQ are
dx
6 3
dy xx x x (1 log x) , ,
2 2 2 2
2 ....(i) 6 ( 3) 2 6 ( 3)2 22
dx 2(1 cot y )
Now when x = 1, x2x – 2xx cot y – 1 = 0, gives 2
1 – 2 cot y – 1 = 0 62 ( 3)2 22
cot y = 0
from equation (i), at x = 1 and cot y = 0, 6 –3 2
we get = , ,
7 7 7
1 1 (1 0) 52. (b) We have
y ' (1) 1
2(1 0) 3
cos ( – ) + cos ( – ) + cos ( – ) =
49. (b) Given that roots of the equation 2
bx2 + cx + a = 0 are imaginary 2 [cos ( – ) + cos ( – ) + cos ( – )] + 3 = 0
c2 – 4ab < 0 ....(i)
2 [cos ( – ) + cos ( – ) + cos ( – )]
Let y = 3b2x2 + 6 bc x + 2c2
3b2x2 + 6 bc x + 2c2 – y = 0 + sin2 + cos2 + sin2 + cos2 sin2 + cos2
S-204 The Pattern Target AIEEE
[sin2 + sin2 + sin2 + 2 sin sin + 2 sin sin +
x2 y2
2 sin sin ] 56. (a) The given ellipse is 1
4 1
+ [cos2 cos2 cos2 2cos cos 2 cos cos So, A = (2, 0) and B= (0, 1)
2cos cos If PQRS is the rectangle in which it is inscribed, then
[sin + sin + sin ]2 + (cos cos cos )2= 0 P = (2, 1).
sin + sin + sin = 0 and cos cos cos =0 x2 y2
Let 1 be the ellipse circumscribing the
A and B both are true. a 2 b2
53. (d) Let A Sum of the digits is 8 rectangle PQRS.
B Product of the digits is 0
Then A = {08, 17, 26, 35, 44}
B = {00, 01, 02, 03, 04, 05, 06, 07, 08, 09, 10, 20, 30, 40,} Q B (0,1) P (2, 1)
A B = { 08 }
1
P ( A B) 50 1
P (A/B) =
P( B ) 14
50
14 A
O (2,0) (4,0)
54. (a) Given that
AP BP CP 1
P (1, 0), Q (– 1, 0) and
AQ BQ CQ 3
S
3AP = AQ R
Let A = (x, y) then
3AP = AQ 9 AP2= AQ2
9 (x – 1)2 + 9y2 = (x + 1)2 + y2 Then it passed through P (2,1 ).
9 x2 – 18x + 9 + 9y2 = x2 +2x +1 + y2 4 1
8x2 – 20x + 8y2 + 8 = 0 1 ....(a)
2
a b2
5 Also, given that, it passes through (4, 0)
x2 + y2 – x +1 = 0 ....(1)
3
16
A lies on the circle given by eq (1). As B and C also 0 1 a 2 16
follow the same condition, they must lie on the same a2
circle. b2 = 4/3 [substituting a2 = 16 in eqn (a)]
centre of circumcircle of ABC
x2 y2
The required ellipse is 1 or x2 + 12y2 =16
5 16 4/3
= Centre of circle given by (1) = ,0
4 57. (a) Given that
d × 100 × 101 4 4 4 4
=1+ Now z z z
101 × 2 z z z z
=1 + 50 d 4
mean deviation from the mean = 255 z 2
z
1
[|1 (1 50d ) | | (1 d ) (1 50d ) | | (1 2d ) z 2
–2 z 4 0
101
(1 50d ) | .... | (1 100d ) (1 50 d ) |] 255 2 20 2 20
z – z – 0
2d [1 2 3 ... 50] 101 255 2 2
50 51 z – (1 5) z – (1 5) 0
2d 101 255
2
5 1 z 5 1
101 255
d 10.1
50 51 z max 5 1
AIEEE-2009 Solved Paper S-205
3 p2 u , v , w pq u , v , w 2q 2 u , v , w g' R2
62. (d) We know that
(3 p2 – pq + 2q2) u , v , w 0
g ( R h)2
3p2 – pq + 2q2 = 0 g /9 R
2
q2 7 q2 g R h
2p2 + p2 – pq + 0
4 4
R 1
2
q 7 2 R h 3
2p2 + p q 0
2 4 h = 2R
p = 0, q = 0, p = q / 2 63. (a) The heat flow rate is given by
This is possible only when p = 0, q = 0
there is exactly one value of (p, q). d kA( 1 )
dt x
60. (c) Let I = [cot x ] dx1 ....(1)
0 x d
1
cot ( x) dx kA dt
=
0
x d
cot x dx 1 kA dt
....(2)
0
Adding two values of I in eqn s (1) & (2), where 1 is the temperature of hot end and is
We get temperature at a distance x from hot end.
The above equation can be graphically represented by
2I = cot x cot x dx
0 option (a) .
= ( 1)dx [ [ x] [ x] 1, if x z WPQ
0 (VQ VP )
64. (c)
and [x] + [–x] = 0, if x z] q
= x 0 WPQ q (VQ VP )
I=– = (– 100 × 1.6 × 10–19) (– 4 – 10)
2
= +2.24 × 10–16J
S-206 The Pattern Target AIEEE
65. (a) The magnetic field at O due to current in DA is 71. (d) Let F be the force between Q and Q. The force between
I q and Q should be attractive for net force on Q to be
o
B1 (directed vertically upwards) zero. Let F be the force between Q and q . The resultant
4 a 6
The magnetic field at O due to current in BC is of F and F is R. For equilibrium
I o Q q
B2 (directed vertically downwards)
4 b 6
The magnetic field due to current AB and CD at O is
zero.
Therefore the net magnetic field is R
F
B B1 B2 (directed vertically upwards)
oI o I
4 a6 4 b 6 Q
q
oI 1 1 oI
F
(b a)
24 a b 24ab F
Q
Increasing
Energy n=5
n
90 –
n=4
n=3 82. (c) P
n=2
n=1
Applying Snell’s law at Q
76. (b) Third bright fringe of known light coincides with the
4th bright fringe of the unknown light. sin 90 1
n
sin(90º ) cos
3(590) D 4 D
d d 1
cos
n
3
590 442.5 nm
4
1 n2 1
sin 1 cos 2 1 ...(1)
77. (a) Given u 3iˆ 4 ˆj , a 0.4iˆ 0.3 ˆj , t = 10 s n2 n
Applying Snell’s Law at P
v u at 3iˆ 4 ˆj (0.4iˆ 0.3 ˆj) 10 7iˆ 7 ˆj sin
n sin n sin n 2 1 ; from (1)
sin
|v | 72 72 7 2 units 2
2 4 1
78. (a) 400 nm, hc = 1240 eV.nm, K.E. =1.68 eV sin 1 1
3 3 3
hc hc
We know that W K .E W K .E 1 1
or sin
3
1240
W 1.68 = 3.1 – 1.68 = 1.42 eV
400
83. (c)
79. (b) Maximum number of beats = ( + 1) – ( – 1) = 2
A Y
2
u=0 a = 2m/s vm
80. (a) Wire (1)
Electric s Motor
siren cycle
v 2m u2 2as 3A Y
v2m 2 2 s
/3
vm 2 s Wire (2)
According to Doppler’s effect
As shown in the figure, the wires will have the same
v vm
v' v Young’s modulus (same material) and the length of the
v wire of area of cross-section 3A will be /3 (same volume
as wire 1).
330 2 s For wire 1,
0.94v v
330 F/A
Y ...(i)
s = 98.01 m x/
For wire 2 ,
81. (d) For A B C , is positive. This is because Eb
F '/ 3 A
Y ...(ii)
for C is greater than the E b for A and B. x /( / 3)
Again for F D E , is positive. This is F F'
From (i) and (ii) , F ' 9F
because Eb for D and E is greater than Eb for F . A x 3A 3 x
S-208 The Pattern Target AIEEE
84. (a) Statement 1 is true.
1 1 1
Statement 2 is true and is the correct explanation of (1) For a convex lens
v u f
85. (d) Here y ( A B ) A.B A B . Thus it is an AND gate
for which truth table is when u , v f
A B y when u f,v
0 0 0
Then u 2 f ,v 2f
0 1 0
1 0 0 f
Also v
1 1 1 f
1
u
86. (a) For an SHM, the acceleration a 2
x where 2 is
a As |u| increases, v decreases for | u | f . The graph
a constant. Therefore is a constant. The time period
x between |v| and |u| is shown in the figure. A straight line
aT passing through the origin and making an angle of
T is also constant. Therefore is a constant. 45°with the x-axis meets the experimental curve at
x
87. (c) P (2f, 2f ).
O
89. (b) We know that a single p-n junction diode connected to
an a-c source acts as a half wave rectifier [Forward
biased in one half cycle and reverse biased in the other
half cycle].
C. M
h
C. M Reference 90. (b)
level for P.E.
R
1 2 x dx
The moment of inertia of the rod about O is m .
3
The maximum angular speed of the rod is when the rod
is instantaneously vertical. The energy of the rod in
1
this condition is I 2 where I is the moment of inertia
2
of the rod about O. When the rod is in its extreme Let us consider a spherical shell of thickness dx and
portion, its angular velocity is zero momentarily. In this
radius x. The volume of this spherical shell = 4 x 2 dx .
case, the energy of the rod is mgh where h is the
maximum height to which the centre of mass (C.M) The charge enclosed within shell
rises
Q.x
1 1 2 = [4 x2dx]
1 2 2
R 4
mgh = I ml
2 2 3
2 2 4Q
h = x 3dx
6g R4
The charge enclosed in a sphere of radius r1 is
88. (d) |v| Experimental
r1 r1
curve 4Q 4Q x 4 Q 4
x3 dx r1
4
R R4 4 0
R4
0
Straight
line The electric field at point p inside the sphere at a
distance r1 from the centre of the sphere is
(2f, 2f)
P Q
r4
1 4 1 1 Q
45° E R r12
|u|
4 0 r12 4 0 R4
AIEEE - 2010
Time : 3 Hours Max. Marks : 432
Section-1 : Mathematics (144 marks) : Question No.1 to 9 and 13 to 21 and 25 to 30 are of 4 marks each and Question No.
10 to12 and 22 to 24 are of 8 marks each.
Section-2 : Chemistry (144 marks) : Question No. 34 to 39 and 43 to 60 are of 4 marks each and Question No. 31 to 33 and
40 to 42 are of 8 marks each.
Section-3 : Physics (144 marks) : Question No. 63 to 69, 70 to 79 and 84 to 90 are of 4 marks each and Question No.
61 to 62 and 80 to 83 are of 8 marks each.
Negative Marking : One fourth (¼) marks will be deducted for indicating incorrect response of each question.
Section - 1 4
4. The equation of the tangent to the curve y x , that
x2
is parallel to the x-axis, is
(a) y = 1 (b) y = 2
4 5
1. Let cos ( ) and sin ( ) , (c) y = 3 (d) y = 0
5 13
5. Solution of the differential equation
where 0 , .
4 cos x dy y (sin x y ) dx, 0 x is
Then tan 2 = 2
56 19 (a) y sec x = tan x + c (b) y tan x = sec x + c
(a) (b) (c) tan x = (sec x + c) y (d) sec x = (tan x + c) y
33 12
6. The area bounded by the curves y = cos x and y = sin x
20 25
(c) (d) 3
7 16
between the ordinates x = 0 and x = is
2. Let S be a non-empty subset of R. Consider the following 2
statement :
P : There is a rational number x S such that x > 0. (a) 4 2 2 (b) 4 2 1
Which of the following statements is the negation of the
(c) 4 2 1 (d) 4 2 2
statement P ?
(a) There is no rational number x S such than x < 0. 7. If two tangents drawn from a point P to the parabola y2 = 4x
(b) Every rational number x S satisfies x < 0. are at right angles, then the locus of P is
(c) x S and x < 0 x is not rational. (a) 2x + 1 = 0 (b) x = – 1
(d) There is a rational number x S such that x < 0.
(c) 2x – 1 = 0 (d) x = 1
3. Let a ˆj kˆ an d c iˆ ˆj kˆ . Then the vector b
8. If the vectors a iˆ ˆj 2kˆ , b 2iˆ 4 ˆj kˆ and
satisfying a b c 0 and a. b 3
c iˆ ˆj kˆ are mutually orthogonal, then ( , )
(a) 2iˆ ˆj 2kˆ (b) iˆ ˆj 2kˆ
(a) (2, –3) (b) (–2, 3)
(c) iˆ ˆj 2kˆ (d) iˆ ˆj 2kˆ
(c) (3, –2) (d) (–3, 2)
S-210 The Pattern Target AIEEE
9. Consider the following relations: 13. Statement -1 : The point A(3, 1, 6) is the mirror image of the
R = {(x, y) | x, y are real numbers and x = wy for some rational point B(1, 3, 4) in the plane x – y + z = 5.
number w}; Statement -2: The plane x – y + z = 5 bisects the line segment
joining A(3, 1, 6) and B(1, 3, 4).
m p
S { , | m,n, p and q are integers such that n, q 0 (a) Statement -1 is true, Statement -2 is true ; Statement -2
n q
is not a correct explanation for Statement -1.
and qm = pn}. (b) Statement -1 is true, Statement -2 is false.
Then (c) Statement -1 is false, Statement -2 is true .
(a) Neither R nor S is an equivalence relation (d) Statement - 1 is true, Statement 2 is true ; Statement -2
(b) S is an equivalence relation but R is not an equivalence is a correct explanation for Statement -1.
relation
10 10 10
(c) R and S both are equivalence relations 14. Let S1 j ( j 1)10C J , S2 j10C j and S3 j 2 10C j .
(d) R is an equivalence relation but S is not an equivalence j 1 j 1 j 1
relation Statement -1 : S3 = 55 × 29.
10. Let f : R R be defined by Statement - 2: S1 = 90 × 28 and S2 = 10 × 28 .
k 2 x, if x 1 (a) Statement -1 is true, Statement -2 is true ; Statement -2
f ( x)
2 x 3, if x 1 is not a correct explanation for Statement -1.
If f has a local minimum at x = – 1 , then a possible value of (b) Statement -1 is true, Statement -2 is false.
k is (c) Statement -1 is false, Statement -2 is true .
(d) Statement - 1 is true, Statement 2 is true ; Statement -2
1
(a) 0 (b) is a correct explanation for Statement -1.
2
15. Let A be a 2 × 2 matrix with non-zero entries and let A2 = I ,
(c) –1 (d) 1 where I is 2 × 2 identity matrix. Define
11. The number of 3 × 3 non-singular matrices, with four entries
Tr(A) = sum of diagonal elements of A and
as 1 and all other entries as 0, is
|A| = determinant of matrix A.
(a) 5 (b) 6
(c) at least 7 (d) less than 4 Statement - 1 : Tr(A) = 0.
Statement -2 : |A| = 1.
Directions Question number 12 to 16 are Assertion–Reason
(a) Statement -1 is true, Statement -2 is true ; Statement -2
type question. Each of these questions contains two
statements is not a correct explanation for Statement -1.
Statement-1 (Assertion ) and Statement-2 (Reason). (b) Statement -1 is true, Statement -2 is false.
Each of these questions also has four alternative choices, (c) Statement -1 is false, Statement -2 is true .
only one of which is the correct answer. You have to select (d) Statement - 1 is true, Statement 2 is true ; Statement -2
the correct choice. is a correct explanation for Statement -1.
16. Let f : R R be a continuous function defined by
12. Four numbers are chosen at random (without replacement)
from the 1
f ( x)
set {1, 2, 3, ...20}. e x
2e x
Statement -1: The probability that the chosen numbers when
1 1
Statement -1 : f (c) = , for some c R.
arranged in some order will form an AP is . 3
85
Statement -2 : If the four chosen numbers form an AP, then 1
the set of all possible values of common difference is Statement -2 : 0 < f (x) , for all x R
2 2
( 1, 2, 3, 4, 5) .
(a) Statement -1 is true, Statement -2 is true ; Statement -2
(a) Statement -1 is true, Statement -2 is true ; Statement -2
is not a correct explanation for Statement -1.
is not a correct explanation for Statement -1
(b) Statement -1 is true, Statement -2 is false.
(b) Statement -1 is true, Statment -2 is false
(c) Statement -1 is false, Statement -2 is true .
(c) Statement -1 is false, Statment -2 is true.
(d) Statement -1 is true, Statement -2 is true ; Statement -2 (d) Statement - 1 is true, Statement 2 is true ; Statement -2
is a correct explanation for Statement -1. is a correct explanation for Statement -1.
AIEEE-2010 Solved Paper S-211
17. For a regular polygon, let r and R be the radii of the inscribed
2 3
and the circumscribed circles. A false statement among the (a) (b)
3 2
following is
(c) 3 (d) 1
r 1 24. Let p(x) be a function defined on R such that p (x)
(a) There is a regular polygon with
R 2 = p (1 – x), for all x [0, 1], p (0) = 1 and p (1) = 41. Then
r 2 1
(b) There is a regular polygon with p ( x ) dx equals
R 3
0
r 3 (a) 21 (b) 41
(c) There is a regular polygon with
R 2 (c) 42 (d) 41
r 1 25. Let f : (–1, 1) R be a differentiable function with f(0) = – 1
(d) There is a regular polygon with and f (0) = 1. Let g(x) = [f (2f (x) + 2)]2. Then g (0) =
R 2
18. If and are the roots of the equation x2 – x + 1 = 0, then (a) –4 (b) 0
2009 + 2009 = (c) –2 (d) 4
(a) –1 (b) 1 26. There are two urns. Urn A has 3 distinct red balls and urn B
has 9 distinct blue balls. From each urn two balls are taken
(c) 2 (d) –2
out at random and then transferred to the other. The number
19. The number of complex numbers z such that of ways in which this can be done is
|z – 1| = |z + 1| = |z – i| equals (a) 36 (b) 66
(a) 1 (b) 2 (c) 108 (d) 3
(c) (d) 0 27. Consider the system of linear equations ;
20. A line AB in three-dimensional space makes angles 45° and x1 + 2x2 + x3 = 3
120° with the positive x-axis and the positive y-axis
2x1 + 3x2 + x3 = 3
respectively. If AB makes an acute angle with the positive
3x1 + 5x2 + 2x3 = 1
z-axis, then equals
The system has
(a) 45° (b) 60°
(a) exactly 3 solutions
(c) 75° (d) 30°
(b) a unique solution
x y (c) no solution
21. The line L given by = 1 passes through the point
5 b (d) infinite number of solutions
(13, 32). The line K is parallel to L and has the equation 28. An urn contains nine balls of which three are red, four are
blue and two are green. Three balls are drawn at random
x y
1 . Then the distance between L and K is without replacement from the urn. The probability that the
c 3 three balls have different colours is
17 2 1
(a) 17 (b) (a) (b)
15 7 21
23 23 2 1
(c) (d) (c) (d)
17 15 23 3
22. A person is to count 4500 currency notes. Let an denote the 29. For two data sets, each of size 5, the variances are given to
number of notes he counts in the nth minute. If a1 = a2 = ... = be 4 and 5 and the corresponding means are given to be 2
a10 = 150 and a10, a11, ... are in an AP with common difference and 4, respectively. The variance of the combined data set
–2, then the time taken by him to count all notes is is
(a) 34 minutes (b) 125 minutes 11
(c) 135 minutes (d) 24 minutes (a) (b) 6
2
23. Let f : R R be a positive increasing function with
13 5
f (3 x) (c) (d)
lim 1 2 2
x f ( x) 2 2
30. The circle x + y = 4x + 8y + 5 intersects the line 3x – 4y = m
f (2 x) at two distinct points if
Then lim (a) – 35 < m < 15 (b) 15 < m < 65
x f ( x)
(c) 35 < m < 85 (d) – 85 < m < – 35
S-212 The Pattern Target AIEEE
Section - 2 37. If sodium sulphate is considered to be completely
dissociated into cations and anions in aqueous solution,
the change in freezing point of water ( Tf), when 0.01 mol
of sodium sulphate is dissolved in 1 kg of water, is
31. The standard enthalpy of formation of NH 3 is (Kf = 1.86 K kg mol–1)
– 46.0 kJ mol–1. If the enthalpy of formation of H2 from its (a) 0.372 K (b) 0.0558 K
atoms is – 436 kJ mol–1 and that of N2 is – 712 kJ mol–1, the (c) 0.0744 K (d) 0.0186 K
average bond enthalpy of N – H bond in NH3 is 38. From amongst the following alcohols the one that would
(a) – 964 kJ mol–1 (b) + 352 kJ mol–1 react fastest with conc. HCl and anhydrous ZnCl 2, is
(c) + 1056 kJ mol –1 (d) – 1102 kJ mol–1 (a) 2-Butanol (b) 2- Methylpropan-2-ol
(c) 2-Methylpropanol (d) 1- Butanol
32. The time for half life period of a certain reaction A 39. In the chemical reactions,
Products is 1 hour. When the initial concentration of the
reactant ‘A’, is 2.0 mol L–1, how much time does it take for its NH2
concentration to come from 0.50 to 0.25 mol L–1 if it is a zero NaNO2 HBF 4
order reaction ? HCl, 278 K
A B
(a) 4 h (b) 0.5 h
(c) 0.25 h (d) 1 h the compounds ‘A’ and ‘B’ respectively are
33. A solution containing 2.675 g of CoCl3. 6 NH3 (molar mass = (a) nitrobenzene and fluorobenzene
267.5 g mol–1) is passed through a cation exchanger. The (b) phenol and benzene
chloride ions obtained in solution were treated with excess (c) benzene diazonium chloride and fluorobenzene
of AgNO3 to give 4.78 g of AgCl (molar mass = 143.5 g
(d) nitrobenzene and chlorobenzene
mol–1). The formula of the complex is
40. 29.5 mg of an organic compound containing nitrogen was
(At. mass of Ag = 108 u)
digested according to Kjeldahl’s method and the evolved
(a) [Co(NH3 )6 ]Cl3 ammonia was absorbed in 20 mL of 0.1 M HCL solution. The
(b) [CoCl2 (NH3 )4 ]Cl excess of the acid required 15 mL of 0.1 M NaOH solution
for complete neutralization. The percentage of nitrogen in
(c) [CoCl3 (NH3 )3 ] the compound is
(d) [CoCl(NH3 )5 ]Cl2 (a) 59.0 (b) 47.4
34. Consider the reaction : (c) 23.7 (d) 29.5
41. The energy required to break one mole of Cl – Cl bonds in
Cl 2 (aq) H 2S(aq) S(s) 2H (aq) 2Cl (aq) Cl2 is 242 kJ mol–1. The longest wavelength of light capable
The rate equation for this reaction is of breaking a single Cl – Cl bond is
rate = k[Cl 2 ][H 2S] (c = 3 × 108 ms–1 and NA = 6.02 × 1023 mol–1).
Which of these mechanisms is/are consistent with this rate (a) 594 nm (b) 640 nm
equation? (c) 700 nm (d) 494 nm
42. Ionisation energy of He is 19.6 × 10–18 J atom–1. The energy
+
A. Cl 2 H 2S H Cl Cl HS (slow) of the first stationary state (n = 1) of Li2+ is
Cl HS H Cl S (fast) (a) 4.41 × 10–16 J atom–1
(b) –4.41 × 10–17 J atom–1
B. H 2S H HS (fast equilibrium) (c) –2.2 × 10–15 J atom–1
(d) 8.82 × 10–17 J atom–1
Cl 2 HS 2Cl H S (Slow)
43. Consider the following bromides :
(a) B only (b) Both A and B
(c) Neither A nor B (d) A only
35. If 10–4 dm3 of water is introduced into a 1.0 dm3 flask at Me Me
Me Br Me
300 K, how many moles of water are in the vapour phase
when equilibrium is established ? Br Br
(Given : Vapour pressure of H2O at 300 K is 3170 Pa; (A) (B)
R = 8.314 J K–1 mol–1) The correct order of SN1 reactivity is
(a) 5.56× 10–3 mol (b) 1.53 × 10–2 mol (a) B > C > A (b) B > A > C
(c) 4.46 × 10–2 mol (d) 1.27 × 10–3 mol (c) C > B > A (d) A > B > C
36. One mole of a symmetrical alkene on ozonolysis gives two 44. Which one of the following has an optical isomer?
moles of an aldehyde having a molecular mass of 44 u. The (a) [Zn(en) (NH3)2]2+ (b) [Co(en)3]3+
alkene is (c) [Co(H2O)4(en)]3+ (d) [Zn(en)2]2+
(a) propene (b) 1-butene (en = ethylenediamine)
(c) 2-butene (d) ethene
AIEEE-2010 Solved Paper S-213
45. On mixing, heptane and octane form an ideal solution. At 51. The correct sequence which shows decreasing order of the
373 K, the vapour pressures of the two liquid components ionic radii of the elements is
(heptane and octane) are 105 kPa and 45 kPa respectively.
Vapour pressure of the solution obtained by mixing 25.0 g (a) Al3 Mg 2 Na F O2
of heptane and 35 g of octane will be
(molar mass of heptane = 100 g mol–1 and of octane = 114 g (b) Na Mg 2 Al3 O2 F
mol–1)
(a) 72.0 kPa (b) 36.1 kPa (c) Na F Mg 2 O2 Al3
(c) 96.2 kPa (d) 144.5 kPa
46. The main product of the following reaction is (d) O2 F Na Mg 2 Al3
conc.H 2SO 4 52. Solubility product of silver bromide is 5.0 × 10–13. The
C6 H 5 CH 2 CH(OH)CH(CH 3 ) 2 ?
quantity of potassium bromide (molar mass taken as 120 g
(a) H 5C 6 H mol–1) to be added to 1 litre of 0.05 M solution of silver
C=C
H CH(CH3)2 nitrate to start the precipitation of AgBr is
(a) 1.2 × 10–10 g (b) 1.2 × 10–9 g
(b) C6H5CH2 CH3
C=C (c) 6.2 × 10–5 g (d) 5.0 × 10–8 g
H CH3
53. The Gibbs energy for the decomposition of Al2O3 at 500°C
(c) H5C6CH2CH2 is as follows :
C = CH 2
H 3C
2 4
(d) C6H5 CH(CH3)2 Al2 O3 Al + O2 , r G 966 kJ mol 1
C=C 3 3
H H The potential difference needed for electrolytic reduction
47. Three reactions involving H2PO4– are given below : of Al2O3 at 500°C is at least
(i) H3PO4 + H2 H3O+ + H2PO4–
(ii) H2PO4 + H2O HPO42– + H3O+
– (a) 4.5 V (b) 3.0 V
(iii) H2PO4– + OH– H3PO4 + O2– (c) 2.5 V (d) 5.0 V
54. At 25°C, the solubility product of Mg(OH)2 is 1.0 × 10–11. At
In which of the above does H 2 PO4 act as an acid ? which pH, will Mg2+ ions start precipitating in the form of
(a) (ii) only (b) (i) and (ii) Mg(OH)2 from a solution of 0.001 M Mg2+ ions?
(c) (iii) only (d) (i) only (a) 9 (b) 10
48. In aqueous solution the ionization constants for carbonic (c) 11 (d) 8
acid are
55. Percentages of free space in cubic close packed structure
K1 = 4.2 × 10–7 and K2 = 4.8 × 10–11.
and in body centered packed structure are respectively
Select the correct statement for a saturated 0.034 M solution
of the carbonic acid. (a) 30% and 26% (b) 26% and 32%
(c) 32% and 48% (d) 48% and 26%
(a) The concentration of CO32 is 0.034 M.
56. Out of the following, the alkene that exhibits optical
(b) The concentration of CO32 is greater than that of isomerism is
(a) 3-methyl-2-pentene (b) 4-methyl-1-pentene
HCO3 . (c) 3-methyl-1-pentene (d) 2-methyl-2-pentene
(c) The concentrations of H + an d HCO3 are 57. Biuret test is not given by
(a) carbohydrates (b) polypeptides
approximately equal.
(c) urea (d) proteins
(d) The concentration of H+ is double that of CO32 .
58. The correct order of E values with negative sign
M2 / M
49. The edge length of a face centered cubic cell of an ionic
substance is 508 pm. If the radius of the cation is 110 pm, the for the four successive elements Cr, Mn, Fe and Co is
radius of the anion is (a) Mn > Cr > Fe > Co (b) Cr < Fe > Mn > Co
(a) 288 pm (b) 398 pm (c) Fe > Mn > Cr > Co (d) Cr > Mn > Fe > Co
(c) 618 pm (d) 144 pm 59. The polymer containing strong intermolecular forces e.g.
50. The correct order of increasing basicity of the given hydrogen bonding, is
conjugate bases (R = CH3) is (a) teflon (b) nylon 6, 6
(a) RCOO HC C R NH 2 (c) polystyrene (d) natural rubber
60. For a particular reversible reaction at temperature T, H and
(b) R HC C RCOO NH 2
S were found to be both +ve. If Te is the temperature at
(c) RCOO NH 2 HC C R equilibrium, the reaction would be spontaneous when
(a) Te > T (b) T > Te
(d) RCOO HC C NH 2 R
(c) Te is 5 times T (d) T = Te
S-214 The Pattern Target AIEEE
Section - 3 64. Statement -1 : When ultraviolet light is incident on a
photocell, its stopping potential is V0 and the maximum
kinetic energy of the photoelectrons is Kmax .When the
ultraviolet light is replaced by X-rays, both V0 and Kmax
61. A rectangular loop has a sliding connector PQ of length l
increase.
and resistance R and it is moving with a speed v as
shown. The set-up is placed in a uniform magnetic field Statement -2 : Photoelectrons are emitted with speeds
going into the plane of the paper. The three currents I1, I2 ranging from zero to a maximum value because of the range
and I are of frequencies present in the incident light.
(a) Statement -1 is true, Statement -2 is true ; Statement -2
is the correct explanation of Statement -1.
l
P (b) Statement -1 is true, Statement -2 is true; Statement -2
is not the correct explanation of Statement -1
(3) Statement -1 is false, Statement -2 is true.
R R v R (d) Statement -1 is true, Statement -2 is false.
I 65. A ball is made of a material of density where
I2 oil water with oil and water representing the
I1 Q
densities of oil and water, respectively. The oil and water are
immiscible. If the above ball is in equilibrium in a mixture of
Blv 2 Blv this oil and water, which of the following pictures represents
(a) I1 I2 , I
6R 6R its equilibrium position ?
Blv 2 Blv
(b) I1 I2 ,I
3R 3R
Blv
(c) I1 I2 I water oil
R
Bl Bl (a) (b)
(d) I1 I2 , I
6R 3R oil
water
62. Let C be the capacitance of a capacitor discharging through
a resistor R. Suppose t1 is the time taken for the energy
stored in the capacitor to reduce to half its initial value and
t2 is the time taken for the charge to reduce to one-fourth its
initial value. Then the ratio t1/ t2 will be
1
(a) 1 (b)
2 water oil
1
(c) (d) 2 (c) (d)
4
oil
Directions: Questions number 3 - 4 contain Statement -1 water
and Statement -2. Of the four choices given after the
statements, choose the one that best describes the two
statements.
63. Statement -1: Two particles moving in the same direction 66. A particle is moving with velocity k ( yiˆ xjˆ) , where k
do not lose all their energy in a completely inelastic collision. is a constant. The general equation for its path is
Statement -2 : Principle of conservation of momentum holds (a) y = x2 + constant
true for all kinds of collisions. (b) y2 = x + constant
(a) Statement -1 is true, Statement -2 is true ; Statement -2 (c) xy = constant
is the correct explanation of Statement -1. (d) y2 = x2 + constant
(b) Statement -1 is true, Statement -2 is true; Statement -2 67. Two long parallel wires are at a distance 2d apart. They
is not the correct explanation of Statement -1 carry steady equal currents flowing out of the plane of the
(c) Statement -1 is false, Statement -2 is true. paper as shown. The variation of the magnetic field B along
(d) Statement -1 is true, Statement -2 is false. the line XX' is given by
AIEEE-2010 Solved Paper S-215
X X
(a)
2
d d x (m)
B
0 2 4 6 8 10 12 14 16
t(s)
X X
(b)
(a) 0.4 Ns (b) 0.8 Ns
(3) 1.6 Ns (b) 0.2 Ns
d d
Directions: Questions number 70-71 are based on the following
B
paragraph.
A nucleus of mass M + m is at rest and decays into two
X X M
daughter nuclei of equal mass each. Speed of light is c.
(c) 2
70. The binding energy per nucleon for the parent nucleus is E1
d d and that for the daughter nuclei is E2. Then
(a) E2 = 2E1 (b) E1 > E2
B (c) E2 > E1 (d) E1 2 E2
V ( R1 R2 ) V
(d) at t = 0 and i
R1 R2 R2 at t = O
S-216 The Pattern Target AIEEE
q q 80. The potential energy function for the force between two
ˆj ˆj
(a) 2 2 (b) 2 2 atoms in a diatomic molecule is approximately given by
4 0r 4 0r
q q a b
ˆj ˆj U(x) = 12 , where a and b are constants and x is the
(c) 2 2 (d)
2 2 2 x x6
2 0r 0r
distance between the atoms. If the dissociation energy of
74. The combination of gates shown below yields
the molecule is D U (x ) U at equilibrium , D is
A
b2 b2
X (a) (b)
2a 12a
B
b2 b2
(c) (d)
(a) OR gate (b) NOT gate 4a 6a
(c) XOR gate (d) NAND gate 81. Two identical charged spheres are suspended by strings of
75. A diatomic ideal gas is used in a Carnot engine as the working equal lengths . The strings make an angle of 30° with each
substance. If during the adiabatic expansion part of the cycle other. When suspended in a liquid of density 0.8g cm–3, the
the volume of the gas increases from V to 32 V, the efficiency angle remains the same. If density of the material of the
of the engine is sphere is 1.6 g cm–3 , the dielectric constant of the liquid is
(a) 0.5 (b) 0.75
(a) 4 (b) 3
(c) 0.99 (d) 0.25
(c) 2 (d) 1
76. If a source of power 4kW produces 1020 photons/second ,
the radiation belongs to a part of the spectrum called 82. Two conductors have the same resistance at 0°C but their
(a) X -rays (b) ultraviolet rays temperature coefficients of resistance are 1 and 2 . The
(c) microwaves (d) -rays respective temperature coefficients of their series and parallel
77. The respective number of significant figures for the numbers combinations are nearly
23.023, 0.0003 and 2.1 × 10–3 are
1 2 1 2
(a) 5, 1, 2 (b) 5, 1, 5 (a) , 1 2 (b) 1 2, ,
2 2
(c) 5, 5, 2 (d) 4, 4, 2
78. In a series LCR circuit R = 200 and the voltage and the
1 2 1 2 1 2
frequency of the main supply is 220V and 50 Hz respectively. (c) 1 2, (d) ,
On taking out the capacitance from the circuit the current 1 2 2 2
lags behind the voltage by 30°. On taking out the inductor 83. A point P moves in counter-clockwise direction on a circular
from the circuit the current leads the voltage by 30°. The path as shown in the figure. The movement of 'P' is such
power dissipated in the LCR circuit is that it sweeps out a length s = t3 + 5, where s is in metres and
(a) 305 W (b) 210 W t is in seconds. The radius of the path is 20 m. The
(c) Zero W (d) 242 W acceleration of 'P' when t = 2 s is nearly.
79. Let there be a spherically symmetric charge distribution with
5 r y
charge density varying as (r ) 0 upto r = R ,
4 R
B
and (r ) 0 for r > R , where r is the distance from the
P(x,y)
origin. The electric field at a distance r(r < R) from the origin
is given by m
20
0r 5 r 4 0r 5 r
x
(a) 4 3 R (2) 3 3 R O A
0 0
84. Two fixed frictionless inclined planes making an angle 30° 87. The initial shape of the wavefront of the beam is
and 60° with the vertical are shown in the figure. Two blocks (a) convex
A and B are placed on the two planes. What is the relative (b) concave
vertical acceleration of A with respect to B ? (c) convex near the axis and concave near the periphery
(d) planar
A 88. The speed of light in the medium is
B (a) minimum on the axis of the beam
(b) the same everywhere in the beam
(c) directly proportional to the intensity I
(d) maximum on the axis of the beam
89. A small particle of mass m is projected at an angle with the
x-axis with an initial velocity in the x-y plane as shown in
60° 30°
0 sin
the figure. At a time t , the angular momentum of
(a) 4.9 ms–2 in horizontal direction g
(b) 9.8 ms–2 in vertical direction the particle is
(c) Zero y
(d) 4.9 ms–2 in vertical direction
v0
85. For a particle in uniform circular motion, the acceleration a
at a point P(R, ) on the circle of radius R is (Here is measured
from the x-axis )
2 2
x
(a) cos iˆ sin ˆj
R R (a) mg 2
cos ˆj
0t
2 2
(b) sin iˆ cos ˆj (b) mg 0 t cos kˆ
R R
1
2 2 (c) mg 0t
2
cos kˆ
2
(c) cos iˆ sin ˆj
R R
1 2
(d) mg 0t cos iˆ
2 2 2
(d) iˆ ˆj
R R where iˆ, ˆj and k̂ are unit vectors along x, y and z-axis
Directions: Questions number 86 – 88 are based on the following respectively.
paragraph. An initially parallel cylindrical beam travels in a medium
of refractive index (I) = 0 + 2 I, where 0 and 2 are positive 90. The equation of a wave on a string of linear mass density
constants and I is the intensity of the light beam. The intensity of 0.04 kg m–1 is given by
the beam is decreasing with increasing radius.
86. As the beam enters the medium , it will t x
y =0.02(m) sin 2 .
(a) diverge 0.04(s ) 0.50(m)
(b) converge The tension in the string is
(c) diverge near the axis and converge near the periphery (a) 4.0 N (b) 12.5 N
(d) travel as a cylindrical beam (c) 0.5 N (d) 6.25 N
S-218 The Pattern Target AIEEE
ANSWER KEY
1 (a) 11 (c) 21 (c) 31 (b) 41 (d) 51 (d) 61 (b) 71 (b) 81 (c)
2 (b) 12 (b) 22 (a) 32 (c) 42 (b) 52 (b) 62 (c) 72 (b) 82 (d)
3 (d) 13 (a) 23 (d) 33 (a) 43 (a) 53 (c) 63 (a) 73 (c) 83 (a)
4 (c) 14 (b) 24 (a) 34 (d) 44 (b) 54 (b) 64 (d) 74 (a) 84 (a)
5 (d) 15 (b) 25 (a) 35 (d) 45 (a) 55 (b) 65 (b) 75 (b) 85 (c)
6 (d) 16 (d) 26 (c) 36 (c) 46 (a) 56 (c) 66 (a) 76 (a) 86 (b)
7 (b) 17 (b) 27 (c) 37 (b) 47 (a) 57 (a) 67 (a) 77 (a) 87 (d)
8 (d) 18 (b) 28 (a) 38 (b) 48 (c) 58 (a) 68 (c) 78 (d) 88 (a)
9 (b) 19 (a) 29 (a) 39 (c) 49 (d) 59 (b) 69 (b) 79 (c) 89 (c)
10 (c) 20 (b) 30 (a) 40 (c) 50 (d) 60 (b) 70 (c) 80 (d) 90 (d)
b1iˆ b2 ˆj b3 kˆ . iˆ ˆj kˆ 0, 1
sec x tan x c
Solution : t(I.F) = (I.F) sec x dx y
where b b1iˆ b2 ˆj b3 kˆ
b1 b2 b3 0 ...(i) 6. (d)
cos x sin x
and a. b 3 ( ˆj kˆ).(b1iˆ b2 ˆj b3kˆ) 3
5 3
b2 b3 3 2
4
From equation (i) 0 2
b1 b2 b3 (3 b3 ) b3 3 2b3 4 2
If b3 2, then b iˆ ˆj 2kˆ 4 2
2 (cos x sin x )dx (sin x cos x) dx sin x dx 4 2 2
4. (c) Since tangent is parallel to x-axis, 0
4 2
dy 8
0 1 3 0 x=2 y=3 7. (b) The locus of perpendicular tangents is directrix
dx x
i.e., x a; x 1
Equation of tangent is y – 3 = 0 (x – 2) y = 3
AIEEE-2010 Solved Paper S-219
a2 bc ab bd
2x + 3 A2
k-2x
ac cd bc d 2
1
a 2 bc 1, bc d 2 1
–1
ab bd ac cd 0
c 0 and b 0 a d 0
This is true where k = – 1
A | ad bc a 2 bc 1
1 ... ...
11. (c) ... 1 ... are 6 non-singular matrices because 6 1 ex
16. (d) f ( x)
... ... 1 ex 2e x
e2 x 2
blanks will be filled by 5 zeros and 1 one. (e 2 x 2) e x 2e 2 x .e x
f '( x )
... ... 1 (e 2 x 2) 2
... 1 ...
Similarly, are 6 non-singular matrices. f '( x) 0 e2 x 2 2e 2 x
1 ... ...
e2 x 2 ex 2
So, required cases are more than 7, non-singular 3 × 3
matrices. 2 1
maximum f (x) =
12. (b) n(S) = 20C4 4 2 2
Statement-1:
1
common difference is 1; total number of cases = 17 0 f ( x) x R
2 2
common difference is 2; total number of cases = 14
common difference is 3; total number of cases = 11 1 1
common difference is 4; total number of cases = 8 Since 0 for some c R
3 2 2
common difference is 5; total number of cases = 5
common difference is 6; total number of cases = 2 1
f ( c)
3
S-220 The Pattern Target AIEEE
17. (b) If O is centre of polygon and AB is one of the side, then b
21. (c) Slope of line L =
r 5
by figure cos
n R 3
O Slope of line K =
c
Line L is parallel to line k.
/n b 3
bc = 15
5 c
R
r (13, 32) is a point on L.
13 32 32 8
1
5 b b 5
A B 3
b 20 c
r 1 1 3 4
, , for
R 2 2 2 Equation of K : y 4 x 3 4x y 3 0
n = 3, 4, 6 respectively. 52 32 3 23
Distance between L and K =
1 1 4 17 17
18. (b) x2 x 1 0 x th
2 22. (a) Till 10 minute number of counted notes = 1500
n
1 3i 3000 2 148 ( n 1)( 2) n 148 n 1
x 2
2
n 2 149 n 3000 0
1 3 2 1 i 3
i n 125, 24
2 2 2 2
But n 125 is not possible
cos i sin , cos i sin total time = 24 + 10 = 34 minutes.
3 3 3 3 23. (d) f(x) is a positive increasing function.
2009 2009
( 2 2009
) ( ) 2009 2
1 0 f ( x ) f (2 x ) f (3 x )
2 1 1
cos
4 I p (1 x)dx ...(ii)
1 0
cos ( being acute)
2 1
on adding (i) and (ii), 2 I (42) dx I 21
3 0
AIEEE-2010 Solved Paper S-221
d SECTON II - CHEMISTRY
25. (a) g '( x ) 2 f (2 f ( x) 2) f (2 f ( x) 2)
dx N2 3H 2 2NH 3 H 2 46.0 kJ mol–1
31. (2)
2 f (2 f ( x) 2) f '(2 f ( x)) 2).(2 f '( x)) Let x be the bond enthalpy of N – H bond then
[Note : Enthalpy of formation or bond formation
g '(0) 2 f (2 f (0) 2). f '(2 f (0) 2).2 f '(0)
enthalpy is given which is negative but the given
4 f (0)( f '(0))2 = 4(–1)(1)2 = – 4 reaction involves bond breaking hence values should
be taken as positive.]
26. (c) Total number of ways = 3 C2 9
C2 H= Bond energies of products – Bond energies
of reactants
9 8
=3× 3 36 108 2 × – 46 = 712 + 3 × (436) – 6x
2
– 92 = 2020 – 6x
1 2 1 3 2 1 6x = 2020 + 92
D 2 3 1 0 D1 3 3 1 0 6x = 2112
27. (c) x = + 352 kJ/mol
3 5 2 1 5 2
32. (c) For the reaction
Given system, does not have any solution. A Product
No solution given t1/ 2 = 1 hour
for a zero order reaction
28. (a) 9
n( S ) C3
A0 initial conc.
n( E ) 3
C1 4
C1 2
C1 tcompletion
k rate constant
3 4 2 24 3! 24 6 2 A0
Probability = 6!
9
C3 9! 9 8 7 7 t1/ 2
2k
2 2
29. (a) x 4, y 5, x 2, y 4 A0 2
or k 1 mol lit –1 hr–1
2 t1/ 2 2 1
1 1
xi2 (2) 2 4; yi2 (4) 2 5 Further for a zero order reaction
5 5
dx change in concentration
xi2 40; yi2 105 xi2 yi2 145 k=
dt time
( xi yi ) 5(2) 5(4) 30 0.50 0.25
1
Variance of combined data time
2 time = 0.25 hr.
1 1
= xi2 yi2 xi yi
10 10 33. (a) CoCl3.6NH3 xCl
2.675g
145 11
9
10 2 xCl AgNO3 x AgCl
4.78g
30. (a) Circle x 2 2
y 4x 8 y 5 0
2.675
Number of moles of the complex = = 0.01 moles
Centre = (2, 4), Radius = 4 16 5 5 267.5
If circle is intersecting line 3 x 4 y m, at two distinct 4.78
Number of moles of AgCl obtained = 0.03
points. 143.5
length of perpendicular from centre to the line < moles
radius No. of moles of AgCl obtained = 3 × No. of moles of
complex
6 16 m
5 10 m 25 0.03
5 n= 3
0.01
–25 < m + 10 < 25 – 35 < m < 15 Hence the formula of the complex is [Co(NH3)6]Cl3
S-222 The Pattern Target AIEEE
34. (d) Since the slow step is the rate determining step hence
if we consider option (1) we find Step 2 : (CH 3 )3 C O H 2 (CH 3 )C H2O
3° Carbocation
Rate= k Cl2 H 2S
Now if we consider option (2) we find Step 3 : (CH3 )3 C Cl (CH 3 )3 C Cl
t Butyl chloride
Rate = k Cl2 HS ...(1) 39. (c) Primary aromatic amines react with nitrous acid to yield
From equation (i) arene diazonium salts.
ArNH2 + NaNO2 + 2HX cold
H HS
k 1° Aromatic amine
H 2S
Ar—N = N+X– + NaX + 2H2O
k H 2S Arenediazonium salt
or HS The diazonium group can be replaced by fluorine by
H
treating the diazonium salt with fluoroboric acid
Substituting this value in equation (1) we find (HBF4). The precipitated diazonium fluoroborate is
H 2S Cl 2 H 2S isolated, dried and heated until decomposition occurs
Rate = k Cl2 K k' to yield the aryl fluoride. This reaction is known as
H H
Balz-Schiemann reaction.
hence only , mechanism (1) is consistent with the given Ar—N2+X– HBF4 Ar—N2+BF4– heat
rate equation.
35. (d) From the ideal gas equation : Ar—F + BF3 + N2
40. (c) Moles of HCl taken = 20 × 0.1 × 10– 3 = 2 × 10–3
PV nRT
Moles of HCl neutralised by NaOH solution
PV 3170 10 3 = 15 × 0.1 × 10–3 = 1.5 × 10–3
or n
RT 8.314 300 Moles of HCl neutralised by ammonia
= 1.27 × 10–3 = 2 × 10–3 – 1.5 × 10–3
36. (c) The given molecular formula suggests that the aldehyde = 0.5 × 10–3
formed will be acetaldehyde hence the alkene will be
1.4 N ×V
CH3CH CH CH3 % of nitrogen = 100
w.t. of Substance
2 butene
3
1.4 0.5 10
O3 H O H = 100 23.7%
C C 29.5 10 3
41. (d) Energy required to break one mole of Cl – Cl bonds in
O O Cl2
242 103 hc 6.626 10 34
3 108
= 23
6.023 10
Zn / H 2O 2CH3CHO H 2O 2
34
6.626 10 3 108 6.023 10 23
37. (b) Sodium sulphate dissociates as =
Na 2SO 4 (s) 2Na SO 4 242 108
= 0.4947 × 10–6 m
hence van’t hoff factor i = 3
= 494.7 nm
Now T f i k f .m
= 3 × 1.86 × 0.01 = 0.0558 K Z2
42. (b) I. E = 13.6 eV ...(i)
38. (b) Tertiary alcohols react fastest with conc. HCl and n2
anhydrous ZnCl2 (lucas reagent) as its mechanism
proceeds through the formation of stable tertiary I1 Z12 n22
carbocation. or I ...(ii)
2 n12 Z 22
Mechanism
Given I1 = – 19.6 × 10–18 , Z1 = 2, n1 = 1 , Z2 = 3 and n2 = 1
CH3 Substituting these values in equation (ii).
| 18
Step 1 : CH3 — C — OH H Cl 19.6 10 4 1
| –
I2 1 9
CH 3
2 Methyl Propan-2-ol 18 9
or I2 19.6 10
4
(CH 3 )3 C — OH 2 Cl = – 4.41 × 10–17 J/atom
AIEEE-2010 Solved Paper S-223
Me ionisation
45. (a) PTotal P A xA P B XB
Me Me
+ Br –
Br
+ = P Heptane X Heptane P Octane X Octane
B
25 /100 35 /114
Me = 105 25 35
45
25 35
ionisation Me –
Me + Br
Be
Me + 100 114 100 114
C 0.25 0.3
= 105 45
0.25 0.3 0.25 0.3
Since S N 1 reactions involve the formation of 105 0.25 45 0.3 26.25 13.5
carbocation as intermediate in the rate determining =
0.55 0.55 0.55
step, more is the stability of carbocation higher will
= 72 kPa
be reactivity of alkyl halides towards SN1 route. Now
we know that stability of carbocations follows the order 46. (a) Whenever dehydration can produce two different
: 3° > 2° > 1°, so SN1 reactivity should also follow the alkenes, major product is formed according to Saytzeff
same order. rule i.e. more substituted alkene (alkene having lesser
number of hydrogen atoms on the two doubly bonded
3° > 2° > 1° > Methyl (SN1 reactivity)
carbon atoms) is the major product.
44. (b) For a substance to be optical isomer following
Such reactions which can produce two or more
conditions should be fulfiled
structural isomers but one of them in greater amounts
(1) A coordination compound which can rotate the than the other are called regioselective ; in case a
plane of polarised light is said to be optically active. reaction is 100% regioselective, it is termed as
(2) When the coordination compounds have same regiospecific.
formula but differ in their abilities to rotate directions In addition to being regioselective, alcohol
of the plane of polarised light are said to exhibit optical dehydrations are stereoselective (a reaction in which
isomerism and the molecules are optical isomers. The a single starting material can yield two or more
optical isomers are pair of molecules which are non- stereoisomeric products, but gives one of them in
superimposable mirror images of each other. greater amount than any other).
(3) This is due to the absence of elements of symmetry
Conc. H SO
in the complex. C6 H5 - CH 2 - CH- CH - CH 3 ¾¾¾ ¾¾¾
2 4®
| |
(4) Optical isomerism is expected in tetrahedral
OH CH3
complexes of the type Mabcd.
Based on this only option (2) shows optical isomerism H H H CH(CH3)2
[Co(en)3]3+ C=C + C=C
C6H5 CH(CH3)2 C6H5 H
en 3+ en 3+
cis trans
(minor) (major)
en Co Co en 47. (1)
en
en
(i) H3PO4 H 2O4 H3O H 2 PO4
d–form Mirror –form acid1 base2 acid 2 base1
en Co en en H 2PO 4 OH H3PO4 O
(iii) acid 2 base2
base1 acid1
48. (c) H 2CO3 (aq) H 2O(l ) HCO3– (aq) H3O+ (aq) 52. (b) AgBr Ag+ + Br-
0.034 x x x
Ksp = [Ag+] [Br–]
[HCO3 ][H3O ] x x For precipitation to occur
K1
[H 2CO3 ] 0.034 x Ionic product > Solubility product
K sp 5´10-13
x2 [Br- ] = = = 10-11
[Ag+ ]
7
4.2 10 x 1.195 10 4 0.05
0.034
As H2CO3 is a weak acid so the concentration of i.e., precipitation just starts when 10–11 moles of KBr
H2CO3 will remain 0.034 as 0.034 >> x. is added to 1 AgNO3 solution
Number of moles of Br– needed from KBr = 10–11
x = [H+] = [ HCO3 ] = 1.195 × 10–4 Mass of KBr = 10–11 × 120 = 1.2 × 10–9 g
53. (c) G = – nFE
Now, HCO3– ( aq ) H 2 O(l ) CO32– (aq ) H3 O ( aq )
x y y y
DG 966´103
or E = = = -2.5 V
As HCO3 is again a weak acid (weaker than H2CO3) -nF 4´96500
with x >> y. The potential difference needed for the reduction
= 2.5 V.
[CO32– ][H3O+ ] y (x y)
K2 54. (b) Mg(OH)2 Mg++ + 2OH-
[HCO3– ] ( x y)
Ksp = [Mg++][OH–]2
Note : [H3O+] = H+ from first step (x) and from second
1.0 × 10–11 = 10–3 × [OH–]2
step (y) = (x + y)
[As x > > y so x + y x and x – y x] 10-11
[OH- ] = = 10-4
y x -3
10
So, K 2 =y
x
pOH = 4
11
K2 4.8 10 y [CO32 ] pH + pOH = 14
pH = 10
So the con centration of [H + ] [ HCO3– ]= 55. (b) Packing fraction is defined as the ratio of the volume
concentrations obtained from the first step. As the of the unit cell that is occupied by the spheres to the
dissociation will be very low in second step so there volume of the unit cell.
will be no change in these concentrations. P.F. for cpp and bcc are 0.74 and 0.68 respectively.
Thus the final concentrations are So, the free space in ccp and bcc are 26% & 32%
respectively.
[H+] = [ HCO3 ] = 1.195 × 10–4 & [CO23 ] 4.8 10 11
56. (c) For a compound to show optical isomerism, presence
49. (d) For an Fcc crystal of chiral carbon atom is a necessary condition.
edge length
rcation ranion H
2 |
508 H 2 C = HC — C* — CH 2 - CH3
110 ranion |
2 CH3
ranion = 254 – 110 = 144 pm 3- methyl-1-pentene
50. (d) The correct order of basicity is
57. (a) Biuret test produces violet colour on addition of dilute
RCOO CH C NH 2 R CaSO4 to alkaline solution of a compound containing
51. (d) All the given species contains 10 e – each i.e. peptide linkage.
isoelectronic. Polypeptides, proteins and urea have - C- NH -
For isoelectronic species anion having high negative ||
O
charge is largest in size and the cation having high
positive charge is smallest. (peptide) linkage while carbohydrates have glycosidic
llinkages. So, test of carbohydtrates should be different
from that of other three.
O F Na Mg Al3
AIEEE-2010 Solved Paper S-225
and E 0 = -0.28 V.
Co 2+ Co
O
OH H2N
HO + NH2
O
Adipic acid Hexamethylenediamine
H2 O
H
O O
H2N N OH
NH2 + HO NH2 + HO
O O
Hexamethylenediamine Adipic acid
H2O H2O
Polymerization
H H
O O
N N
N N
O
H H
Nylon
1 dy
t1 = time for the charge to reduce to of its initial kx ...(2)
2 dt
value From (1) and (2) ,
1 dy x
and t2 = time for the charge to reduce to of its
4 dx y
initial value
We have , y dy x dx
q2 q1e t / CR y2 x 2 constant
67. (a) The magnetic field varies inversely with the distance
q2 t for a long conductor. That is,
ln
q1 CR
1
B
1 t1 d
ln ...(1)
2 CR so, graph (1) is the correct one.
68. (c) At t = 0 , no current will flow through L and R1
1 t2
and ln ...(2) V
4 CR
Current through battery = R
2
By (1) and (2) ,
At t = ,
1
ln R1R2
t1 2 effective resistance, Reff
R1 R2
t2 1
ln
4 Current through battery
V
1 = R
ln eff
1 2 1
= 2 1 = V ( R1 R2 )
2ln 4
2 = R1R2
63. (a) In completely inelastic collision, all energy is not lost 69. (b) In each impulse, x varies from 0 to 2m and again from
(so, statement -1 is true) and the principle of 2 m to 0 during the time interval of 4s. We have ,
conservation of momentum holds good for all kinds of Impulse = Change in momentum
collisions (so, statement -2 is true) . Statement -2
explains statement -1 correctly because applying the = m v2 v1
principle of conservation of momentum, we can get the = 2 × 0.4 × (1 – 0) = 0.8 NS
common velocity and hence the kinetic energy of the
70. (c) In nuclear fission, the binding energy per nucleon of
combined body.
daughter nuclei is always greater than the parent
64. (d) We know that nucleus.
eV0 K max h 71. (b) By conservation of energy,
where, is the work function . 2.M 2 1 2M 2
M m c2 c . v ,
Hence, as increases (note that frequency of X-rays 2 2 2
is greater than that of U.V. rays), both V0 and Kmax where v is the speed of the daughter nuclei
increase. So statement - 1 is correct
M 2
65. (b) Oil will float on water so, (2) or (4) is the correct option. mc 2 v
But density of ball is more than that of oil,, hence it will 2
sink in oil.
2 m
v c
66. (a) v k ( y iˆ x ˆj ) M
x-component of v = ky 72. (b) As a result of emission of 1 -particle, the mass number
decreases by 4 units and atomic number decreases by
dx
ky ...(1) 2 units. And by the emission of 1 positron the atomic
dt number decreases by 1 unit but mass number remains
y - component of v = kx constant.
AIEEE-2010 Solved Paper S-227
Mass number of final nucleus = A – 12 74. (a) The final boolean expression is,
Atomic number of final nucleus = Z – 8
Number of neutrons = (A – 12) – (Z – 8) X A.B
=A–Z –4
= A B
Number of protons = Z – 8
=A+B OR gate
A Z 4
Required ratio =
Z 8
73. (c) Let us consider a differential element dl. charge on this 75. (b)
P T1
element. (V, T1)
j
(32 V, T2)
+
+
+ dl
+
+ d
+ T2
+ dE
+
cos
i V
O
dE We have
dE sin
TV r 1
constant
q
dq dl T1V r 1
T2 (32V )r 1
r
q T1 (32)r 1.T2
= ( rd ) ( dl rd )
r For diatomic gas ,
q 7
= d r
5
Electric field at O due to dq is
2
r 1
1 dq 5
dE .
4 0 r2 2
1 q T1 (32) 5 .T2
= 4 . d
2
0 r T1 4T2
The component dE cos will be counter balanced by
T2
another element on left portion. Hence resultant field Now, efficiency = 1 T
at O is the resultant of the component dE sin only. 1
q T2
E dE sin
2 2
sin d = 1 4T
04 r 0 2
q 1
cos =1
2 2 0 4
4 r 0
3
q = 0.75.
= ( 1 1) 4
2 2
4 r 0 76. (a) Power, P nh , n = no, of photons per second
q P
= 2 2
2 r 0 nh
The direction of E is towards negative y-axis.
4 103
q =
E ˆj 10 20 6.63 10 34
2 2
2 r 0 = 6 × 1016 Hz
S-228 The Pattern Target AIEEE
77. (a) Number of significant figures in 23.023= 5
Number of significant figures in 0.0003 = 1
Number of significant figures in 2.1 × 10–3 = 2
dx
So, the radiation belongs to X-rays part of the spectrum. x
78. (d) When capacitance is taken out, the circuit is LR.
L
tan
R
1 200
L = R tan 200
3 3
5 r3 1 r4
Again , when inductor is taken out, the circuit is CR. =4 0 . .
4 3 R 4
1
tan
CR 5 r 3
= 0r
3 R
1 1 200
R tan 200 Electric field at r,
c 3 3
1 q
E . 2
4 0 r
2 3
1 1 0r 5 r
Now, Z R2 L = .
C 4 2 3 R
0 r
2 0r 5 r
200 200 = 4
= (200) 2 0 3 R
3 3 80. (d) At equilibrium ;
= 200 dU ( x )
0
Power dissipated = Vrms I rms cos dx
12a 6b
Vrms R R
= Vrms . . cos 11
x x5
Z Z Z
1
V 2rms R 2a 6
x
= 2 b
Z
a b b2
(220)2 200 U at equilibrium
2a
2 2a = 4a
=
(200)2 b b
In liquid , = R0 2 ( 1 2) t
Fe ' T 'sin 30 ...(A)
1 2
mg FB T 'cos 30 = 2 R0 1 t
2
But FB Buoyant force
1 2
eq
2
In Parallel ,
30° 1 1 1
FB T R R1 R2
T cos 30° 1 1
= R 1 t R0 1 t
0 1 2
Fe
T sin 30°
1 1 1
R0 R0 (1 1 t) R0 (1 2 t)
(1 eq t)
mg 2
2(1 eq t) (1 1 t )(1 2 t)
= V (d )g
V (1.6 0.8) g 1 2
eq
= 0.8 Vg 2
m 0.8 mg mg 83. (a) s t3 5
= 0.8 g
d 1.6 2
ds
velocity , v 3t 2
mg dt
mg T ' cos 30
2
dv
mg Tangential acceleration a t = 6t
T 'cos 30 ...(B) dt
2
From (A) and (B), v2 9t 4
Radial acceleration ac =
2 Fe'
R R
tan 30 At t = 2s,
mg
From (1) and (2) at 6 2 12 m/s2
Fe 2 Fe' 9 16
ac 7.2 m/s2
(2) 20
mg mg
Resultant acceleration
Fe 2 Fe'
= at2 ac2
If K be the dielectric constant, then
Fe
Fe' = (12) 2 (7.2) 2
K
= 144 51.84
2 Fe
Fe
K = 195.84
K=2 = 14 m/s2
S-230 The Pattern Target AIEEE
84. (1) mg sin = ma
c0
a = g sin c=
where a is along the inclined plane
vertical component of acceleration is g sin2 c0
relative vertical acceleration of A with respect to B =
0 2 (I )
is
As I is decreasing with increasing radius, it is maximum
2 2 g on the axis of the beam. Therefore, c is minimum on the
g (sin 60 sin 30] 4.9 m/s 2 in vertical
2 axis of the beam.
direction
89. (3) L m( r v )
85. (c) Clearly
a ac cos ( iˆ) ac sin ( ˆj ) L m v0 cos t iˆ (v0 sin t
1 2 ˆ
gt ) j
2
v2 v2
= cos iˆ sin ˆj v0 cos iˆ (v0 sin gt ) ˆj
R R
Y 1
= mv0 cos t gt kˆ
2
P( R, ) 1
= mgv0 t 2 cos kˆ
2
R
X t x
O 90. (d) y 0.02(m)sin 2
0.04( s) 0.50(m)
Comparing this equation with the standard wave
equation
86. (b) In the medium, the refractive index will decrease from y a sin(wt kx)
the axis towards the periphery of the beam. we get
Therefore, the beam will move as one move from the
axis to the periphery and hence the beam will converge. 2
0.04
1
25 Hz
0.04
Decreasing
2
Axis k 0.5 m
0.50
velocity, v
= 25 × 0.5 m/s = 12.5 m/s
Velocity on a string is given by
87. (d) Initially the parallel beam is cylindrical . Therefore, the T
wavefront will be planar. v
88. (a) The speed of light (c) in a medium of refractive index
( ) is given by T v2 = (12.5)2 × 0.04 = 6.25 N
c0
, where c0 is the speed of light in vacuum
c
AIEEE - 2011 (Held on 1st May 2011)
Time : 3 Hours • Each correct answer has + 4 marks • Each wrong answer has – 1 mark. Max. Marks : 360
(a) 1.7 × 105 (b) 2.7 × 106 27. A water fountain on the ground sprinkles water all around
(c) 3.3 × 107 (d) 1.3 × 104 it. If the speed of water coming out of the fountain is v, the
21. A Carnot engine operating between temperatures T1 and total area around the fountain that gets wet is :
1 v4 v4
T2 has efficiency . When T2 is lowered by 62 K its (a) 2 (b) 2 g2
6 g
1
efficiency increases to . Then T1 and T2 are, respectively: v2 v2
3 (c) 2 (d)
g g
(a) 372 K and 330 K (b) 330 K and 268 K
28. 100g of water is heated from 30°C to 50°C. Ignoring the
(c) 310 K and 248 K (d) 372 K and 310 K
slight expansion of the water, the change in its internal
22. If a wire is stretched to make it 0.1% longer, its resistance energy is (specific heat of water is 4184 J/kg/K):
will : (a) 8.4 kJ (b) 84 kJ
(a) increase by 0.2% (b) decrease by 0.2% (c) 2.1 kJ (d) 4.2 kJ
(c) decrease by 0.05% (d) increase by 0.05% 29. The half life of a radioactive substance is 20 minutes. The
23. This question has a paragraph followed by two statements, approximate time interval (t2 – t1) between the time t2 when
Statement – 1 and Statement – 2. Of the given four 2 1
alternatives after the statements, choose the one that of it had decayed and time t1 when of it had decayed is :
3 3
describes the statements. (a) 14 min (b) 20 min
A thin air film is formed by putting the convex surface of a (c) 28 min (d) 7 min
plane-convex lens over a plane glass plate. With 30. This question has Statement – 1 and Statement – 2. Of the
monochromatic light, this film gives an interference pattern four choices given after the statements, choose the one
due to light reflected from the top (convex) surface and the that best describes the two statements.
bottom (glass plate) surface of the film. Statement – 1: A metallic surface is irradiated by a
Statement – 1 : When light reflects from the air-glass plate monochromatic light of frequency v > v0 (the threshold
interface, the reflected wave suffers a phase change of . frequency). The maximum kinetic energy and the stopping
potential are Kmax and V0 respectively. If the frequency
Statement – 2 : The centre of the interference pattern is
incident on the surface is doubled, both the Kmax and V0
dark.
are also doubled.
(a) Statement – 1 is true, Statement – 2 is true, Statement – 2 Statement – 2 : The maximum kinetic energy and the
is the correct explanation of Statement – 1. stopping potential of photoelectrons emitted from a surface
(b) Statement – 1 is true, Statement – 2 is true, Statement – 2 are linearly dependent on the frequency of incident light.
is not the correct explanation of Statement – 1. (a) Statement–1 is true, Statement–2 is true, Statement – 2
(c) Statement – 1 is false, Statement – 2 is true. is the correct explanation of Statement – 1.
(d) Statement – 1 is true, Statement – 2 is false. (b) Statement–1 is true, Statement–2 is true, Statement – 2
24. A car is fitted with a convex side-view mirror of focal length is not the correct explanation of Statement – 1.
20 cm. A second car 2.8 m behind the first car is overtaking (c) Statement – 1 is false, Statement – 2 is true.
the first car at a relative speed of 15 m/s. The speed of the (d) Statement – 1 is true, Statement – 2 is false.
image of the second car as seen in the mirror of the first
one is : Section - 2
1
(a) m/s (b) 10 m/s
15
31. The presence or absence of hydroxyl group on which
1 carbon atom of sugar differentiates RNA and DNA?
(c) 15 m/s (d) m/s
10
(a) 1st (b) 2nd
25. Energy required for the electron excitation in Li++ from the (c) 3rd (d) 4 th
first to the third Bohr orbit is :
32. Among the following the maximum covalent character is
(a) 36.3 eV (b) 108.8 eV
shown by the compound
(c) 122.4 eV (d) 12.1 eV
(a) FeCl2 (b) SnCl2
26. The electrostatic potential inside a charged spherical ball is
given by = ar2 + b where r is the distance from the centre (c) AlCl3 (d) MgCl2
and a, b are constants. Then the charge density inside the 33. Which of the following statement is wrong?
ball is : (a) The stability of hydride increases from NH3 to BiH3
(a) –6a 0r (b) –24 a 0 in group 15 of the periodic table.
(c) –6a 0 (d) –24 a r (b) Nitrogen cannot form d – p bond.
S-234 The Pattern Target AIEEE
(c) Single N – N bond is weaker than the single P – P 42. A gas absorbs a photon of 355 nm and emits at two
bond. wavelengths. If one of the emissions is at 680 nm, the
(d) N2O4 has two resonance structures. other is at:
34. Phenol is heated with a solution of mixture of KBr and (a) 1035 nm (b) 325 nm
KBrO3. The major product obtained in the above reaction is : (c) 743 nm (d) 518 nm
(a) 2-Bromophenol
43. Which of the following statements regarding sulphur is
(b) 3-Bromophenol
incorrect?
(c) 4-Bromophenol
(a) S2 molecule is paramagnetic.
(d) 2, 4, 6-Tribromophenol
(b) The vapour at 200°C consists mostly of S8 rings.
35. A 5.2 molal aqueous solution of methyl alcohol, CH3OH, is
supplied. What is the mole fraction of methyl alcohol in (c) At 600°C the gas mainly consists of S2 molecules.
the solution? (d) The oxidation state of sulphur is never less than +4 in
(a) 0.100 (b) 0.190 its compounds.
(c) 0.086 (d) 0.050 44. The entropy change involved in the isothermal reversible
expansion of 2 mole of an ideal gas from a volume of 10
36. The hybridization of orbitals of N atom in NO3– , NO2 and
dm3 to a volume of 100 dm3 at 27°C is:
NH4 are respectively : (a) 38.3 J mol–1 K–1 (b) 35.8 J mol–1 K–1
(a) sp, sp2, sp3 (b) sp2, sp, sp3 (b) 32.3 J mol–1 K–1 (d) 42.3 J mol–1 K–1
(c) sp, sp3, sp2 (d) sp2, sp3, sp
45. Which of the following facts about the complex
37. Ethylene glycol is used as an antifreeze in a cold climate. [Cr(NH3)6]Cl3 is wrong?
Mass of ethylene glycol which should be added to 4 kg of
(a) The complex involves d2sp3 hybridisation and is
water to prevent it from freezing at –6°C will be : (Kf for
octahedral in shape.
water = 1.86 K kg mol–1, and molar mass of ethylene glycol
= 62 g mol–1) (b) The complex is paramagnetic.
(a) 804.32 g (b) 204.30 g (c) The complex is an outer orbital complex
(c) 400.00 g (d) 304.60 g (d) The complex gives white precipitate with silver nitrate
solution.
38. The reduction potential of hydrogen half-cell will be
negative if : 46. The structure of IF7 is
(a) p(H2) = 1 atm and [H+] = 2.0 M (a) square pyramid
(b) p(H2) = 1 atm and [H+] = 1.0 M (b) trigonal bipyramid
(c) p(H2) = 2 atm and [H+] = 1.0 M (c) octahedral
(d) p(H2) = 2 atm and [H+] = 2.0 M (d) pentagonal bipyramid
39. Which of the following reagents may be used to distinguish 47. The rate of a chemical reaction doubles for every 10°C rise
between phenol and benzoic acid? of temperature. If the temperature is raised by 50°C, the
(a) Aqueous NaOH (b) Tollen’s reagent rate of the reaction increases by about :
(c) Molisch reagent (d) Neutral FeCl3 (a) 10 times (b) 24 times
40. Trichloroacetaldehyde was subjected to Cannizzaro’s (c) 32 times (d) 64 times
reaction by using NaOH. The mixture of the products 48. The strongest acid amongst the following compounds is :
contains sodium trichloroacetate and another compound. (a) CH3COOH
The other compound is : (b) HCOOH
(a) 2, 2, 2-Trichloroethanol (c) CH3CH2CH(Cl)CO2H
(b) Trichloromethanol (d) ClCH2CH2CH2COOH
(c) 2, 2, 2-Trichloropropanol 49. Identify the compound that exhibits tautomerism :
(d) Chloroform (a) 2-Butene (b) Lactic acid
41. Which one of the following orders presents the correct (c) 2-Pentanone (d) Phenol
sequence of the increasing basic nature of the given
50. A vessel at 1000 K contains CO2 with a pressure of 0.5 atm.
oxides?
Some of the CO2 is converted into CO on the addition of
(a) Al2O3 < MgO < Na2O < K2O
graphite. If the total pressure at equilibrium is 0.8 atm, the
(b) MgO < K2O < Al2O3 < Na2O value of K is :
(c) Na2O < K2O < MgO < Al2O3 (a) 1.8 atm (b) 3 atm
(d) K2O < Na2O < Al2O3 < MgO
(c) 0.3 atm (d) 0.18 atm
AIEEE-2011 Solved Paper S-235
51. In context of the lanthanoids, which of the following 60. Silver Mirror test is given by which one of the following
statements is not correct? compounds?
(a) There is a gradual decrease in the radii of the members (a) Acetaldehyde (b) Acetone
with increasing atomic number in the series.
(c) Formaldehyde (d) Benzophenone
(b) All the members exhibit +3 oxidation state.
(c) Because of similar properties the separation of Section - 3
lanthanoids is not easy.
(d) Availability of 4f electrons results in the formation of
compounds in +4 state for all the members of the
series. 61. The lines L1 : y – x = 0 and L2 : 2x + y = 0 intersect the line
52. ‘a’ and ‘b’ are van der Waals’ constants for gases. Chlorine L 3 : y + 2 = 0 at P an d Q respectively.
is more easily liquefied than ethane because The bisector of the acute angle between L1 and L2 intersects
(a) a and b for Cl2 > a and b for C2H6 L3 at R.
(b) a and b for Cl2 < a and b for C2H6
Statement-1: The ratio PR : RQ equals 2 2 : 5
(c) a for Cl2 < a for C2H6 but b for Cl2 > b for C2H6
(d) a for Cl2 > a for C2H6 but b for Cl2 < b for C2H6 Statement-2: In any triangle, bisector of an angle divides
53. The magnetic moment (spin only) of [NiCl4]2– is : the triangle into two similar triangles.
(a) Statement-1 is true, Statement-2 is true; Statement-2
(a) 1.82 BM (b) 5.46 BM
is not a correct explanation for Statement-1.
(c) 2.82 BM (d) 1.41 BM (b) Statement-1 is true, Statement-2 is false.
54. In a face centred cubic lattice, atom A occupies the corner (c) Statement-1 is false, Statement-2 is true.
positions and atom B occupies the face centre positions. (d) Statement-1 is true, Statement-2 is true; Statement-2
If one atom of B is missing from one of the face centred is a correct explanation for Statement-1.
points, the formula of the compound is : 62. If A = sin2 x + cos4x, then for all real x :
(a) A2 B (b) AB2
13
(c) A2B3 (d) A2B5 (a) A 1 (b) 1 A 2
16
55. The outer electron configuration of Gd (Atomic No. : 64) is :
(a) 4f 3 5d5 6s2 (b) 4f8 5d0 6s2 3 13 3
(c) A (d) A 1
(c) 4f 4 5d4 6s2 (d) 4f7 5d1 6s2 4 16 4
56. Boron cannot form which one of the following anions? 63. The coefficient of x7 in the expansion of
(a) BF63– (b) BH4 – (1– x – x2 + x3 )6 is
(a) –132 (b) –144
(c) B (OH)–4 (d) BO2–
(c) 132 (d) 144
57. Ozonolysis of an organic compound gives formaldehyde
as one of the products. This confirms the presence of : 1 cos{2( x 2)}
64. lim
(a) two ethylenic double bonds x 2 x 2
(b) a vinyl group
(a) equals 2 (b) equals – 2
(c) an isopropyl group
(d) an acetylenic triple bond 1
58. Sodium ethoxide has reacted with ethanoyl chloride. The (c) equals (d) does not exist
2
compound that is produced in the above reaction is :
65. Statement-1: The number of ways of distributing 10
(a) Diethyl ether (b) 2-Butanone
identical balls in 4 distinct boxes such that no box is empty
(c) Ethyl chloride (d) Ethyl ethanoate
is 9C3 .
59. The degree of dissociation ( ) of a weak electrolyte, AxBy Statement-2: The number of ways of choosing any 3 places
is related to van’t Hoff factor (i) by the expression from 9 different places is 9C3 .
i –1 i –1 (a) Statement-1 is true, Statement-2 is true; Statement-2
(a) (b) is not a correct explanation for Statement-1.
(x y –1) x y 1
(b) Statement-1 is true, Statement-2 is false.
x y –1 x y 1 (c) Statement-1 is false, Statement-2 is true.
(c) (d)
i –1 i –1 (d) Statement-1 is true, Statement-2 is true; Statement-2
is a correct explanation for Statement-1.
S-236 The Pattern Target AIEEE
72. A man saves Rs. 200 in each of the first three months of his
d2x service. In each of the subsequent months his saving
66. equals :
dy 2 increases by Rs. 40 more than the saving of immediately
previous month. His total saving from the start of service
1 2 will be Rs. 11040 after
d2y dy
3
d2y dy
(a) (b) (a) 19 months (b) 20 months
dx 2 dx dx 2 dx
(c) 21 months (d) 18 months
3 1
d2y dy d2y 1
(c) (d) 73. The domain of the function f ( x) is
dx 2 dx dx 2
x x
(c) ~ ( P ~ R) Q 1 3 1 3
(c) p,q (d) p ,q
2 2 2 2
(d) ~ P (Q ~ R)
89. The area of the region enclosed by the curves
83. The shortest distance between line y – x =1 and curve x =
y2 is 1
y = x, x = e, y = and the positive x-axis is
x
3 2 8
(a) (b) 3
8 3 2 (a) 1 square unit (b) square units
2
4 3
(c) (d) 5 1
3 4 (c) square units (d) square unit
2 2
84. If the mean deviation about the median of the numbers a,
x
2a,.......,50a is 50, then | a | equals 5
90. For x 0, , define f ( x) t sin t dt . Then f has
(a) 3 (b) 4 2 0
(c) 5 (d) 2
(a) local minimum at and 2
85. Statement-1: The point A(1, 0, 7)) is the mirror image of the
(b) local minimum at and local maximum at 2
x y 1 z 2 (c) local maximum at and local minimum at 2
point B(1, 6, 3) in the line :
1 2 3 (d) local maximum at and 2
S-238 The Pattern Target AIEEE
ANSWER KEY
1 (a) 16 (c) 31 (b) 46 (d) 61 (b) 76 (d)
2 (a) 17 (b) 32 (c) 47 (c) 62 (d) 77 (a)
3 (b) 18 (a) 33 (a) 48 (c) 63 (b) 78 (c)
4 (c) 19 (a) 34 (d) 49 (c) 64 (d) 79 (a)
5 (c) 20 (b) 35 (c) 50 (a) 65 (a) 80 (a)
6 (a) 21 (d) 36 (b) 51 (d) 66 (c) 81 (a)
7 (c) 22 (a) 37 (a) 52 (d) 67 (d) 82 (a)
8 (d) 23 (b) 38 (c) 53 (c) 68 (a) 83 (a)
9 (c) 24 (a) 39 (d) 54 (d) 69 (d) 84 (b)
10 (a) 25 (b) 40 (a) 55 (d) 70 (c) 85 (a)
11 (a) 26 (c) 41 (a) 56 (a) 71 (b) 86 (a)
12 (a) 27 (a) 42 (c) 57 (b) 72 (c) 87 (a)
13 (d) 28 (a) 43 (d) 58 (d) 73 (b) 88 (b)
14 (c) 29 (b) 44 (a) 59 (a) 74 (d) 89 (b)
15 (c) 30 (c) 45 (c) 60 (a,c) 75 (d) 90 (c)
SECTION 1 : PHYSICS 4. (c) W = 2T V
W = 2T4 [(52) – (3)2] × 10–4
1. (a) Given wave equation is
= 2 × 0.03 × 4 [25 – 9] × 10–4 J
y(x,t) = e ax 2 bt 2 2 ab xt = 0.4 × 10–3 J = 0.4 mJ
5. (c) As insect moves along a diameter, the effective mass
[( ax )2 ( b t ) 2 2 a x . b t ] and hence the M.I. first decreases then increases so
=e
from principle of conservation of angular momentum,
2
b
angular speed, first increases then decreases.
( ax bt ) 2 x t 6. (a) Let, x1 = A sin t and x2 = A sin ( t + )
=e =e a
q2 x2 tan
x
sin =
2l
For small , sin tan
q2 x 3
l l
Tcos
Tsin
From Snell's law, 2 sin i 3 sin r
r = 45°
d
13. (d) Current in a small element, dI I
On solving t LC k
4 V A
12. (a) Angle of incidence is given by M
t
1 1
d 4t t 2 dt E = 13.6 (3)2 = 108.8 eV
0 0 12 32
26. (c) Electric field
t3
2t 2 (as = 0 at t = 0, 6s)
3 d
E = – 2ar ....(i)
dr
6 3
t
d 2t 2 dt By Gauss's theorem
3
0 0
1 q
= 36 rad E ....(ii)
4 0 r2
36
n 6 From (i) and (ii),
2
20. (b) We have, V = V0 (1 – e–t/RC) Q = –8 0ar3
dq = – 24 0ar2 dr
t / RC
120 = 200 1 e
dq
t = RC in (2.5) Charge density, = – 6 0a
4 r 2dr
R = 2.71 × 106
AIEEE-2011 Solved Paper S-241
27. (a) Total area around fountain 33. (a) The ease of formation and stability of hydrides
2 decreases rapidly from NH3 to BiH3. This is evident
A Rmax
from their dissociation temperature which decreases
v 2 sin v 2 sin 90 v2 from NH3 to BiH3. As we go down the group the size of
Where Rmax = central atom increases and thus metal-hydrogen bond
g g g
becomes weaker due to decreased overlap between
v4 the large central atom and small hydrogen atom.
A NH3 > PH3 > AsH3 > SbH3 > BiH3
g2
(most stable) (least stable)
28. (a) U = Q = mc T = 100 × 10 –3 × 4184 34. (d) 5KBr + KBrO3 + 3H2O 3Br2 + 6KOH
(50 – 30) 8.4 kJ
29. (b) Number of undecayed atom after time t2 ; OH
OH Br Br
N0 t2
N0e ...(i)
3
+ Br2
Number of undecayed atom after time t1;
2N0 t1 Br
N0e ...(ii)
3 35. (c) 5.2 molal solution means 5.2 moles of methyl alcohol in
t2 1 1000
From (i), e 1000 gm water or in mole of water..
3
18
1 mole fraction of methyl alcohol
– t2 = loge ...(iii)
3 moles of methyl alcohol
= moles of methyl alcohol + moles of water
2
From (ii) – e t2 =
3
5.2
= 0.086
2 1000
– t1 = loge 3 ...(iv) 5.2+
18
Solving (iii) and (iv), we get 36. (b) The formula to find the hybridisation of central atom is
t2 – t1 = 20 mm
1
30. (c) By Einstein photoelectric equation, Z= [Number of valence electrons on central atom +
Kmax = eV0 = hv – hv0 2
When v is doubled, Kmax and V0 become more than No. of monovalent atom altached to it + negative
double. charge if any – positive charge if any]
SECTION-2 : CHEMISTRY 1
For NO3– Z [5 0 1 – 0] 3
31. (b) RNA has D (–) - Ribose and the DNA has 2–Deoxy 2
D (–) – ribose as the carbohydrate unit. O
||
O OH 5 O N sp2
HOCH2 HOCH2 OH
O O–
H H 4 H H 1
For NO2+
H H H H
3 2 1
OH OH OH H Z= [5 0 0 –1] 2
2
ribose 2-deoxy ribose
From the structures it is clear that 2nd carbon in DNA O N O sp
do not have OH group.
32. (c) The proportion of covalent character in an ionic bond For NH4
is decided by polarisability of the metal cation as well 1
as the electrongativity of both elements involved in Z= [5 4 0 – 1] 4
2
bonding. Polarisability is further decided by the density
of positive charge on the metal cation. AlCl3 is H +
considered to show maximum covalent character among |
the given compounds. This is because Al 3 + bears 3 N
unit of positive charge and shows strong tendency to H | H sp3
distort the electron cloud, thus the covalent character H
in Al-Cl bond dramatically increases.
S-242 The Pattern Target AIEEE
37. (a) Given Kf = 1.86 K kg mol–1 43. (d) Oxidation of sulphur varies from – 2 to + 6 in its various
Tf = 0 – (– 6) = 6ºC compounds.
As we know that 44. (a) Entropy change for an isothermal reversible process is
Tf = Kf × molality given by
K f ×1000× mass of solute V2
= 100
molar mass of × mass of solvent S = nR ln V = 2 × 8.314 × 2.303 log
solute in kg 1 10
= 38.3 J mol–1 K–1
Substituting given values in formula 45. (c) [Cr (NH3)6]Cl3 is an inner orbital complex, because in
1.86 ×1000×w this complex d-orbital used is of lower quantum number
6= ; w = 0.8 kg = 800 gm ie (n – 1). It results from d 2 sp3 (inner orbital)
62 × 4
hybridization.
1 46. (d) Pentagonal bipyramidal shape.
38. (c) H+ + e– H2
2
1 F
0.059 PH 2
E = Eº – log 2
1 [H ] F
F
Now if PH2 2 atm and [H+] = 1M 3 3
F I sp d
0.059 21 / 2 F
then E = 0 – log –2
1 1
F
39. (d) Phenol gives a violet colour with neutral ferric chloride
solution whereas benzoic acid does not give this test.
F
40. (a) CCl3CHO + NaOH CCl3CH 2 OH CCl3COONa
2, 2, 2 trichloro ethanol 47. (c) Since for every 10ºC rise in temperature rate doubles
In cannizzaro’s reaction the compounds which do not for 50ºC rise in temp increase in reaction rate = 25 = 32
contain -hydrogen atoms undergo oxidation and times
reduction simultaneously i.e undergo disproportion 48. (c) The electron withdrawing (–I ) group – Cl withdraws
ation and form one molecule of sodium salt of carboxylic electrons from O – H bond and thus helps the cleavage
acid as oxidation product and one molecule of alcohol of the O – H bond releasing hydrogen as H+.
as reduction product.
O
41. (a) On moving across a period ionisation energy increases ||
hence the electropositive nature of metals decreases 49. (c) H3C – C – CH 2 – CH2 – CH3
therefore the ease of formation of ion also decreases Keto form
and hence the basic character decreases. Further basic
O–H
character of alkali metals oxides increases from Li2O to |
Tautomerism
Cs2O. Hence the correct order will be H 3C – C = CH – CH 2 – CH 3
enol form
Al2O3 < MgO < Na2 O < K2 O
42. (c) Energy of absorbed photon
= Sum of the energies of emitted photon 50. (a) CO 2 C(grapnite) 2CO
hc hc hc Pinitial 0.5atm 0
1 2 Pfinal (0.5 – x)atm 2x atm
1 1 1
– 6 (2 0.3) 2 (0.6) 2
2 355 10–9 680 10 –9 1.346 10 =
(0.5 – 0.3) (0.2)
= 1.8 atm
or 6 –9
2= 1/1.346 × 10 = 743 × 10 m = 743 nm
51. (d) The most stable oxidation states of lanthanides is + 3.
AIEEE-2011 Solved Paper S-243
52. (d) The value of a is a measure of the magnitudeof the Total no. of moles (i) = 1 – + x + y
attractive forces between the molecules of the gas. i – 1 = x + y – = (x + y – 1)
Greater the value of ‘a’, larger is the attractive
intermolecular force between the gas molecules. i –1
=
The value of b related to the effective size of the gas (x y –1)
molecules. It is also termed as excluded volume. 60. (a,c) Both formaldehyde and acetaldehyde give silver mirror
The gases with higher value of a and lower value of b with Tollen’s reagent.
are more liquefiable, hence for Cl2 “a” should be greater
than for C2H6 but for it b should be less than for C2H6. SECTION-3 : MATHEMATICS
61. (b)
53. (c) [NiCl4]2– (d8)
L3
L1
sp3 hybridisation
i.e the number of unpaired electrons in [NiCl4]2– is 2. 0
x= P(–2, –2)
y–
n n(n 2) 2(4) = 2 2
R(–1, –2)
= 2 × 1.41 = 2. 8 2 BM
O
1 (0, 0) 2x
54. (d) No. of atoms in the corners (A) = 8 1 +y Q
8 =0
1
No. of atoms at face centres (B) = 5 × L2
2
= 2.5
Formula is = AB2.5 or A2B5 L1 : y – x = 0
55. (d) The configuration of Gd is 4f 7 5d1 6s2. L2 : 2x + y = 0
L3 : y + 2 = 0
56. (a) Boron cannot form BF63– due to non-availability of d-
On solving the equation of line L1 and L2 we get their
orbitals.
point of intersection (0, 0) i.e., origin O.
57. (b) Compound must contain a vinyl group (– C CH 2 ) On solving the equation of line L1 and L3,
|
we get P = (– 2, – 2).
in order to give formaldehyde as one of the product.
Similarly, we get Q = (– 1, – 2)
We know that bisector of an angle of a triangle, divide
the opposite side the triangle in the ratio of the sides
R R O H including the angle [Angle Bisector Theorem of a
C = CH2 + O3 C C Triangle]
R R H
O O PR OP ( 2)2 ( 2) 2 2 2
RQ OQ ( 1) 2
( 2) 2 5
y x–
Now 0 sin 2 (2 x) 1
59. (a) A x By xA yB
1 2 1
t=0 1 0 0 0 sin (2 x )
4 4
teq 1– x y
S-244 The Pattern Target AIEEE
68. (a) Let for statement 1 xRy = x – y I . As xRx is an integer
1 2 1
1 1 sin (2 x ) 1 and yRx as well as xRz (for xRy and yRz) is also an
4 4 integer.
3 Hence equivalence.
1 A
4 1
Similarly as x = y hence =1 for reflexive and
63. (b) (1 – x – x2 + x3)6 = [(1– x) – x2 (1 – x)]6
= (1– x)6 (1 – x2)6 being a rational for symmetric for some non zero and
= (1 – 6x + 15x2 – 20x3 + 15x4 – 6x5 + x6) product of rationals also being rational equivalence
× (1 – 6x2 + 15x4 – 20x6 + 15x8 – 6x10 + x12) But not symmetric because of = 0 case
Coefficient of x7 = (– 6) (– 20) + (– 20)(15) Both relations are equivalence but not the correct
+ (– 6) (–6) explanation.
= – 144 1
8log(1 x )
69. (d) I dx
1 cos{2( x 2)} 2 sin( x 2) 1 x2
64. (d) lim lim 0
x 2 x 2 x 2 x 2
Put x = tan ,
L.H.L = 2 sin( x 2) dx
(at x = 2)
lim 1 sec 2 dx sec 2 d
x 2 ( x 2) d
/4
2 sin( x 2) log(1 tan )
R.H.L lim 1 I 8 .sec 2 d
(at x = 2) = ( x 2) 2
x 2
0 1 tan
Thus L.H.L R.H.L /4
(at x = 2) (at x = 2)
I 8 log(1 tan )d ...(i)
0
1 cos{2( x 2)}
Hence, lim does not exist.
x 2 x 2 /4
65. (a) The number of ways of distributing 10 identical balls in =8 log 1 tan
4
d
4 distinct boxes such that no box empty is same as the 0
dx dy / dx dy /4
I 8.(log 2)[ x]0 / 4 8 log(1 tan )d
1 d2y 1 1 d y2 0
. . =–
dy
2
dx 2 dy dy
3
dx 2
dx I 8. .log 2 I [From equation (i)]
dx dx 4
2I 2 log 2
dy dy
67. (d) y 3 dx I log 2
dx y 3
70. (c) Since both the roots lie in the line Re z = 1 i.e., x = 1,
n y 3 x c hence real part of both the roots are 1.
Let both roots be 1 + i and 1 – i
Since y (0) = 2, n5 c Product of the roots, 1 + 2 =
n y 3 x n5 2
1 1
When x = n 2 , then n |y + 3| = n 2 + n5
1
n | y + 3 | = n 10
y + 3 = ±10 y = 7, – 13 (1, )
AIEEE-2011 Solved Paper S-245
1 1 2 2 3 Now, b c b d
sin
12 22 2
. 12 22 32 a (b c ) a (b d )
(a . c )b (a . b )c (a . d )b (a . b )d
5 3
2 (a.b )d (a.c )b (a.b )c
14. 5
a.c
2 d c b
(5 3 ) a.b
cos 1
14(5 2
) 79. (a) If the two circles touch each other, then they must
touch each other internally.
a a
1 (5 3 )2 So, c a c
cos 1 2 2 2
14(5 )
80. (a) In this case
5 C P(C D) P (C )
But it is given that cos 1 P
14 D P( D ) P ( D)
Where, 0 P( D) 1 ,
2
(5 3 ) 5 2 C
1 2 hence P P (C )
14(5 14 3 D
S-246 The Pattern Target AIEEE
81. (a) =0 joining points A and B.
86. (a) A =A
4 k 2 B =B
k 4 1 0 Now (A(BA)) = (BA) A = (A B )A = (AB)A = A(BA)
Similarly ((AB)A) = (AB)A
2 2 1
So, A(BA) and (AB)A are symmetric matrices.
4(4 2) k (k 2) 2(2k 8) 0 Again (AB) = B A = BA
Now if BA = AB, then AB is symmetric matrix.
8 k2 2k 4k 16 0 87. (a) (1 + )7 = A + B
(– 2)7 = A + B
k2 6k 8 0 – 2=A+B
(k 4)(k 2) 0, k 4, 2 1 + =A + B
A = 1, B = 1.
82. (a) Suman is brilliant and dishonest if and only if Suman is
rich is expressed as 88. (b) L.H .L lim f ( x)
( at x 0) x 0
Q ( P ~ R)
sin{( p 1)( h)} sinh
Negation of it will be ~ (Q ( P ~ R)) lim
h 0 h
83. (a) Shortest distance between two curve occurred along = p+1+1=p+2
the common normal, so – 2t = – 1
1 1
t = 1/2 R.H .L lim f ( x) lim
y ( at x 0) x h 0 1 1 2
3 1
f (0) q ,q p
2 2
(t2, t) 89. (b) Area of required region AOBC
1 e
1 1 3
x = xdx dx 1 sq. units
O 0 1
x 2 2
y x=e
y=x
3 2
So shortest distance between them is
8
84. (b) Median is the mean of 25th and 26th observation (1, 1)
A
25a 26a
M 25.5a 1
2 B e, e
O
xi M x
M .D (M ) C
N
1
50 [2 a (0.5 1.5 2.5 ....24.5)] 90. (c) f '( x) x sin x
50
f '( x) 0
25
2500 2 a (25) a 4 x 0 or sin x 0
2
x 2 ,
85. (a) The direction ratio of the line segment joining points
A(1, 0, 7) and B(1, 6, 3) is 0, 6, –4. 1
f ''( x) x cos x sin x
The direction ratio of the given line is 1, 2, 3. 2 x
Clearly 1 × 0 + 2 × 6 + 3 × (–4) = 0 1
So, the given line is perpendicular to line AB. =
(2 x cos x sin x)
2 x
Also , the mid point of A and B is (1, 3, 5) which lies on
the given line. At x = , f ''( x ) 0
So, the image of B in the given line is A, because the Hence, local mixima at x =
given line is the perpendicular bisector of line segment At x 2 , f ''( x ) 0
Hence local minima at x = 2
(Rescheduled Paper Held on 11 May-2011)
AIEEE - 2011
(Rescheduled Paper, held on 11 May 2011)
Time : 3 Hours • Each correct answer has + 4 marks • Each wrong answer has – 1 mark. Max. Marks : 360
Gm Gm mv3
(c) (d) (a) zero (b)
2R R 2g
4. The minimum force required to start pushing a body up
rough (frictional coefficient ) inclined plane is F1 while
3 mv3 3 mv 2
the minimum force needed to prevent it from sliding down (c) (d)
16 g 2 g
is F2. If the inclined plane makes an angle from the
F1 9. The specific heat capacity of a metal at low temperature
horizontal such that tan 2 then the ratio is (T) is given as
F2
3
(a) 1 (b) 2 T
C p (kJK 1kg 1 ) 32
(c) 3 (d) 4 400
S-248 The Pattern Target AIEEE
A 100 gram vessel of this metal is to be cooled from 20ºK 15. Combination of two identical capacitors, a resistor R and a
to 4ºK by a special refrigerator operating at room
dc voltage source of voltage 6V is used in an experiment
temperature (27°C). The amount of work required to cool
the vessel is on a (C-R) circuit. It is found that for a parallel combination
(a) greater than 0.148 kJ of the capacitor the time in which the voltage of the fully
(b) between 0.148 kJ and 0.028 kJ charged combination reduces to half its original voltage is
(c) less than 0.028 kJ 10 second. For series combination the time for needed for
(d) equal to 0.002 kJ reducing the voltage of the fully charged series
10. A wooden cube (density of wood ‘d’) of side ‘ ’ floats in combination by half is
a liquid of density ‘ ’ with its upper and lower surfaces
(a) 10 second (b) 5 second
horizontal. If the cube is pushed slightly down and
released, it performs simple harmonic motion of period ‘T’ (c) 2.5 second (d) 20 second
16. A beaker contains water up to a height h1 and kerosene of
d
(a) 2 (b) 2 height h2 above water so that the total height of ( water +
g dg kerosene) is (h1 + h2). Refractive index of water is 1and
d that of kerosene is 2. The apparent shift in the position of
(c) 2 (d) 2 the bottom of the beaker when viewed from above is
d g d g
11. A container with insulating walls is divided into equal parts 1 1
by a partition fitted with a valve. One part is filled with an (a) 1 h1 1 h2
1 2
ideal gas at a pressure P and temperature T, whereas the
other part is completly evacuated. If the valve is suddenly
opened, the pressure and temperature of the gas will be : 1 1
(b) 1 h1 1 h2
P T 1 2
(a) , (b) P, T
2 2
T P 1 1
(c) P, (d) ,T (c) 1 h2 1 h1
2 2 1 2
12. In a Young’s double slit experiment, the two slits act as
coherent sources of wave of equal amplitude A and
1 1
wavelength . In another experiment with the same (d) 1 h2 1 h1
arrangement the two slits are made to act as incoherent 1 2
sources of waves of same amplitude and wavelength. If the 17. A metal rod of Young’s modulus Y and coefficient of
intensity at the middle point of the screen in the first case is
thermal expansion is held at its two ends such that its
I1 length remains invariant. If its temperature is raised by t°C,
I1 and in the second case is I 2 , then the ratio is
I2 the linear stress developed in it is
(a) 2 (b) 1
Y
(c) 0.5 (d) 4 (a) (b) Y t
13. The output of an OR gate is connected to both the inputs t
of a NAND gate. The combination will serve as a:
(a) NOT gate (b) NOR gate 1 t
(c) (d)
(c) AND gate (d) OR gate (Y t ) Y
14. Two positive charges of magnitude ‘q’ are placed, at the
18. A travelling wave represented by y = A sin ( t – kx) is
ends of a side (side 1) of a square of side ‘2a’. Two negative
charges of the same magnitude are kept at the other corners. superimposed on another wave represented by
Starting from rest, if a charge Q moves from the middle of y = A sin ( t + kx). The resultant is
side 1 to the centre of square, its kinetic energy at the (a) A wave travelling along + x direction
centre of square is (b) A wave travelling along – x direction
(a) zero (c) A standing wave having nodes at
1 2 qQ 1
(b) 1 n
4 0 a 5 x ,n 0,1, 2....
2
1 2 qQ 2 (d) A standing wave having nodes at
(c) 1
4 0 a 5
1 2 qQ 1 1
(d) 1 x n ; n 0,1, 2....
4 0 a 5 2 2
AIEEE-2011 Rescheduled Solved Paper S-249
19. A thin circular disc of radius R is uniformly charged with 26. If a spring of stiffness ‘k’ is cut into parts ‘A’ and ‘B’ of
density 0 per unit area. The disc rotates about its axis length : B 2 : 3, then the stiffness of spring ‘A’ is
A
with a uniform angular speed .The magnetic moment of given by
the disc is
3k 2k
R4 (a) (b)
(a) R 4
(b) 5 5
2
5
4 (c) k (d)
R 2k
(c) (d) 2 R 4
4 27. Statement - 1 : A nucleus having energy E1 decays by –
20. An aluminium sphere of 20 cm diameter is heated from 0°C emission to daughter nucleus having energy E2, but the
– rays are emitted with a continuous energy spectrum
to 100°C. Its volume changes by (given that coefficient of
having end point energy E1 – E2.
linear expansion for aluminium 23 10 6/ C )
A Statement - 2 : To conserve energy and momentum in –
(a) 2.89 cc (b) 9.28 cc decay at least three particles must take part in the
(c) 49.8 cc (d) 28.9 cc transformation.
21. Two mercury drops (each of radius ‘r’) merge to form bigger (a) Statement-1 is correct but statement-2 is not correct.
drop. The surface energy of the bigger drop, if T is the (b) Statement-1 and statement-2 both are correct and
surface tension, is : statement-2 is the correct explanation of statement-1.
(a) 4 r 2T (b) 2 r 2T (c) Statement-1 is correct, statement-2 is correct and
statement-2 is not the correct explanation of
(c) 28 / 3 r 2T (d) 25/ 3 r 2T statement-1
22. If a ball of steel (density = 7.8 g cm–3) attains a terminal (d) Statement-1 is incorrect, statement-2 is correct.
velocity of 10 cm s–1 when falling in water (Coefficient of
28. When monochromatic red light is used instead of blue
viscosity water × 10–4 Pa.s), then, its terminal light in a convex lens, its focal length will
velocity in glycerine ( = 1.2 g cm–3, Pa.s) would (a) increase
be, nearly (b) decrease
(a) 6.25 × 10–4 cm s–1 (c) remain same
(b) 6.45 × 10–4 cm s–1 (d) does not depend on colour of light
(c) 1.5 × 10–5 cm s–1 29. Statement - 1: On viewing the clear blue portion of the sky
(d) 1.6 × 10–5 cm s–1 through a Calcite Crystal, the intensity of transmitted light
23. A horizontal straight wire 20 m long extending from east to varies as the crystal is rotated.
west falling with a speed of 5.0 m/s, at right angles to the
Statement - 2: The light coming from the sky is polarized
horizontal component of the earth’s magnetic field 0.30 ×
due to scattering of sun light by particles in the atmosphere.
10–4 Wb/m2. The instantaneous value of the e.m.f. induced
The scattering is largest for blue light.
in the wire will be
(a) 3 mV (b) 4.5 mV (a) Statement -1 is true, statement-2 is false.
(c) 1.5 mV (d) 6.0 mV (b) Statement-1 is true, statement-2 is true, statement-2
24. After absorbing a slowly moving neutron of mass mN is the correct explanation of statement-1
(momentum 0) a nucleus of mass M breaks into two (c) Statement-1 is true, statement-2 is true, statement-2
nuclei of masses m1 and 5m1 (6m1 = M + mN) respectively. is not the correct explanation of statement-1
If the de Broglie wavelength of the nucleus with mass m1 (d) Statement-1 is false, statement-2 is true.
is , the de Broglie wavelength of the nucleus will be 30. Statement - 1 : Two longitudinal waves given by
(a) 5 (b) /5 equations : y1 ( x, t ) 2a sin( t kx ) and y2 ( x, t ) = a
(c) (d) 25 sin (2 t 2kx) will have equal intensity..
25. Which of the following four alternatives is not correct ? Statement - 2 : Intensity of waves of given frequency in
We need modulation : same medium is proportional to square of amplitude only.
(a) to reduce the time lag between transmission and (a) Statement-1 is true, statement-2 is false.
reception of the information signal (b) Statement-1 is true, statement-2 is true, statement-2
(b) to reduce the size of antenna is the correct explanation of statement-1
(c) to reduce the fractional band width, that is the ratio (c) Statement-1 is true, statement-2 is true, statement-2
of the signal band width to the centre frequency is not the correct explanation of statement-1
(d) to increase the selectivity (d) Statement-1 is false, statement-2 is true.
S-250 The Pattern Target AIEEE
Section - 2 38. The value of enthalpy change ( H) for the reaction
C2 H5OH( ) 3O2 (g) 2CO2 (g) 3H 2 O( )
–1
at 27° C is – 1366.5 k J mol . The value of internal energy
31. Identify the incorrect statement from the following : change for the above reaction at his temperature will be :
(a) Ozone absorbs the intense ultraviolet radiation of the (a) – 1369.0 kJ (b) – 1364.0 kJ
sun.
(c) – 1361.5 kJ (d) – 1371.5 kJ
(b) Depletion of ozone layer is because of its chemical
reactions with chlorofluoro alkanes. 39. Thermosetting polymer, Bakelite is formed by the reaction
(c) Ozone absorbs infrared radiation. of phenol with :
(d) Oxides of nitrogen in the atmosphere can cause the (a) CH3CHO (b) HCHO
depletion of ozone layer. (c) HCOOH (d) CH3CH2CHO
32. When r, P and M represent rate of diffusion, pressure and 40. Ozonolysis of an organic compound 'A' produces acetone
molecular mass, respectively, then the ratio of the rates of and propionaldhyde in equimolar mixture. Identify 'A' from
diffusion (rA / rB ) of two gases A and B, is given as : the following compounds:
(a) 1 – Pentene
(a) ( PA / PB ) ( M B / M A )1/ 2 (b) 2 – Pentene
(c) 2 – Methyl – 2 – pentene
(b) ( PA / PB )1/ 2 ( M B / M A )
(d) 2 – Methyl – 1 – pentene
(c) ( PA / PB ) ( M A / M B )1/ 2 41. Consider the reaction :
4NO2 (g) O 2 (g) 2N 2 O5 (g), rH = – 111 kJ.
(d) ( PA / PB )1/ 2 ( M A / M B )
If N2O5(s) is formed instead of N2O5(g) in the above
33. Consider thiol anion (RS ) and alkoxy anion (RO ) . reaction, the rH value will be :
(given, H of sublimation for N2O5 is 54 kJ mol–1)
Which of the following statements is correct ?
(a) + 54 kJ (b) + 219 kJ
(a) RS is less basic but more nucleophilic than RO (c) – 219 J (d) – 165 kJ
(b) RS is more basic and more nucleophilic than RO 42. An acid HA ionises as
RS is less basic and less nucleophilic than RO The pH of 1.0 M solution is 5. Its dissociation constant
(d)
would be :
34. The change in the optical rotation of freshly prepared
8
solution of glucose is known as: (a) 5 (b) 5 10
(a) racemisation (b) specific rotation (c) 1 10 5 (d) 1 10 10
(c) mutarotation (d) tautomerism 43. The correct order of electron gain enthalpy with negative
35. The molality of a urea solution in which 0.0100 g of urea, sign of F, Cl, Br and I, having atomic number 9, 17, 35 and
[(NH2)2CO] is added to 0.3000 dm3 of water at STP is : 53 respectively, is :
(a) F > Cl > Br > I (b) Cl > F > Br > I
(a) 5.55 10 4 m (b) 33.3 m (c) Br > Cl > I > F (d) I > Br > Cl > F
(c) 3.33 × 10–2 m (d) 0.555 m 44. The frequency of light emitted for the transition n = 4 to
36. The molecular velocity of any gas is : n = 2 of the He+ is equal to the transition in H atom
corresponding to which of the following ?
(a) inversely proportional to absolute temperature. (a) n = 2 to n = 1 (b) n = 3 to n = 2
(b) directly proportional to square of temperature. (c) n = 4 to n = 3 (d) n = 3 to n = 1
(c) directly proportional to square root of temperature. 45. A 5% solution of cane sugar (molar mass 342) is isotonic
(d) inversely proportional to the squar e root of with 1% of a solution of an unknown solute. The molar
temperature. mass of unknown solute in g/mol is :
(a) 171.2 (b) 68.4
37. The correct order of acid strength of the following
(c) 34.2 (d) 136.2
compounds :
(A) Phenol (B) p–Cresol 46. In view of the sings of rG for the following reactions :
(C) m–Nitrophenol (D) p–Nitrophenol PbO2 Pb 2PbO, rG <0
(a) D > C > A > B (b) B > D > A > C
SnO2 Sn 2SnO, rG >0
(c) A > B > D > C (d) C > B>A> D
AIEEE-2011 Rescheduled Solved Paper S-251
which oxidation states are more characteristics for lead 53. The number of types of bonds between two carbon atoms
and tin ? in calcium carbide is :
(a) For lead + 2, for tin + 2 (a) One sigma, One pi (b) Two sigma, one pi
(b) For lead + 4, for tin + 4 (c) Two sigma, two pi (d) One sigma, two pi
(c) For lead + 2, for tin + 4 54. A reactant (A) froms two products :
(d) For lead + 4, for tin + 2 k1
A B, Activation Energy Ea1
47. The Ksp for Cr(OH)3 is 1.6 × 10–30. The more solubility of
k2
this compound in water is : A C, Activation Energy Ea2
30
If Ea2 = 2Ea1, then k1 and k2 are related as :
(a) 30 (b)
4 1.6 10 4 1.6 10 / 27
(a) k2 k1e Ea1 / RT (b) k2 k1e Ea2 / RT
(c) 1.6 10 30/ 27 (d) 2 1.6 10 30
(c) k1 Ak 2 e Ea1 / RT (d) k1 2k 2 e Ea2 / RT
48. The products obtained on heating LiNO2 will be :
55. Copper crystallises in fcc lattice with a unit cell edge of
(a) Li 2 O NO2 O2 (b) Li 3 N O2
361 pm. The radius of copper atom is :
(c) Li 2 O NO O 2 (d) LiNO3 O2 (a) 108 pm (b) 128 pm
(c) 157 pm (d) 181 pm
49. Resistance of 0.2 M solution of an electrolyte is 50 . The
56. The mass of potassium dichromate crystals required to
specific conductance of the solution is 1.3 S m–1 . If
oxidise 750 cm3 of 0.6 M Mohr's salt solution is :
resistance of the 0.4 M solution of the same electrolyte is
(Given molar mass potassium dichromate = 294, Mohr's
260 , its molar conductivity is :
salt = 392)
(a) 6.25 × 10–4 S m2 mol–1
(a) 0.45 g (b) 22.05 g
(b) 625 × 10–4 S m2 mol–1
(c) 2.2 g (d) 0.49 g
(c) 62.5 S m2 mol–1 57. What is the best description of the change that occurs
(d) 6250 S m2 mol–1 when Na2O(s) is dissolved in water ?
50. Among the ligands NH3, en, CN– and CO the correct order (a) Oxide ion accepts sharing in a pair of electrons
of their increasing field strength, is : (b) Oxide ion donates a pair of electrons
(a) NH 3 en CN CO (c) Oxidation number of oxygen increases
(d) Oxidation number of sodium decreases
(b) CN NH3 CO en 58. Which of the following has maximum number of lone
pairs associated with Xe ?
(c) en CN NH3 CO (a) XeF4 (b) XeF6
(c) XeF2 (d) XeO3
(d) CO NH3 en CN
59. In the chemical reactions :
51. Consider the following reaction :
NH2
C2H5OH + H2SO4 Product
Among the following, which one cannot be formed as a NaNO
2 CuCN
A B,
product under any conditions ? HCl, 278K
(a) Ethylene
(b) Acetylene
the compounds A and B respectively are :
(c) Diethyl ether
(a) Benzene diazonium chloride & benzonitrile
(d) Ethyl-hydrogen sulphate
(b) Nitrobenzene and chlorobenzene
52. The non aromatic compound among the following is :
(c) Phenol and bromobenzene
(d) Fluorobenzene and phenol
(a) (b) 60. Which one of the following complex ions has geometrical
S isomers ?
(a) [Ni(NH3)5Br]+
(b) [Co(NH3)2(en)2]3+
(c) (d)
(c) [Cr(NH3)4(en)2]3+
– (d) [Co(en)3]3+
(en ethylenediamine)
S-252 The Pattern Target AIEEE
Section - 3 66. The lines x y a and ax – y = 1 interesect each other in
the first quadrant. Then the set of all possible values of a
is the interval :
2
61. Let f be a function defined by f x x 1 1, x 1 . (a) 0, (b) [1, )
Statement - 1 :
(c) 1, (d) 1,1
1
The set x : f x f x 1, 2 . 67. If the vector
Statement - 2: pi j k , i q j k and iˆ ˆj rkˆ p q r 1
1
f is a bijection and f x 1 x 1, x 1.
are coplanar, then the value of pqr p q r is
(a) Statement-1 is true, Statement-2 is true; Statement-2
is a correct explanation for Statement-1. (a) 2 (b) 0
(b) Statement-1 is true, Statement-2 is true; Statement-2 (c) – 1 (d) –2
is NOT a correct explanation for Statement-1. 68. The distance of the point (1, – 5, 9) from the plane x – y +
(c) Statement-1 is true, Statement-2 is false. z = 5 measured along a straight x = y = z is
(d) Statement-1 is false, Statement-2 is true.
(a) 10 3 (b) 5 3
62. If 1is the complex cube root of unity and matrix
0 (c) 3 10 (d) 3 5
H , then H70 is equal to –
0 69. Let a, b, c be three non-zero vectors which are pairwise
(a) 0 (b) –H
non-collinear. If a 3b is collinear with c and b 2c is
(c) H2 (d) H
63. Let [.] denote the greatest integer function then the value collinear with a , then a 3b 6c is :
1.5 (a) (b)
a c
of x x 2 dx is :.
0
(c) 0 (d) a c
3 70. If A (2, – 3) and B (– 2, 1) are two vertices of a triangle and
(a) 0 (b)
2 third vertex moves on the line 2 x 3 y 9, then the locus
of the centroid of the triangle is :
3 5
(c) (d) (a) x y 1 (b) 2x 3 y 1
4 4
64. The curve that passes through the point (2, 3), and has the (c) 2x 3 y 3 (d) 2x 3 y 1
property that the segment of any tangent to it lying between
the coordinate axes is bisected by the point of contact is 71. There are 10 points in a plane, out of these 6 are collinear.
given by : If N is the number of triangles formed by joining these
points. Then :
6
(a) 2 y 3x 0 (b) y (a) N 100 (b) 100 N 140
x
(c) 140 N 190 (d) N 190
2 2
x y
(c) x2 y2 13 (d) 2 72. Define f (x) as the product of two real functions
2 3
65. A scientist is weighing each of 30 fishes. Their mean weight 1
sin , if x 0
worked out is 30 gm and a standarion deviation of 2 gm. f1 x x, x R, and f 2 x x
Later, it was found that the measuring scale was misaligned 0, if x 0
and always under reported every fish weight by 2 gm. The
correct mean and standard deviation (in gm) of fishes are as follows :
respectively :
(a) 32, 2 (b) 32, 4 f1 x . f 2 x , if x 0
f ( x)
(c) 28,2 (d) 28, 4 0 if x 0
AIEEE-2011 Rescheduled Solved Paper S-253
(d) 2a f a a2 f ' a
AIEEE-2011 Rescheduled Solved Paper S-255
ANSWER KEY
1 (c) 16 (b) 31 (c) 46 (b) 61 (c) 76 (a)
2 (a) 17 (b) 32 (d) 47 (a) 62 (d) 77 (a)
3 (a) 18 (d) 33 (d) 48 (d) 63 (c) 78 (b)
4 (c) 19 (c) 34 (c) 49 (d) 64 (b) 79 (c)
5 (a) 20 (d) 35 (d) 50 (c) 65 (d) 80 (d)
6 (a) 21 (c) 36 (a) 51 (c) 66 (c) 81 (d)
7 (c) 22 (a) 37 (c) 52 (a) 67 (b) 82 (d)
8 (c) 23 (a) 38 (b) 53 (b) 68 (a) 83 (b)
9 (d) 24 (c) 39 (c) 54 (d) 69 (d) 84 (b)
10 (a) 25 (a) 40 (d) 55 (a) 70 (d) 85 (c)
11 (d) 26 (d) 41 (c) 56 (b) 71 (b) 86 (d)
12 (a) 27 (b) 42 (d) 57 (b) 72 (b) 87 (d)
13 (b) 28 (a) 43 (c) 58 (a) 73 (d) 88 (c)
14 (d) 29 (b) 44 (a) 59 (d) 74 (c) 89 (b)
15 (c) 30 (a) 45 (b) 60 (a) 75 (a) 90 (c)
SECTION-1 : PHYSICS 3. (a) Here, centripetal force will be given by the gravitational
force between the two particles.
1. (c) K.E. t or K. E. = ct
1 2 Gm2 2
mv ct m R
2 2R
2 R
p2 Gm 2 m m
ct ( p = mv) 3
2m 4R
p 2ctm Gm
4 R3
1
F= 2 cm If the velocity of the two particles with respect to the
2 t centre of gravity is v then v = R
1
F Gm Gm
t v 3
R=
4R 4R
2. (a) Path difference at p
x1 0 4. (c) N1 F1
Phase difference at P
1 = 0°
Intensity at p
I1 I 0 I 0 2 I0 cos 0 4I0
mg sin
Path difference at Q f1 mg cos
mg
x2
4
Phase difference at Q N2
2
F
2 f2
2
.
4 2
Intensity at Q.
mg sin
I2 I0 I0 2 I0 cos 2I0 mg cos
2 mg
I1 4I0 2
Thus, For the upward motion of the body
I2 2I0 1
mg sin + f1 F1
S-256 The Pattern Target AIEEE
or, F1 = mg sin + mg cos 10. (a)
For the downward motion of the body,
mg sin – f 2 F2
d
or F2 = mg sin – mg cos l
F1 sin cos
=
F2 sin cos
tan 2 3 l0
3
tan 2
5. (a)
R R R R At equilibrium
R = 100 5 Fb = mg
4R = 400 20
Thus, tolerance of combination is also 5%. A 0g dA g ....(i)
6. (a) Lorentz force acting on the particle
ˆi ˆj kˆ
F = q E + v× B = q 3i j 2k 3 4 1 d
1 1 3
=q 3i j 2kˆ i 12 1 j 9 1 k 3 4
x
= q 3iˆ j 2k 13i 10 j k = q 10i 11j k 0
Fy 11qjˆ
Thus, the y component of the force. Restoring force,
7. (c) Potential gradient F = mg – Fb'
V IR I I F mg A 0 x g
x
A A dA a dA g A 0g gAx
7
0.2 4 10 0.8 g
x 7
0.1 V/m a x
8 10 8 d
Therefore, wooden cube performs S.H.M.
8. (c)
g d
T 2
d g
v
v cos 11. (d)
P, T
H Vacuum
A 1 Water h1
B (A+B)
(A+B)
1
14. (d) Potential at point A, Apparent shift due to water = h1 1
1
2kq 2kq
VA
a a 5 1
Apparent shift due to kerosene = h2 1
2
kq
(potential due to each q and potential due
a 1 1
Thus, total apparent shift : h1 1 h2 1
kq 1 2
to each – q = )
a 5 17. (b) For linear expansion
2a
q q . T
A
F/A F/A
. T and Y
/ . T
2a
B
–q –q So, F = AY t ( T t)
Potential at point. B,
F
Thermal stress Y t.
VB 0 A
( Point B is equidistant from all the four charges) 18. (d) y = A sin ( t – kx) + A sin ( t + kx)
Using work energy theorem, y = 2A sin t cos kx
WAB Q VA VB For standing wave nodes
electric
cos kx = 0
2 kqQ 1 1 2Qq 1 2
= 1 = 1 .x (2n 1)
a 5 4 0 a 5 2
15. (c) Time constant for parallel combination = 2RC 2n 1
x ,n 0,1, 2,3,...........
RC 4
Time constant for series combination =
2 q Magnetic dipole moment
In first case : 19. (c)
2m Angular momentum
t1
2 RC
V0
V V0 e ...(1)
2
In second case :
t2
( RC / 2) V0
V V0 e ....(2)
2
From (1) and (2)
t1 t2
Magnetic dipole moment (M)
2 RC RC / 2
q mR 2 1
t1 10 M . . . R4 .
t2 2.5 sec. 2m 2 4
4 4
S-258 The Pattern Target AIEEE
20. (d) V V0 . T , but 3 . Speed of electro-magnetic waves will not change due
to modulation. So there will be time lag between
V V0 (3 ) T transmission and reception of the information signal.
4 3 2 3
10 3 23 10 6 100 26. (d) Here, A , B
3 5 5
V = 28. 9 cc
k kA A kB B
21. (c) Sum of volumes of 2 smaller drops
= Volume of the bigger drop 2
k kA
4 4 3 5
2. r 3 R R 21/ 3 r
3 3
5k 5k
T .4 R 2 kA kB .
Surface energy 2 3
= T 4 22 / 3 r 2 = T .2
8/3
r2. 27. (b) Statement-1: Energy of -particle from 0 to maximum
( ')
v' ' v 1 1 1
and ( 1)
( ) f R1 R2
( ') v 1 1
v' fR fB .
( ) ' fB fR
4 29. (b) When viewed through a polaroid which is rotated then
(7.8 1.2) 10 8.5 10
= the light from a clear blue portion of the sky shows a
(7.8 1) 13.2 rise and fall of intensity.
v' 6.25 10 4 cm/s
23. (a) Incident sunlight
W E (unpolarised)
ind Bv
= 0.3 10 4
5 20 = 3 × 10–3 V = 3 mV.. Sun
24. (c) pi 0 Scattered light
(polarised)
pf p1 p2
pi pf
0 p1 p2
to observer
p1 p2
30. (a) Since, I A2 2
h
1
p1 I1 (2a)2 2
h I2 a 2 (2 ) 2
2
p2
I1 I 2
1 2
Intensity depends on frequency also.
. At x = , f ''( x ) 0
1 2
25. (a) Low frequencies cannot be transmitted to long Hence, local mixima at x =
distances. Therefore, they are super imposed on a high At x 2 , f ''( x ) 0
frequency carrier signal by a process known as Hence local minima at x = 2
modulation.
AIEEE-2011 Rescheduled Solved Paper S-259
2RT
(ii) Average velocity v
M
(iii) Root mean square velocity in all three cases
3RT
v
M
S-260 The Pattern Target AIEEE
40. (c) From the products formed it is clear that the compound For same frequency,
has 5 carbon atoms with a double bond and methyl
group on 2nd carbon atom. 1 1 1 1
z2 2 2
CH3 2 4 n12 n22
| O3 / Zn, H2 O
CH3 C CH CH 2 CH3 Since, z = 2
(2–Methyl–2–pentene) 1 1 1 1
(A) n12 n22 12
22
CH3 O n1 = 1 & n2 = 2
| 45. (b) For isotonic solutions
CH3 C O CH3 CH 2 C
1 2
H C1 C2
Acetone Propionadhyde 5 / 342 1/ M
41. (d) 0.1 0.1
5 1
342 M
342
M 68.4 gm/mol
– 111 – 54 = H' 5
H' = – 165 kJ 46. (c) Negative r G value indicates that + 2 oxidation state
42. (d) pH = 5 means is more stable for Pb2+. Also it is supported by inert
[H+] = 10–5 pair effect that + 2 oxidation state is more stable for Pb
HA H A 1 and + 4 oxidation state is more stable for Sn.
i.e. Sn++ < Pb++, Sn4+ > Pb4+
t=0 c 0 0 47. (b) Cr(OH)3 (s) Cr3 (aq.) 3OH (aq.)
teq c (1 – ) c c
[H ][A ] (c )2 [H ]2 27 S 4 K sp
Ka
[HA] c (1– ) c [H ] 1/ 4 1/ 4
K sp 1.6 10 30
But, [H+] << C S
27 27
Ka = (10–5)2 = 10–10
43. (b) As we move down in a group electron gain enthalpy 48. (a) 4LiNO3 2Li 2 O 4NO2 O2
becomes less negative because the size of the atom
increases and the distance of added electron from the 1
49. (a) k
nucleus increases. Negative electron gain enthalpy of R A
F is less than Cl. This is due to the fact that when an
1
electron is added to F, the added electron goes to the 1.3
smaller n = 2 energy level and experiences significant 50 A
repulsion from the other electrons present in this level. 1
In Cl, the electron goes to the larger n = 3 energy level 65m
A
and consequently occupies a larger region of space
leading to much less electron–electron repulsion. So k 1000
the correct order is molarity
Cl > F > Br > I. [molarity is in moles/litre but 1000 is used to convert
44. (a) For He+ liter into cm3]
1 1 1 1
v RH Z 2 65 m 1 1000 cm 3
2 2 260
2 4
For H 0.4 moles
650 m 1 1
1 1 1 m3
v RH 260 4 mol 1000
n12 n22
= 6.25 × 10–4 S m2 mol–1
AIEEE-2011 Rescheduled Solved Paper S-261
50. (a) Ligands can be arranged in a series in the orders of 56. (b) Potassium dichromateoxidise Fe++ to Fe+++ as
increasing field strength as given below : K 2 Cr2 O 7 7H 2SO 4 6FeSO 4
Weak field ligands : K 2SO4 Cr2 (SO 4 )3 3Fe 2 (SO4 )3 7H 2O
I Br S SCN Cl N3 , F i.e. change in o.s. of Cr per atom = 6
< Urea, OH– < oxalate 294
Therefore the eq.wt. of K 2 Cr2 O7
Strong field ligands 6
Now, since
O H 2O NCS EDTA Py, NH3 < Normality = Gram Eq. / volume of solution in litre
en = SO 3 – < bi py, Phen NO2 CH3 W
or N × V =
E
C6 H 5 CN CO W
Such a series is termed as spectrochemical series. It is 0.6 0.75 1 6
294
an experimentally determined series based on the W = 22.05 g
absorption of light by complexes with different ligands.
57. (b) O 2 (base) H 2O(acid)
51. (b) C2 H5 - OH + H2SO4 433K CH2 = CH2 OH (C.B.) OH (C.A.)
ethylene
413 K O2– acts as Lewis base.
CH3 - CH2 - O - CH 2 - CH3
di ethyl ether
383 K
CH3 CH 2 HSO4 + H2 O
ethyl hydrogen sulphate
361 2
r 127.6 128pm
4 Hence XeF2 has maximum no. of lone pairs of electrons.
S-262 The Pattern Target AIEEE
NH 2 N +2 Cl- 2
0
62. (d) H2
2
59. (a) 0
NaNO2
HCl, 278K k 1
k
0 0
Diazotization Benzene diazonium
If H k then Hk + 1 k 1
chloride 0
(A) So by principal of mathematical induction,
C=N 70 69
0
H 70
70 69
CuCN 0
0
H
Benzonitrile
(B)
1.5
Sandmayer reaction
63. (c) x x 2 dx
0
3+
en
1 2 1.5
NH3
x[ x 2 ]dx x x 2 dx x x 2 dx
60. (b) Co3+ 0 1 2
1 2 1.5
en NH3
x.0 dx xdx 2xdx
0 1 2
cis-
2
x2 1.5
NH 3 =0 x2
3+ 2 1
2
1 1 1 1 3
2 1 2.25 2 0.25
en en 2 2 2 4 4
Co3+
dy
64. (b) Y – y = X x
dx
Y
NH 3
trans- B (0, y-xdy/dx)
SECTION-3 : MATHEMATICS
2
61. (a) f x x 1 1, x 1
Since f is a bijective function
f :[1, ) [1, ) (x, y)
2 2
y x 1 1 x 1 y 1
1
x 1 y 1 f y 1 y 1
1
f x 1 x 1 x 1 X
X´ O
Hence statement-2 is correct A (x – y)/( dy/dx), 0)
Y´
1
Now f x f x
y
2
X-intercept x
f x x x 1 1 x dy / dx
x2 3 x 2 0 x 1, 2 xdy
Y-intercept = y –
Hence statement-1 is correct dx
AIEEE-2011 Rescheduled Solved Paper S-263
–1
1 a a a2 1 a
y a 0 0 G
1 a 1 a (h, k)
a2 1 a2 1
0 0
a 1 a 1 B(–2, 1) C( )
S-264 The Pattern Target AIEEE
3h 7
n 1 n7 1 n7 n is divisible by 7
2 3k
7
n 1 n 7 1 is divisible by 7
3k 2
Hence both Statements 1 and 2 are correct and
Third vertex , lies on the line Statement 2 is the correct explanation of Statement -1.
74. (b) Circle whose diametric end points are (1,0) and (0,1)
2x 3 y 9 will be of smallest radius. Equation of this smallest circle
is (x – 1) (x – 0) + (y – 0) (y – 1) = 0
2 3 9
x2 + y2 – x – y = 0
2 3h 3 3k 2 9 75. (b) ae = 2
e=2
2h 3k 1 a =1
2x 3 y 1
b2 a 2 e2 1
10 6
71. (a) Number of required triangles = C3 C3
b2 14 1
10 9 8 6 5 4
120 20 100 b2 3
6 6
x2 y2
x sin 1/ x , x 0 Equation of hyperbola, 1
72. (c) f x at x 0 a2 b2
0 ,x 0
x2 y2
1
1 1 3
LHL = lim– h sin
h 0 h 3 x2 y2 3
= 0 × a finite quantity betwen – 1 and 1 76. (a) x ky z 0
=0
kx 3 y kz 0
1
lim
RHL = h 0 h sin = 0 3x y z 0
h
The given system of equations will have non trivial
Also, f (0) = 0 solution, if
Thus LHL = RHL = f 0 1 k 1
f x is continuous on R. k 3 k 0
f 2 x is not continuous at x = 0 3 1 1
From (i) and (ii), roots of the correct equation 82. (d) sin 4 + 2sin 4 cos 3 = 0
2
x 0 are 6 and 1.
7x 6 sin 4 1 2 cos 3 0
78. (b) Let A.P. be a, a d , a 2d ,.........
1
a2 a4 ........... a200 = sin 4 0 or cos 3
2
100 4 n ;n I
2 a d 100 1 d ....(i)
2
2
and a1 a3 a5 ......... a199 or 3 2n ,n I
3
100 3
2a + (100 – 1) d = ....(ii) , ,
2 4 2 4
On solving (i) and (ii), we get 2 8 4
or , , [ , (0, )]
d 9 9 9
100
dx x 1 83. (b)
79. (c) dy y2 y3
1 1
dy –
y2 y
I.F. = e e
1 1
1 –
y y
So x.e 3
e dy
y
1
Let t
y
1
dy dt
y2
4
1 1 x2
1 Area 2 x dx
I tet dt et tet e y
e y
c 0
4
y
1 1 1 4
1 x3/ 2 x3 4 64
xe y
e y
e y
c 2 8
y 3/ 2 12 3 12
0
1 32 16
x 1 c.e1/ 16 sq. units
y 3 3
Since y 1 1
tan x
, x 0
1 84. (b) f ( x) x
c
e 1, x 0
1 1 1/ y In right neighbourhood of ‘0’
x 1 .e
y e tan x > x
f x
2
9 tan x
1 Y
80. (d) lim =0 x
x 5 x 5
y = tan x
lim [(f (x) )2 – 9] = 0 lim f (x) = 3
x 5 x 5
81. (d) Statement-1 : Determinant of skew symmetric matrix of y=x
X´ X
odd order is zero. O
T
Statement-2 : det A = det (A).
det (– A) = – (– 1)n det (A).
where A is a n n order matrix.
Y´
S-266 The Pattern Target AIEEE
In left neighbourhood of ‘0’
at x = 0, f x 1
tan x < x
x 0 is the point of minima
tan x
1 as tan x 0 So Statement 1 is true.
x Statement 2 obvious.
85. (c) A B A B A B A (A B) A (A B) A B A (A B) [A (A B) B] [B [A (A B)]
T F T F T T F F T T
F T T F F F T F T F
T T T T T T T T T T
F F F F F F T F T T
f x g x As A, B R for matrix P
ax2 bx c a1x 2 b1 x c1 A P 1 BP
1
2
PAP B
a a1 x b b1 x c c1 0. 1
B PAP
It has only one solution, x = – 1 1
b b1 a a1 c c1 ...(1) B P –1 A (P–1)
R E E
(c) (d)
Rr Rr
(a) 2 SY T (b) SY T 4. An electromagnetic wave in vacuum has the electric and
(c) SY T (d) 2SY T
magnetic field E and B , which are always perpendicular
2. The figure shows an experimental plot discharging of a
capacitor in an RC circuit. The time constant of this to each other. The direction of polarization is given by X
circuit lies between : and that of wave propagation by k . Then
25
(a) X || B and k || B E
Potential difference
(b) X || E and k || E B
20
V in volts
(c) X || B and k || E B
15
(d) X || E and k || B E
10 5. If a simple pendulum has significant amplitude (up to a
factor of 1/e of original) only in the period between t = 0s
5
to t = s, then may be called the average life of the
0
50 100 150 200 250 300 pendulum. When the spherical bob of the pendulum suffers
a retardation (due to viscous drag) proportional to its
Time in seconds velocity with b as the constant of proportionality, the
(a) 150 sec and 200 sec average life time of the pendulum is (assuming damping is
(b) 0 sec and 50 sec small) in seconds :
2012- 2 The Pattern Target AIEEE
0.693
(a) (b) b
0)
0)
b
loge ( –
loge ( –
1 2
(c) (d)
b b
6. Hydrogen atom is excited from ground state to another (a) (b)
state with principal quantum number equal to 4. Then the 0
0 t
number of spectral lines in the emission spectra will be : t
(a) 2 (b) 3
(c) 5 (d) 6
0)
7. A coil is suspended in a uniform magnetic field, with the
0)
plane of the coil parallel to the magnetic lines of force.
loge ( –
loge ( –
When a current is passed through the coil it starts
oscillating; It is very difficult to stop. But if an aluminium
(c) (d)
plate is placed near to the coil, it stops. This is due to :
(a) developement of air current when the plate is placed 0
(b) induction of electrical charge on the plate 0 t
t
(c) shielding of magnetic lines of force as aluminium is a
paramagnetic material. 12. A particle of mass m is at rest at the origin at time
(d) Electromagnetic induction in the aluminium plate t = 0. It is subjected to a force F(t) = F0e–bt in the x direction.
giving rise to electromagnetic damping. Its speed v(t) is depicted by which of the following curves?
8. The mass of a spaceship is 1000 kg. It is to be launched F0 F0
from the earth's surface out into free space. The value of g
mb mb
and R (radius of earth) are 10 m/s2 and 6400 km respectively.
The required energy for this work will be : (a) v(t) (b) v(t)
(a) 6.4 × 1011Joules (b) 6.4 × 108 Joules
9 t t
(c) 6.4 × 10 Joules (d) 6.4 × 1010 Joules
9. Helium gas goes through a cycle ABCDA (consisting of F0 F0
two isochoric and isobaric lines) as shown in figure mb mb
Efficiency of this cycle is nearly : (Assume the gas to be (c) (d) v(t)
v(t)
close to ideal gas)
t t
13. Two electric bulbs marked 25W – 220 V an d
B C 100W – 220V are connected in series to a 440 V supply.
2P0
Which of the bulbs will fuse?
(a) Both (b) 100 W
P0 D (c) 25 W (d) Neither
A
14. Resistance of a given wire is obtained by measuring the
current flowing in it and the voltage difference applied
2V0 across it. If the percentage errors in the measurement of
V0
the current and the voltage difference are 3% each, then
(a) 15.4 % (b) 9.1 % error in the value of resistance of the wire is :
(c) 10.5% (d) 12.5 % (a) 6% (b) zero
10. In Young's double slit experiment, one of the slit is wider (c) 1% (d) 3%
than other, so that amplitude of the light from one slit is 15. A boy can throw a stone up to a maximum height of 10 m.
double of that from other slit. If Im be the maximum The maximum horizontal distance that the boy can throw
intensity, the resultant intensity I when they interfere at the same stone up to will be :
phase difference is given by :
(a) 20 2 m (b) 10 m
Im Im
(a) (4 5 cos ) (b) 1 2 cos 2
9 3 2 (c) 10 2 m (d) 20 m
523K 1700K
Ti(s) 2I 2 (g) TiI 4 (g) Ti(s) 2I 2 (g)
2n 2 h 2 (m1 m2 )n2 h2
(c) (d) (a) Zone refining (b) Cupellation
(m1 m2 )r 2 2m1m2 r 2
(c) Polling (d) Van Arkel
28. A spectrometer gives the following reading when used to 33. Lithium forms body centred cubic structure. The length of
measure the angle of a prism. the side of its unit cell is 351 pm. Atomic radius of the lithium
Main scale reading : 58.5 degree
will be :
Vernier scale reading : 09 divisions
Given that 1 division on main scale corresponds to 0.5 (a) 75 pm (b) 300 pm
degree. Total divisions on the Vernier scale is 30 and match (c) 240 pm (d) 152 pm
with 29 divisions of the main scale. The angle of the prism 34. The molecule having smallest bond angle is :
from the above data : (a) NCl3 (b) AsCl3
(a) 58.59 degree (b) 58.77 degree (c) SbCl3 (d) PCl3
(c) 58.65 degree (d) 59 degree 35. Which of the following compounds can be detected by
29. This questions has statement-1 and statement-2. Of the Molisch's Test ?
four choices given after the statements, choose the one (a) Nitro compounds (b) Sugars
that best describe the two statements. (c) Amines (d) Primary alcohols
An insulating solid sphere of radius R has a uniformly
36. The incorrect expression among the following is :
positive charge density . As a result of this uniform charge
distribution there is a finite value of electric potential at Gsystem
the centre of the sphere, at the surface of the sphere and (a) T
Stotal
also at a point out side the sphere. The electric potential at
infinite is zero. Vf
Statement -1 When a charge q is take from the centre of (b) In isothermal process, wreversible = –nRT n
the surface of the sphere its potential energy changes by Vi
q Hº T Sº
(c) ln K (d) K = e– Gº/RT
3 0 . RT
Statement -2 The electric field at a distance r (r <R) from 37. The density of a solution prepared by dissolving 120 g of
urea (mol. mass = 60 u) in 1000 g of water is 1.15 g/mL. The
r
the centre of the sphere is . molarity of this solution is :
3 0 (a) 0.50 M (b) 1.78 M
(a) Statement 1 is true, Statement 2 is true; Statement 2 is (c) 1.02 M (d) 2.05 M
not the correct explanation of statement 1. 38. The species which can best serve as an initiator for the
(b) Statement 1 is true Statement 2 is false. cationic polymerization is :
(c) Statement 1 is false Statement 2 is true.
(a) LiAlH4 (b) HNO3
(d) Statement 1 is true, Statement 2 is true, Statement 2 is
the correct explanation of Statement 1 (c) AlCl3 (d) BaLi
AIEEE-2012 Solved Paper 2012- 5
39. Which of the following on thermal decomposition yields a 48. Very pure hydrogen (99.9) can be made by which of the
basic as well as acidic oxide ? following processes ?
(a) NaNO3 (b) KClO3 (a) Reaction of methane with steam
(c) CaCO3 (d) NH4NO3 (b) Mixing natural hydrocarbons of high molecular weight
40. The standard reduction potentials for Zn 2+ /Zn, (c) Electrolysis of water
Ni2+/Ni and Fe2+/Fe are –0.76,–0.23 and –0.44 V respectively. (d) Reaction of salts like hydrides with water
49. The electrons identified by quantum numbers n and :
The reaction X +Y 2 X 2+ + Y will be spontaneous
(A) n = 4, = 1 (B) n = 4, = 0
when :
(C) n = 3, = 2 (D) n = 3, = 1
(a) X = Ni, Y = Fe (b) X = Ni, Y = Zn
can be placed in order of increasing energy as :
(c) X= Fe, Y = Zn (d) X= Zn, Y = Ni
(a) (C) < (D) < (B) < (A) (b) (D) < (B) < (C) < (A)
41. According to Freundlich adsorption isotherm which of the
(c) (B) < (D) < (A) < (C) (d) (A) < (C) < (B) < (D)
following is correct?
50. For a first order reaction (A) products the concentration
x 0 of A changes from 0.1 M to 0.025 M in 40 minutes.
(a) p
m The rate of reaction when the concentration of A is 0.01 M
is :
x (a) 1.73 × 10–5 M/min (b) 3.47 × 10–4 M/min
(b) p1
m –5
(c) 3.47 × 10 M/min (d) 1.73 × 10–4 M/min
51. Iron exhibits +2 and + 3 oxidation states. Which of the
x
(c) p1/n following statements about iron is incorrect ?
m (a) Ferrous oxide is more basic in nature than the ferric
(d) All the above are correct for different ranges of pressure oxide.
42. The equilibrium constant (K c ) for the reaction (b) Ferrous compounds are relatively more ionic than the
N2(g) + O2(g) 2NO(g) at temperature T is 4 × 10–4. The corresponding ferric compounds.
value of Kc for the reaction (c) Ferrous compounds are less volatile than the
1 1 corresponding ferric compounds.
NO( g) N2 ( g ) O 2 ( g ) at the same temperature is: (d) Ferrous compounds are more easily hydrolysed than
2 2
the corresponding ferric compounds.
(a) 0.02 (b) 2.5 × 102
–4 52. The pH of a 0.1 molar solution of the acid HQ is 3. The value
(c) 4 × 10 (d) 50.0 of the ionization constant, Ka of the acid is :
43. The compressibility factor for a real gas at high pressure is : (a) 3 × 10–1 (b) 1 × 10–3
RT (c) 1 × 10 –5 (d) 1 × 10–7
(a) 1 + (b) 1
pb 53. Which branched chain isomer of the hydrocarbon with
molecular mass 72u gives only one isomer of mono
pb pb substituted alkyl halide ?
(c) 1 + (d) 1 – (a) Tertiary butyl chloride (b) Neopentane
RT RT
44. Which one of the following statements is correct? (c) Isohexane (d) Neohexane
(a) All amino acids except lysine are optically active 54. Kf for water is 1.86 K kg mol–1. If your automobile radiator
(b) All amino acids are optically active holds 1.0 kg of water, how many grams of ethylene glycol
(c) All amino acids except glycine are optically active (C2H6O2) must you add to get the freezing point of the
(d) All amino acids except glutamic acids are optically solution lowered to –2.8ºC ?
active (a) 72 g (b) 93 g
(c) 39 g (d) 27 g
45. Aspirin is known as :
55. What is DDT among the following ?
(a) Acetyl salicylic acid (b) Phenyl salicylate
(a) Greenhouse gas
(c) Acetyl salicylate (d) Methyl salicylic acid
(b) A fertilizer
46. Ortho-Nitrophenol is less soluble in water than p- and m-
(c) Biodegradable pollutant
Nitrophenols because :
(d) Non-biodegradable pollutant
(a) o-Nitrophenol is more volatile steam than those of m-
56. The increasing order of the ionic radii of the given
and p-isomers.
isoelectronic species is :
(b) o-Nitrophenol shows intramolecular H-bonding
(a) Cl–, Ca2+ , K+, S2– (b) S2–, Cl–, Ca2+ , K+
(c) o-Nitrophenol shows intermolecular H-bonding 2+ + – 2–
(c) Ca , K , Cl , S (d) K+, S2–, Ca2+, Cl–
(d) Melting point of o-Nitrophenol is lower than those of 57. 2-Hexyne gives trans-2-Hexene on treatment with :
m- and p-isomers. (a) Pt/H2 (b) Li / NH3
47. How many chiral compounds are possible on (c) Pd/BaSO4 (d) Li AlH4
monochlorination of 2- methyl butane ? 58. Iodoform can be prepared from all except :
(a) 8 (b) 2 (a) Ethyl methyl ketone (b) Isopropyl alcohol
(c) 4 (d) 6 (c) 3-Methyl 2-butanone (d) Isobutyl alcohol
2012- 6 The Pattern Target AIEEE
59. In which of the following pairs the two species are not 65. The negation of the statement
isostructural ? "If I become a teacher, then I will open a school", is :
(a) CO23 and NO3 (b) PCl 4 and SiCl4
(a) I will become a teacher and I will not open a school.
(c) PF5 and BrF5 (d) AlF63
and SF6
(b) Either I will not become a teacher or I will not open a
60. In the given transformation, which of the following is the
school.
most appropriate reagent ?
(c) Neither I will become a teacher nor I will open a school.
CH CH COCH3 (d) I will not become a teacher or I will open a school.
5 tan x
Reagent 66. If the dx x a ln sin x 2 cos x k , then a is
HO tan x 2
CH CH CH2 CH3 equal to :
(a) –1 (b) –2 (c) 1 (d) 2
67. Statement-1 : An equation of a common tangent to the
HO
parabola y 2 16 3 x and the ellipse 2 x 2 y2 4 is
(a) NH2 NH2 , OH (b) Zn – Hg/ HCl
(c) Na, Liq NH3 (d) NaBH4 y 2x 2 3
Section - 3 4 3
Statement-2 : If the line y mx , m 0 is a common
m
tangent to the parabola y 2 16 3 x and the ellipse 2x2 +
y2 = 4, then m satisfies m4 + 2m2 = 24
61. The equation esinx – e–sinx – 4 = 0 has :
(a) Statement-1 is false, Statement-2 is true.
(a) infinite number of real roots
(b) Statement-1 is true, statement-2 is true; statement-2 is
(b) no real roots
a correct explanation for Statement-1.
(c) exactly one real root (d) exactly four real roots
(c) Statement-1 is true, statement-2 is true; statement-2 is
62. Let a and b be two unit vectors. If the vector s not a correct explanation for Statement-1.
(d) Statement-1 is true, statement-2 is false.
c aˆ 2 bˆ and d 5 aˆ 4bˆ are perpendicular to each other,,
1 0 0
then the angle between â and b̂ is :
68. Let A 2 1 0 . If u1 and u2 are column matrices such
(a) (b) 3 2 1
6 2
1 0
(c) (d)
3 4 that Au1 0 and Au2 1 , then u1 + u2 is equal to :
63. A spherical balloon is filled with 4500 cubic meters of helium 0 0
gas. If a leak in the balloon causes the gas to escape at the
rate of 72 cubic meters per minute, then the rate (in meters
1 1 1 1
per minute) at which the radius of the balloon decreases 49
minutes after the leakage began is : 1 1 1 1
(a) (b) (c) (d)
9 7 0 1 0 1
(a) (b)
7 9 2n 2n
69. If n is a positive integer , then 3 1 3 1 is :
2 9
(c) (d) (a) an irrational number
9 2
(b) an odd positive integer
64. Statement-1: The sum of the series 1 + (1 + 2 + 4) + (4 + 6
(c) an even positive integer
+ 9) + (9 + 12 + 16) + .... + (361 + 380 + 400) is 8000.
(d) a rational number other than positive integers
n
3
70. If 100 times the 100th term of an AP with non zero common
Statement-2: k3 k 1 n3 , for any natural difference equals the 50 times its 50th term, then the 150th
k 1 term of this AP is :
number n. (a) – 150 (b) 150 times its 50th term
(a) Statement-1 is false, Statement-2 is true. (c) 150 (d) Zero
(b) Statement-1 is true, statement-2 is true; statement-2 is 71. In a PQR, If 3 sin P + 4 cos Q = 6 and 4 sin Q + 3
a correct explanation for Statement-1. cos P = 1, then the angle R is equal to :
(c) Statement-1 is true, statement-2 is true; statement-2 is 5 3
not a correct explanation for Statement-1. (a) (b) (c) (d)
6 6 4 4
(d) Statement-1 is true, statement-2 is false.
AIEEE-2012 Solved Paper 2012- 7
72. A equation of a plane parallel to the plane 79. If f : R R is a function defined by f (x) = [x]
x – 2y + 2z –5 = 0 and at a unit distance from the origin is : 2x 1
cos , where [x] denotes the greatest integer
2
(a) x – 2y + 2z – 3 = 0 (b) x – 2y + 2z + 1 = 0
(c) x – 2y + 2z – 1 = 0 (d) x – 2y + 2z + 5 = 0 function, then f is .
(a) continuous for every real x.
73. If the line 2x + y = k passes through the point which divides
(b) discontinuous only at x = 0
the line segment joining the points (1,1) and (2,4) in the ratio
3 :2, then k equals : (c) discontinuous only at non-zero integral values of x.
ANSWER KEY
1 (d) 16 (c) 31 (b) 46 (b) 61 (b) 76 (b)
2 (d) 17 (d) 32 (d) 47 (c) 62 (c) 77 (c)
3 (c) 18 (a) 33 (d) 48 (d) 63 (c) 78 (d)
4 (b) 19 (a) 34 (c) 49 (b) 64 (b) 79 (a)
5 (d) 20 (a) 35 (b) 50 (b) 65 (a) 80 (c)
6 (d) 21 (a) 36 (c) 51 (d) 66 (d) 81 (b)
7 (d) 22 (c) 37 (d) 52 (c) 67 (b) 82 (a)
8 (d) 23 (b) 38 (c) 53 (b) 68 (d) 83 (c)
9 (a) 24 (c) 39 (c) 54 (b) 69 (a) 84 (b, c)
10 (d) 25 (a) 40 (d) 55 (d) 70 (d) 85 (a)
11 (a) 26 (d) 41 (d) 56 (c) 71 (b) 86 (b)
12 (c) 27 (d) 42 (d) 57 (b) 72 (a) 87 (d)
13 (c) 28 (c) 43 (c) 58 (d) 73 (c) 88 (c)
14 (a) 29 (c) 44 (c) 59 (c) 74 (d) 89 (c)
15 (d) 30 (b) 45 (a) 60 (a) 75 (a) 90 (b)
R R 1 200 200
. T … (ii) or
R R T 10 log e 10 log e 2
log e
From equation (i) and (ii) 2
F 200 200
Y F Y.S. T
S. T = 124.300
2.302 0.693 1.609
The ring is pressing the wheel from both sides, Which lies between 100 and 150
Thus
Fnet = 2F = 2YS T 3. (c) E inside the charged sphere RP
2. (d) The discharging of a capacitor is given as Ein 0 …(i) O P Q
q q 0 exp t / RC
E on the surface of the charge
RC = time constant = sphere
q q 0 e t/ 1 q 1
Es 2
i.e., Es nˆ …(ii)
If e is the capacitance of the capacitor 4 0 R R2
q CV and q CV0 E on any point away from the uniformly
charged sphere is given
Thus, CV = CV0 e t /
1 q
V V0 e t/ …(i) E nˆ
4 0 r2
2012- 10 The Pattern Target AIEEE
1
6. (d) For ground state, the principal quantum no
E nˆ … (iii) (n) = 1. There is a 3rd excited state for principal
r2 quantum number.
R is the radius of the sphere, which is constant,
thus E is maximum and constant at the surface of 4 3rd excited state
Energy states
the surface of the uniformally charged sphere.
4. (b) The E.M. wave are transverse in nature i.e., 3 2nd excited state
k E
= H …(i)
2 1st excited state
2 2 k b P0V0 2
0 where 0 ,
m 2 13 13
P0V0
1 2 2
The average life =
b 2 200
Efficiency in % 100 15.4%
13 13
AIEEE-2012 Solved Paper 2012- 11
10. (d) Let a1 = a, I1 = a12 = a2 The current flowing through the circuit
a2 = 2a, I2 = a22 = 4a2 B1 B2
I2 = 4I1
Ir = a12 + a22 + 2a1a2 cos
= I1 I 2 2 I1 I2 cos
R1 r2
Ir = I 1 4 I 1 2 4 I12 cos
Ir 5I1 4 I1 cos … (1) 440V
Now, Imax 2 2 2
( a1 a2 ) (a 2 a) 9a 440
I=
Imax Reff
Imax 9I1 I1
9
Reff = R1 R2
Substituting in equation (1)
5Imax 4I max V12 (220)2 V22 (220)2
Ir cos R1 ; R2
9 9 P1 25 P 100
Imax
Ir 5 4 cos 440 440
9 I=
(220)2 (220)2 (220) 21 1
Imax 25 100 25 100
Ir 5 8 cos2 4
9 2
40
I= Amp
Imax 220
Ir 1 8 cos 2
9 2
25 40 100
11. (a) Newton's law of cooling I1 A I A I2 A
220 220 200
d d
k( 0) kdt Thus the bulb marked 25W-220 will fuse.
dt ( 0)
V V V
Integrating 14. (a) R R R
I I I
log( 0) kt c
Which represents an equation of straight line.
Thus the option (a) is correct. R V 1 V /V
bt R 1 = I I
12. (c) Given that F(t) = F0 e R 1
I
dv bt
m = F0 e
dt R V I
= = (3 + 3)% = 6%
dv F0 R V I
bt
= e
dt m
u2 sin 2 u2 sin 2
v t
15. (d) R ,H
F0 bt
g 2g
dv = e dt
0 m Hmax at 2 =90
0
t u2
F e bt F bt 0
Hmax =
v= 0 = 0 e e 2g
m b mb
0
u2
F 10 u2 10 g 2
v = 0 1 e bt 2g
mb
13. (c) The current upto which bulb of marked 25W -220V,
u2 sin 2 u2
W1 25 R= Rmax
Amp (9) g
will not fuse I1 = V 220
1
10 g 2
W2 100 Rmax = 20 meter
Similarly, I2 = V Amp g
2 220
2012- 12 The Pattern Target AIEEE
16. (c) 22. (c) The efficiency of the engine is given as
17. (d) The surface tension of the liquid film is given as
T2
F 1 100
T T1
, where F is the force, and l = length of the
2l
For first case
slider. T1 = 500 K; = 40
F 2lT T2
At equilibrium, F = W 40 = 1 100
500
2Tl = mg
40 T2
mg 1.5 10 2 1.5 =1
T 0.025 N/m = 0.025Nm 100 500
2 30 10 2
2l 60
T2 60
18. (a) The magnetic field due a disc is given as = T2 = 300 K
500 100
0Q 1 For second case :
B i.e., B
2 R R
60 300
19. (a) = 1 T
A
100 2
Y2 = A.AB 300 40
Y = A.AB B.AB T2 = 100
Y1 = AB 100 300
T2 = T2 = 750 K
40
B
Y3 = B.AB 1 2
23. (b) w= kx
By expanding this Boolen expression 2
Y A.B B.A 1 1 2
w1 = k1x 2 ; w2 = k2 x
Thus the truth table for this expression should be 2 2
(1). Since w1 > w2 Thus (k1 > k2)
20. (a) Let d is the maximum distance, upto it condetict v12 v22
the objects C 24. (c) a1 = … (1) a2 = … (2)
r1 r2
From AOC
2 r1 2 r2
OC 2 AC 2 AO2 h t1 and t2
d v1 v2
(h R)2 d2 R2 B
Putting v1 and v2 in equations (1) and (2) in
A terms of t.
d2 ( h R )2 R2 R
2 2
R 4 r1 4 r2
2 2 a1 ; a2
d (h R) R O t t
d h2 2hR
a1 r1
Now, a r2
2
d 5002 2 6.4 10 6 = 80 km 25. (a) The fundamental frequency of open tube
21. (a) 1 1 0 v
0n 1H 1e Q
0
2l0 … (i)
The mass defect during the process
where l is the length of the tube
m mn mH me
v = speed of sound
= 1.6725 × 10–27 – (1.6725 × 10–27+ 9 × 10–31kg) That of closed pipe
= – 9 × 10–31 kg
v
The energy released during the process c … (ii)
E = mc2 4lc
E = 9 × 10–31× 9 × 1016 = 81 × 10–15 Joules l0
According to the problem lc
81 10 15 2
E= 0.511MeV
1.6 10 –19 v v
Thus c c … (iii)
l0 /2 2l
AIEEE-2012 Solved Paper 2012- 13
Wolf-kishner
54. (b) Tf = i × Kf × m Reduction
HO
Given Tf = 2.8, Kf = 1.86 K kg mol–1 i = 1
(ethylene glygol is a non- electrolyte) CH CH CH2 CH3
wt. of solvent = 1 kg
Let of wt of solute = x HO
Mol. wt of ethylene glycol = 62 –OH group and alkene are acid-sensitive groups
so clemmensen reduction can not be used.Acid
x 2.8 62 sensitive substrate should be reacted in the Wolf-
2.8 = 1 × 1.86 × or x = = 93 gm
62 1 1.86 Kishner reduction which utilise strongly basic
55. (d) DDT is a non-biodegradable pollutant. conditions.
56. (c) Among isoelectronic species ionic radii increases
as the charge increases. SECTION-3 : MATHEMATICS
Order of ionic radii Ca2+ < K+ < Cl– < S2– 61. (b) Given equation is esinx – e–sinx – 4 = 0
The number of electrons remains the same but Put esin x = t in the given equation, we get
nuclear charge increases with increase in the t2 – 4t – 1 = 0
atomic number causing decrease in size.
4 16 4 4 20 4 2 5
57. (b) Anti addition of hydrogen atoms to the triple bond t 2 5
2 2 2
occurs when alkynes are reduced with sodium
(or lithium) metal in ammonia, ethylamine, or e sin x 2 5 t e sin x
alcohol at low temperatures. This reaction called,
a dissolving metal reduction, produces an (E)- or
e sin x 2 5 and e sin x 2 5
trans-alkene.
Sodium in liq. NH3 is used as a source of electrons e sin x 2 5 0 and sin x ln 2 5 1
in the reduction of an alkyne to a trans alkene.
So rejected So, rejected
CH3 CH2 CH2 C C CH3
2-Hexyne
Hence given equation has no solution.
The equation has no real roots.
Li/NH 3
Birch reduction
CH3 CH2 CH2 H 62. (c) Let c aˆ 2 bˆ and d 5 aˆ 4bˆ
C C
H CH3 Since c and d are perpendicular to each other
Trans-2-Hexene
dr dr 2 dt
72 4 9 9 I2 2 2 ln t C = 2 ln (sin x–2cos x) + C
dt dt 9 t
64. (b) nth term of the given series Hence, I1 I 2 dx 2 ln sin x 2 cos x c
2 2
Tn n 1 n 1 n n
= x + 2ln |(sin x – 2 cos x)| + k a =2
3 67. (b) Given equation of ellipse is 2x + y2 = 4
2
n 1 n3
3 3
n n 1 2x2 y2 x2 y2
n 1 n 1 1
4 4 2 4
n
3
Sn k3 k 1 =n x2 y2
k 1 Equation of tangent to the ellipse 1 is
2 4
n = 20 which is a natural number.
Now, put n =1,2,3,....20 y mx 2 m2 4 ...(1)
T1 = 13 – 03
T2 = 23 – 13 x2 y2
( equation of tangent to the ellipse 1
a2 b2
T20 = 203 – 193
Now, T1 + T2 + --- + T20 = S20 is y = mx + c where c a2 m2 b2 )
S20 = 203 – 03 = 8000
Now, Equation of tangent to the parabola
Hence, both the given statement is true.
65. (a) Let p : I become a teacher. 4 3
q : I will open a school y2 16 3 x is y mx ...(2)
m
Negation of p q is ~ (p q) = p ^ ~q
( equation of tangent to the parabola y2 = 4ax is
i.e. I will become a teacher and I will not open a
school. a
y mx )
sin x m
5 On comparing (1) and (2), we get
5 tan x cos x dx
66. (d) dx
tan x 2 sin x
2 4 3
cos x 2m 2 4
m
5 sin x cos x Squaring on both the sides, we get
dx
cos x sin x 2 cos x 16 (3) = (2m2 + 4) m2
AIEEE-2012 Solved Paper 2012- 17
900 p t 2
2 ln t 1
1
1
1
50 3.y 2 1 y2
2 .
t 1 2 1 1
1
900 p t 50 e 2 2 2 0
t
p t 900 50e 2 3 3 2
2 1 2 2
let p (t1) = 0 2 3.y 2 .y
3 2 3
t1 0
t1 = 2ln 18
0 900 50 e 2
3 3 3 2
76. (b) Given, f x ln x bx 2 ax 2 2y 2
1 2
y 2
5
y 2
3 3
1 0
f' x 2bx a
x
At x = –1, f x –1– 2b + a =0 5 20 2
2. 2 2
a – 2b = 1 ...(i) 3 3
1 78. (d) Number of white balls = 10
At x = 2, f x 4b a 0 Number of green balls = 9
2
and Number of black balls = 7
1 Required probability = (10 + 1) (9 + 1) ( 7 + 1)
a + 4b = ...(ii)
2 – 1 = 11.10.8 –1 = 879
1 1 [ The total number of ways of selecting one or
On solving (i) and (ii) we get a ,b more items from p identical items of one kind, q
2 4
identical items of second kind; r identical items of
1 x 1 2 x2 x third kind is ( p + 1) (q + 1) (r + 1) –1 ]
Thus, f ' x
x 2 2 2x
2x 1
79. (a) Let f x x cos
2 2
x2 x 2 x x 2 x 1 x 2
Doubtful points are x = n, n I
2x 2x 2x
AIEEE-2012 Solved Paper 2012- 19
2x 1 z z z2 z z 0
L.H.L = lim x cos
x n 2
2
2n 1 Either z z 0 or z z z 0
n 1 cos 0
2 Either z z real axis
( [x] is the greatest integer function) or z 2 z z zz z z 0
2x 1 2n 1 represents a circle passing through origin.
R.H.L = lim x cos n cos 0
x n 2 2 83. (c) Given P3 = Q3 ...(1)
Now, value of the function at x = n is and P2Q = Q2P ...(2)
f (n) = 0 Subtracting (1) and (2), we get
Since, L.H.L = R.H.L. = f(n) P3 – P2 Q = Q3 – Q2 P
2x 1 P2 (P–Q) + Q2 (P – Q) = 0
f (x) = [x] cos is continuous for every (P2 + Q2) (P–Q) = 0
2
If |P2 + Q2| 0 then P2 + Q2 is invertible.
real x.
P–Q=0 P=Q
x 1 y 1 z 1 Which gives a contradiction ( P Q)
80. (c) Given lines are
2 3 4 Hence |P2 + Q2| = 0
x 3 y k z x
and
1 2 1 84. (b, c) g x cos 4t dt
Thus, a , b , c and d are given as 0
a c,b,d 0
g x cos 4t dt
Now, a c = (3 – 1, k + 1, 0 –1) 0
= (2, k + 1, –1) (from graph of cos4t, it is clear that
2 k 1 1 x
2 3 4 0 cos 4t dt cos 4t dt
1 2 1 x 0
2 (3 – 8) – k + 1 (2 – 4) – 1 (4 – 3) = 0 = g (x g ( g (x – g (
2 (–5) – ( k +1) (–2) – 1 (1) = 0 ( from graph of cos4t, g (
9 85. (a) Let centre of the circle be (1,h)
– 10 + 2k + 2 – 1 = 0 k
2 Y [ circle touches x-axis at (1,0)]
81. (b) Given sample space = {1,2,3,.....,8}
Let Event
A : Maximum of three numbers is 6.
B : Minimum of three numbers is 3. (1,h) (2,3)
This is the case of conditional probability C B
We have to find P (minimum) is 3 when it is given
that P (maximum) is 6. A(1,0) X
2
B P B A C1 2 1
P 5 Let the circle passes through the point B (2,3)
A P A C2 10 5
CA = CB (radius)
82. (a) Since we know z z iff z is real. CA2 = CB2
(1 – 1)2 + (h – 0)2 = (1 – 2)2 + ( h – 3)2
z2 z2 h2 = 1 + h2 + 9 – 6h
Therefore,
z 1 z 1
10 5
2 2 h
zzz z2 z. z . z z 2 z .z z2 z .z z 2 6 3
2 10
z z z z z z z 0
Thus, diameter is 2h .
3
2012- 20 The Pattern Target AIEEE
86. (b) Let X = {1,2,3,4,5} 22 2
Total no. of elements = 5 Also, OP x 0 1 0 0 x
m
Each element has 3 options. Either set Y or set Z or
none. ( Y Z = Similarly, point Q (0,y) will satisfy equation of PQ
So, number of ordered pairs = 35 y –2 = m (x– 1)
87. (d) Equation of circle is (x – 1)2 + y2 =1 y – 2 = m (–1) y = 2 – m and OQ = y = 2–m
radius = 1 and diameter = 2 1 1 2
Length of semi-minor axis is 2. Area of POQ OP OQ 1 2 m
2 2 m
Equation of circle is x2 + (y – 2)2 = 4 = (2)2
radius = 2 and diameter = 4 1
( Area of base height )
Length of semi major axis is 4 2
We know, equation of ellipse is given by
1 4 1 4
2 m 2 4 m
x2 y2 2 m 2 m
2 2
1
Major axis Minor axis m 2
2
2 m
x2 y2 x2 y2
1 1
4 2
2 2
16 4 Q
x2 + 4y2 =16
x 2 , x 2 0 (1,2)
88. (c) f x x 2
2 x , x 2 0
x 2 , x 2 P
O
2 x , x 2
m 2
x 5 , x 5 Let Area = f (m) = 2
Similarly, f x x 5 2 m
5 x , x 5
1 2
f x x 2 x 5 x 2 5 x 3, 2 x 5 Now, f ' m
2 m2
Thus f (x) = 3 , 2 x 5 Put f (m) = 0 m2 = 4 m= ±2
f (x) = 0 , 2 x 5 4
f (4) = 0 Now, f '' m
m3
Y
1
f '' m m 2 0
2
1
f '' m m 0 2
2
Area will be least at m = –2
X Hence, slope of PQ is –2.
2 5 90. (b) Let ABCD be a parallelogram such that
Section - 1 æ 1 2ö
maximum energy loss possible is given as f çè mv ÷ø then
2
æ m ö
1. A uniform cylinder of length L and mass M having cross- f= ç ÷.
sectional area A is suspended, with its length vertical, from èM + mø
a fixed point by a massless spring such that it is half Statement - II: Maximum energy loss occurs when the
submerged in a liquid of density s at equilibrium position. particles get stuck together as a result of the collision.
The extension x0 of the spring when it is in equilibrium is: (a) Statement - I is true, Statment - II is true, Statement - II
is the correct explanation of Statement - I.
Mg Mg æ LAs ö (b) Statement-I is true, Statment - II is true, Statement - II is
(a) (b) ç1 – ÷
k k è M ø not the correct explanation of Statement - II.
(c) Statement - I is true, Statment - II is false.
Mg æ LAs ö Mg æ LAs ö (d) Statement - I is false, Statment - II is true.
(c) ç1 – ÷ (d) ç1 + ÷
k è 2M ø k è M ø 4. Let [ Î0 ] denote the dimensional formula of the permittivity of
2. A metallic rod of length ‘l’ is tied to a string of length 2l and vacuum. If M = mass, L = length, T = time and A = electric
made to rotate with angular speed w on a horizontal table current, then:
with one end of the string fixed. If there is a vertical magnetic
(a) Î0 = [M–1 L–3 T2 A] (b) Î0 = [M1 L3 T5 A2]
field ‘B’ in the region, the e.m.f. induced across the ends of
the rod is (c) Î0 = [M1 L2 T1 A2] (d) Î0 = [M1 L2 T1 A]
2 Bwl 2 5. A projectile is given an initial velocity of (iˆ + 2 ˆj ) m/s, where
(a)
2
iˆ is along the ground and ĵ is along the vertical. If g = 10
2
(b)
3Bwl m/s2 , the equation of its trajectory is :
2
(a) y = x - 5x2 (b) y = 2 x - 5x2
2
4 Bwl
(c) (c) 4 y = 2 x - 5x2 (d) 4 y = 2 x - 25 x 2
2
6. The amplitude of a damped oscillator decreases to 0.9 times
5 Bwl 2 its original magnitude in 5s. In another 10s it will decrease to
(d)
2 a times its original magnitude, where a equals
3. This question has statement I and statement II. Of the four (a) 0.7 (b) 0.81
choices given after the statements, choose the one that best (c) 0.729 (d) 0.6
describes the two statements. 7. Two capacitors C1 and C2 are charged to 120 V and 200 V
Statement - I: Apoint particle of mass m moving with speed respectively. It is found that connecting them together the
u collides with stationary point particle of mass M. If the potential on each one can be made zero. Then
2013- 2
1 A 2 gP0 1 MV0
Signal C R (c) (d)
2p MV0 2p AgP0
18. If a piece of metal is heated to temperature q and then allowed
(a) 10.62 MHz (b) 10.62 kHz to cool in a room which is at temperature q0, the graph between
(c) 5.31 MHz (d) 5.31 kHz the temperature T of the metal and time t will be closest to
13. Abeam of unpolarised light of intensity I0 is passed through
a polaroidAand then through another polaroid B which is
T T
oriented so that its principal plane makes an angle of 45°
q0
relative to that of A. The intensity of the emergent light is (a) (b)
(a) I0 (b) I0/2
(c) I0/4 (d) I0/8 O t O t
14. The supply voltage to room is 120V. The resistance of the
lead wires is 6W. A 60 W bulb is already switched on. What
is the decrease of voltage across the bulb, when a 240 W T T
heater is switched on in parallel to the bulb? q0 q0
(c) (d)
(a) zero (b) 2.9 Volt
(c) 13.3 Volt (d) 10.04 Volt O t O t
JEE Main-2013 Solved Paper 2013- 3
19. This questions has Statement I and Statement II. Of the four 23. The anode voltage of a photocell is kept fixed. The wavelength
choices given after the Statements, choose the one that best l of the light falling on the cathode is gradually changed. The
describes into two Statements. plate current I of the photocell varies as follows :
Statement-I : Higher the range, greater is the resistance of
ammeter.
I I
Statement-II : To increase the range of ammeter, additional
shunt needs to be used across it.
(a) Statement-I is true, Statement-II is true, Statement-II is (a) (b)
the correct explanation of Statement-I.
(b) Statement-I is true, Statement-II is true, Statement-II is O l O l
not the correct explanation of Statement-I.
(c) Statement-I is true, Statement-II is false.
I I
(d) Statement-I is false, Statement-II is true.
20. In an LCR circuit as shown below both switches are open
initially. Now switch S1 is closed, S2 kept open. (q is charge (c) (d)
on the capacitor and t = RC is Capacitive time constant).
Which of the following statement is correct ? O l O l
O V O V
C S2
L V O
I R
(a) Work done by the battery is half of the energy
Y I
dissipated in the resistor (c) (d) G
(b) At t = t, q = CV/2 B
O V
(c) At t = 2t, q = CV (1 – e–2)
(d) At t = 2 t, q = CV (1 – e–1) 25. Assume that a drop of liquid evaporates by decrease in its
21. Two coherent point sources S1 and S2 are separated by a surface energy, so that its temperature remains
small distance 'd' as shown. The fringes obtained on the unchanged.What should be the minimum radius of the drop
screen will be for this to be possible? The surface tension is T, density of
liquid is r and L is its latent heat of vaporization.
27. The graph between angle of deviation (d) and angle of Section - 2
incidence (i) for a triangular prism is represented by
O A B (a) II > IV > I > III (b) I > II > III > IV
L L (c) III > I > II > IV (d) IV > III > I > II
38. For gaseous state, if most probable speed is denoted by C*,
Q 3Q average speed by C and mean square speed by C, then for
(a) 8pe 0 L (b) 4pe L a large number of molecules the ratios of these speeds are :
0
(a) C* : C : C = 1.225 : 1.128 : 1
Q Q ln 2 (b) C* : C : C = 1.128 : 1.225 : 1
(c) (d) 4pe L (c) C* : C : C = 1 : 1.128 : 1.225
4pe 0 L ln 2 0 (d) C* : C : C = 1 : 1.225 : 1.128
JEE Main-2013 Solved Paper 2013- 5
39. The rate of a reaction doubles when its temperature changes 48. Synthesis of each molecule of glucose in photosynthesis
from 300 K to 310 K. Activation energy of such a reaction involves :
will be: (R = 8.314 JK–1 mol–1 and log 2 = 0.301) (a) 18 molecules of ATP (b) 10 molecules of ATP
(a) 53.6 kJ mol –1 (b) 48.6 kJ mol–1 (c) 8 molecules of ATP (d) 6 molecules of ATP
(c) 58.5 kJ mol –1 (d) 60.5 kJ mol–1 49. The coagulating power of electrolytes having ions Na+, Al3+
40. A compound with molecular mass 180 is acylated with and Ba2+ for arsenic sulphide sol increases in the order :
CH3COCl to get a compound with molecular mass 390. The (a) Al3+ < Ba2+ < Na+ (b) Na+ < Ba2+ < Al3+
2+
(c) Ba < Na < Al + 3+ (d) Al3+ < Na+ < Ba2+
number of amino groups present per molecule of the former
compound is : 50. Which of the following represents the correct order of
(a) 2 (b) 5 increasing first ionization enthalpy for Ca, Ba, S, Se and Ar ?
(a) Ca < S < Ba < Se < Ar (b) S < Se < Ca < Ba < Ar
(c) 4 (d) 6
(c) Ba < Ca < Se < S < Ar (d) Ca < Ba < S < Se < Ar
41. Which of the following arrangements does not represent
the correct order of the property stated against it ? -18
æ Z2ö
51. Energy of an electron is given by E = – 2.178 × 10 Jç 2 ÷ .
èn ø
(a) V2+ < Cr2+ < Mn2+ < Fe2+ : paramagnetic behaviour
(b) Ni2+ < Co2+ < Fe2+ < Mn2+ : ionic size Wavelength of light required to excite an electron in an
(c) Co3+ < Fe3+ < Cr 3+ < Sc3+ : stability in aqueous hydrogen atom from level n = 1 to n = 2 will be :
(h = 6.62 × 10 –34 Js and c = 3.0 × 108 ms–1)
solution
(a) 1.214 × 10–7 m (b) 2.816 × 10.–7 m
(d) Sc < Ti < Cr < Mn : number of oxidation states
(c) 6.500 × 10–7 m (d) 8.500 × 10–7 m
42. The order of stability of the following carbocations :
52. Compound (A), C8H9Br, gives a white precipitate when
Å warmed with alcoholic AgNO3. Oxidation of (A) gives an
CH 2 acid (B), C8H6O4. (B) easily forms anhydride on heating.
Identify the compound (A).
Å Å
CH 2 = CH - C H 2 ; CH3 - CH 2 - CH 2 ; is : CH2Br C2H5
I II (a) (b)
III Br
CH3
(a) III > II > I (b) II > III > I
(c) I > II > III (d) III > I > II CH2Br
43. Consider the following reaction :
CH2Br
z
xMNO4 - + yC2O4 2- + zH + ® xMn 2+ + 2yCO2 + H 2O (c) (d)
2 CH3
The value’s of x, y and z in the reaction are, respectively : CH3
(a) 5, 2 and 16 (b) 2, 5 and 8 53. Four successive members of the first row transition elements
(c) 2, 5 and 16 (d) 5, 2 and 8 are listed below with atomic numbers. Which one of them is
0
44. Which of the following is the wrong statement expected to have the highest E 3+ 2+ value ?
M /M
(a) ONCl and ONO– are not isoelectronic. (a) Cr(Z = 24) (b) Mn(Z = 25)
(b) O3 molecule is bent (c) Fe(Z = 26) (d) Co(Z = 27)
(c) Ozone is violet-black in solid state 54. How many litres of water must be added to 1 litre an aque-
(d) Ozone is diamagnetic gas. ous solution of HCl with a pH of 1 to create an aqueous
45. A gaseous hydrocarbon gives upon combustion 0.72 g of solution with pH of 2 ?
water and 3.08 g. of CO2. The empirical formula of the (a) 0.1 L (b) 0.9 L
hydrocarbon is : (c) 2.0 L (d) 9.0 L
(a) C2H4 (b) C3H4 55. The first ionisation potential of Na is 5.1 eV. The value of
(c) C6H5 (d) C7H8 electron gain enthalpy of Na+ will be :
46. In which of the following pairs of molecules/ions, both the (a) – 2.55 eV (b) – 5.1 eV
species are not likely to exist ? (c) – 10.2 eV (d) + 2.55 eV
56. An organic compound A upon reacting with NH3 gives B.
(a) H +2 , He22 - (b) H -2 , He 22 - On heating B gives C. C in presence of KOH reacts with Br 2
to given CH3CH2NH2. A is :
(c) H 22 + , He 2 (d) H -2 , He22 +
(a) CH3COOH (b) CH3CH2CH2COOH
47. Which of the following exists as covalent crystals in the
solid state ? (c) CH3 - CH - COOH (d) CH3CH2COOH
|
(a) Iodine (b) Silicon
CH3
(c) Sulphur (d) Phosphorus
2013- 6
ANSWER KEY
1 (c) 16 (c) 31 (c) 46 (c) 61 (c) 76 (d)
2 (d) 17 (c) 32 (a,b) 47 (b) 62 (c) 77 (a)
3 (d) 18 (c) 33 (c) 48 (a) 63 (c) 78 (c)
4 (b) 19 (d) 34 (d) 49 (b) 64 (c) 79 (b)
5 (b) 20 (c) 35 (a) 50 (c) 65 (c) 80 (c)
6 (c) 21 (d) 36 (a) 51 (a) 66 (d) 81 (a)
7 (b) 22 (b) 37 (c) 52 (d) 67 (c) 82 (b)
8 (b) 23 (d) 38 (c) 53 (d) 68 (b) 83 (c)
9 (a) 24 (a) 39 (a) 54 (d) 69 (b) 84 (a)
10 (c) 25 (d) 40 (b) 55 (b) 70 (a) 85 (b)
11 (a) 26 (d) 41 (a) 56 (d) 71 (c) 86 (a)
12 (b) 27 (c) 42 (d) 57 (b) 72 (a) 87 (b)
13 (c) 28 (a) 43 (c) 58 (b) 73 (b) 88 (a)
14 (d) 29 (b) 44 (a) 59 (a) 74 (c) 89 (b)
15 (b) 30 (d) 45 (d) 60 (b) 75 (d) 90 (d)
sLAg C2 [AT]2
Mg - Hence, e 0 = = = [M -2 L-3T 4 A 2 ]
Þ x0 = 2 = Mg æ 1 - LAs ö N.m 2 -2 2
MLT .L
ç ÷
k è 2M ø
k r
5. (b) From equation, v = iˆ + 2 ˆj
Hence, extension of the spring when it is in equilibrium
Þ x=t … (i)
Mg æ LAs ö
is, x 0 = ç1 - ÷
k è 2M ø 1
y = 2t - (10t 2 ) … (ii)
w l 2
2. (d) Here, induced e.m.f. 2l
dx From (i) and (ii), y = 2 x - 5x2
3l x
[(3l) – (2l) 2 ]
2
e = ò (wx) Bdx = Bw bt
2l
2 -
(c) Q A = A0 e
2m
6. (where, A0 = maximum amplitude)
5 Bl 2 w
= According to the questions, after 5 second,
2
2 2
P P
3. (d) Maximum energy loss = - -
b(5)
2m 2(m + M) 0.9A 0 = A 0 e 2m … (i)
C1 C2 -GMm
+ – – + Ei = +K
7. (b) R
120 V 200 V Ei = E f
For potential to be made zero, after connection
5GMm
Therefore minimum required energy, K =
é qù 6R
120 C1 = 200 C2 êQ C = v ú Given : Resistance R = 100 kilo ohm = 100 × 10 3 W
ë û 12. (b)
Þ 3C1 = 5C2 Capacitance C = 250 picofarad = 250 × 10–12F
t = RC = 100 × 103 × 250 × 10–12 sec
8. (b) Fundamental frequency,
= 2.5 × 107 × 10–12 sec
v 1 T 1 T é T mù = 2.5 × 10–5 sec
f = = = êQ v = and m = ú
2l 2l m 2l Ar ë m lû The higher frequency whcih can be detected with
tolerable distortion is
Tl T Y Dl 1 gDl
Also, Y = Þ = Þ f = ....(i)
ADl A l 2l lr f =
1
=
1
Hz
Dl 2pma RC 2p ´ 0.6 ´ 2.5 ´ 10 -5
l = 1.5 m, = 0.01, r = 7.7 × 103 kg/m3 (given)
l
g = 2.2 × 1011 N/m2 (given) 100 ´ 104 4
= Hz = ´ 104 Hz = 10.61 KHz
25 ´ 1.2p 1.2 p
Dl
Putting the value of l, , r and g in eqn. (i) we get, This condition is obtained by applying the condition
l that rate of decay of capacitor voltage must be equal or
2 103
f = ´ less than the rate of decay modulated singnal voltage
7 3 for proper detection of mdoulated signal.
or, f » 178.2 Hz 13. (c) Relation between intensities
9. (a) As we know, Magnetic flux, f = B. A
45° B
I0 (I0/2)
m 0 (2)(20 ´ 10 -2 ) 2
´ p (0.3 ´ 10 -2 )2 (unpolarised) IR
2[(0.2) + (0.15) ]
2 2
A
on solving
= 9.216 × 10–11 æI ö I 1 I
= 9.2 × 10–11 weber I r = ç 0 ÷ cos 2 (45°) = 0 ´ = 0
è 2ø 2 2 4
Velocity of light in vacuum
10. (c) \ n =
Velocity of light in medium 6W
3
\ n= (Lead) Bulb
2 R = 3cm
32 + (R – 3mm)2 = R2 14. (d)
Þ 32 + R2 – 2R(3mm) + (3mm)2 = R2 3mm
Þ R » 15 cm
1 æ 3 öæ 1 ö
= ç –1 ÷ç ÷ Þ f = 30 cm 120 V
f è 2 øè 15 ø
Power of bulb = 60 W (given)
11. (a) As we know,
120 ´ 120 é V2 ù
- GMm Resistance of bulb = = 240W êQ P = ú
Gravitational potential energy = 60 ë R û
r
Power of heater = 240W (given)
and orbital velocity, v0 = GM / R + h
120 ´ 120
Resistance of heater = = 60W
1 GMm 1 GM GMm 240
E f = mv02 - = m -
2 3R 2 3R 3R Voltage across bulb before heater is switched on,
GMm æ 1 ö -GMm 240
= ç - 1÷ = V1 = ´ 120 = 117.73 volt
3R è 2 ø 6R 246
2013- 10
Voltage across bulb after heater is switched on, 21. (d) It will be concentric circles.
48 22. (b) From question,
V2 = ´ 120 = 106.66 volt B0 = 20 nT = 20 × 10–9T
54
Hence decrease in voltage (Q velocity of light in vacuum C = 3 × 10 8 ms–1)
r r r
V1 – V2 = 117.073 – 106.66 = 10.04 Volt (approximately) E0 = B0 ´ C
15. (b) Heat is extracted from the source in path DA and AB is r r r
| E 0 |=| B | . | C |= 20 ´ 10 -9 ´ 3 ´ 108
3 æ P0V0 ö 5 æ 2 P0V0 ö
DQ = + R
2 çè R ÷ø 2 çè R ÷ø
R = 6 V/m.
23. (d) As l is increased, there will be a value of l above
which photoelectrons will be cease to come out so
3 5 æ 13 ö
Þ P0V0 + 2 P0V0 = ç ÷ P0V0 photocurrent will become zero. Hence (d) is correct
2 2 è 2ø
answer.
r 24. (a) For same value of current higher value of voltage is
required for higher frequency hence (1) is correct
16. (c) o o
answer.
25. (d) When radius is decrease by DR,
From conservation of angular momentum about any fix
point on the surface, 4pR 2 DRrL = 4pT[R 2 - (R - DR) 2 ]
mr 2 w 0 = 2mr 2 w Þ rR 2 DRL = T[R 2 - R 2 + 2RDR - DR 2 ]
w0 r Þ rR 2 DRL = T2RDR [ DR is very small]
Þ w = w0 / 2 Þ v =
2
[Q v = rw ]
2T
Mg ÞR=
17. (c) = P0 P0V0 g = PV g rL
A 26. (d) DE = hv
Mg = P0A … (1) P0 Ax0 g = PA( x0 - x) g DE é 1 1 ù k(2n - 1)
n= = kê - 2ú= 2
ë (n - 1) n û n (n - 1) 2
h 2
P0 x0g
P=
( x0 - x) g 2k
» 3
Piston
Let piston is displaced by distance x n
æ P xg ö 1
0 0 x or, n µ 3
Mg - ç ÷ A = Frestoring n
çè ( x - x) g ÷ø
0 x0 Cylinder 27. (c) For the prism as the angle of incidence (i) increases,
containing the angle of deviation (d) first decreases goes to
æ x0g ö ideal gas
minimum value and then increases.
P0 A ç 1 - ÷ = Frestoring [ x0 - x » x0 ] 28. (a)
g
çè ( x - x) ÷ø y
0
gP0 Ax
F=- F F
x0 x
q q
\ Frequency with which piston executes SHM. a a
29. (b) Given : N 34. (d) Higher the value of standard reduction potential,
stronger is the oxidising agent, hence MnO4– is the
M1 = 1.20 Am 2 BH
strongest oxidising agent.
B1
M 2 = 1.00 Am2 S B2 S 35. (a) Process is isothermal reversible expansion, hence
DU = 0, therefore q = – W.
20 O
r= cm = 0.1m N N Since q = + 208 J, W = – 208 J
2 r r 37. (c) Electron withdrawing substituents like –NO2 , Cl
Bnet = B1 + B2 + BH increase the acidity of phenol while electron releasing
S substituents like – CH3, – OCH3 decreases acidity.
m 0 ( M1 + M 2 )
Bnet = + BH hence the correct order of acidity will be
4p r3
OH OH OH OH
-7
10 (1.2 + 1)
= 3
+ 3.6 ´ 10 -5 = 2.56 ´ 10 -4 wb/m2
(0.1) > > >
30. (d) L L
NO2 Cl CH3 OCH3
O A B
dx
III I II IV
Electric potential is given by, (–M, –I) (–I > +M) (+I, + HC ) (+ M)
æ qö 2RT
1 çè L ÷ø
2L 2L dx q 38. (c) Most probable speed (C*) =
kdq =
V= ò x
= ò 4pe 0 x 4 pe 0L
ln(2) M
L L
8RT
Average Speed (C) =
pM
SECTION-2 : CHEMISTRY
31. (c) Octahedral coordination entities of the type Ma3b3 3RT
Root mean square velocity (C) =
exhibit geometrical isomerism. The compound exists M
both as facial and meridional isomers, both contain
plane of symmetry 2RT 8RT 3RT
C*: C : C = : :
NH3 NH3 M pM M
Cl NH NH3 Cl 4 3
= 1: = 1:1.128 :1.225
3
:
Co Co p 2
NH3 Cl Cl 41. (a)
Cl
Cl NH3 (a) V = 3d 3 4s 2 ; V2+ = 3d 3 = 3 unpaired electrons
Cr = 3d 5 4s 1 ; Cr2+ = 3d 4 = 4 unpaired electrons
fac- mer
32. (a,b) The molecular orbital structures of C2 and N2 are Mn = 3d 5 4s2 ; Mn2+ = 3d 5 = 5 unpaired electrons
Fe = 3d 6 4s 2 ; Fe2+ = 3d 6 = 4 unpaired electrons
N2 = s1s2 s *1s2 s2s2 s *2s2 s2 px2p2 py2p2 pz2 Hence the correct order of paramagnetic behaviour
V2+ < Cr2+ = Fe2+ < Mn2+
C2 = s1s 2 s *1s 2 s 2s 2 s * 2s 2 p 2 py 2 p 2 Pz2 (b) For the same oxidation state, the ionic radii generally
Both N2 and C2 have paired electrons, hence they are decreases as the atomic number increases in a particular
diamagnetic. transition series. hence the order is
33. (c) Carbocations are planar hence can be attacked on Mn++ > Fe++ > Co++ > Ni++
either side to form racemic mixture. (c) In solution, the stability of the compound depends upon
electrode potentials, SEP of the transitions metal ions are
Å given as
Cl - CH- CH3 ¾¾¾¾ 5 ®Ph - CH - CH + SbCl- ¾¾
SbCl
®
Toluene 3 6 Co3+ / Co = + 1.97, Fe3+ / Fe = + 0.77 ;
| (carbocation)
Cr3+ / Cr2+ = – 0.41, Sc 3+ is highly stable as it does not
Ph
(-) show + 2 O. S.
(d) Sc – (+ 2), (+ 3)
Ti – (+ 2), (+ 3), (+ 4)
Ph - CH - CH3 + SbCl5
| Cr – (+ 1), (+ 2), (+ 3), (+ 4), (+ 5), (+ 6)
Cl Mn – (+ 2), (+ 3), (+ 4), (+ 5), (+ 6), (+ 7)
(d+l) mixture i.e. Sc < Ti < Cr = Mn
2013- 12
2 = s1s s*1s
(c) H2+
42. (d) Higher stability of allyl and aryl substituted methyl 46. 0 0
carbocation is due to dispersal of positive charge due
1
to resonance Bond order for H2+
2 =
(0 - 0) = 0
2
+ + He2 = s1s2s*1s2
CH 2 = CH - C H 2 ¬¾® CH 2 - CH = CH 2
Resonating structures of allyl carbocation 1
Bond order for He2 = (2 - 2) = 0
2
+
CH2 CH2 CH2 CH2 so both H22+ and He2 does not exist.
Å Å 50. (c) On moving down a group size increases hence
ionisation enthalpy decreases, hence Se < S and Ba <
Å
Ca. Further, Ar being an inert gas has maximum IE.
Resonating structures of benzyl carbocation
æ 1 1 ö hc
51. (a) DE = 2.178 ´ 10-18 ç 2 - 2 ÷ =
whereas in alkyl carbocations dispersal of positive è1 2 ø l
charge on different hydrogen atoms is due to
3 hc 6.62 ´ 10-34 ´ 3 ´ 108
hyperconjugation. Hence the correct order of stability Þ 2.178 ´ 10-18 ´ = =
will be 4 l l
Å
6.62 ´ 10-34 3 ´ 108 ´ 4
CH2 l= = 1.214 × 10–7m
2.178 ´ 10-18 ´3
Å Å
> CH 2 = CH - C H 2 > CH3 - CH 2 - CH 2
Allyl, I Propyl, II CH2Br
Benzyl, III 52. (d) Alcholic
AgBr¯
43. (c) On balancing the given equations, we get CH3 AgNO3
A
2MnO 4 - + 5C 2O 4 2- + 16H + ¾¾
® 2Mn ++ Oxidation
+10CO 2 + 8H 2O
COOH CO
So, x = 2, y = 5 & z = 16 D
O
44. All options are correct, CO
COOH
(a) + 7 + 17 = 32e - ü not Acid (B) Phthalic Anhydride
ý isoelectronic
ONO- = 8 + 7 + 8 + 1 = 24e - þ
54. (d) Q pH = 1 ; H+ = 10–1 = 0.1 M
.. pH = 2 ; H+ = 10–2 = 0.01 M
(b) 7 8Å O 1.278A°
1. 2 The central atom is sp2 \ M1 = 0.1 V1 = 1
O 116.8° O hybridized with one lone pair. M2 = 0.01 V2 = ?
(c) It is a pale blue gas. At – 249.7°, it forms violet black From
crystals. M1V1 = M2V2
(d) It is diamagnetic in nature due to absence of unpaired 0.1 × 1 = 0.01 × V2
electrons. V2 = 10 litres
\ Volume of water added = 10 – 1 = 9 litres
45. (d) Q 18 gm, H2O contains = 2 gm H
\ 0.72 gm H2O contains 55. ® Na + + e -
(b) Q For Na ¾¾ IE1 = 5.1 eV
2 \ For Na + + e - ¾¾ ® Na EF = -5.1 eV
= ´ 0.72 gm = 0.08 gm H
18 (because the reaction is reverse)
Q 44 gm CO2 contains = 12 gm C 57. (b) Li2 = s1s2 s*1s2 s2s2
\ 3.08 gm CO2 contains 1
\ Bond order = (4 , 2) < 1
2
12
= ´ 3.08 = 0.84 gm C Li2+ = s1s2 s*1s2 s2s1
44
1
0.84 0.08 B.O. < (3 , 2) < 0.5
\C:H= : 2
12 1
Li2– = s1s2 s*1s2s2s2s*2s1
= 0.07 : 0.08 = 7 : 8
1
\ Empirical formula = C7H8 B.O. = (4 , 3) < 0.5
2
JEE Main-2013 Solved Paper 2013- 13
The bond order of Li2+ and Li2– is same but Li2+ is more 65. (c) We have,
stable than Li2– because Li2+ is smaller in size and has 2 uuur uuur uuur
electrons in antibonding orbitals whereas Li2– has 3 AB + BC + CA = 0
electrons in antibonding orbitals. Hence Li2+ is more uuur uuur uuur
Þ BC = AC - AB
stable than Li2–.
uuur uuur
58. (b) Reaction of alcohols with Lucas reagent proceeds uuuur AC - AB
through carbocation formation, SN1 mechanism. Now, BM =
2
Further 3° carbocations (from tertiary alcohols) are uuur
highly stable thus reaction proceeds through SN1 æ uuuur BC ö
çQ BM =
mechanism. è 2 ÷ø
60. (b) For one mole of the oxide
Moles of M = 0.98 Also, we have
Moles of O2– = 1 uuur uuuur uuuur
Let moles of M3+ = x AB + BM + MA = 0
\ Moles of M2+ = 0.98– x uuur uuur
uuur AC - AB uuuur
On balancing charge Þ AB + = AM
(0.98 – x) × 2 + 3x – 2 = 0 2
x = 0.04 uuur uuur
uuuur AB + AC
0.04 Þ AM = = 4iˆ - ˆj + 4kˆ
\ % of M3+ = ´100 < 4.08% 2
0.98 uuuur
SECTION-3 : MATHEMATICS Þ AM = 33
é 1 11 111 ù
dP = 7ê + + 3 +¼... + up to 20 terms ú
62. (c) Given, Rate of change is = 100 - 12 x ë10 100 10 û
dx
Multiply and divide by 9
Þ dP = (100 – 12 x ) dx
7 é 9 99 999 ù
By integrating = ê + + + ...¼ up to 20 terms ú
9 ë10 100 1000 û
ò dP = ò (100 - 12 x ) dx
P = 100x – 8x3/2 + C éæ 1ö æ 1 ö æ 1 öù
7 êç 1 - ÷ + ç 1 - 2 ÷ + ç 1 - 3 ÷ ú
Given when x = 0 then P = 2000 = ê è 10 ø è 10 ø è 10 øú
Þ C = 2000 9
ëê +¼...up to 20 terms ûú
Now when x = 25 then
P = 100 × 25 – 8 × (25)3/2 + 2000
= 4500 – 1000 é 1 æ æ 1ö öù
20
Þ P = 3500 ê ç 1 - çè ÷ø ÷ ú
7ê 10 è 10 ø ú
63. (c) Given = ê 20 - ú
9ê 1
n(A) = 2, n(B) = 4, n(A × B) = 8 1- ú
Required number of subsets = ê 10 ú
8C + 8C + . . . . . . + 8C = 28 – 8C – 8C – 8C ë û
3 4 8 0 1 2
= 256 – 1 – 8 – 28 = 219
7 é179 1 æ 1 ö ù
64. (c) Given lines will be coplanar 20
7
= ê + ç ÷ ú = [179 + (10)–20]
-1 1 1 9 ëê 9 9 è 10 ø ûú 81
If 1 1 -k = 0 68. (b) Suppose B(0, 1) be any point on given line and
k 2
co-ordinate of A is ( 3, 0). So, equation of
Þ –1(1 + 2k) – (1 + k2) + 1(2 – k) = 0
Þ k = 0, –3
2013- 14
2sin 2 x 3 + cos x x
A (3, 0) = lim · ·
x ®0 x2 1 tan 4 x
sin 2 x x
A = 2 lim 2
· lim 3 + cos x · lim
x ®0 x x®0 x ® 0 tan 4 x
(1, –2)
1 4x 1
= 2.4 lim = 2.4. = 2
4 x®0 tan 4 x 4
JEE Main-2013 Solved Paper 2013- 15
p /3
dx
76. (d) Let I = ò
p/61+ tan x
p /3 p /3
dx tan x dx
= ò = ò …(i)
p/6 æp ö
1 + tan ç - x÷ p/61+ tan x
è2 ø
Also, Given, I
p /3
tan x dx
= ò …(2)
p/61+ tan x
By adding (1) and (2), we get Now, x-co-ordinate of incentre is given as
p/3 ax1 + bx2 + cx3
2I= ò dx a+b+c
p/6
2 ´ 0 + 2 2.0 + 2.2
1 ép pù p Þ x-coordinate of incentre =
Þ I = ê - ú = , statement-1 is false 2+ 2+2 2
2 ë 3 6 û 12
2
b b = = 2- 2
2+ 2
ò f ( x)dx = ò f (a + b - x) dx
80. (c) Given expression can be written as
a a
It is fundamental property. 10
æ 1/ 3 æ x + 1ö ö æ 1 ö
10
77. (a) From the given equation of ellipse, we have ç ( x + 1) - ç ÷÷ = ç x1/ 3 + 1 - 1 - ÷
è è x øø è xø
9
a = 4, b = 3, e = 1 - = (x1/3 – x–1/2)10
16 General term = Tr+1 = 10Cr (x1/3)10–r(–x–1/2)r
10 - r 10 - r - r
-r
=
10
Cr x 3 · ( -1) r · x 2 = 10Cr ( -1) r · x 3 2
10 - r r
Term will be independent of x when - =0
3 2
Þ r=4
So, required term = T5 = 10C4 = 210
2 1 11
= 5· + =
ò {(2 y + 3) - y } dy
3
5 5
3 3 35 Required area = 2
a = 2, b = 2 2, c = 2 3
y3
x1 = 0, x2 = 0, x3 = 2 = y2 + 3y – = 9.
3 0
2013- 16
( )
y = ± x + 5 , both statements are correct as m = ±1
satisfies the given equation of statement-2.
88. (a) Let y = sec(tan–1 x) and tan–1 x = q.
Þ x = tan q
Thus, we have y = sec q
AB p2 + q2 Þ y = 1 + x2 (Q sec2q = 1 + tan2q)
=
sinq sin(p - (q + a ))
dy 1
Þ = · 2x
p + q sin q
2 2
dx 2 1 + x2
AB =
sin q cos a + cos q sin a At x = 1,
( p 2 + q 2 ) sin q dy 1
=
= .
q sin q + p cos q dx 2
89. (b) Given expression can be written as
æ q p ö
ççQ cos a = and sina = ÷÷ sin A sin A cos A cos A
è p +q
2 2
p + q2
2
ø ´ + ´
cos A sin A - cos A sin A cos A - sin A
85. (b) | P | = 1(12 – 12) – a(4 – 6) + 3(4 – 6) = 2a – 6
Now, adj A = P Þ | adj A | = | P | æ sin A ö
Q tan A = and
Þ | A |2 = | P | ç cos A ÷
Þ | P | = 16 ç ÷
ç cos A ÷
Þ 2a – 6 = 16 cot A =
è sin A ø
Þ a = 11
1 ìï sin 3 A - cos3 A üï
x = í ý
sin A - cos A îï cos A sin A þï
86. (a) Since, y = ò t dt, x ÎR
0
sin 2 A + sin A cos A + cos 2 A
dy = = 1 + sec A cosec A
therefore = x sin A cos A
dx
å ( xi - x )2
dy
But from y = 2x, =2 90. (d) If initially all marks were xi then s12 = i
dx N
Þ |x|=2 Þ x=±2 Now each is increased by 10
±2
å ( xi - x )
2
Points y = ò t dt = ± 2 å [( xi +10) -( x +10)]2
0 s12 = i
= i
= s12
\ equation of tangent is N N
y – 2 = 2(x – 2) or y + 2 = 2(x + 2) Hence, variance will not change even after the grace
Þ x-intercept = ± 1. marks were given.
JEE Main
Section - 1 2g g
(a) (b)
3 2
5g
(c) (d) g
1. The current voltage relation of a diode is given by 6
I e1000V T 1 mA, where the applied voltage V is in volts
4. A block of mass m is placed on a surface with a vertical cross
and the temperature T is in degree kelvin. If a student makes
an error measuring 0.01 V while measuring the current of x3
section given by y . If the coefficient of friction is 0.5,
5 mA at 300 K, what will be the error in the value of current in 6
mA? the maximum height above the ground at which the block
(a) 0.2 mA (b) 0.02 mA can be placed without slipping is:
(c) 0.5 mA (d) 0.05 mA 1 2
2. From a tower of height H, a particle is thrown vertically (a) m (b) m
6 3
upwards with a speed u. The time taken by the particle, to hit
the ground, is n times that taken by it to reach the highest 1 1
(c) m (d) m
point of its path. The relation between H, u and n is: 3 2
(a) 2gH n2u 2 (b) gH n 2
2 2
u d 5. When a rubber-band is stretched by a distance x, it exerts
restoring force of magnitude F = ax + bx2 where a and b are
(c) 2gH nu 2 n 2 (d) gH n 2 u2 constants. The work done in stretching the unstretched
rubber-band by L is:
3. A mass ‘m’ is supported by a massless string wound around
a uniform hollow cylinder of mass m and radius R. If the 1
(a) aL2 bL3 (b) aL2 bL3
string does not slip on the cylinder, with what acceleration 2
will the mass fall or release?
aL2 bL3 1 aL2 bL3
(c) (d) 2 2 3
2 3
6. A bob of mass m attached to an inextensible string of length
l is suspended from a vertical support. The bob rotates in a
horizontal circle with an angular speed rad/s about the vertical.
R About the point of suspension:
m (a) angular momentum is conserved.
(b) angular momentum changes in magnitude but not in
direction.
(c) angular momentum changes in direction but not in
magnitude.
m (d) angular momentum changes both in direction and
magnitude.
2014- 2
1 sin
(a) P
1 sin
1 cos 600 k
(b) C
1 cos A
d2 400 K
1 tan
(c) V
1 tan
(a) The change in internal energy in whole cyclic process
1 sin d1 is 250 R.
(d)
1 cos (b) The change in internal energy in the process CA is 700 R.
10. On heating water, bubbles being formed at the bottom of the (c) The change in internal energy in the process AB is -350
vessel detach and rise. Take the bubbles to be spheres of R.
radius R and making a circular contact of radius r with the (d) The change in internal energy in the process BC is
bottom of the vessel. If r << R and the surface tension of – 500 R.
water is T, value of r just before bubbles detach is: 13. An open glass tube is immersed in mercury in such a way
(density of water is ) that a length of 8 cm extends above the mercury level. The
w
open end of the tube is then closed and sealed and the tube
is raised vertically up by additional 46 cm. What will be
length of the air column above mercury in the tube now?
(Atmospheric pressure = 76 cm of Hg)
(a) 16 cm (b) 22 cm
(c) 38 cm (d) 6 cm
14. A particle moves with simple harmonic motion in a straight
line. In first s, after starting from rest it travels a distance a,
and in next s it travels 2a, in same direction, then:
(a) amplitude of motion is 3a
R
(b) time period of oscillations is 8
(c) amplitude of motion is 4a
2r (d) time period of oscillations is 6
JEE Main-2013 Solved Paper 2014- 3
15. A pipe of length 85 cm is closed from one end. Find the 21. In the circuit shown here, the point ‘C’ is kept connected to
number of possible natural oscillations of air column in the point ‘A’ till the current flowing through the circuit becomes
pipe whose frequencies lie below 1250 Hz. The velocity of constant. Afterward, suddenly, point ‘C’ is disconnected
sound in air is 340 m/s. from point ‘A’ and connected to point ‘B’ at time t = 0. Ratio
(a) 12 (b) 8
of the voltage across resistance and the inductor at t = L/R
(c) 6 (d) 4
will be equal to:
16. Assume that an electric field E 30x 2 ˆi exists in space.
Then the potential difference VA VO , where VO is the A C R
potential at the origin and VA the potential at x = 2 m is:
(a) 120 J/C (b) -120 J/C L
(c) -80 J/C (d) 80 J/C
17. A parallel plate capacitor is made of two circular plates B
separated by a distance 5 mm and with a dielectric of dialectric
constant 2.2 between them. When the electric field in the
dielectric is 3 104 V m the charge density of the positive
plate will be close to: e
(a) (b) 1
7 2 7 2 1 e
(a) 6 10 C m (b) 3 10 C m
(c) 3 104 C m 2 (d) 6 104 C m 2 1 e
(c) –1 (d)
e
18. In a large building, there are 15 bulbs of 40 W, 5 bulbs of 100
W, 5 fans of 80 W and 1 heater of 1 kW. The voltage of 22. During the propagation of electromagnetic waves in a
electric mains is 220 V. The minimum capacity of the main medium:
fuse of the building will be: (a) Electric energy density is double of the magnetic energy
(a) 8 A (b) 10 A density.
(c) 12 A (d) 14 A (b) Electric energy density is half of the magnetic energy
19. A conductor lies along the z-axis at 1.5 z 1.5 m and density.
carries a fixed current of 10.0 A in â z direction (see figure). (c) Electric energy density is equal to the magnetic energy
density.
For a field B 3.0 10 4 e 0.2x aˆ y T, find the power required
(d) Both electric and magnetic energy densities are zero.
to move the conductor at constant speed to x = 2.0 m, y = 0
m in 5 10 3 s. Assume parallel motion along the x-axis. 3
23. A thin convex lens made from crown glass has
2
z focal length f. When it is measured in two different liquids
1.5 4 5
I having refractive indices and , it has the focal lengths
3 3
f1 and f2 respectively. The correct relation between the focal
lengths is:
B y
2.0 (a) f1 = f2 < f
(b) f1 > f and f2 becomes negative
x (c) f2 > f and f1 becomes negative
–1.5 (d) f1 and f2 both become negative
24. A green light is incident from the water to the air - water
interface at the critical angle ( ). Select the correct statement.
(a) 1.57 W (b) 2.97 W
(a) The entire spectrum of visible light will come out of the
(c) 14.85 W (d) 29.7 W
water at an angle of 90º to the normal.
20. The coercivity of a small magnet where the ferromagnet gets
(b) The spectrum of visible light whose frequency is less
demagnetized is 3 103 Am 1. The current required to be
than that of green light will come out to the air medium.
passed in a solenoid of length 10 cm and number of turns
100, so that the magnet gets demagnetized when inside the (c) The spectrum of visible light whose frequency is more
solenoid, is: than that of green light will come out to the air medium.
(a) 30 mA (b) 60 mA (d) The entire spectrum of visible light will come out of the
water at various angles to the normal.
(c) 3 A (d) 6 A
2014- 4
35. Resistance of 0.2 M solution of an electrolyte is 50 . The rate law for the formation of C is:
The specific conductance of the solution is 1.4 S m –1 .
The resistance of 0.5 M solution of the same electrolyte is 280 dc dc 2
(a) k A B (b) k A B
. The molar conductivity of 0.5 M solution of the electrolyte dt dt
in S m2 mol–1 is:
(a) 5 × 10–4 (b) 5 × 10–3 dc 2 dc
3 (c) k A B (d) k A
(c) 5 × 10 (d) 5 × 102 dt dt
36. For complete combustion of ethanol,
41. Among the following oxoacids, the correct decreasing order
C2 H5OH l 3O2 g 2CO2 g 3H 2 O l , of acid strength is:
the amount of heat produced as measured in bomb calorimeter, (a) HOCl HClO2 HClO3 HClO4
is 1364.47 kJ mol–1 at 25ºC. Assuming ideality the enthalpy of
combustion, cH, for the reaction will be: (b) HClO4 HOCl HClO 2 HClO3
(R = 8.314 kJ mol–1) (c) HClO4 HClO3 HClO2 HOCl
1 1
(a) 1366.95 kJ mol (b) 1361.95 kJ mol
(d) HClO2 HClO 4 HClO3 HOCl
(c) 1460.95 kJ mol 1 (d) 1350.50 kJ mol 1
42. The metal that cannot be obtained by electrolysis of an
37. The equivalent conductance of NaCl at concentration C and aqueous solution of its salts is:
at infinite dilution are C and , respectively. The correct (a) Ag (b) Ca
relationship between C and is given as: (c) Cu (d) Cr
(Where the constant B is positive) 43. The octahedral complex of a metal ion M 3+ with four
monodentate ligands L1, L2, L3 and L4 absorb wavelengths
(a) C B C (b) C B C in the region of red, green, yellow and blue, respectively. The
increasing order of ligand strength of the four ligands is:
(c) C B C (d) C B C
38. Consider separate solutions of 0.500 M C 2H5OH(aq), (a) L4 L3 L2 L1 (b) L1 L3 L2 L4
0.100 M Mg3 (PO4)2 (aq), 0.250 M KBr(aq) and 0.125 M (c) L3 L2 L4 L1 (d) L1 L2 L4 L3
Na3PO4(aq) at 25 C. Which statement is true about these
44. Which one of the following properties is not shown by NO?
solutions, assuming all salts to be strong electrolytes?
(a) They all have the same osmotic pressure. (a) It is diamagnetic in gaseous state
(b) 0.100 M Mg3(PO4)2(aq) has the highest osmotic pressure. (b) It is neutral oxide
(c) 0.125 M Na3PO4(aq) has the highest osmotic pressure. (c) It combines with oxygen to form nitrogen dioxide
(d) 0.500 M C2H5OH(aq) has the highest osmotic pressure. (d) It’s bond order is 2.5
1 45. In which of the following reactions H2O2 acts as a reducing
39. For the reaction SO2 g O SO3 , if agent?
2 2 g g
x
where the symbols have usual meaning (a) H 2 O 2 2H 2e 2H 2 O
KP KC RT
then the value of x is (assuming ideality): (b) H 2 O 2 2e O2 2H
1
(a) –1 (b) (c) H 2 O 2
2 2e 2OH
1
(c) (d) 1 (d) H 2 O 2 2OH 2e O2 2H 2 O
2
40. For the non - stoichimetre reaction 2A B C D, the 46. The correct statement for the molecule, CsI3 is:
following kinetic data were obtained in three separate (a) It is a covalent molecule.
experiments, all at 298 K.
(b) It contains Cs+ and I3 ions.
48. Given below are the half-cell reactions: 53. The most suitable reagent for the conversion of
Mn 2 2e Mn ; E 0 1.18 V R CH 2 OH R CHO is:
(a) KMnO4
2 Mn3 e Mn 2 ; E0 1.51V (b) K2Cr2O7
(c) CrO3
The E0 for 3Mn 2 Mn 2Mn 3 will be: (d) PCC (Pyridinium Chlorochromate)
(a) –2.69 V; the reaction will not occur 54. The major organic compound formed by the raction of
(b) –2.69 V; the reaction will occur 1, 1, 1-trichloroethane with silver powder is:
(c) –0.33 V; the reaction will not occur (a) Acetylene (b) Ethene
(d) –0.33 V; the reaction will occur (c) 2 - Butyne (d) 2 - Butene
55. Sodium phenoxide when heated with CO2 under pressure at
49. Which series of reactions correctly represents chemical
125ºC yields a product which on acetylation produces C.
reactions related to iron and its compound?
dil. H 2SO 4 H 2SO 4 , O 2 ONa 125 H
(a) Fe FeSO 4 CO 2 B C
5 Atm Ac2 O
heat
Fe 2 SO4 3
Fe The major product C would be
(a) X (b) Y
(c) I B (d) I
(c) N (d) Y – X 67. If the coefficents of x 3 and x 4 in the expansion of
62. If z is a complex number such that z 2, then the minimum 1 ax bx 2 1 2 x
18
in powers of x are both zero, then
1 (a, b) is equal to:
value of z :
2 272 272
(a) 14, (b) 16,
3 3
5
(a) is strictly greater than 251 251
2 16, 14,
(c) (d)
3 3
3 5
(b) is strictly greater than but less than 68. If 10
9
2 11 108
1
3 11 10
2 7
........
2 2
9 9
5 10 11 k 10 , then k is equal to:
(c) is equal to
2 (a) 100 (b) 110
(d) lie in the interval (1, 2) 121 441
(c) (d)
63. If a R and the equation 10 100
2
69. Three positive numbers form an increasing G. P. If the middle
3 x x 2 x x a2 0 term in this G.P. is doubled, the new numbers are in A.P. then
the common ratio of the G.P. is:
(where [x] denotes the greatest integer x ) has no integral
(a) 2 3 (b) 2 3
solution, then all possible values of a lie in the interval:
(c) 2 3 (d) 3 2
(a) 2, 1 (b) , 2 2,
sin cos 2 x
(c) 1, 0 0,1 (d) 1, 2 70. lim is equal to:
x 0 x2
64. Let and be the roots of equation px 2 qx r 0, (a) (b)
1 1 (c) (d) 1
p 0. If p, q, r are in A.P. and 4, then the value of 2
1
is: 71. If g is the inverse of a function f and f ' x , then
1 x5
34 2 13 g x is equal to:
(a) (b)
9 9 1 5
(a) 5
(b) 1 g x
61 2 17 1 g x
(c) (d)
9 9 (c) 1 + x5 (d) 5x4
2014- 8
72. If f and g are differentiable functions in [0, 1] satisfying 80. The locus of the foot of perpendicular drawn from the centre
f (0) = 2 = g(1), g(0) = 0 and f (1) = 6, then for some c ]0,1[ of the ellipse x2 + 3y2 = 6 on any tangent to it is
(a) f (c) = g (c) (b) f (c) = 2g (c) 2
(a) x2 y2 6 x2 2 y2
(c) 2f (c) = g (c) (d) 2f (c) = 3g (c)
73. If x = –1 and x = 2 are extreme points of
2
f x log x x 2
x then (b) x2 y2 6 x2 2 y2
1 1 2
(a) 2, (b) 2, (c) x2 y2 6x2 2 y2
2 2
1 1 2
(c) 6, (d) 6, (d) x2 y2 6 x2 2 y2
2 2
1
81. Let C be the circle with centre at (1, 1) and radius = 1. If T is the
1 x circle centred at (0, y), passing through origin and touching
74. The integral 1 x e x dx is equal to
x the circle C externally, then the radius of T is equal to
1 1 1 1
(a)
x
(b) x (a) (b)
x 1 e x c xe x c 2 4
x
1 1 3 3
(c) x 1 e x c (d) xe x x c
(c) (d)
2 2
1 1 1
86. Let A and B be two events such that P A B , (a) (b)
6 4 12
1 1
P A B and P A , where A stands for the
4 4 1 1
complement of the event A. Then the events A and B are (c) (d)
6 3
(a) independent but not equally likely.
89. A bird is sitting on the top of a vertical pole 20 m high and its
(b) independent and equally likely. elevation from a point O on the ground is 45 . It flies off
(c) mutually exclusive and independent. horizontally straight away from the point O. After one second,
(d) equally likely but not independent. the elevation of the bird from O is reduced to 30 . Then the
87. The variance of first 50 even natural numbers is speed (in m/s) of the bird is
437 (a) 20 2 (b) 20 3 1
(a) 437 (b)
4
833
(c) (d) 833 (c) 40 2 1 (d) 40 3 2
4
1 90. The statement p q is:
88. Let f k x sin k x cosk x where x R and k 1.
k
(a) a tautology (b) a fallacy
Then f 4 x f 6 x equals (c) eqivalent to p q (d) equivalent to p q
2014- 10
ANSWER KEY
PHYSICS
1 (a) 6 (c) 11 (c) 16 (c) 21 (c) 26 (b)
2 (c) 7 (d) 12 (d) 17 (a) 22 (c) 27 (c)
3 (b) 8 (a) 13 (a) 18 (c) 23 (b) 28 (a)
4 (a) 9 (c) 14 (d) 19 (b) 24 (b) 29 (d)
5 (c) 10 (N) 15 (c) 20 (c) 25 (d) 30 (b)
CHEMISTRY
31 (a) 36 (a) 41 (c) 46 (b) 51 (b) 56 (a)
32 (b) 37 (c) 42 (b) 47 (b) 52 (d) 57 (d)
33 (c) 38 (a) 43 (b) 48 (a) 53 (d) 58 (a)
34 (b) 39 (b) 44 (a) 49 (c) 54 (c) 59 (a)
35 (a) 40 (d) 45 (d) 50 (b) 55 (a) 60 (c)
MATHEMATICS
61 (b) 66 (d) 71 (b) 76 (c) 81 (b) 86 (a)
62 (d) 67 (b) 72 (b) 77 (c) 82 (c) 87 (d)
63 (c) 68 (a) 73 (a) 78 (d) 83 (c) 88 (b)
64 (b) 69 (b) 74 (d) 79 (a) 84 (c) 89 (b)
65 (a) 70 (b) 75 (b) 80 (a) 85 (b) 90 (c)
6. (c) Torque working on th e bob of mass m is, ( R cos R sin )d 2 g ( R cos R sin )d1 g
mg sin . (Direction parallel to plane of rotation
d1 cos sin 1 tan
of particle)
d 2 cos sin 1 tan
10. None of the given option is correct.
When the bubble gets detached,
l l Bouyant force = force due to surface tension
m mg
As is perpendicular to L , direction of L changes
but magnitude remains same.
R
Mv 2
7. (d) 2 F cos 45 F'
R
r
GM 2 GM 2 T×dl
Where F and F '
( 2 R) 2 4R 2 Force due to excess pressure = upthrust
2T
Access pressure in air bubble =
F R
M M
F'
2T 4 R3
F ( r2 ) wg
R R 3T
o 2 R4 w g 2 wg
r2 r R2
3T 3T
M 11. (c) Rate of heat flow is given by,
M
KA( 1 2)
Q=
l
2 GM 2 GM 2 Mv 2
Where, K = coefficient of thermal conductivity
2( R 2)2 4 R2 R
l = length of rod and A = area of cross-section of rod
GM 2 1 1 100°C
Mv 2
R 4 2 Copper
Gm 2 4 1 Gm T
v (1 2 2) B
R 4 2 2 R Brass
Steel
0°C 0°C
stress
8. (a) Young's modulus Y = If the junction temperature is T, then
strain
stress = Y strain QCopper = QBrass + QSteel
Stress in steel wire = Applied pressure
0.92 4(100 T ) 0.26 4 (T 0)
Pressure = stress = Y × strain = +
46 13
L
Strain = T (As length is constant) 0.12 4 (T 0)
L
= 2 × 1011 × 1.1 × 10–5 × 100 = 2.2 × 108 Pa 12
9. (c) Pressure at interface A must be same from both the 200 – 2T = 2T + T
sides to be in equilibrium. T = 40°C
0.92 4 60
QCopper 4.8 cal/s
46
R d2 Rsin 12. (d) In cyclic process, change in total internal energy is
zero.
Rcos R Ucyclic = 0
Rsin – Rsin
A 5R
d1 UBC = nCv T 1 T
2
2014- 12
VA VO [10 x3 ]20 80 J / C
Length of the air column above mercury in the tube is, 17. (a) Electric field in presence of dielectric between the two
plates of a parallel plate capaciator is given by,
P + x = P0
P = (76 – x) E
8 × A × 76 = (76 – x) × A × (54 – x) K 0
x = 38 Then, charge density
Thus, length of air column = 54 – 38 = 16 cm. K 0E
14. (d) In simple harmonic motion, starting from rest,
= 2.2 × 8.85 × 10–12 × 3 × 104 6 × 10–7 C/m2
At t = 0 , x = A 18. (c) Total power consumed by electrical appliances in the
x = Acos t ..... (i) building, Ptotal = 2500W
When t = , x = A – a Watt = Volt × ampere
When t = 2 , x = A –3a 2500 = V × I 2500 = 220 I
From equation (i) 2500
A – a = Acos ......(ii) I 11.36 12A
220
A – 3a = A cos2 ......(iii) (Minimum capacity of main fuse)
As cos2 = 2 cos2 –1 …(iv) 19. (b) Work done in moving the conductor is,
From equation (ii), (iii) and (iv) 2 2
2
W Fdx 3.0 10 4 e 0.2 x 10 3dx
A 3a A a 0 0
2 1
A A
A 3a 2 A2 2a 2 4 Aa A2 l=3m
A A 2 I = 10 A
z
A2 – 3aA = A2 + 2a2 – 4Aa
2a2 = aA x
A = 2a
3 2 0.2 x
a 1 = 9 10 e dx
0
A 2
3
Now, A – a = A cos 9 10 0.2 2 4 0.2 x
= [ e 1] B 3.0 10 e
0.2
A a
cos (By exponential function)
A
9 10 3
1 2 = [1 e 0.4 ]
cos or 0.2
2 T 3
= 9 × 10–3 × (0.33) = 2.97 × 10–3J
T= 6 Power required to move the conductor is,
15. (c) Length of pipe = 85 cm = 0.85m
W
Frequency of oscillations of air column in closed organ P=
pipe is given by, t
3
(2n 1) 2.97 10
f P 2.97 W
3
4L (0.2) 5 10
JEE Main-2013 Solved Paper 2014- 13
1 Bo 2
28. (a) P n
Magnetic energy density = B
2 0 For forward bias, p-side must be at higher potential
Thus, = B than n-side. V ( )Ve
Energy is equally divided between electric and magnetic 29. (d)
field (1) Infrared rays are used to treat muscular strain because
23. (b) By Lens maker's formula for convex lens these are heat rays.
(2) Radiowaves are used for broadcasting because these
1 2 waves have very long wavelength ranging from few
1
f L R centimeters to few hundred kilometers
(3) X-rays are used to defect fracture of bones because
4 they have high penetrating power but they can't
for, L1 , f1 4R
3 penetrate through denser medium like dones.
(4) Ultraviolet rays are absorbed by ozone of the
5
for L2 , f2 5R atmosphere.
3 30. (b) Measured length of rod = 3.50 cm
f2 = (–) ve For vernier scale with 1 Main Scale Division = 1 mm
24. (b) For critical angle c, air 9 Main Scale Division = 10 Vernier Scale Division,
Least count = 1 MSD –1 VSD = 0.1 mm
1 Water
sin c SECTION-2 : CHEMISTRY
For greater wavelength or lesser frequency is less. 31. (a) The electronic configuration of Rubidium (Rb = 37) is
So, critical angle would be more. So, they will not suffer
1s 2 2s 2 2 p 6 3s 2 3 p6 3d 10 4s 2 4 p 6 5s1
reflection and come out at angles less then 90°.
25. (d) According to malus law, intensity of emerging beam is Since last electron enters in 5s orbital
given by, 1
I = I0cos2 Hence n = 5, l = 0, m = 0, s
2
2014- 14
a 1 2 1 2
PV RT 50 1.4 10 70 10
V 280 280
PV a = 2.5 × 10–3 S cm–1
1
RT VRT 1000 2.5 10 3 1000
Now, m
a M 0.5
Hence, Z 1 = 5 S cm mol = 5 × 10–4 S m2 mol–1
2 –1
VRT
36. (a) C 2 H5 OH( ) 3O2 ( g ) 2CO2 ( g ) 3H 2 O( )
33. (c) Bomb calorimeter gives U of the reaction
Cl – Cl –
Given, U = –1364.47 kJ mol–1
Cl – Cl ng = – 1
1 8.314 298
+ H = U + ngRT = 1364.47
Cs 1000
= – 1366.93 kJ mol–1
37. (c) According to Debye Huckle onsager equation,
Cl – Cl
– C B C
– Cl
Cl 38. (a) = i CRT
Relation between radius of cation, anion and edge C2 H5 OH 1 0.500 R T 0.5 RT
length of the cube
Mg3 (PO 4 )2 5 0.100 R T 0.5 RT
2r 2r 3a
Cs Cl
KBr 2 0.250 R T 0.5 RT
3a
r r 4 0.125 RT 0.5 RT
Cs Cl 2 Na 3PO4
1.4 meq. of acid Since the osmotic pressure of all the given solutions is
34. (b) % of N = equal. Hence all are isotonic solution.
mass of organic compound
1
M 39. (b) SO 2 (g) O 2 (g) SO3 (g)
meq. of H2SO4 = 60 2 = 12 2
10
KP KC ( RT ) x
M
meq. of NaOH = 20 2 where x ng number of gaseous moles in product
10
meq. of acid consumed = 12 – 2 = 10 – number of gaseous moles in reactant
1.4 10 1 3 1
% of N = 10% =1 1 1
1.4 2 2 2
JEE Main-2013 Solved Paper 2014- 15
45. (d) The reducing agent loses electron during redox reaction
d [C ]
40. (d) Let rate of reaction = = k[A]x [B]y i.e. oxidises itself.
t
1 2
Now from the given data (a) H 2 O 2 2H 2e 2H 2 O (Re d.)
1.2 × 10 – 3 = k [0.1]x[0.1]y .....(i)
1.2 × 10 – 3 = k [0.1]x[0.2]y .....(ii) 1 0
2.4 × 10 – 3 = k [0.2]x[0.1]y .....(iii) (b) H 2 O2 O 2 2H 2e (Ox.)
Dividing equation (i) by (ii)
1 2
3
1.2 10 k[0.1]x [0.1] y (c) H 2 O2 2e 2 OH (Re d.)
3
1.2 10 k [0.1]x [0.2] y
0
We find, y = 0 (d) H 2 O 2 1 2OH O 2 H 2 O 2e (Ox.)
Now dividing equation (i) by (iii)
46. (b) CsI3 dissociates as CsI3 Cs+ + I3–
3 x y
1.2 10 k[0.1] [0.1] w
3 47. (b) Number of moles of O 2
2.4 10 k [0.2]x [0.1] y 32
We find, x = 1 4w w
Number of moles of N 2
d [C ] 28 7
Hence k[ A]1 [ B ]0
dt w w
Ratio : 7 : 32
41. (c) Acidic strength increases as the oxidation number of 32 7
central atom increases.
Hence acidic strength order is 48. (a) (a) Mn 2 2e Mn; E 0 1.18V ; ... (i)
(+7) (+5) (+3) (+1) (b) Mn3 e Mn 2 ; E 0 1.51V ; ... (ii)
HClO4 > HClO3 > HClO2 > HClO Now multiplying equation (ii) by two and subtracting
42. (b) On electrolysis of aqueous solution of s-block elements from equation (i)
H2 gas discharge at cathode.
3Mn 2 Mn 2Mn 3 ;
1
At cathode: H 2O e H 2 OH E 0 EOx. ERed. = – 1.18 + (– 1.51) = – 2.69 V
2
(–ve value of EMF (i.e. G = +ve) shows that the
43. (b) B reaction is non-spontaneous)
49. (c) In equation (i) Fe2(SO4)3 and in equation (ii) Fe2(SO4)3
V G on decomposing will form oxide instead of Fe.
The correct sequence of reactions is
O2 ,heat Co,600 C
Fe Fe3O 4
R Y
O Fe2 (SO 4 )3 Fe
For a given metal ion, weak field ligands create a complex 50. (b) The complex [CoCl(NH3 )5 ] decomposes under
with smaller , which will absorbs light of longer and acidic medium, so
thus lower frequency. Conservely, stronger field ligands
create a larger , absorb light of shorter and thus
[CoCl(NH3 )5 ] 5H Co 2 5NH 4 Cl
higher v i.e. higher energy. 51. (b) Steric congestion around the carbon atom undergoing
the inversion process will slow down the SN2 reaction,
Red Yellow Green Blue hence less congestion faster will the reaction. So, the
= 650 nm 570 nm 490 nm 450 nm
order is
So order of ligand strength is CH3Cl > (CH3)CH2 – Cl > (CH3)2CH – Cl > (CH3)3CCl
L1 < L3 < L2 < L4 52. (d) R – CH2 – NH2 + CHCl3 + 3KOH (alc)
44. (a) Nitric oxide is paramagnetic in the gaseous state Carbyl amine reaction
because of the presence of one unpaired electron in its R – CH2 – NC + 3KCl + 3H2O
outermost shell. Alkyl isocynide
The electronic configuration of NO is 53. (d) An excellent reagent for oxidation of 1° alcohols to
aldehydes is PCC.
2 *2 2 *2 2 2 2 *1
1s 1s 2 s 2s 2 pz 2 px 2 p y 2 px PCC
R CH 2 OH R CHO
2014- 16
Cl HO – CH2 – CH2 – OH
|
54. (c) 2Cl C CH3 + 6Ag Ethylene glycol
CH3C CCH3 6AgCl
[ ]
| 2 butyne
Cl CO CH 2 CH 2 O
n
1, 1, 1-trichloroethane
Dacron (Polyester)
ONa OH 59. (a) DNA contains ATGC bases
So quinoline is not present in DNA.
55. (a) + CO2
COONa 60. (c) CH3 COOH
LiA1H 4
CH3 CH 2 OH
Sodium
(A)
Phenoxide
PCl5
OH
H2SO4
CH3CH2Cl
COOH (B)
Salicylic acid Alc. KOH
Terephthalic acid
JEE Main-2013 Solved Paper 2014- 17
71. (b) Since f (x) and g(x) are inverse of each other /3
x x
1 x 4cos 4 cos x
g'( f (x)) = 2 0 2 /3
f '( x )
3 3
1 4 4 0 4 4 3 4
3 =
5 f ( x)
g '( f ( x)) 1 x 5 3 2 2 3
1 x
76. (c) Given curves are x2 + y2 = 1 and y2 = 1 –x. Intersecting
Here x = g(y)
points are x = 0, 1
5 5 Area of shaded portion is the required area.
g ( y) 1 g ( y) g ( x) 1 g ( x)
So, Required Area = Area of semi-circle
72. (b) Since, f and g both are continuous function on [0, 1]
+ Area bounded by parabola
and differentiable on (0, 1) then c (0,1) such that
1
r2
f (1) f (0) 6 2 = 2 1 xdx
f (c ) 4 2
1 1 0
g (1) g (0) 2 0 1
and g (c ) 2
= 2 1 x dx ( radius of circle = 1)
1 1 2
0
Thus, we get f (c) 2 g (c)
73. (a) Let f (x) = log | x | + x2 + x 3 1
(1 x ) 2 4 4
= 2 = ( 1) = Sq. unit
Differentiate both side, f ( x ) 2 x 1 2 3 2 3 2 3
x 2 0
Since x = –1 and x = 2 are extreme points therefore 77. (c) Given differential equation is
f ( x ) 0 at these points.
dp(t ) 1
p (t ) 200
Put x = –1 and x = 2 in f ( x ) , we get dt 2
– –2 + 1 = 0 +2 = 1 ...(i) By separating the variable, we get
4 1 0 1
+8 = –2 ...(ii) dp(t) = p(t ) 200 dt
2 2
On solving (i) and (ii), we get
dp(t )
1 dt
6 3 =2 1
2 p (t ) 200
2
1 x 1
74. (d) Let I 1 x e x dx Integrate on both the sides,
x
d ( p (t ))
x 1 1 x 1 dt
= e x dx + x e x dx 1
x p (t ) 200
2
x 1x 1 x 1x 1 x 1
= x.e x 1 e dx x e x dx 1 dp(t )
x 2 x Let p(t ) 200 = s ds
2 2
1
x 1 x 1 1 x 1
d p (t )
= x.e x x e x dx x e x dx
So, dt
x x 1
p (t ) 200
1 2
x
= xe x C
2ds
dt
x x x s
75. (b) Let I = 1 4 sin 2 4sin dx = 2 sin 1 dx
2 2 2
0 0 p (t )
2 log s = t + c 2log 200 t c
/3 2
x x
= 1 2 sin dx 2 sin 1 dx
2 2 1
0 /3 p(t )
200 e 2 k
x 1 x x 5 5 2
sin x , x
2 2 2 6 3 2 6 3 Using given condition p(t) = 400 – 300 et/2
JEE Main-2013 Solved Paper 2014- 19
( x2 y 2 )2 a2 x2 b2 y 2
( x2 y 2 )2 6 x2 2 y2
2 1 2
Now, slope of PS =
13 9 Equation of circle C ( x 1)2 ( y 1)2 1
2
2 Radius of T = | y |
Since, required line is parallel to PS therefore slope of T touches C externally therefore,
required line = slope of PS Now, eqn of line passing Distance between the centres = sum of their radii
2 (0 1)2 ( y 1)2 1 | y |
through (1, –1) and having slope is
9 (0 – 1)2 + (y –1)2 = (1 + |y|)2
2 1 + y2 + 1 – 2y = 1 + y2 + 2| y |
y ( 1) ( x 1) 2 | y | = 1 – 2y
9
9y + 9 = –2x + 2 2x + 9y + 7 = 0 1
If y > 0 then 2y = 1 – 2y y=
79. (a) Given lines are 4
4ax + 2ay + c = 0 If y < 0 then –2y = 1 – 2y 0 = 1
5bx + 2by + d = 0 (not possible)
The point of intersection will be 1
y
x y 1 4
= = 82. (c) Given parabolas are
2 ad 2bc 4ad 5bc 8ab 10ab
y2 = 4x ...(i)
2(ad bc) bc ad x2 = –32y ...(ii)
x
2ab ab Let m be slope of common tangent
Equation of tangent of parabola (1)
5bc 4ad 4ad 5bc
y 1
2ab 2ab y mx ...(i)
m
Point of intersection is in fourth quadrant so x is
positive and y is negative. Equation of tangent of parabola (2)
y = mx + 8m2 ...(ii)
Also distance from axes is same
(i) and (ii) are identical
So x = – y ( distance from x-axis is –y as y is negative)
1
bc ad 5bc 4ad 8m2 m3 m
3bc – 2ad = 0 m 8 2
ab 2ab
ALTERNATIVE METHOD:
x2 y2 1
80. (a) Given equation of ellipse can be written as 1 Let tangent to y2 = 4x be y mx
6 2 m
a2 = 6, b2 = 2 Since this is also tangent to x2 = – 32y
Now, equation of any variable tangent is
1 32
2 2 2 x2 32 mx x2 + 32mx + =0
y mx a m b ...(i) m m
where m is slope of the tangent Now, D = 0
So, equation of perpendicular line drawn from centre to 32
tangent is (32) 2 4 0
m
x
y ...(ii) 4 1
m m3 m
32 2
2014- 20
a 1 b 3 c 4 We know, P ( A B) P( A) P ( B ) P ( A B)
83. (c) (let)
2 1 1 5 3 1 1
a=2 +1 A (1, 3, 4) P( B ) P( A B)
6 4 4 4
b=3–
c=4+ 1
3i j 5k P( B)
3
a 1 b 3 c 4 P P ( A) P ( B ) so they are not equally likely.
P , ,
2 2 2 3i j 5k
3 1 1
A Also P ( A) P( B ) P( A B)
6 8 4 3 4
= 1, , (a, b, c) So A & B are independent.
2 2
87. (d) First 50 even natural numbers are 2, 4 , 6 ....., 100
6 8
2( 1) 3 0 xi2
2 2 Variance = ( x) 2
3 +6=0 = –2 N
2
a = –3, b = 5, c = 2 2 22 42 ... 100 2 2 4 ... 100
x 3 y 5 z 2 50 50
Required line is
3 1 5 4(12 22
32 .... 50 2 )
84. (c) Given, l + m + n = 0 and l 2 = m2 + n2 (51)2
50
Now, (–m –n)2 = m2 + n2
50 51 101
mn = 0 4 (51)2 = 3434 – 2601 2 = 833
m = 0 or n = 0 50 6
If m = 0 then l = –n 1
88. (b) Let f k ( x) (sin k x cosk x )
1 k
We know l 2 + m2 + n2 = 1 n Consider
2
1 1
1 1 f 4 ( x) (sin 4 x cos 4 x)
f 6 ( x) (sin 6 x cos6 x)
i.e. (l1, m1, n1) , 0, 4 6
2 2 1 1
[1 2 sin 2 x cos 2 x] [1 3sin 2 x cos 2 x]
If n = 0 then l = –m 4 6
l 2 + m2 + n2 = 1 2m2 = 1 1 1 1
1 4 6 12
m 89. (b) Let the speed be y m/sec.
2
Let AC be the vertical pole of height 20 m.
1 Let O be the point on the ground such that AOC = 45°
Let m Let OC = x
2 A B
1 Time t = 1 s
l and n = 0 From AOC, 20
45° 20
2
20 30°
tan 45 ...(i) O x C y D
1 1 x
(l2, m2, n2 ) , ,0
2 2 20
and from BOD, tan 30 ...(ii)
1 x y
cos
2 3 1 20
From (i) and (ii), we have x = 20 and
85. (b) L.H.S = (a b).[(b c) (c a)] 3 x y
1 20
= (a b).[(b c. a )c (b c.c )a ] 20 y 20 3
3 20 y
= (a b).[[b c a ] c] [ b c. c 0] So, y 20( 3 1)
2
= [a b c].(a b. c ) [a b c ] i.e., speed 20( 3 1) m/s
= [a b b c c a ] [a b c ]2 90. (c) p q ~q p ~ q ~ (p ~ q)
F F T F T
So = 1
F T F T F
1 1 5
86. (a) Given, P ( A B) P( A B) 1 T F T T F
6 6 6
T T F F T
1 1 3 Clearly equivalent to p q
P ( A) P ( A) 1
4 4 4
1
Time : 3 hrs. INSTRUCTIONS Max. Marks : 360
Negative Marking : One fourth (¼) marks will be deducted for indicating incorrect response of each question.
. . . Best of Luck
:
H O : : H O : :
| || | ||
A. H–C –C B. H–C –C
| | +
: :
O– H O– H
B
H B
H
:
CH3
:
H O : :
| |
1. IUPAC name of is C. H–C –C
CH2CH3 | +
: :
O– H
B
H
(a) 1-methyl-3 ethyl cyclohexane (a) A (b) B
(b) 1-ethyl-3 methyl benzene (c) C (d) All have same energy
(c) 1-ethyl-3 methyl cyclo hexane 3. The hybridization of atomic orbitals of nitrogen in
(d) Cyclo hexane-1-ethyl-3-methyl NO 2 , NO 2 and NH 4 are
2. Which of the following resonance structure is lowest in (a) sp2, sp3 and sp2 respectively
energy? (b) sp, sp2 and sp3 respectively
(c) sp2, sp and sp3 respectively
(d) sp2, sp3 and sp respectivley
FT-2 The Pattern Target AIEEE
4. In a compound AOH, electronegativity of ‘A’ is 2.1, the Ka
10. Given, HF H 2 O H 3O F— ;
compound would be
(a) Acidic K
F— H 2O b HF OH — .
(b) Neutral towards acid & base Which relation is correct
(c) Basic
1
(d) Amphoteric (a) Kb = Kw (b) Kb
Kw
5. Which of the following orders is wrong?
(a) Electron affinity– N < O < F < Cl Ka
(c) Ka × Kb = Kw (d) Kw
(b) Ist ionisation potential – Be < B < N < O Kb
(c) Basic property– MgO < CaO < FeO < Fe2O3 11. In an amino acid, the carboxyl group ionises at pK a1 = 2.34
(d) Reactivity–Be < Li < K < Cs Cl
and ammonium ion at pK a 2 = 9.60. The isoelectric point of
6. The dipole moment of chlorobenzene is 1.5 D. The the amino acid is at pH
(a) 5.97 (b) 2.34
Cl (c) 9.60 (d) 6.97
Cl 12. AB, A2 and B2 are diatomic molecules. If the bond enthalpies
dipole moment of is of A2, AB and B2 are in the ratio 1:1 :0.5 and enthalpy of
Cl Cl formation of AB from A2 and B2 is –100 kJ mol–1 . What is
(a) 2.86D (b) 2.25D the bond energy of A2 :
(c) 1.5D (d) 0D (a) 200 kJ mol–1 (b) 100 kJ mol–1
7. Following substances are in solid state : (c) 300 kJ mol–1 (d) 400 kJ mol–1
(A) Methane (B) Cesium chloride 13. Equal volume of 0.1 M urea and 0.1 M glucose are mixed.
(C) Ice (D) Lithium The mixture will have
Which non-conductive solid when melts converts into (a) Lower osmotic pressure
conductive liquid? (b) Same osmotic pressure
(a) C, D (b) Only C (c) Higher osmotic pressure
(c) Only B (d) A, B and C (d) None of these
14. If the following half cells have the E° values as
8. On applying pressure to the equilibrium
ice water, which phenomenon will happen Fe 3 e – Fe 2 ; E° = + 0.77V and Fe 2 2e Fe ;
(a) More ice will be formed E° = – 0.44V. The E° of the half cell Fe 3 3e – Fe will be
(b) More water will be formed (a) 0.33 V (b) 1.21 V
(c) Equilibrium will not be disturbed (c) 0.04 V (d) 0.605 V
(d) Water will evaporate 15. The oxidation states of sulphur in the anions
9. The energy levels for ( Z 1) can be given by SO 32 , S2 O 24 and S2 O 62 follow the order
ZA
(a) En for A(+Z–1) = Z2 × En for H (a) SO 32 S 2O 24 S2 O62
(b) En for A(+Z–1) = Z × En for H
(b) S 2O 24 S2 O 62 SO 32
1
(c) En for A(+Z–1) = × En for H
Z2 (c) S 2 O 62 S2 O 24 SO 32
1
(d) En for A(+Z–1) = × En for H (d) S 2O 24 SO32 S2 O62
Z
FT-3
16. Which of the following statements is not correct? 22. An aqueous solution of colourless metal sulphate M gives
(a) C – Cl bond in vinyl chloride is less polar than in CH3Cl a white precipitate with NH4OH. This was soluble in excess
(b) C – Cl bond in vinyl chloride is stronger than in CH3Cl of NH4OH. On passing H2S through this solution a white
(c) C – Cl bond in vinyl chloride is shorter than in CH3Cl ppt. is formed. The metal M in the salt is
(d) Vinyle chloride undergo nucleophilic substitution more (a) Ca (b) Ba
readily than CH3Cl. (c) Al (d) Zn
17. Which of the following will not be soluble in sodium
carbonate solution? 23. A laboratory reagent imparts green colour to the flame. On
heating with solid K2Cr2O7 and conc. H2SO4 it evolves a red
OH gas. Identify the reagent
O2N NO2 COOH
(a) CaCl2 (b) BaCl2
(a) (b) (c) CuCl2 (d) None of these
24. Which of the following compound can not used in
NO2 preparation of iodoform?
(a) CH3CHO (b) CH3COCH3
OH SO2OH (c) HCHO (d) 2-propanol
NO2 25. The correct priorities for the substituents shown below,
(c) (d) according to the E-Z sequence rule is
I. – CN II. – CBr (CH 3 ) 2
18. Although Al has a high oxidation potential it resists corrosion
because of the formation of a tough, protective coat of O
||
(a) Al(NO3)2 (b) AlN III. – COOH IV. – CH 2 C OCH 3
(c) Al2O3 (d) Al2(CO3)2
19. A metal which is not affected by conc. H2SO4, HNO3 or O
||
alkalis forms a compound X. This compound X can be used V. – C H
to give a complex which finds its application for toning in
photography? The metal is (a) II, III, V, I, IV (b) V, II, I, IV, III
(a) Au (b) Ag (c) III, IV, I, II, V (d) II, V, I, IV, III
(c) Hg (d) Cu 26. Identify X in the sequence given :
20. If Cl2 gas is passed into aqueous solution of KI containing
some CCl4 and the mixture is shaken, then”. NH2
(a) Upper layer becomes violet CHCl3 HCl
(Y) X + methanoic acid
(b) Lower layer becomes violet KOH (300 K)
(c) Homogenous violet layer is formed
(d) None Cl
21. In Lassaigne’s test, the organic compound is fused with a
piece of sodium metal in order to (a) NH2 (b) C N Cl
Cl
(a) increase the ionisation of the compound
(b) decrease the melting point of the compound
(c) increase the reactivity of the compound
(d) convert the covalent compound into a mixture of ionic (c) N C Cl (d) CH3–NH Cl
compounds
FT-4 The Pattern Target AIEEE
27. Select the rate law that corresponds to the data shown for
the following reaction A B C
Expt. No. (1) (2) Initial Rate
1 0.012 0.035 0.10
31. A d.c. supply of 120V is connected to a large resistance X. A
2 0.024 0.070 0.80
voltmeter of resistance 10k placed in series in the circuit
3 0.024 0.035 0.10
reads 4V. What is the value of X?
4 0.012 0.070 0.80
(a) Rate = K[B]3 (b) Rate = K [B]4 (a) 190 k (b) 90 k
(c) Rate = K [A] [B]3 (d) Rate = K [A]2 [B]2 (c) 290 k (d) 390 k
28. An alkene upon ozonolysis yield 32. An insect trapped in a circular groove of radius 12 cm. moves
CHO – CH2– CH2– CH2 – CHO only. The alkene is along the groove steadily and completes 7 revolutions in
(a) CH2= CH – CH2 –– CH2 –– CH2 –– CH2 –– CH3 100 sec. What is the linear speed of the motion
(a) 2.3 cm/sec. (b) 5.3 cm/sec.
(b)
(c) 0.44 cm/sec. (d) None of these
33. The period of the satellite of the earth orbiting very near to
the surface of the earth is T0. What is the period of the
(c) geostationary satellite in terms of T0
T0
(a) (B) 7T0
7
(d)
(c) 7T0 (d) 7 7T0
29. 1 mol of N 2 and 3 mol of H 2 are placed in a closed container 34. In “Al” and “Si”. If temperature is changed from normal
at a pressure of 4 atm. The pressure falls to 3 atm at the same temperature to 70 K then
temperature when the following equilibrium is attained (a) The resistance of Al will increase and that of Si will
decrease
N 2 (g ) 3H 2 (g) 2 NH 3 (g ) .
(b) The resistance of Al will decrease and that of Si will
The K p for the dissociation of NH 3 is increase
(c) Resistance of both decrease
3 3 (d) Resistnance of both increase
(a) atm 2 (b) 0.5 (1.5) 3 atm 2
0.5 (1.5)3 35. For Doppler’s effect, which of the following is a correct
statement
(1.5) 3 (I) During observer motion both frequency and
0.5 (1.5)3 2
(c) atm 2 (d) atm wavelength will change
3 3 0.5
(II) During source motion both frequency and wavelength
30. 0.5 g mixture of K2Cr2O7 and KMnO4 was treated with excess will change
of KI in acidic medium. I2 liberated required 100 cm3 of 0.15N. (III) During source motion only frequency will change
Na2S2O3 solution for titration. The percentage amount of (IV) During observer motion only frequency will change
K2Cr2O7 in the mixture is (a) I, II (b) I, II
(a) 85.36 % (b) 14.64 % (c) II, III (d) II, IV
(c) 58.63 % (d) 26.14 %
FT-5
36. When white light passes through a dispersive medium, it 41. The circular head of a screw gauge is divided into 200
breaks up into various colours. Which of the following divisions and move 1 mm ahead in one revolution. If the
statements is true? same instrument has a zero error of –0.05 mm and the reading
(a) Velocity of light for violet is greater than the velocity of on the main scale in measuring diameter of a wire is 6 mm
light for red colour. and that on circular scale is 45. The diameter of the wire is
(b) Velocity of light for violet is less than the velocity of
(a) 6.275 mm (b) 6.375 mm
light for red.
(c) 5.75 mm (d) 5.50 mm
(c) Velocity of light is the same for all colours
(d) Velocity of light for differnet colours has nothing to do 42. Ammeter and voltmeter readings were recorded as 0.25 A
with the phenomenon of dispersion and 0.5 V during the experiment to determine the resistance
37. An engine has an efficiency of 1/6. When the temperature of a given wire using Ohm's law. The correct value of the
of sink is reduced by 62°C, its efficiency is doubled. resistance is
Temperatures of source and sink are (a) 0.125 (b) 2
(a) 99°C, 37°C (b) 124°C, 62°C (c) 0.75 (d) 0.25
(c) 37°C, 99°C (d) 62°C, 124°C 43. A wheel is rotating at 900 r.p.m. about its axis. When power
38. Two isothermals are shown in figure at temperatures T1 and is cut off it comes to rest in 1 minute. The angular retardation
T2. Which of the following relations is correct? in rad/s2 is
(a) /2 (b) /4
(c) /6 (d) /8
44. Choose the wrong statement from the following
P (a) a metre bridge can be used to determine the emf of a
primary cell
(b) a metre bridge works on the principle of Wheatstone
T1
bridge
T2 (c) a potentiometer is used to compare the emfs of two
primary cells
V (d) a multimeter can be used as an ohmmeter
45. The adjacent figure shows the position graph of one
(a) T1 > T2 (b) T1 < T2 dimensional motion of a particle of mass 4 kg. The impulse
1 at t =0 s and t = 4 is given respectively as:
(c) T1 = T2 (d) T1 = T
2 2
39. During the experiment to determine the resistivity of a wire x(m)
by a metre bridge, the jockey is moved gently along the wire
from left to right to 3
(a) find a deflection in the galvanometer towards left
(b) find a deflection in the galvanometer towards right
(c) find no deflection in the galvanometer
(d) find a maximum deflection in the galvanometer 0 4 t(s)
40. The radius of curvature of a thin plano-convex lens is 10 cm
(of curved surface) and the refractive index is 1.5. If the (a) 0,0
plane surface is silvered, then it behaves like a concave
(b) 0, – 3 kg ms – 1
mirror of focal length
(a) 10 cm (b) 15 cm (c) + 3 kg ms–1, 0
(c) 20 cm (d) 5 cm (d) + 3 kg ms– 1, – 3 kg. ms–1
FT-6 The Pattern Target AIEEE
46. In case of a p-n junction diode at high value of reverse bias,
the current rises sharply. The value of reverse bias is known
as 10kg
(a) cut off voltage (b) zener voltage
(c) inverse voltage (d) critical voltage µ=0.6
47. The following observations were taken while comparing 15 kg
the e.m.f.s of two primary cells using a potentiometer.
Balance point when E 1 (Leclanche cell) in the circuit F = 80N
( 1 in cm) = 350
Balance point when E2 (Daniel cell) is the circuit ( 2 in cm) Smooth
= 275
The ratio (E1/E2) of emfs is approximately 51. The magnitude of acceleration of the 10 kg block is
(a) 1.30 (b) 1.17 (a) 3.2 m/s² (b) 2.0 m/s²
(c) 1.27 (d) 1.5 (c) 1.6 m/s² (d) 0.8 m/s²
48. A photon materializes into an electron-positron pair. The 52. If applied force F = 120 N, then magnitude of acceleration of
kinetic energy of the electron is found to be 0.19 MeV. What 15 kg block will be –
was the energy of the photon? (a) 8 m/s² (b) 4 m/s²
(a) 0.38 MeV (b) 0.70 MeV (c) 3.2 m/s² (d) 4.8 m/s²
(c) 1.40 MeV (d) None
49. If 10% of a radioactive material decays in 5 days, then the ASSERTION & REASON
amount of the original material left after 20 days is DIRECTIONS (Qs. 53 and 54) : Each of these questions
approximately contains two statements: Statement-1(Assertion) and
(a) 60% (b) 65% (c) 70% (d) 75% Statement-2 (Reason). Choose the correct answer (ONLY
50. Lights of two different frequencies, whose photons have ONE option is correct) from the following -
energies 1 eV and 2.5 eV respectively, successively illuminate (a) Statement-1 is false, Statement-2 is true.
a metal whose work function is 0.5 eV. The ratio of the (b) Statement-1 is true, Statement-2 is true; Statement-2
maximum speeds of the emitted electrons will be is a correct explanation for Statement-1.
(a) 1 : 5 (b) 1 : 4 (c) 1 : 2 (d) 1 : 1 (c) Statement-1 is true, Statement-2 is true; Statement-2
is not a correct explanation for Statement-1.
DIRECTIONS (Qs. 51 and 52) : Read the following passage (d) Statement-1 is true, Statement-2 is false.
and answer the questions that follows :
53. Statement-1 : When two semiconductor of p and n type are
A block of mass 15 kg is placed over a frictionless horizontal brought in contact, they form p-n junction which act like a
surface. Another block of mass 10 kg is placed over it, that is rectifier.
connected with a light string passing over two pulleys fastened Statement-2 : A rectifier is used to convert alternating current
to the 15 kg block. A force F = 80N is applied horizontally to the into direct current.
free end of the string. Friction coefficient between two blocks is 54. Statement 1 : A nucleus at rest splits into two nuclear parts
0.6. The portion of the string between 10 kg block and the upper having radii in the ratio 1 : 2. Their velocities will be in the
pulley is horizontal. Pulley, string and connecting rods are ratio 8 : 1.
massless. (Take g = 10 m/s²) Statement 2 : The radius of a nucleus is proportional to the
cube root of its mass number.
FT-7
55. A plate of mass (M) is placed on a horizontal frictionless 59. Two long parallel wires P and Q are both perpendicular to
surface and a body of mass (m) is placed on this plate. The the plane of the paper with distance of 5 m between them. If
coefficient of dynamic friction between this body and the P and Q carry current of 2.5 amp and 5 amp respectively in
plate is . If a force 3 mg is applied to the body of mass (m) the same direction, then the magnetic field at a point half-
along the horizontal, the acceleration of the plate will be way between the wires is
m mg 3 0
(a) g (b) 0
M M m (a) (b)
2
mg mg
(c) (d) 3 0 0
M M m (c) (d)
2 2
56. Two rods of length d1 and d2 and coefficients of thermal
conductivities K1 and K2 are kept touching each other. Both 60. In a circuit L, C and R are connected in series with an
have the same area of cross-section the equivalent of thermal alternating voltage source of frequency f. The current leads
conductivity is the voltage by 45°. The value of C is
(a) K1 + K2 (b) K1d1 + K2d2
1 1
d1K1 d 2 K 2 d1 d 2 (a) f (2 fL R )
(b) 2 f (2 fL R )
(c) (d)
d1 d 2 (d1 / K1 d 2 / K 2 )
1 1
57. Two masses A and B of 10 kg and 5 kg respectively are (c) f (2 fL R )
(d) 2 f (2 fL R )
connected with a string passing over a frictionless pulley
fixed at the corner of a table (as shown in figure). The
coefficient of friction between the table and the block is 0.2.
The minimum mass of C that may be placed on A to prevent
it from moving is equal to
C 61. Area of triangle formed by the vertices (0, 0), (6, 0), (4, 3) is
A (a) 6 (b) 9
(c) 18 (d) 24
62. In a box containing 100 bulbs, 10 are faulty. The probability
B that from a sample of 5 bulbs none are defective.
5 5
(a) 15 kg (b) 10 kg 1 9
(a) (b)
(c) 5 kg (d) 0 kg 10 10
58. A particle starts S.H.M. from the mean position. Its amplitude
is a and total energy E. At one instant5 its kinetic energy is 9 1
(c) (d)
3 E/4, its displacement at this instant is 10 5 5
(a) 16 (b) 17
C7 C7 41 30
(c) (d)
17 16 91 91
(c) C8 (d) C8
73. The eccentricity of the ellipse represented by
d5y 25x2 + 16y2 – 150x – 175 = 0 is
67. If y = A sin t then
dt 5 2 3
(a) (b)
5 5 5 5
(a) A cos t (b) A sin t
2 2
4 1
(c) (d)
5 5
5 5
(c) A cos t (d) A sin t
2 2 74. If f(x) = | x – 2| and g(x) = f (f(x)) then for x > 10, g'(x) equal
(a) –1 (b) 0
68. for which x 2 y 4 z 8 and x 1 y 2 z 3
(c) 1 (d) 2x – 4
3 7 3
75. If a, b, c are in A.P., b, c, d are in G.P. and c, d, e are in H.P. then
are perpendicular equals
a, c, e are in
(a) 5 (b) 6 (c) 7 (d) 8
(a) A.P. (b) G.P.
x 1 y 3 z 4 (c) H.P. (d) None
69. The angle between the two lines &
2 2 1
15
1
x 4 y 4 z 1 76. The coefficient of x10 in the expansion of 3x 2 is
is x2
1 2 2
1 4 1 3 15! 10 15! 10
(a) cos (b) cos (a) 3 (b) – 3
9 9 10! 5! 10! 5!
1 2 1 1 15! 35 15! 8
(c) cos (d) cos (c) (d) 3
9 9 10! 5! 7! 8!
FT-9
77. The mean of discrete observations y1, y2, y3,...,yn is given Statement-2 : ln x is an increasing function.
by
n n 81. Statement–1 : Range of f(x) = 4 x 2 is [0, 2]
yi yi
(a) i 1 (b) i 1 Statement–2 : f(x) is increasing for 0 x 2 and decreasing
n
n
i
for – 2 x 0.
i 1 82. Let x, y, z are three integers lying between 1 and 9 such that
n n x 51, y 41 and z 31 are three digit numbers.
y if i y if i
(c) i 1 (d) i 1 Statement-1 : The value of the determinant
n n
fi
i 1 5 4 3
dx x51 y41 z31
78. is is zero.
(x ) (x ) x x y z
3 3 1
(c) (d)
2 2 3
2
Time : 3 hrs. INSTRUCTIONS Max. Marks : 360
Negative Marking : One fourth (¼) marks will be deducted for indicating incorrect response of each question.
. . . Best of Luck
COOEt
O
O
EtONa
O OH
(c) (d)
COOEt
(a) Knovengel reaction
(b) Perkin reaction
(c) Reformatsky reaction OH
(d) Dieckmann’s condensation. 16. To obtain silver from silver amalgam it is heated in vessel
12. Choose the correct options – which is made of –
(a) The change of atmospheric temperature with altitude (a) Cu (b) Fe
is called the lapse rate. (c) Ni (d) Zn
(b) The gases responsible for greenhouse effects are CO2, 17. Which of the following compounds does not give nitrogen
Water vapour, CH4 and ozone. gas on heating –
(c) The order of contribution to the acid rain is (a) NH4NO2 (b) (NH4)2SO4
H2SO4 < HNO3 < HCl
(c) NH4ClO4 (d) (NH4)2Cr2O7
(d) Both (a) and (b)
FT-13
18. Choose the correct options – 22. Choose the incorrect option –
(a) The size of the colloidal particles is in between the size (a) When a dilute solution of an acid is added to a dilute
of the molecule and the size of the particle of a coarse solution of a base, neutralization reaction takes place.
suspension. (b) In acid-base titrations, at the end point, the amount of
(b) White portion of the egg is condensed colloidal solution acid becomes chemically equivalent to the amount of
whereas the sol prepared from this substance is a dilute base present.
colloidal solution. (c) In the case of a strong acid and a strong base titration,
(c) Dialysis is a slow process. at the end point the solution becomes acidic.
(d) All of these (d) In acid-base titrations the end point is determined by
19. Choose the correct option for the hydrogen ion concentration of the solution.
23. Light of wavelength strikes a metal surface with intensity
Cl
‘x’ and the metal emits ‘y’ electrons per second of maximum
energy ‘z’ . What will happen to ‘y’ and ‘z’ if ‘x’ is halved?
HBr (a) y will remain same (b) y will be doubled
I
HBr (c) z will remain the same (d) z will halved
II
Perdoxide
NO2
(a) I and II are identical
(b) I and II are different 24.
Br2 CH3ONa
(c) Mechanism of formation of I and II are same X Y
Fe, CH3OH,
(d) None of these
HgSO4 dil. OH
20. HC CH (A) (B) Cl
H2SO 4 5ºC
Product (Y) of this reaction is –
Give the IUPAC name of “B” is –
(a) 2-butanal (b) 3-hydroxy butanal NO2 NO2
(c) 3-formyl-2-propanol (d) 4-oxo-2-propanol
(a) (b)
Br2 / h Alcoholic
21. Major (X) OCH3 Br
KOH /
Cl OCH3
H Br
Major (Y) Major (Z) NO2 NO2
Peroxide
Major final product (Z) is –
(c) (d)
Br
OCH3
(a) (b)
OCH3
Br
(i) NaOI SOCl 2 C6 H5 NH 2
Br 25. A B C
(ii) H NaOH
Br Conc. HNO3
D E (Major) E can be –
(c) (d) Conc. H 2SO4
(1 equivalent)
FT-14 The Pattern Target AIEEE
O
(c) t 1/2 (d) t 3/4
NH
(a)
[A] 0 [A] 0
NO2 28. This question contains Statement-1 and Statement-2. Of the
four choices given after the statements, choose the one that
NO2 best describes the two statements.
O Statement 1 : Anhydrous copper (II) chloride is a covalent
while anhydrous copper (II) fluoride is ionic in nature.
NH Statement 2 : In halides of transition metals, the ionic
(b) character decreases with increase in atomic mass of the
halogen.
(a) Statement-1 is false, Statement-2 is true.
(b) Statement-1 is true, statement-2 is true and statement-
2 is correct explanation for statement-1
O
(c) Statement-1 is true, statement-2 is true and statement-
2 is NOT correct explanation for statement-1
NH
(c) (d) Statement-1 is true, Statement-2 is false.
NO2 29. At 300K, the vapour pressure of an ideal solution containing
3 mole of A and 2 mole of B is 600 torr. At the same
temperature, if 1.5 mole of A and 0.5 mole of C (non-volatile)
H
are added to this solution the vapour pressure of solution
(d) O2N NH2 increases by 30 torr. What is the value of PB0 ?
(a) 940 (b) 405
Cl
(c) 90 (d) None of these
26. The fH°(N2O5, g) in kJ/mol on the basis of the following
30. The standard redox potentials Eº of the following systems
data is –
are :
2NO(g) + O2(g) 2NO2(g) ; rH° = –114 kJ/mol
4NO2(g) + O2(g) 2N2O5(g) ; rH° = –102.6 kJ/mol System Eº (volts)
H°(NO, g) = 90.2 kJ/mol (i) MnO4– + 8H+ + 5e Mn2+ + 4H2O 1.51
f
(a) 15.1 (b) 30.2 (ii) Sn4+ + 2e Sn2+ 0.15
(c) –36.2 (d) None of these (iii) Cr2O7 + 14H+ + 6e
2– 2Cr3+ + 7H2O 1.33
27. Which graph represents zero order reaction [A(g) B(g)] (iv) Ce4+ + e Ce3+ 1.61
The oxidising power of the various species decreases in the
order-
(a) Ce4+ > Cr2O72– > Sn4+ > MnO4–
(a) [B] (b) d[B] (b) Ce4+ > MnO4– > Cr2O72– > Sn4+
dt (c) Cr2O72– > Sn4+ > Ce4+ + MnO4–
(d) MnO4– > Ce4+ > Sn4+ > Cr2O72–
t t
FT-15
(b) 10 4 13 1
5 B 2 He 7 N 1H
(c) 10 1 11
(a) mg (b) mg/2 5 B 0n 5 B
(d) 14 1 12
(c) mg / 3 (d) 2mg 7 N 1H 6 C
FT-16 The Pattern Target AIEEE
38. For the determination of the focal length of a convex mirror, 42. An electron and a proton each travels with equal speed
a convex lens is required because around circular orbits in the same uniform magnetic field as
(a) it is not possible to obtain the image produced by a indicated (not to scale) in the diagram. The field is into the
convex mirror on the screen page on the diagram. The electron travels ............... around
(b) a convex lens has high resolving power so it helps to the ............ circle and the proton travels ............... around the
measure the focal length correctly ............... circle.
(c) a convex mirror always forms a real image which is
diminished by the convex lens
(d) none of these
39. Choose the wrong statement regarding the experiment to
identify a diode, an LED, a resistor and a capacitor B
(a) when pointer moves in one way when voltage is applied
and doesn't move when reversed and there is light
emission, the item is LED (a) clockwise, smaller, counter-clockwise, larger
(b) when pointer moves in one way when voltage is applied (b) counter-clockwise, smaller, clockwise, larger
and doesn't move when reversed and there is no (c) clockwise, larger, counter-clockwise, smaller
emission of light, the item is a diode (d) counter-clockwise, larger, clockwise, smaller
(c) when pointer moves in one way when voltage is applied 43. In the screw gauge, the screw should be moved in the same
and not when reversed, the item is a resistor direction to avoid which of the following error?
(d) none of these (a) Zero error (b) Parallel error
40. The activity of a radioactive substance drops to 1/32 of its (c) Back lash error (d) Index error
initial value in 7.5 h. Find the half life. q
44. Consider Gauss’s law E.dA . Which of the following
(a) 2 hours (b) 1.5 hours 0
(c) 1 hours (d) 2.5 hours is true ?
41. A ring of mass M and radius R lies in x-y plane with its
centre at origin as shown. The mass distribution of ring is
+ – +
nun-uniform such that at any point P on the ring, the mass
per unit length is given by = 0 cos2 (where 0 is a
positive constant). Then the moment of inertia of the ring
about z-axis is – (a) E must be the electric field due to the enclosed charge
y
(b) If net charge inside the Gaussian surface = 0, then E
must be zero everywhere over the Gaussian surface.
M P (c) If the only charge inside the Gaussian surface is an
R electric dipole, then the integral is zero.
x (d) E is parallel to dA everywhere over the Gaussian
surface.
45. Six resistors of 10 each are connected as shown. The
equivalent resistance between points X and Y is –
X
1
(a) MR2 (b) MR 2
2
Y
1M 1 M (a) 20 (b) 5
(c) R (d) R
2 0 0 (c) 25/3 (d) 10
FT-17
46. What happens when we do not stir the mixture continuously Statement 1 : The temperature of gas continuously
in the experiment to determine the specific heat capacity of increases.
a given solid ?
Statement 2 : According to first law of thermodynamics dQ
(a) Temperatue of mixture is increased
= dU + dW where symbols have their usual meaning.
(b) Temperature of mixture is not constant
(c) Temperature of mixture is constant 49. Statement 1 : In YDSE, as shown in figure, central bright
(d) None of these fringe is formed at O. If a liquid is filled between plane of
slits and screen, the central bright fringe is shifted in upward
ASSERTION & REASON direction.
Hg 76cm.
51. The electric potential V is given as a function of distance x
by V = (5x2 + 10x – 9) volt. Value of electric field at x = 1m is
(a) –20 V/m (b) 6 V/m
76cm.
(c) 11 V/m (d) +20 V/m
FT-18 The Pattern Target AIEEE
56. A charged particle of mass m and having a charge Q is placed
DIRECTIONS (Qs. 52 - 54) : Read the following passage in an electric field E which varies with time as E = E0 sin t.
and answer the questions that follows : What is the amplitude of the S.H.M. executed by the particle?
(a) 5 kJ (b) 25 kJ
m M (c) 15 kJ (d) 20 kJ
37° 59. A body is executing simple harmonic motion. At a
///////////////////////////////////////// displacement x from mean position, its potential energy is
E1 = 2J and at a displacement y from mean position, its
3 4
(a) m (b) m potential energy is E2 = 8J. The potential energy E at a
5 5 displacement (x + y) from mean position is
6 3 (a) 10J (b) 14J
(c) m (d) m
5 2 (c) 18J (d) 4J
FT-19
60. A thin convex lens of focal length 10cm and refractive index
2 y z
1.5 is cut vertically into two equal pieces. They are placed as
shown with a liquid of refractive index 3 between them. What 64. If x 2, y 2, z 2 and x 2 z 0 , then the value of
is the focal length of the combination ? x y 2
2 y z
2 x 2 y 2 z
(a) 1 (b) 0
(c) 3 (d) 4
65. Box contains 2 one rupee, 2 five rupee, 2 ten rupee and 2 twenty
(a) –10 cm. (b) –10/4 cm.
rupee coin. Two coins are drawn at random simultaneously.
(c) –10/3 cm. (d) None The probability that their sum is Rs. 20 or more, is
(a) 1/4 (b) 1/2
(c) 3/4 (d) 1/8
66. The equation (5x – 1)2 + (5y – 2)2 = ( 2 – 4 + 4) (3x + 4y – 1)2
61. Equation of straight line ax + by + c = 0 where 3a+4b+ c = 0, represents an ellipse if
which is at maximum distance from (1, –2), is (a) (0, 1] (b) (–1, 2)
(a) 3x + y – 17 = 0 (b) 4x + 3y – 24 = 0 (c) (2, 3) (d) (–1, 0)
(c) 3x + 4y – 25 = 0 (d) x + 3y – 15 = 0
2
d 501
67. The value of the definite integral,
1 tan K
10 x 2 if 3 x 3 1
x2 – (k – 2) x + (k2 + 3k + 5) = 0 (k R) .
ASSERTION & REASON
80. Statement-1 : If | z1 | = 30, | z2 – (12 + 5i) | = 6, then maximum 85. Consider the following statements :
value of | z1 – z2 | is 49.
S1 : Number of integrals values of ‘a’ for which the roots of
Statement-2 : If z1, z2 are two complex numbers, then the equation x2 + ax + 7 = 0 are imaginary with positive real
| z1 – z2 | | z1 | + | z2 | and equality holds when origin, z1 and parts is 5.
z2 are collinear and z1, z2 are on the opposite side of the S2 : Let , are roots x2 – (a + 3) x + 5 = 0 and , a, are in
origin. A.P. then roots are 2 and 5/2
81. Consider the family of straight lines 2x sin 2 + y cos2 = 2 S3 : Solution set of logx (2 + x) logx (6 – x) is (1, 2]
cos 2
State, in order, whether S1, S2, S3 are true or false.
Statement-1 : All the lines of the given family pass through
(a) FFT (b) TFT
the point (3, –2).
(c) TFF (d) TTT
Statement-2 : All the lines of the given family pass through
a fixed point. 86. If the substitution x = tan –1 (t) transforms the differential
/4 d2 y dy
dx equation 2
xy sec 2 x 0 into a differential
82. Consider I = dx dx
/4
1 sin x
d2 y dy
equation (1 t 2 ) 2
(2t y tan 1 (t)) k then k is
Statement-1 : I = 0 dt dt
equal to
because
(a) –2 (b) 2
a
(c) –1 (d) 0
Statement-2 : f (x) dx 0 , wherever f (x) is an odd
a 87. Let f : R R and fn (x) = f (fn–1 (x)) n 2, n N,
function. the roots of equation f3(x) f2(x) f (x) – 25f2(x) f (x) + 175 f (x)
83. Statement-1 : Let f : R R be a function such that = 375. Which also satisfy equation f (x) = x will be
3 2
f (x) = x + x + 3x + sin x. Then f is one-one. (a) 5 (b) 15
Statement-2 : f (x) neither increasing nor decreasing function. (c) 10 (d) Both (a) and (b)
84. Statement-1 : If the lengths of subtangent and subnormal at 88. A triangle ABC satisfies the relation 2 sec 4C + sin2 2A +
point (x, y) on y = f (x) are respectively 9 and 4. Then x = 6 sin B = 0 and a point P is taken on the longest side of the
triangle such that it divides the side in the ratio 1 : 3. Let Q
Statement-2 : Product of sub tangent and sub normal is and R be the circumcentre and orthocentre of ABC. If PQ :
square of the ordinate of the point. QR : RP = 1 : : , then the value of 2 + 2.
(a) 9 (b) 8
(c) 6 (d) 7
FT-22 The Pattern Target AIEEE
Negative Marking : One fourth (¼) marks will be deducted for indicating incorrect response of each question.
. . . Best of Luck
1.7
CHO CO2 K
1.6
1.5 (a)
1.4
CH2OH (b) CH2OH
Concentration (M)
1.3
1.2
1.1
1.0
0.9
0.8 Experiment-1
0.7
CH2OH CO2 K
Experiment-2
0.6 (c) (d)
10 515 20
time (min.) CH2OH CO2 K
(a) 0.072 M min–1 (b) 0.036 M min–1
(c) 0.1296 M min –1 (d) 1 M min–1
FT-26 The Pattern Target AIEEE
26. Precautions to be taken in the study of reaction rate for the 29. In the given molecule choose the correct order of their angle.
reaction between potassium iodate (KIO3) and sodium
O
sulphite (Na2 SO3) using starch solution as indicator at
different concentrations and temperature – y C x
(a) The concentration of sodium thiosulphate solution H z O–H
should always be less than the concentration of the
potassium iodide solution. (a) x > y > z (b) x < y < z
(c) x = y > z (d) Can’t be predicted
(b) Freshly prepared starch solution should be used
30. What would be the reduction potential of an electrode at
(c) Experiments should be performed with the fresh 298 K, which originally contained 1M K2Cr2O7 solution in
solutions of H2O2 and KI. acidic buffer solution of pH = 1.0 and which was treated
(d) All of these with 50% of the Sn necessary to reduce all Cr 2O72– to Cr3+.
Assume pH of solution remains constant.
27. Calculate the millimoles of SeO32– in solution on the basis of
following data : Given : E 0 1.33 V , log2 = 0.3,
Cr2O72 /Cr3 ,H
M
70 ml of solution of KBrO3 was added to SeO32– solution. 2.303RT
60 0.06
The bromine evolved was removed by boiling and excess of F
M
(a) 1.285 V (b) 1.193 V
KBrO3 was back titrated with 12.5 mL of solution of (c) 1.187 V (d) None of these
25
NaAsO2.
The reactions are given below.
32. The speed of sound is measured by a resonance tube at the 36. A particle falls freely near the surface of the earth. Consider
room temperature once by filling the tube with water and a fixed point O (not vertically below the particle) on the
then with glycerine as v1 and v2. Which of the following ground. Then pickup the incorrect alternative or alternatives.
relation relations is/are true in this context?
(a) The magnitude of angular momentum of the particle
(a) v1 = 2v2
about O is increasing
(b) v1 > v2
(c) v1 < v2 (b) The magnitude of torque of the gravitational force on
(d) v1 and v2 does not depend on the nature of the liquid the particle about O is decreasing
taken in the tube. (c) The moment of inertia of the particle about O is
33. In making an Ohm's law circuit, which of the following decreasing
connection is correct?
(a) Voltmeter in series and ammeter in parallel (d) The magnitude of angular velocity of the particle about
(b) Voltmeter in parallel and ammeter in series. O is increasing
(c) Voltmeter and ammeter both are in parallel 37. Two spherical bubbles are in contact with each other
(d) Voltmeter and ammeter both are in series. internally as shown. The radius of curvature of the common
34. Two potentiometers A and B having 4 wires and 10 wires, surface is R, then –
each having 100 cm in length are used to compare e.m.f. of 2
cells. Which one will give a larger balancing length?
(a) Balancing length doesn't depend on the total length of R1
the wire. R2 Common
(b) Both A and B will give same balancing length surface
(c) Potentiometer B
(d) Potentiometer A
35. The mass of block is m1 and that of liquid with the vessel is
m2. The block is suspended by a string (tension T) partially (a) R > R1 (b) R1 > R > R2
in the liquid. (c) R < R2 (d) R = R1
/////////////// 38. Two masses A and B are connected with two inextensible
string to write constraint relation between vA and vB.
Student A : vA cos = vB.
Student B : vB cos = vA.
//////// ////////
(a) 6.25 A, 3.75 V (b) 3.00 A, 5 V (c) y = 0.4 sin t (d) y = 0.2 sin 2t
4 2
(c) 3.75 A, 3.75 V (d) 6.25 A, 6.25 V
56. The fundamental frequency of a sonometer wire of length
51. Consider an optical communication system operating at l ~ is n0. A bridge is now introduced at a distance of (<< )
800 nm. Suppose, only 1% of the optical source frequency from the centre of the wire. The lengths of wire on the two
is the available chann el bandwidth for optical sides of the bridge are now vibrated in their fundamental
communication. How many channels can be accommodated modes. Then, the beat frequency nearly is –
for transmitting audio signals requiring a bandwidth of (a) n0 / (b) 8n0 /
8 kHz ? (c) 2n0 / (d) n0 /2
57. A rectangular loop of wire with dimensions shown is coplanar
(a) 4.8 × 108 (b) 48
with a long wire carrying current I. The distance between
the wire and the left side of the loop is r. The loop is pulled
(c) 6.2 × 108 (d) 4.8 × 105 to the right as indicated. What are the directions of the
induced current in the loop and the magnetic forces on the
DIRECTIONS (Qs. 52 - 54) : Read the following passage left and right sides of the loop as the loop is pulled?
and answer the questions that follows :
A monoatomic ideal gas is filled in a nonconducting r
container. The gas can be compressed by a movable
nonconducting piston. The gas is compressed slowly to 12.5% of I b
its initial volume.
52. Find final temperature of the gas if it is T0 initially – a
(a) 4T0 (b) 3T0
Induced current Force on left Force on right
(c) 2/3 T0 (d) T0
side side
53. The initial adiabatic bulk modulus of the gas is Bi and the final
(a) Counterclockwise To the left To the left
value of the adiabatic bulk modulus of the gas is Bf then –
(a) 32Bi = Bf (b) Bi = Bf (b) Counterclockwise To the right To the left
(c) Bi = 32Bf (d) Bi = 4Bf (c) Clockwise To the right To the left
(d) Clockwise To the left To the right
FT-31
Q Q
/// (a) (b)
/////// plane 16 0 4 0
//// d
/////// cline
/ n
//// 2 i Q
//// µ=1/ (c) 8 0 (d) None of these
30°
horizontal level
mg 3
(a) (b) mg
2 4
(c) mg (d) 0 61. Let L1 be a straight line passing through the origin and L2 be
the straight line x + y = 1. If the intercepts made by the circle
59. The circuit shown has been operating for a long time. The
x2 + y2 – x + 3y = 0 on L1 and L2 are equal, then which of the
instant after the switch in the circuit labeled S is opened,
following equation can represent L1 ?
what is the voltage across the inductor VL and which labeled
point (A or B) of the inductor is at a higher potential ? (a) x + 7y = 0 (b) x – y = 0
(c) x – 7y = 0 (d) Both (a) and (b)
Take R1 = 4.0 , R2 = 8.0 and L = 2.5 H.
L S 3 5 12 5
62. Let P and Q then incorrect about
7 12 7 3
A B
the matrix (PQ)–1 –
R1 R2 (a) nilpotent (b) idempotent
= 12V (c) involutory (d) symmetric
63. The equation sin x + x cos x = 0 has at least one root in –
48 44 logically equivalent.
(a) (b)
91 91 Statement-2 : The end columns of the truth table of both
statements are identical.
88 24
(c) (d) 84. Statements-1 : Period of f(x) = sin 4 {x} + tan [x], where,
91 91
[x] & {x} denote the G.I.F. & fractional part respectively is 1.
ASSERTION & REASON
Statements-2: A function f(x) is said to be periodic if there
exist a positive number T independent of x such that
DIRECTIONS (Qs. 80 - 84) : Each of these questions
contains two statements: Statement-1(Assertion) and f(T + x) = f(x). The smallest such positive value of T is called
Statement-2 (Reason). Choose the correct answer the period or fundamental period.
(ONLY ONE option is correct) from the following - 85. The value of the expression
(a) Statement-1 is false, Statement-2 is true.
(b) Statement-1 is true, Statement-2 is true; Statement-2 1 1 1 1
1 1 2
2 2 2
is a correct explanation for Statement-1.
(c) Statement-1 is true, Statement-2 is true; Statement-2
is not a correct explanation for Statement-1. 1 1 1 1
3 3 2
.... n n 2
(d) Statement-1 is true, Statement-2 is false.
where is an imaginary cube root of unity, is –
80. Statement-1 : If a, b, c are non real complex and , are the
roots of the equation ax2 + bx + c = 0 then Im ( ) 0. n (n 2 2) n (n 2 2)
(a) (b)
because 3 3
Statement-2 : A quadratic equation with non real complex
coefficient do not have root which are conjugate of each n (n 2 1)
(c) (d) None of these
other. 3
FT-34 The Pattern Target AIEEE
86. If s, s' are the length of the perpendicular on a tangent from 88. Set of values of m for which two points P and Q lie on the
the foci, a, a' are those from the vertices is that from the
line y = mx + 8 so that APB = AQB = where
x2 y2 2
centre and e is the eccentricity of the ellipse, 1, A (– 4, 0), B (4, 0) is –
a2 b2
(a) ( , 3) ( 3, ) { 2, 2}
ss c2
then 2
aa c (b) [ 3, 3] { 2, 2}
(a) e (b) 1/e (c) ( , 1) (1, ) { 2, 2}
(c) 1/e2 (d) e2
(d) { 3, 3}
87. One percent of the population suffers from a certain disease.
There is blood test for this disease, and it is 99% accurate, in 89. The trace Tr(A) of a 3 × 3 matrix A = (aij ) is defined by the
other words, the probability that it gives the correct answer relation Tr(A) = a11 + a22 + a33 (i.e., Tr(A) is sum of the main
is 0.99, regardless of whether the person is sick or healthy. A diagonal elements). Which of the following statements
person takes the blood test, and the result says that he has cannot hold ?
the disease. The probability that he actually has the disease, (a) Tr(kA) = kTr(A) (k is a scalar)
is – (b) Tr(A + B) = Tr(A) + Tr(B)
(a) 0.99% (b) 25% (c) Tr(I3) = 3
(c) 50% (d) 75% (d) Tr(A2) = Tr(A)2
/2 n
an
90. Let a n (1 sin t) sin 2t dt then nlim
n
n is equal
1
0
to
(a) 1/2 (b) 1
(c) 4/3 (d) 3/2
4
Time : 3 hrs. INSTRUCTIONS Max. Marks : 360
Negative Marking : One fourth (¼) marks will be deducted for indicating incorrect response of each question.
. . . Best of Luck
20. Point out the incorrect statment among the following : equilibrium constant is 3.8 10–6 and pH = 6.0. What would
(a) The oxidation state of oxygen is +2 in OF2. be the ratio of concentration of bicarbonate ion to carbon
(b) Acidic character follows the order H2O < H2S < H2Se dioxide?
< H2Te. (a) 3.8 10–12 (b) 3.8
(c) The tendency to form multiple bonds increases in (c) 6 (d) 13.4
moving down the group from sulphur to tellurium 27. Amongst the following the compound that is both
(towards C and N) paramagnetic and coloured is
(d) Sulphur has a strong tendency to catenate while (a) K2Cr2O7 (b) (NH4)2[TiCl6]
oxygen shows this tendency to a limited extent. (c) CoSO4 (d) K3[Cu(CN)4]
21. Removal of Fe, Cu, W from Sn metal after smelting is by 28. A reaction rate constant is given by
............... because ............. K = 1.2 1010 e 2500/RT . It means
(a) Poling; of more affinity towards oxygen for impurities (a) log K vs T will give a straight line
(b) Selective oxidation; of more affinity towards oxygen (b) log K vs 1/T gives a straight line with a slope
for impurities – 2500/2.303 R
(c) Electrolytic refining; impurities undissolved in elec- (c) half life of reaction will be more at higher temperature
trolyte (d) log K vs 1/T gives a straight line with a slope 2500/R
(d) Liquation; Sn having low melting point compared to 29. The correct statement among the following is :
impurities. (a) The alkali metals when strongly heated in oxygen form
22. Among KO 2 , AlO2 , BaO2 and NO 2 , unpaired superoxides.
(b) Caesium is used in photoelectric cells.
electron is present in
(c) NaHCO3 is more soluble in water than KHCO3.
(a) NO 2 and BaO 2 (b) KO 2 and AlO-2 (d) The size of hydrated ions of alkali metals increases
(c) KO 2 only (d) BaO 2 only from top to bottom.
23. If a 0.1 M solution of glucose (Mol. wt 180) and 0.1 molar 30. The e.m.f. of a Daniell cell, Zn ZnSO 4 CuSO4 Cu , at
solution of urea (Mol. wt. 60) are placed on two sided (0.01M) (1.0M)
semipermeable membrane to equal heights, then it will be
correct to say that 298 K is E1. When the concentration of ZnSO4 is 1.0 M and
(a) there will be no net movement across the membrane that of CuSO4 is 0.01 M, the e.m.f. changed to E 2. What is
(b) glucose will flow across the membrane into urea solution the relationship between E1 and E2?
(a) E1 < E2 (b) E1 = E2
(c) urea will flow across the membrane into glucose
(c) E2 = 0 E1 (d) E1 > E2
solution
(d) water will flow from urea solution to glucose solution
24. Which reagent can be used to identify nickel ion?
(a) Resorcinol (b) Dimethyl glyoxime
(c) Diphenyl benzidine (d) Potassium ferrocyanide
31. A bus is moving with a velocity of 10m/s on a straight road.
25. When pink [Co(H2O)6 ]2 is dehydrated the colour changes A scooterist wishes to overtake the bus in 100 seconds. If
to blue. The correct explanation for the change is : the bus is at a distance of 1 km from the scooterist, at what
(a) The octahedral complex becomes square planar. velocity should the scooterist chase the bus?
(b) A tetrahedral complex is formed. (a) 50 m/sec (b) 40 m/sec
(c) Distorted octahedral structure is obtained. (c) 30 m/sec (d) 20 m/sec
(d) Dehydration results in the formation of polymeric species. 32. The length of an elastic string is x when the tension is 5N. Its
26. When CO2 dissolves in water, the following equilibrium is length is y when the tension is 7N. What will be its length,
established when the tension is 9N?
CO2 + 2H2 O H3 O + + HCO3 – ; for which the (a) 2y + x (b) 2y – x
(c) 7x – 5y (d) 7x + 5y
FT-38 The Pattern Target AIEEE
33. The masses of the blocks A and B are 0.5 kg and 1 kg respectively. 38. Two points of a rod move with velocities 3v and v
These are arranged as shown in the figure and are connected perpendicular to the rod and in the same direction, separated
by a massless string. The coefficient of friction between all by a distance r. Then the angular velocity of the rod is
contact surfaces is 0.4. The force, necessary to move the block (a) 3v/r (b) 4v/r
B with constant velocity, will be (g = 10m/s2) (c) 5v/r (d) 2v/r
39. For hydrogen gas C p – C v = a and for oxygen gas
A Cp – Cv = b. So, the relation between a and b is given by
F B
(a) a = 16 b (b) 16a = b
(c) a = 4b (d) a = b
40. A non-conducting partition divides a container into two
(a) 5N (b) 10N equal compartments. One is filled with helium gas at 200 K
(c) 15N (d) 20N and the other is filled with oxygen gas at 400 K. The number
34. A rod of length L is placed on x – axis between of molecules in each gas is the same. If the partition is
x = 0 and x = L. The linear density i.e., mass per unit length removed to allow the gases to mix, the final temperature will
denoted by , of this rod, varies as, = a + bx. What should be be
the dimensions of b? (a) 350 K (b) 325 K
(a) M2L1T 0 (b) M1 L–2 T 0 (c) 300 K (d) 275 K
(c) M–1 L3T 1 (d) M–1 L2T 3
41. Statement-1 : If collision occurs between two elastic bodies
35. A body is thrown vertically upwards from the surface of their kinetic energy decreases during the time of collision.
earth in such a way that it reaches upto a height equal to
Statement-2 : During collision intermolecular space
10Re. The velocity imparted to the body will be
decreases and hence elastic potential energy increases.
(a) 10.6 km/s (b) 0.106 km/s
(a) Statement-1 is true, Statement-2 is true; Statement-2 is
(c) 1.06 km/s (d) zero
a correct explanation for Statement-1.
36. A wheel is rolling on a plane road. The linear velocity of
centre of mass is v. Then velocities of the points A and B on (b) Statement-1 is true, Statement-2 is true; Statement-2 is
circumference of wheel relative to road will be not a correct explanation for Statement-1.
(c) Statement-1 is true, Statement-2 is false.
B
(d) Statement-1 is false, Statement-2 is true.
v 42. A bucket full of hot water is kept in a room and it cools from
75oC to 70oC in T1 minutes, from 70oC to 65oC in T2 minutes
A and from 65oC to 60oC in T3 minutes. Then
(a) vA = v, vB = 0 (b) vA = vB = 0 (a) T1= T2 = T3 (b) T1 < T2 < T3
(c) vA = 0, vB = v (d) vA = 0, vB = 2v (c) T1 > T2 > T3 (d) T1< T3< T2
37. A metallic wire of density d is lying horizontal on the surface 43. The equivalent capacity of the network, (with all capacitors
of water. The maximum length of wire so that it may not sink having the same capacitance C)
will be
A
2Tg 2 T
(a) (b) dg
d
B
2T (a) (b) zero
(c) dg
(d) any length
(c) C[( 3 – 1) / 2] (d) C[( 3 + 1) / 2]
FT-39
44. The current I vs voltage V graphs for a given metallic wire at 47. A straight section PO of a circuit lies along the x-axis from
two different temperatures T1 and T2 are shown in the figure. x = –a/2 to x = +a/2, and carries a steady current ‘I’. The
It is concluded that magnitude of magnetic field due to the section PO at a point
to y = + a will be
T1
(a) proportional to a (b) proportional to a2
T2 (c) proportional to 1/a (d) equal to zero
I 48. A transformer is used to light a 140 W, 24 V bulb from a
240 V a.c. mains. The current in the main cable is 0.7 A. The
efficiency of the transformer is
V (a) 63.8 % (b) 83.3 %
(c) 16.7 % (d) 36.2 %
(a) T1 > T2 (b) T1 < T2
49. In the given circuit, the current drawn from the source is
(c) T1 = T2 (d) T1 = 2T2
45. In the circuit shown the effective resistance between B and
V 100x sin(100 t )
C is
20
10
~
20
XL
XC
R
(a) 20 A (b) 10 A
(c) 5 A (d) 5 2 A
50. Two circular coils X and Y, having equal number of turns, carry
equal currents in the same sense and subtend same solid angle
at point O. If the smaller coil X is midway between O and Y, then
if we represent the magnetic induction due to bigger coil Y at O
(a) 3 (b) 4 as BY and due to smaller coil X at O as BX then
(c) 4/3 (d) 3/4
Y d
46. Statement-1 : Orbital velocity of a satellite is greater than d/2
its escape velocity.
2r
Statement-2 : Orbit of a satellite is within the gravitational r O
field of earth whereas escaping is beyond the gravitational
X
field of earth.
(a) Statement-1 is true, Statement-2 is true; Statement-2 is
a correct explanation for Statement-1. BY BY
(a) 1 (b) 2
(b) Statement-1 is true, Statement-2 is true; Statement-2 is BX BX
not a correct explanation for Statement-1.
(c) Statement-1 is true, Statement-2 is false. BY 1 BY 1
(c) BX 2 (d) BX 4
(d) Statement-1 is false, Statement-2 is true.
FT-40 The Pattern Target AIEEE
51. A flat plate P of mass ‘M’ executes SHM in a horizontal electron in the first Bohr orbit of the hydrogen atom?
plane by sliding over a frictionless surface with a frequency (a) 1 : 1 (b) 1 : 2
V. A block ‘B’ of mass ‘m’ rests on the plate as shown in (c) 1 : 4 (d) 2 : 1
figure. Coefficient of friction between the surface of B and P
55. In frequency modulation
is . What is the maximum amplitude of oscillation that the
plate block system can have if the block B is not to slip on (a) the amplitude of modulated wave varies as frequency
of carrier wave
the plate :
(b) the frequency of modulated wave varies as amplitude
of modulating wave
(c) the amplitude of modulated wave varies as amplitude
of carrier wave
(d) the frequency of modulated wave varies as frequency
of modulating wave
g g
(a) (b) 56. The radioactivity of a sample is R1 at a time T1 and R2 at a
2 2 2
4 V 4 V time T2. If the half life of the specimen is T, the number of
atoms that have disintegrated in the time (T 2 –T 1) is
g
(c) (d) proportional to
4 2 V 2g 2 2
V2
(a) (R1T1 – R2T2) (b) (R1 – R2)
52. A glass slab has the left half of refractive index n1, and the
right half of n2=3n 1. The effective refractive index of the (c) (R1 – R2)/T (d) (R1– R2) × T
whole slab is 57. The figure of merit of a galvanometer is defined as
(a) the voltage required to produce unit deflection in the
n1
(a) (b) 2n galvanometer
2
(b) the current required to produce unit deflection in the
3n 1 2n 1 galvanometer
(c) (d) (c) the power required to produce unit deflection in the
2 3
galvanometer
53. In the arrangement shown L1, L2 are slits and S1, S2 two
independent sources on the screen, interference fringes (d) none of these
screen 58. P–V plots for two gases during adiabatic processes are
L1 shown in the figure. Plots 1 and 2 should correspond
respectively to
× ×
S1 S2
L2
(a) will not be there P
(b) will not be there if the intensity of light reaching the 1
screen from S1 and S2 are equal.
2
(c) will be there under all circumstances
(d) we will have only the central fringe V
54. What is the ratio of the circumference of the first Bohr orbit
for the electron in the hydrogen atom to the de Brogile (a) He and Ar (b) He and O2
wavelength of electrons having the same velocity as the (c) O2 and N2 (d) O2 and He
FT-41
59. Two identical thin rings, each of radius R metres, are coaxially a,b,c,d are in
placed at a distance R metres apart. If Q1 coulomb and Q2 (a) A.P. (b) G.P.
coulomb are respectively, the charges uniformly spread on
(c) H.P. (d) satisfy ab = cd
the two rings, the work done in moving a charge q from the
centre of one ring to that of the other is 63. Which of the following is correct?
2 1
R log( a 2b ) (log a log b )
q 2
A
B
log x log y log z
R (b) If , then x a .y b .zc abc
b c c a a b
R
1 1 1
(a) zero (c) 2
log xy xyz log yz xyz log zx xyz
(b) q(Q1–Q2) ( 2 –1)/ 2 4 0R
(d) All are correct
(c) q 2 (Q1 + Q2) / 4 0R
64. Statement-1: If n is an odd prime, then integral part of
(d) q(Q1+Q2) ( 2 +1)/ 2 4 0R
( 5 2) n is divisible by 20n.
60. Statement-1 : Wein's distribution law fails at longer
wavelengths. Statement-2 : If n is prime, then nc1, nc2, nc3 . . . ncn – 1 must
Statement-2 : Intensity of radiations of very short be divisible by n.
wavelength is small. (a) Statement-1 is true, Statement-2 is true; Statement-2 is
(a) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
a correct explanation for Statement-1. (b) Statement-1 is true, Statement-2 is true; Statement-2 is
(b) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
not a correct explanation for Statement-1. (c) Statement-1 is true, Statement-2 is false.
(c) Statement-1 is true, Statement-2 is false. (d) Statement-1 is false, Statement-2 is true.
(d) Statement-1 is false, Statement-2 is true. 65. The value of cos360 cos420 cos780 is
5 1 5 1
Given : sin18 and cos 36
4 4
61. If one root of the equation x2 + px + 12 = 0 is 4 while the (a) 1/4 (b) 1/8
equation x2 + px + q = 0 has equal roots, the value of q is (c) 1/16 (d) [( 5 – 1)/4]2
66. If x =1/5, the value of cos (cos x + 2 sin–1 x) is
–1
49
(a) (b) 4 24 24
4 (a) (b)
25 25
(c) 3 (d) 12
62. If a, b, c, d and p are distinct non zero real numbers such that 1 1
(a2 + b2 + c2) p2 –2(ab + bc + cd)p + (b2 + c2+ d2) 0 then (c) (d)
5 5
FT-42 The Pattern Target AIEEE
72. Which of the following is correct?
67. If 0 < < /2 such that + + = and cot , cot , (a) If A and B are square matrices of order 3 such that
2
cot are in arithmetic progression, then the value of cot | A | = –1, | B | = 3, then the determinant of 3 AB is equal
cot is to 27.
(a) 1 (b) 3 (b) If A is an invertible matrix, then det (A–1) is equal to
(c) cot2 (d) cot + cot det (A)
68. Statement-1 : The straight line 2x + 3y = 4 intersects the (c) If A and B are matrices of the same order, then (A + B) 2
hyperbola 4x2 – 9y2 = 36 in exactly one point. = A2 + 2AB + B2 is possible if AB = I
Statement-2 : The line is parallel to an asymptote of the (d) None of these
hyperbola.
73. If the solution of the linear equations x – 2y + z = 0;
(a) Statement-1 is true, Statement-2 is true; Statement-2 is
2x – y + 3z = 0 and x + y – z = 0 is trivial then the value of
a correct explanation for Statement-1.
is given by
(b) Statement-1 is true, Statement-2 is true; Statement-2 is
not a correct explanation for Statement-1.
4 4
(c) Statement-1 is true, Statement-2 is false. (a) =– (b) –
5 5
(d) Statement-1 is false, Statement-2 is true.
69. If (c) =2 (d) 2
1, 2 are the solutions of the equation
2tan2 – 4tan + 1 = 0, then tan ( 1 + 2) is equal to 74. Let f(x) = | x – 1 |. Then
(a) –4 (b) 4 (a) f(x2) = (f(x))2 (b) f(x + y) = f(x) + f(y)
(c) 1 (d) 2
(c) f(| x |) = | f(x) | (d) None of these
70. Let z1 and z2 be complex number such that
75. Statement-1 : If 1/ 2 x 1 , then
z1 z2 | z1 | | z 2 |
1 1 x 3 3x 2
cos x cos is equal to /3
z1 2 2
Statement-1 : arg z 0
2
76. If AB = 0, then for the matrices (b) Statement-1 is true, Statement-2 is true; Statement-2 is
not a correct explanation for Statement-1.
A= cos 2 cos sin (c) Statement-1 is true, Statement-2 is false.
cos sin sin 2 (d) Statement-1 is false, Statement-2 is true.
x x 1
cos 2 cos sin 81. Let f ( x) ; where {x} is the fractional part of
and B = , – is x x 2
cos sin sin 2
x, then lim f ( x )
x 1/ 3
(a) an odd number of (a) has value 0 (b) has value 1
2
(c) has vlaue – (d) has value
(b) an odd multiple of
82. The order of the differential equation
(c) an even multiple of
2 2 3/ 2 5
dy d2 y
(d) 0 1 5 11 is
dx dx 2
2
77. Let f (x) , g(x) cos x and h(x) x 3 then the (a) 1 (b) 2
x 1
range of the composite function fogoh, is (c) 3 (d) 4
(a) R+ (b) R – {0} 1
83. dx log(f (x))2 C, then f(x) is
(c) [1, ) (d) R+ – {1} f (x)
78. The set of points where f(x) = (x – 1)2 (x + | x –1 | ) is thrice
(a) x + (b) 2x +
differentiable, is
(a) R (b) R – {0} x
(c) (d) x2
(c) R – {1} (d) R–{0,1} 2
79. Let f (x) = 1/(x – 1) and g (x) = 1/(x2 + x – 2). Then the set of 2a f (x)
points where (gof)(x) is discontinuous, is 84. Value of dx is
0 f (x) f (2a x)
(a) {1} (b) {–2,1}
(c) {1/2, 1, 2} (d) {1/2 , 1} (a) 0 (b)
4
(n 2 )! (c) a (d) none of these
80. Statement-1 : is a natural number for all n N.
(n!)n
85. The value of
Statement-2 : The number of ways of distributing mn things
/4
1 a b
8 xyz
89. In a ABC, if 1 c a 0, then (d)
[(1 x )(1 y 2 )(1 z 2 )]
2
1 b c
5
Time : 3 hrs. INSTRUCTIONS Max. Marks : 360
Negative Marking : One fourth (¼) marks will be deducted for indicating incorrect response of each question.
. . . Best of Luck
16. The number of electrons present in 6.4 g of calcium carbide 22. An organic compound A (C4H10O) has two enantiomeric
is – (NA = Avagadro’s number) forms and on dehydration it gives B(major product) and C
(minor product). B and C are treated with HBr/ Peroxide and
(a) 4 NA (b) 0.4 NA
the compounds so produced were subjected to alkaline
(c) 0.1 NA (d) 0.2 NA hydrolysis then-
17. An organic compound ‘X’ on ozonolysis followed by (a) B will give an isomer of A
O
|| (b) C will give an isomer of A
reduction with Zn/H2O gives 2 moles of H C H and (c) Neither of them will give isomer of A
O O (d) Both B and C will give isomer of A
|| ||
H C CH 2 C H . ‘X’ is 23. A reaction is found to be second order w.r.t. one of the
reactants & has rate constant of 0.5 mol–1 dm3 min–1. If initial
(a) CH2 = CH – CH2 – CH = CH2
concentration is 0.2 mol dm–3 then t1/2 of reaction is
(b) CH2 = CH–CH2–CH2–CH = CH2
(a) 5 min (b) 10 min
(c) H – C C–C C–H
(c) 15 min (d) 20 min
(d) CH2 = CH – CH = CH2
18. The rate of SN1 reaction is fastest in the hydrolysis of CH 2 O
which of the following halides ? 24. O CH 2
(a) C6H5CH2Br (b) CH3Br CH 2 O
(c) (CH3)2CHBr (d) (CH3)3CBr
19. Two elements A & B form compounds having molecular The above shown polymer is obtained when a carbonyl
formulae AB2 and AB4. When dissolved in 20.0 g of benzene compound is allowed to stand. It is a white solid. The polymer
1.00g of AB2 lowers f.p. by 2.30C whereas 1.00g of AB4 is
lowers f.p. by 1.30C. The molal depression constant for (a) trioxane (b) formose
benzene in 1000g is 5.1. The atomic masses of A and B are
(c) paraformaldehyde (d) metaldehyde.
(a) 52, 48 (b) 42, 25
25. Concentration of NH4Cl and NH4OH in a buffer solution is
(c) 25, 42 (d) None in the ratio of 1 : 1, Kb for NH4OH is 10–10. The pH of the
20. To detect iodine in presence of bromine, the sodium extract buffer is
is treated with NaNO 2 + glacial acetic acid + CCl 4 . Iodine (a) 4 (b) 5
is detected by the appearance of (c) 9 (d) 11
(a) yellow colour of CCl4 layer 26. Aniline when diazotized in cold and when treated with
(b) purple colour of CCl4 dimethyl aniline gives a coloured product. Its structure
(c) brown colour in the organic layer of CCl4 would be
(d) deep blue colour in CCl4 (a) CH3NH N=N NHCH3
21. An element (atomic mass =100g/mol) having bcc structure
has unit cell edge 400pm. The density (in g/cm3) of the (b) CH3 N=N NH2
element is
(c) (CH3)2N N=N
(a) 10.376 (b) 5.19
(c) 7.289 (d) 2.144 (d) (CH3)2N NH
FT-48 The Pattern Target AIEEE
NH2
(i) NaNO2HCl
27. A. Compound ‘A’ is
(ii) CuCl
31. A bullet comes out of a gun with a velocity of 600 m/s. At
what height from the target should the gun be aimed in
Cl order to hit the target situated at a distance of 200 m from the
NO2
gun ?
(a) (b)
(a) 0.55 m (b) 55 m
(c) 5.5 m (d) 55 10–4m
32. A gramophone record is revolving with an angular velocity
Cl NO2
. A coin is placed at a distance r from the center of the
(c) (d) record. The static coefficient of friction is . The coin will
revolve with the record if
g g
28. The standard reduction potential of (a) r only
2
(b) r 2
only
Li+/Li, Ba2+/Ba, Na+/Na and Mg2+/Mg are –3.05, –2.73, –2.71
g g
and –2.37 volts respectively. Which one of the following is (c) r only (d) r only
2 2
strongest oxidising agent?
(a) Na+ (b) Li+ 33. A pendulum consists of a wooden bob of mass m and length
‘ ’. A bullet of mass m1 is fired towards the pendulum with a
(c) Ba2+ (d) Mg2+ speed v1. The bullet emerges out of the bob with a speed
29. Phospholipids are esters of glycerol with v1/3, and the bob just completes motion along a vertical
(a) two carboxylic acid residues and one phosphate group circle. Then v1 is
35. A particle of mass m1 collides head-on with another 40. Statement-1 : Two different gas molecules having same
stationary particle of mass m2 (m2 > m1). The collision is temperature will have the same K.E.
perfectly inelastic. The fraction of kinetic energy which is Statement -2 : KE of gas molecules follows the Boltzmann's
converted into heat in this collision is law.
(a) m2/(m1 + m2) (b) m1/(m1 + m2) (a) Statement-1 is true, Statement-2 is true; Statement-2 is
(c) m1/(m1 – m2) (d) m2/(m1 – m2) a correct explanation for Statement-1.
36. The correct graph between the gravitational potential (Vg) (b) Statement-1 is true, Statement-2 is true; Statement-2 is
due to a hollow sphere and distance from its centre will be not a correct explanation for Statement-1.
(c) Statement-1 is true, Statement-2 is false.
r (d) Statement-1 is false, Statement-2 is true.
41. 1/2 mole of helium gas is contained in a container at S.T.P.
Vg
(a) (b) V The heat energy needed to double the pressure of the gas,
g
r keeping the volume constant (heat capacity of the gas
= 3 Jg–1 K–1) is
(a) 3276 J (b) 1638 J
+Vg (c) 819 J (d) 409.5 J
+Vg r = Re r r = Re r 42. A sphere, a cube and a thin circular plate all made of the
same material and having the same mass, are initially heated
to a temperature of 200oC. Which of these objects will cool
(c) (d)
slowest when left in air at room temperature?
–Vg –Vg (a) the sphere (b) the cube
(c) the circular plate (d) all will cool at same rate
37. A beaker containing a liquid of density moves up with an 43. A system changes from the state (P1, V1) to (P2, V2) as
acceleration a. The pressure due to the liquid at a depth h shown in the figure below. What is the work done by the
below the free surface of the liquid is system?
(a) h g (b) h (g – a)
(P2, V2)
g a 5 105 B
(c) h (g + a) (d) 2h g
Pressure in N/ m2
g a 5
4 10
38. A wooden block of volume 1000 cc is suspended from a 3 105
spring balance. It weighs 12 N in air. It is then held suspended 2 105
in water with half of it inside water. What would be the A
reading in spring balance now? 1 105 (P1, V1)
(a) 10 N (b) 9 N
(c) 8 N (d) 7 N
D
39. A square wire frame of size L is dropped in a liquid. On 1 2 3 4 5
taking out, a membrane is formed. If the surface tension of Volume in metre 3
the liquid is T, force acting on the frame will be
(a) 2 T. L (b) 4 T. L (a) 7.5 × 105 joule (b) 7.5 × 105 ergs
(c) 8 T. L (d) 10 T. L (c) 12 × 105 joule (d) 6 × 105 joule
FT-50 The Pattern Target AIEEE
44. A cylinder of radius R made of a material of thermal capacitance across terminals AB will be
conductivity K1 is surrounded by a cylindrical shell of inner
radius R & outer radius 2R made of a material of thermal C1
A B
conductivity K2. The two ends of the combined system are
maintained at two different temperatures. There is no loss of
heat across the cylindrical surface and the system is in steady
state. The effective thermal conductivity of the system is C2
(a) K1 + K2 (b) K1K2/ (K1+ K2)
(c) (K1+ 3K2)/4 (d) (3K1+ K2)/4 C1C 2
(a) C1 + C2 (b)
45. Figure shows a closed surface which intersects a conducting C1 C 2
sphere. If a positive charge is placed at the point P, the flux (c) C1 (d) C2
of the electric field through the closed surface 48. In the figure battery B supplies 12 V. Take C1 = 1 F, C2 = 2
F, C3 = 3 F, C4 = 4 F. Charge on capacitor C1 when only
S1 is closed is
•P C1 C3
conducting S2
closed sphere
surface
C2 C4
+ – S1
(a) will remain zero (b) will become positive
(c) will become negative (d) will become undefined B
46. In the circuit shown, the total current supplied by the battery
is (a) 3 C (b) 9 C
(c) 6 C (d) 12 C
B 49. The instantaneous values of current and voltage in an A.C.
3 3
1.5
(c) (d)
6 6
D
50. An alternating voltage E(in volt) 200 2 sin 100 t is
(a) 2 ampere (b) 4 ampere
connected to one microfarad capacitor through an A.C.
(c) 1 ampere (d) 6 ampere ammeter. The reading of the ammeter shall be
47. Parallel plate capacitors C1 & C2 have the same area with C1
having half the plate separation as C2. C1 is inserted wholly (a) 20 2 mA (b) 20 mA
inside C2 as shown with their plates joined as shown. The (c) 40 mA (d) 80 mA
FT-51
51. In fig, what is the magnetic field induction at point O? 55. Statement 1 : A current flows in a conductor only when
there is an electric field within the conductor.
Statement 2 : The drift velocity of electron in presence of
i
electric field decreases.
(a) Statement-1 is true, Statement-2 is true; Statement-2 is
a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is true; Statement-2 is
r not a correct explanation for Statement-1.
O
(c) Statement-1 is true, Statement-2 is false.
(d) Statement-1 is false, Statement-2 is true.
56. Two periodic waves of intensities I1 and I2 pass through a
region at the same time in the same direction. The sum of the
0i 0i 0i
(a) (b) maximum and minimum intensities is
4 r 4r 2 r
(a) 2 (I1 + I2) (b) I1 + I2
0i 0i 0i 0i
(c) (d) (c) ( I1 I2 )2 (d) ( I1 I2 )2
4r 4 r 4r 4 r
57. Range of frequencies alloted for commercial FM radio
52. A proton, a deutron and an particle accelerated through
broadcast is
the same potential difference enter a region of uniform
(a) 88 to 108 MHz (b) 88 to 108 kHz
magnetic field, moving at right angles to B. What is the ratio
of their K.E.? (c) 8 to 108 MHz (d) 88 to 108 GHz
(a) 1 : 2 : 2 (b) 2 : 2 : 1 (c) 1 : 2 : 1 (d) 1 : 1 : 2 58. A hydrogen-like atom has one electron revolving around a
stationary nucleus. The energy required to excite the electron
53. A simple pendulum with a bob of mass ‘m’ oscillates from A to
from the second orbit to the third orbit is 47.2 eV. The atomic
C and back to A such that PB is H. If the acceleration due to
number of the atom is
gravity is ‘g’, then the velocity of the bob as it passes through
B is (a) 3 (b) 4 (c) 5 (d) 6
P
A C 59. In Millikan oil drop experiment, a charged drop of mass
H 1.8 × 10–14 kg is stationary between its plates. The distance
B between its plates is 0.90 cm and potential difference is 2.0
(a) zero (b) 2 gH kilovolts. The number of electrons on the drop is
(c) mgH (d) 2g H (a) 500 (b) 50 (c) 5 (d) 0
60. For the determination of the focal length of a convex mirror,
54. A ray of light is incident at an angle of 60° on one face of a a convex lens is required because
prism of angle 30°. The ray emerging out of the prism makes
(a) it is not possible to obtain the image produced by a
an angle of 30° with the incident ray. The emergent ray is
convex mirror on the screen
(a) Normal to the face through which it emerges
(b) a convex lens has high resolving power so it helps to
(b) Inclined at 30° to the face through which it emerges measure the focal length correctly
(c) Inclined at 60° to the face through which it emerges (c) a convex mirror always forms a real image which is
(d) Inclined at 90° to the normal at face through which it diminished by the convex lens
emerges (d) none of these
FT-52 The Pattern Target AIEEE
(b) Statement-1 is true, Statement-2 is true; Statement-2 is
not a correct explanation for Statement-1.
(c) Statement-1 is true, Statement-2 is false.
61. Let f(x) = [(tanx + sinx )2]. Then (d) Statement-1 is false, Statement-2 is true.
(c) P 5, Q 6, R 3 (d) P 3, Q 2, R 3 aX b
69. If the S.D. of a variable X is , then the S.D. of
c
65. Statement-1 : The area of the ellipse 2x2 + 3y2 = 6 is more
(a,b,c are constant) is
than the area of the circle x2 + y2 – 2x + 4y + 4 = 0.
Statement-2 : If the length of the major axis of an ellipse is a a
more than the diameter of a circle then the area of the ellipse (a) (b)
c c
will be more than the area of the circle.
(a) Statement-1 is true, Statement-2 is true; Statement-2 is a
c c
correct explanation for Statement-1. (c) (d)
a a
FT-53
70. Statement 1 : The determinant of a matrix (a) Statement-1 is true, Statement-2 is true; Statement-2 is a
correct explanation for Statement-1.
0 p q p r (b) Statement-1 is true, Statement-2 is true; Statement-2 is
q p 0 q r not a correct explanation for Statement-1.
is zero.
r p r q 0 (c) Statement-1 is true, Statement-2 is false.
(d) Statement-1 is false, Statement-2 is true.
Statement 2 : The determinant of a skew symmetric 76. Two finite sets have m and n elements. The total number of
matrix of odd order is zero. subsets of the first set is 56 more than the total number of
(a) Statement-1 is true, Statement-2 is true; Statement-2 is a subsets of the second set. Then :
correct explanation for Statement-1. (a) m = 3, n = 6 (b) m = 6, n = 3
(b) Statement-1 is true, Statement-2 is true; Statement-2 is (c) m = 5, n = 6 (d) None of these
not a correct explanation for Statement-1. 77. A set A has 3 elements and another set B has 6 elements.
(c) Statement-1 is true, Statement-2 is false. Then
(d) Statement-1 is false, Statement-2 is true. (a) 3 n (A B) 6 (b) 3 n (A B) 9
(c) 6 n (A B) 9 (d) 0 n (A B) 9
71. If A and B are two sets, then A (A B )' equals :
78. If g = {(1, 1), (2, 3), (3, 5), (4, 7)} is a function described by the
(a) A (b) B formula, g (x) = x+ then what values should be assigned
(c) (d) None to and ?
72. f (x) = x2 [x] (a) = 1, =1 (b) = 2, =–1
(a) increases in (0,1) (b) decreases in (0 , 1)
(c) = 1, =–2 (d) = – 2, =–1
(c) increases in (–1,0) (d) none of these
73. If y = ex + sin x , then d2x/dy2 is equal to 79. The roots and of the quadratic equation px 2 qx r 0
(a) ex – sin x are real and of opposite signs. The roots of
(b) –(ex + cos x)–2 (x )2 (x )2 0 are
(c) –(ex – sin x) (ex + cos x)–2 (a) positive (b) negative
(d) (sin x – ex) (cos x + ex)–3 (c) of opposite signs (d) non real
74. The sum to n terms of the series 80. The value of x for which sin (cot –1 (1+ x)) = cos (tan–1 x) is
(a) 1/2 (b) 1
1 3 7 15 (c) 0 (d) –1/2
+ + + + .............. is
2 4 8 16 81. The equation k sin + cos 2 = 2k – 7 possesses a solution if :
(a) 2 k 6 (b) k > 2
(a) n – 1 – 2 – n (b) 1 (c) k > 6 (d) k < 2
1 2x 2x dy
83. If y sin , then which of the following is not correct? 87. If y tan 1 , then at x 0 is :
2
1 x 1 2 2x 1 dx
dy 2 dy 2
(a) for | x | 1 (b) for | x | 1 3 2 3
dx 1 x 2 dx 1 x2 (a) – log 2 (b) log 2 (c) log 2 (d) None
5 5 2
dy dy
(c) 2 for x 1 (d) does not exist at | x | = 1 a a2 1 a
dx dx
2 3
88. If b b 1 b = 0 and the vectors A = (1, a, a2) ;
1
84. If f (x ) , the number of points of discontinuity of c c2 1 c3
1 x
f{f [f(x)]} is :
(a) 2 (b) 1 B = (1, b, b2) ; C = (1,c, c2) are non-coplanar then the
(c) 0 (d) infinite product abc =
85. Statement -1 : The period of (a) 0 (b) 1
1 (c) –1 (d) None
f (x) = sin 2x cos [2x] – cos 2x sin [2x] is 89. A and B are two independent witnesses (i.e. there is no
2
collusion between them) in a case. The probability that A
Statement -2 : The period of x – [x] is 1
will speak the truth is x and the probability that B will speak
(a) Statement-1 is true, Statement-2 is true; Statement-2 is a
the truth is y. A and B agree in a certain statement. The
correct explanation for Statement-1.
probability that the statement is true is
(b) Statement-1 is true, Statement-2 is true; Statement-2 is
not a correct explanation for Statement-1. x–y xy
(c) Statement-1 is true, Statement-2 is false. (a) (b)
x y 1 x y xy
(d) Statement-1 is false, Statement-2 is true.
86. If the slope of the tangent at (x, y) to a curve passing through x–y xy
(c) (d)
y y 1 – x – y 2 xy 1 – x – y 2xy
1, is given by cos 2 , then the equation of the
4 x x 90. Complex number z satisfies | z – a + ia | = 1 and has the least
curve is absolute value. Its absolute value is
ANSWER KEY
1 (d) 16 (d) 31 (c) 46 (b) 61 (b) 76 (b)
2 (a) 17 (c) 32 (b) 47 (d) 62 (b) 77 (d)
3 (b) 18 (c) 33 (d) 48 (c) 63 (c) 78 (a)
4 (b) 19 (a) 34 (b) 49 (b) 64 (d) 79 (c)
5 (b) 20 (b) 35 (d) 50 (c) 65 (b) 80 (a)
6 (c) 21 (d) 36 (b) 51 (a) 66 (b) 81 (d)
7 (c) 22 (d) 37 (a) 52 (b) 67 (d) 82 (d)
8 (b) 23 (b) 38 (a) 53 (c) 68 (c) 83 (d)
9 (a) 24 (c) 39 (c) 54 (a) 69 (a) 84 (b)
10 (c) 25 (a) 40 (c) 55 (a) 70 (c) 85 (a)
11 (a) 26 (a) 41 (a) 56 (d) 71 (a) 86 (c)
12 (d) 27 (a) 42 (b) 57 (a) 72 (d) 87 (b)
13 (b) 28 (b) 43 (a) 58 (b) 73 (b) 88 (d)
14 (c) 29 (b) 44 (a) 59 (a) 74 (c) 89 (d)
15 (d) 30 (b) 45 (d) 60 (d) 75 (b) 90 (a)
CH3
3 Cl 120
2 1
1. (d) Cl
1 CH CH Ethyl group comes first in alphab- 6. (c)
6 2 Dipole moments of 2Cl and 5Cl
2 3 3
C2H5 Cl 5 4 Cl
3
are vectorically cancelled.
atical order but IUPAC name of 2 Br , 2-bromo-
1 2
It is due 1 Cl and 3 Cl
CH3
2 2
4-ethyl-1-methyl cyclohexane. It follows lowest sum = 1 2 2 1 2 cos
rule
(i.e., lowest set of locant is preferred.) = (1.5) 2 (1.5) 2 2 1.5 1.5 cos 120 = 1.5 D
Lowest sum of locant is = 1 + 2+ 3 = 6 7. (c) CsCl is ionic solid.
2. (a) The order of stability of resonating structures carrying 8. (b) Volume of ice > volume of water & thus increase in
no charge > carrying minimum charge and each atom pressure favours forward reaction showing decrease
having octet complete. in volume.
9. (a) EHe+ = EH × 22; ELi2+ = EH × 32 ..... up to Z
1
3. (b) NO = [5 + 0 + 0 – 1] = 2 sp;
2 [H 3 O ][F – ] [HF][OH – ]
10. (c) Ka and K b .
[HF][H 2 O] [F – ][H 2 O]
1
NO – = [5 + 0 + 1 – 0]= 3 sp2;
2 Therefore, Ka × Kb = [H3O+] [OH–] = Kw.
11. (a) Isoelectric point (pH)
1
NH = [5 + 4 + 0 – 1] = 4 sp3 pKa1 pK a 2
2 = 2.34 9.60
5.97
4. (b) In A — O — H, if EN of ‘A’ is 2.1 then it will be neutral, 2 2
as XA – X0 = X0 – XH. (where X is EN) 12. (d) Let bond energy of A2 be x then bond energy of AB is
5. (b) Correct order is B < Be < O < N. also x and bond energy of B2 is x/2.
Enthalpy of formation of AB is – 100 KJ/mole:
FT-56 The Pattern Target AIEEE
23. (b) The reagent is BaCl2 which imparts green colour to
1 1
A2 B2 2 AB; A2 B2 AB; 100KJ flame. BaCl2 forms chromyl chloride (which is red in
2 2
colour), when treated with K2Cr2O7 and conc. H2SO4.
x x 2x x 4x
or 100 x 100 x 400 KJ 2BaCl2 K 2Cr2O7 3H 2SO4
2 4 4
K 2SO 4 2BaSO 4 2CrO 2Cl 2 3H 2O
(n 1 n 2 ) Chromyl chloride
13. (b) M total st , (red gas)
v1 v 2
24. (c) Formaldehyde can not produce iodoform, as only those
n1 = n2 = 0.1, V1 = V2 = 1litre M total 0.1 M compound which contains either CH3 CH group
|
0.77 2( 0.44) 0.77 0.88 0.11 OH
14. (c) E0
3 3 3 or CH3 CH group on reaction with potassium
||
0.04
O
15. (d) The chemical bond method gives the O.N.
iodide and sod. hypochlorite yield iodoform.
– O O O O
O – – – – N C CH 3
– S = O: O – S – S – O : O–S– S–O | | |
O +4 +3 25. (a) (i) C N, (ii) C CH 3 ,
+5 O O
| | |
+ – + N C Br
–
16. (d) CH2 = CH– Cl CH2 – CH = Cl
Vinyl chloride
OH H O C
17. (c) o-Nitrophenol is not sufficiently strong acid so as to | | |
(iii) C O, (iv) C C OCH 3 ,
react with NaHCO3. | |
| |
18. (c) Because the layer of Al2O3 (oxide) is inert, insoluble O C H O
and impervious.
19. (a) Au, the gold is not attacked by acids and alkalis. It O C
forms AuCl3.AuCl3 further reacts with HCl to form | |
(v) C O
H[Au(Cl)4] which is used in photography. |
H
AuCl3 HCl H[Au(Cl)4 ]
Complex Arrange (NNN), (BrCC), (OOO), (CHH), (OOH) in
increasing atomic number. The order is ii, iii, v, i, iv.
20. (b) 2KI Cl 2 2KCl I 2
CHCl3
I2 CCl 4 Violet Colour 26. (a) Cl
KOH
Note: The excess of Cl2 should be avoided. The layer
may become colorless due to conversion of
HCl
I 2 to HIO 3 Cl NC [Y]
300 K
I 2 5Cl 2 6H 2 O 2HIO 3 10HCl
In case of Br 2 :
Cl + HCOOH
Br2 2H 2 O Cl 2 2HBrO 2HCl
21. (d) To convert covalent compounds into ionic compounds [X]
such as NaCN, Na2S, NaX, etc. 27. (a) Let the rate law be r = [A]x[B]y
2
CH2– C – H 33. (d) T2
2
r 3 law; T0 R 3 & Ts ( 7R)3 ;
O CH2– CH
Ozonolysis 2
28. (b) CH2 CH2 Ts
7 3 ; Ts 7 7 T0
O CH2– CH T0
CH2– C – H
34. (b) T (300K to 70K )
29. (b) N 2 (g ) 3H 2 (g) 2 NH 3(g) T R metal R semi - conductor
1–x 3 – 3x 2x at equilibrium (Al) (Si)
Total moles,
35. (d) (Source) motion , n both change, (Observer) motion
1 – x + 3 – 3x + 2x = 4 – 2x = 3 (given) only n change.
(Since, 4 moles = 4 atm given) 36. (b) As v > r therefore, vv < vr.
x = 0.5
T2 T2 1 5
37. (a) From =1 , 1 =1 ...(i)
3 T1 T1 6 6
p N2 p H2
K p for dissociation of NH 3
2
p NH 3 T2 62 2 2
In 2nd case : 1 ' 1 .....(ii)
T1 6 3
3
1 0 .5 3 3 0 .5
3 3 2 2 6 4
3 3 Using (i), T2 – 62 = T1 T2 T2
= 3 3 5 5
2
2 0 .5 3
3 or 1 T2 62, T2 = 310 K = 310 – 273 = 37°C
5
1 1 1 1 1 –1
( –1) – ( – 1) –
f1 R1 R 2 R R
10k
V
1 2( –1) 1 2( –1) R
F
32. (b) This of example of uniform circular motion. F R R 2( – 1)
2 7 Here R = 20 cm,
= 2 n 2 0.44 rad / sec . ;
T 100 20
F 20cm
V = R = 0.44 × 12 = 5.3 cm/sec. 2(1.5 –1)
FT-58 The Pattern Target AIEEE
41. (a) Pitch = 1 mm 50. (c) Einstein equation KEmax = E – Work function;
Number of divisions on circular scale = 200
1
Pitch mv 2 E W
L.C = 2
Number of divisions on circular scale Using this concept,
1 mm
= = 0.005 mm = 0.0005 cm 1
200
mV12 max
Diameter of the wire = (Main scale reading + Circular 1 .5 1
2 or V1 max 1
scale reading × L.C.) – zero error 1 2 2.5 .5 4 V2 max 2
V2 max
= 6 mm + 45 × 0.005 – (– 0.05) 2
= 6 mm + 0.225 mm + 0.05 mm = 6.275 mm
51. (a) First, let us check upto what value of F, both blocks
V 0.5 move together. Till friction becomes limiting, they will
42. (b) R 2 .
I 0.25 be moving together. Using the FBDs
43. (a) Angular retardation,
F a1
2 (n 2 n1 ) 2 (0 900 / 60)
2 1
rad / s 2 . f
t t 60 2
F
44. (a) A metre bridge is used to determine the resistance of a f
F
given wire and not to determine the emf of a cell.
15 kg
3 F
45. (d) t < 0, vi = 0 and t > 0, vf = m / s
4
a2
3
Impulse = m (vf –vi) 4 0 10 kg block will not slip over the 15 kg block till
4
acceleration of 15 kg block becomes maximum as it is
= 3 kg ms –1
created only by friction force exerted by 10 kg block on
3 it.
Little before 4 second v m / s, little after 4 second,
4 a1 a 2(max)
velocity becomes zero
Therefore, Impulse F f f
for limiting condition as f maximum is 60N.
3 10 15
m(vf vi ) 4 0 3kg ms 1
4 F = 100 N
46. (b) In reverse bias on p-n junction when high voltage is Therefore, for F = 80N, both will move together.
applied, electric break down of junction takes place, Their combined acceleration, by applying NLM using
resulting large increase in reverse current. This high both as system, F = 25a
voltage applied is called zener voltage.
80
47. (d) fc 9 Nm 9 9 1010 a 3.2 m / s 2
25
2.7 106 Hz 2.7 MHz 52. (b) If F = 120N, then there will be slipping, so using FBDs
48. (c) Conservation of Energy, of both (friction will be 60 N)
Photon e e energy For 10 kg block, 120 – 60 = 10a a = 6 m/s².
For 15 kg block, 60 = 50a a = 4 m/s².
EP E E K.E. of both e e 53. (c) Study of junction diode characteristics shows that the
re re
junction diode offers a low resistance path, when
We know that rest mass energy of e e is .5 MeV
forward biased and high resistance path when reverse
Hence E P .5 .5 .19 2 = 1.40MeV biased. This feature of the junction diode enables it to
49. (b) Let initial amount be 100 gm. be used as a rectifier.
disintegrated Left
a a
5 days 100 10 54. (a) µd 6, µ 3,
100 gm 10 90
100 a d
µd ? ; µd µ µa 1
Next 5 days 90 10
90 9 81
100 d 1
6 µ 1
Next 5 days 81 10 3
81 8.1 73
100
73 10 d 3 1
73
Next 5 days
7.3 65 µ ; µd 2
100 6 2
FT-59
is true.
5m
mg
55. (a) a = Magnetic field at half way due to Q wire
M
0I 2 0
Body BQ [upward × ]
m 3 mg
5 2
2
fr 2
M plate
fr [where I 2 2.5Amp. ]
Net magnetic field at half way
fr = mg = Ma
0 0 3 0
56. (d) When two rods are connected in series B BP BQ = (downward )
2 2
A(T1 T2 )t A(T1 T2 ) t 3
Q Hence net magnetic field at midpoint
0
d1 d 2 (d 1 d 2 ) / K 2
K1 K1 60. (d) From figure,
d1 d 2 d1 d2 (d 1 d 2 )
; K
K K1 K2 d1 d 2
K1 K 2
71. (a) Let the co-ordiante of other ends are (x, y, z).
63. (c) 2 tan 2 1 1
The centre of sphere is C(3, 6, 1)
cos 2 tan sin cos
x 2
Therefore, 3 x 4
sin2 = cos2 = sin (90-2 ) + = 2
4
64. (d) Sum of 7 can be obtained when (2,6), (3,5) (3, 6), (4, 4), y 3 z 5
6 y 9 and 1 z 3
(4,5), (4, 6)(5, 3)(5, 4) (5, 5)(5, 6)(6, 2)(6, 3)(6, 4)(6,5)(6, 6) 2 2
15 5
Probability of sum > 7 =
36 12 72. (d) P(E E) P(E E E E E )
4 8
n 5 1 5 1 5 1
Cr r. n r 1. n r 1 n r 1 ...
65. (b) n
. 6 6 6 6 6 6
Cr 1 r. n r n n r
3
5 5 30
1 ......
n r 1 36 6 91
n r 1
n r
73. (b) The ellipse
5
n (n 1) (n 2) (n 3) (n 4)
( x 3) 2 y2 16 3
r 1 1 1 e2 e .
5n 10 5(n 2) 16 25 25 5
74. (c) For x > 10, f (x) = x – 2.
3
66. (b) 14 17 i 14 14 15 16 Therefore, g(x) = x – 2 – 2 = x – 4
C7 C6 C7 C6 C6 C6
i 1
g (x) = 1.
15
C7 15
C6 16
C6 16
C7 16
C6 17
C7 75. (b) a, b, c in A.P. a + c = 2b; b, c, d in G.P..
2ce
dy bd = c2; c, d, e in H.P. d=
67. (d) y = A sin t. A cos t c e
dx
a c 2ce
d2 y 2 c2 (a c)e (c e)c c2 ae.
2
A sin t 2 c e
dx
Therefore, in G.P.
3
d y 3
A cos t 15
dx3 2 1
76. (b) 3x
x2
4
d y 4
4
A sin t
dx 1
r
15
Tr 1 C r (3x 2 )15 r
d5y x2
A 5 cos t A 5 sin t 15
dx 5 2 C r 315 r ( 1) r x 30 2r 2r
dx 5 4 3
78. (a) I = (x ) (x )( x) 1 1 1
[R2 R2 – 100R3 – 10R1]
x y z
Put x = sin 2 cos 2
which is zero provided x, y, z are in A.P.
[see the standard substitutions]
a
dx = 2( ) sin cos d 83. (d) y mx
m
Also, ( x ) ( ) cos 2
9/ 4
10 4m 1 16m 2 40m 9 0
(x ) ( ) sin 2 m
40 5 9
2( ) sin cos d m1 m 2 ; m1m2 =
I 16 2 16
( ) sin 2 ( ) sin cos
1 9
m1 , m2
2 d 2 2 4 4
cos ec d
sin 2 every parabola is symmetric about its axis only
Statement 1 is true.
2 2
( cot ) C cot C 1 1
84. (b) x 1, x 2 1,
x x2
Now , x = sin 2 cos 2 1 1 1
x3 2, x 4 x ,
x3 x4 x
x cos ec 2 cot 2
1 1
x5 1, x 6 2 , etc.
x (1 cot 2
) cot 2 x5 x6
2 3 2
1 1 1
x 2 x x x2 x3
cot ; I C x x2 x3
x x
2 2
79. (c) As the given spheres both have same cenre and 1 1
x4 x5
different radii therefore they are concentric and they x4 x5
do not have any point in common. Hence they do not 2
2 2
intersect. 1 1 3 9 1
x6 x7 +........+ ( x )
80. (a) Statement – II is true, as if f (x) = ln x, then x6 x7 (x 3 )9
1 = (1 + 1 + 4) + (1 + 1 + 4) + (1 + 1 + 4) + ......... 9 times
f (x) 0 (as x > 0, so that f (x) is defined) = 6 × 9 = 54.
x
85. (a) We have,
Statement – I is not true as 0 < ln x < 1, x (1, e) and
hence (ln x)n decreases as n is increasing. So that I n is
cos 2 cos sin cos 2 cos sin
a decreasing sequence. AB 2
cos sin sin cos sin sin 2
x
81. (d) f (x)
4 x2 cos 2 cos 2 cos cos sin sin
2
f (x) is increasing for – 2 x 0 and decreasing for cos sin cos sin 2 cos sin
0 x 2.
cos 2 cos sin cos sin sin 2
5 4 3 cos cos sin sin sin 2 sin 2
82. (d) x51 y41 z31
x y z cos cos cos sin
cos( )
sin cos sin sin
5 4 3 Since, AB = 0, cos( ) 0
100x 51 100y 41 100z 31
x y z is an odd multiple of
2
FT-62 The Pattern Target AIEEE
86. (c) When x > 3, f '(x) > 0; when 2 < x < 3, f '(x) < 0; when 1 <
dV
x < 2, f '(x) > 0; when x < 1, f '(x) < 0. = . 4 cub.cm./min.
dt
87. (b) Required area
We have to find out the rate of increase of water-level
1 1 1
dy
| y | dx | x 5 | dx 2 | x 5 | dx i.e. .
dt
1 1 0
1 6
1 Differentiating (1) with respect to t, we get
5 x 2 1
2 x dx 2 dV dy dy dy 16
6 6 3 .3y 2 . ; 4 y2. ; .
0 0 dt 12 dt 4 dt dt y2
88. (d) Let y be the level of water at time t and x the radius of dy 16 4
the surface and V, the volume of water. When y = 6 cm, cub.cm. / min .
dt 2 9
6
We know that the volume of cone
1 89. (d) Slope of the equations 4x2 – 9y2 = 36
(radius)2 × height
3
dy dy 4x 4x
8x 18y 0 or m1 =
D dx dx 9y 9y
C 5cm B
5
Slope of the straight line, 5x + 2y – 10 = 0 is m 2
2
Therefore, for the perpendicularity, m1m2 = –1
R
M 10 cm 4x 5 10x
Now, 1 y .
9y 2 9
y 10 x
Putting y in 4 x 2 9 y 2 36 gives imaginary
9
A x roots resulting in no tangents.
log 1 2x
V
1 2
x y. Let BAD 90. (a) e 2 x sec 2 e dx
3 log 2 3
tan
BD 5 1
. Put e 2 x t 2 e 2 x dx dt
AD 10 2
Again, from right angled AMR, we have 2 log 2 log 2
When x = log 2, t e =e
2
MR x 1 x y
tan ; ; x . When x = log , t = e 2 log
AR y 2 y 2
2 1 1 t
1 1 y 1 1
V .x 2 y . .y y2 ......(1). sec 2 t dt .
1
tan .
3 3 2 12 2 3 2 3 2
2 3
By question, the rate of change of volume
3 3 1
tan tan 3 3
2 3 6 2 3
FT-63
Ag Hg
Fe vessel
Ag Hg 24. (b) NO2 NO2
Silver amalgam
CH3ONa
Vessel made of other metal will form amalgam with
liberated mercury. X=
SNAr
Br Br
17. (b) (a) NH4NO2 N2 + 2H2O
Cl OCH3
(b) (NH4)2SO4 NH3 + H2SO4 25. (b)
(c) 2NH4ClO4 N2 + Cl2 + 2O2 + 4H2O O O
|| NaOH
|| SOCl2
(d) (NH4)2Cr2O7 N2 + Cr2O3 + 4H2O X C CH 3 X C OH
18. (d) (a) The size of the colloidal particles is in between the H
size of the molecule and the size of the particle of O O
a coarse suspension. || Ph NH 2
||
(b) White portion of the egg is condensed colloidal X C Cl X C NH Ph
solution whereas the sol prepared from this O
substance is a dilute colloidal solution.
(HNO3 H 2SO4 )
(c) Dialysis is a slow process. It may some times take X – C – NH NO2
several days for its completion.
FT-65
1 1 34. (c)
26. (a) N 2 (g) O 2 (g) NO(g) ; fH° = 90.2 mg = T cos 30° 30° T
2 2 R
R N
N 2 (g) O2 (g) 2NO(g) ; fH° = 90.2 × 2 ...... (1)
N = T sin 30°
2NO2 (g) ;
mg
2NO(g) O2 (g) fH° = –114....... (2) N
3
1 mg
2NO 2 (g) O2 (g) N 2 O5 (g) ;
2 35. (b) S1 : No work is done by net force, it only changes
102.6 direction of momentum of particle. Hence S1 is false.
fH° =
51.3 ....... (3) S2 : True by definition.
2
Eq. (1) + (2) + (3) S3 : Nothing is said about acceleration of both particles.
Hence angle between velocity and acceleration of
5 centre of mass may not be zero. Consequently centre
N 2 (g) O 2 (g) N 2 O5 (g)
2 of mass may not move along a straight line. Hence S3 is
fH° (N2O5, g) = 15.1 kJ/mol false.
x 1 3 m1v1 m 2 v 2 ..... mn v n Fnet
27. (d) k ; t x. ; t 3/4 .a S4 : Vcm
t k 4k m1 m 2 ..... m n (m1 m2 ..... m n )
28. (b) Anhydrous copper (II) chloride is a covalent while
anhydrous copper (II) fluoride is ionic in nature because Direction at Pnet is fixed so Vcm is also constant in the
in halides of transition metals, the ionic character direction. So path of CM will be straight line.
decreases with increase in atomic mass of the halogen. 36. (a) The linear relationship between v and x is
Fluorine being the most electronegative, forms ionic v = – mx + C where m and C are positive constants.
salt with copper.
dv
29. (c) P PAo x A PBo x B Acceleration, a v m ( mx C)
dx
3 2 a = m2x – mC
600 PAo PBo ; 3PA 2PB 3000
3 2 2 3 a
4.5 2
630 PAo PBo Hence the graph relating a to x is x
4.5 2 0.5 4.5 2 0.5
4.5 PAo 2 PBo 4410
1.5PA 1410 ; PA 940 and PB 90
30. (b) The redox system having the highest redox potential in 37. (a) Charge conservation is violated in [2, 3, 4], nucleon
the most powerful oxidising agent followed by the one conservation is violated in (4), (1) works.
with the lower redox potential. 38. (a) A convex mirror always forms a virtual image which
31. (a) p mv m 2gh can't be cast on a screen. Hence, a convex lens is used
to find the focal lenght of a convex mirror.
32. (a) Least count of vernier callipers
39. (c) For a resistor the pointer moves in both ways when
= value of one division of main scale – value of one
voltage is applied.
division of vernier scale
Now, N × a = (N + 1) a' t /T 7.5/T
A 1 1 1
(a' = value of one division of vernier scale) 40. (b) Using or
A0 2 32 2
N
a a 5 7.5/T
N 1 1 1 7.5
or or 5 i.e., T = 1.5 hours
a 2 2 T
Least count = a a
N 1 41. (a) Divide the ring into infinitely small lengths of mass
33. (b) Here I = 4A dm1. Even though mass distribution is non-uniform,
c c each mass dm1 is at same distance R from origin.
30 MI of ring about z-axis is
30
180 6 = dm1R2 + dm2 R2 + ..............+ dmnR2 = MR2
I 4 4 6 7 2 6 7 mv
Now, k 42. (a) r r m (v = same for both charges)
22 11 qB
6 43. (c) Back lash error is occurred when the screw is rotated
both in clockwise and anticlockwise direction.
84
7.6 A / rad Q net
11 44. (c) For dipole Qnet = 0 =0
0
FT-66 The Pattern Target AIEEE
second, the particle is at rest at extreme position. Hence
45. (b) D
instantaneous power at x = 2 at t = 1.25 sec is zero.
55. (a) As long as the block of mass m remains stationary, the
block of mass M released from rest comes down by
X Y
2Mg
(before coming it rest momentarily again). Thus
K
the maximum extension in the spring is
C
2Mg
x .......... (1)
Equivalent resistance = 5 k
46. (b) If temp of mixture (T) is not constant then according to For block of mass m to just move up the incline
calculation we find wrong value of specific heat. If we kx = mg sin + µ mg cos .......... (2)
stir continuously then temperature of whole mixture
becomes same. 3 3 4 3
2Mg mg mg or M m
5 4 5 5
47. (b) T ; red yellow, r y 56. (a) For a particle undergoing SHM with an amplitude A and
48. (a) Initial condition angular frequency , the maximum acceleration = 2A
Volume = 76cm × A Here the maximum force on the particle = QE0
Pressure = (76 + 76) dg = 152dg QE 0 2
Final condition maximum acceleration = A
Volume = 152 cm × A m
Pressure = 76dg QE 0
P1V1 = P2V2 A
m 2
Both points lie on the same isothermal line
i.e. both have same T. 1 hc
57. (d) mv 2
49. (a) Filling liquid will increase the optical path length by 2
same amount.
1 hc 4hc 4
CBF will not shift. m '2 . Clearly, v v
2 (3 / 4) 3 3
m 2
50. (c) I ; 58. (c) Process1 2 and Process3 4 are isochoric processes.
12 W12 = 0, W34 = 0, W23 = nR (T3 – T2) = 3R (1600 – 400)
d2 = 3600 R
Restoring troque = qE sin qE W41 = nR (T1 – T4) = 3R (200 – 800) = –1800 R
dt 2
W = (3600 – 1800) R = 1800R = 15 kJ
qE qE 12 2 1 2 1 2
I 2 T 59. (c) E1 kx , E 2 ky ,
m 2 2
m 1
T E k(x y) 2 E1 E 2 2 E1E 2
3qE 2
2 8 2 16 18J
dV
51. (a) E = V = 5x2 + 10x –9, 1 1 1
dx ( 1)
60. (c) f R1 R2
d
E= (5x2 + 10x – 9) = – (10x + 10) R = 10 cm.
dx
On putting value of x in it we get, 1 1 10
E = –(10 × 1 + 10) = –20 V/m f (3 1)
10 10 4
52. (c) The equation of wave moving in negative x-direction, f=20 f' f=20
assuming origin of position at x = 2 and origin of time 1 1 4 1 2 4
(i.e. initial time) at t = 1 sec. f eq 20 10 20 20 10
y = 0.1 sin (2 t + 4x)
Shifting the origin of position to left by 2m, that is, to x 10
= 0. Also shifting the origin of time backwards by 1 sec, f eq cm
3
that is to t = 0 sec. 61. (d) It passes through a fixed point (3, 4)
y = 0.1 sin (2 (t – 1) + 4 (x – 2)) Slope of line joining (3, 4) and (1, –2) is –6/–2 = 3
53. (a) As given the particle at x = 2 is at mean position at Slope of required line = –1/3
t = 1 sec.
Its velocity v = A = 2 × 0.1 = 0.2 m/s. 1
Equation is y – 4 = (x 3)
2 2 3
54. (d) Time period of oscillation T 1 sec. x + 3y – 15 = 0
2
Hence, t = 1.25 sec, that is, at T/4 seconds after t = 1 f (3 h) f (3)
62. (b) f (3 ) lim
h 0 h
FT-67
(2 e h ) 1 eh 1 2 d
lim lim 1 and also I
h 0 h h 0 h 1 1 tan
2
f (3 h) f (3) 1002 501
f (3 ) lim 2I d 2 1 I
h 0 h 2008 2008
1
10 (3 h)2 1 1 (6h h 2 ) 1 Hence, K = 2008.
lim lim
h 0 h h 0 h y am
68. (c) y2 4a i.e., my2 – 4ay – 4a2m = 0
m
6h h 2
m 0; 16a2 + 16a2m2 > 0 which is true m.
= hlim0
h( 1 6h h 2 1) m R – {0}
n
h (h 6) 6 Ck 1 n! (k 1)!(n k 1)! 1
lim 3 69. (a) n
2 Ck 2 k! (n k)! n! 2
h( 1 6h h 2 1)
h 0 1
1 0 1 xi
their mean x i 1
8
6
2 y z 2 y z
0 0 6 6
x 2 2 y 2 z 2 x 2 y 2 z xi 8 6 xi 48
65. (b) Let A be the event such that sum is Rs. 20 or more i 1 i 1
P (1) = 1 – P (Total value is < 20) On multiplying each observation by 3, we get the new
6 2 observations as 3x1, 3x2, 3x3, 3x4, 3x5 and 3x6.
C2 C2 14 1 1
=1 8
1 1 6 6
C2 28 2 2 3x i 3 xi
i 1 i 1
1
2
2
2
3x 4y 1
2 Now, their mean = x
x y ( 2
4 4) 6 6
66. (c)
5 5 5 3 48
x 24
2 2 6
1 2 3x 4y 1 Variance of new observations
i.e., x y | 2|
5 5 5 6 6
is an ellipse. (3x i 24) 2 32 (x i 8) 2
If 0 < | – 2 | < 1 i.e., (1, 2) (2, 3) = i 1 i 1
6 6
67. (d) 1 2 9
2 = Variance of old observations = 9 × 42 = 144
1
2
d 2
tan d Thus, standard deviation of new observations
I = Variance
1 tan 144 12
11 tan 1
2
FT-68 The Pattern Target AIEEE
71. (b) p (p q) is equivalent to p. 78. (b) n = 3, P (success) = P (HT or TH) = 1/2
3
x cos x sin x 1
esin x p = q = and r = 2
72. (a) dx 2
cos2 x
2
1 1 3
= e sin x
x cos x dx – e sin x
tan x sec x dx P (r = 2) = 3C2 .
2 2 8
Excess of
1 1
1. (c) RcZ2 NaOH
[Al(OH)4 ]
n12 n 22 soluble
1 1 5. (b)
1 RcZ2 RcZ2 H2(g) + CO2(g) CO(g) + H2O (g)
12 2
At eqm 0.25–x 0.25–x x x
1 1 3RcZ 2 x2 x
2 RcZ2 2 2 Kp = 0.16 = 0.4 0.1 0.4x x
1 2 4 (0.25 x) 2 0.25 x
1 1 RcZ2 x = 0.0714
3 RcZ2 2 2
2 4 0.0714
Mole% of CO (g) = 100 14.28
1 2 3 0.50
2. (a)
3° 3° Anhydrous HI
6. (a) (I)
O SN2 I
+ HO
(2 H)
N N
2°
Conc. HI
T.S. having carbanion character so Hoffmann alkene is (II)
major product. SN2 I
OH
7. (c) By intermittently filled lime the blown air in the sinterer
3. (d) X (g) X (g) ; Energy = IE + IA = 16.8 eV gets diffused and PbS particles come under limited
IA EA 16.8 supply of air therefore formation of PbSO4 is prevented.
EN pauling 3 3
5.6 5.6 PbS +O PbO + SO2
4. (b) K2SO4.Al2 (SO4 )3.24H2O 2 2
K2SO4 Al2 (SO4 )3 24H2O
swells PbO + SiO2 PbSiO3
Amorphous powder gangue
K 2SO4 .Al2 (SO 4 )3 .24H 2 O 8NaOH 2KOH 2Al(OH)3 4Na 2SO4 PbSiO3 + CaO CaSiO3 + PbO
soluble (white ppt.) soluble more basic less basic
FT-71
OH P X
15. (c) PNH3 PH 2S
2 2
Polymerisation
8. (a) + HCHO Bakellite X
2
X
G RT ln K p RT ln = 2RT ln
2 2
1000 = – 2RT (ln X – ln 2)
9. (a) K (NH 4 )2 S2
N 16. (c)(a) CdS (NH 4 ) 2 S CdS CdS
Yellow Yellow Insoluble
1000
1.50 × 10–4 × 104 = 9 × 10–6 × 10–2 × (b) SnS2 (NH 4 )2 S SnS2
(NH4 )2 S2
(NH4 ) 2 SnS3
N
Yellow Soluble
N = 6 × 10–5
( NH 4 )2 S2
5 (c) Cd 2 (NH 4 )2 S CdS CdS
N 6 10 5 Yellow Insoluble
S M 2 10 mol / L
nf 3 (NH 4 ) 2 S2
(d) Sn 2 (NH 4 ) 2 S SnS (NH4 ) 2 SnS3
Ag3PO4 = 3Ag+ + PO4 Brown (Amm. thiostannate)
Soluble
3S S
Ksp = (3S)3.S = 27S4 = 27 × (2 × 10–5)4 = 4.32 × 10–18 17. (c)
10. (a) (Mg powder + BaO2) acts as ignition mixture e e f f
A f A f A e A e
Fe, Cr are obtained in molten state
Enthalpy of formation of Al2O3 is negative. M M M M
B D B C B D B C
11. (d)
C D C D
(a) For tetrahedral d6 ion, e e f f
B f B f B e B e
4 unpaired electrons M M M M
A D A C A D A C
(b) For [Co(H2O)6 ]3+, C D C D
0 unpaired electrons e e
A C A D
(c) For square planar d7 ion, , M M
B D B C
1 unpaired electrons f f
(d) B.M. = n(n 2) , n = unpaired electrons 18. (b) A (g) + 2B (g) C (g) + D (g)
At eqn 0.5–x 1 – 2x 0.5+x 3.5+x
5.92 B.M. = n(n 2) y 2y 1 4
n = 5 unpaired electrons (due to very high value of equilibrium constant)
12. (c) Ag (1) Ag+ (aq) + A– (aq) 1 4
1012 ; y = 10– 4; [B(g)] = 2y 2 × 10– 4
x x–y y (2y) 2
Ag (1) + H2O ( ) HA (aq) + OH– (aq)
x–y y y 19. (b) SO32 2Al 2OH 3H 2O S2 2 [Al(OH) 4 ]
Ksp = x (x – y) H2S does not evolve as liberated H2S is neutralized by
NaOH and Na2S is formed.
Kw y2 10 14
(10 5 ) 2
Kh ; 20. (b)
Ka (x y) 10 (x y)
10
COOK
x – y = 10–6 COOH
aq. KOH
x = 10–5 + 10–6 Me – CH = CH2 + CHCl3 CH3CH – CH3
x = 1.1 × 10–5 CCl2
CH3 – CH – CH2
33. (b) Voltmeter measures voltage across two points so it is 42. (a) First, the length of wire goes on increasing i.e., area
connected in parallel and ammeter measures current so decreases and finally at breaking stress the wire breaks.
it is connected in series.
34. (c) Larger the length, lesser will be the potential gradient,
so more balancing length will be required.
//////////// 43. (d) For circular wire ×
T
N = (m1g + m2g) – T
L
if T = 0 N = (m1g + m2g) I
if T > 0 N < m1g + m2g I I
and T cannot be negative.
36. (b) If student will use angular momentum = mvr. As it can be easily seen by the direction of I that Q is
He/she may conclude answer (1) as r is decreasing decreasing thus, energy of capacitor is decreasing and
angular momentum must decrease hence (1) is incorrect. 1 2
hence, energy of inductance is increasing or LI
m 2
v
gives that I is increasing.
mg r 45. (b) Student having misconcepts in refraction will mark
option (a)
/////////////////////////////////////////
No effect of refraction on light coming from centre,
O because it is along normal.
d
The magnitude of angular momentum of particle about 46. (d) Voltage gets distributed among all capacitors.
O = mvd 47. (b) Both statements are true and statement-2 is a correct
Since speed v of particle increases, its angular explanation for statement-1.
momentum about O increases. 48. (b) Since F inside is zero so work done to move the charge
Magnitude of torque of gravitational force about is also zero hence potential remains constant.
O = mgd constant 49. (a) E j as area and current are same j is same but are
Moment of inertia of particle about O = mr2 different. E is different.
Hence MI of particle about O decreases.
v sin R 1R 2
Angular velocity of particle about O 50. (b) Req = ( Parallel combination);
r R1 R 2
v and sin increases and r decreases. Req = R1+R2(Series combination)
angular velocity of particle about O increases.
37. (c) Pressure difference is largest between atmosphere and 5 15 125 75 5
Req of the circuit = = =5
smaller bubbles. Hence radius of curvature (R) is smallest. 5 15 100 20 4
38. (a) If student will use vector
R1 = 5
component (only mathe-matically) I1
he/she may conclude both are //////// //////// R 2 = 15 4A
A
R 3 = 1.25
correct. Hence the option (c). vB
Correct concept : Velocity of both 4A 4 – I1
ends of the string along the length vAcos vA
of the string should be same. vB
vB = vA cos . vB
E
//////// We know that I = = 20 = 4 A
39. (d) µI < µII < µIII < µIV R eq 5
vB
I > II > III > IV Potential difference across R1 and R2 are same
Blue < Green < Yellow < Red
40. (a) Physically possible explosions are those in which both (Parallel combination) I1R1 = (4 – I1)R2
particles move in opposite directions. i.e. signs of 5I1 = (4 – I1) 15 I1 = 12–3I1 I1 = 3A
velocities are opposite II, IV & V Thus reading of ammeter = 3A
41. (a) The surface of the calorimeter is exposed to the air. So Voltage across 1.25 = I R = 4 1.25 = 5V (Reading
heat is lost from the surface by radiation. of voltmeter)
FT-74 The Pattern Target AIEEE
51. (a) Optical source frequency 59. (d) Current in the inductor before opening ‘S’
L I S
c
f 3 108 /(800 10 9 ) 3.8 1014 Hz 12 9
I 4.5A
Bandwidth of channel (1% of above) = 3.8 × 1012Hz 4 8 2 R1 =4 R2 =8
12V
4 8
Number of channels = (Total bandwidth of channel)
/ (Bandwidth needed per channel)
(a) Number of channels for audio signal Since current in inductor does not
change instantly, therefore, just
L
(3.8 1012 ) /(8 103 ) ~ 4.8 108 after opening ‘S’, A B I
Bi T0 V T0 V0 / 8 1 1 3
61. (d) Centre of the circle is , .
Bf V0 T V0 4T0 32 2 2
54. (d) For adiabatic process dQ = 0 Its distance from the line x + y – 1 = 0 is 2
dU + dW = 0 Let the required line be mx – y = 0
55. (b) A = 0.1m, m = 0.1 kg, KEmax = 18 × 10–3 J, m 3
4
2 2 2 m = 1, –1/7
3
36 10 k 3.6 m2 1
k 3.6 ; 6 rad / s
(0.1) 2 m 0.1
The lines are x – y = 0, x + 7y = 0
Eqn. y = 0.1 sin 6t 1 0
4 62. (b) PQ
0 1
v v v
56. (b) n0 , n1 , n2 63. (b) Let f (x) = sin x + x cos x
2 2( /2 ) 2( /2 ) x
Beat frequency = n 1 – n 2 Consider g (x) = (sin t t cos t) dt t sin t]0x x sin x
1 1 0
( 2 ) ( 2 )
v = v g (x) = x sin x which is differentiable
2 2 2 2
4 Now, g (0) = 0 and g ( ) = 0, using Rolles Theorem
4 8 v 8 n0 atleast one c (0, ) such that g '(c) = 0
=v 2 2 i.e. c cos c + sin c = 0 for atleast one c (0, )
4 2 r
A B y
57. (d) As flux decreases to maintain flux, 64. (b) Required area = 2 y dx
current in loop is clockwise. Force 3
on DA due to long wire is towards I v =2 12 x dx 24
A S(3,0) x
left while on wire BC 0
towards right. D C
58. (d) If student use, f = µ mg cos
65. (a) Number of digits are 9
1 3 Select 2 places for the digit 1 and 2 in 9C2 ways
= mg cos 30 = mg from the remaining 7 places select any two places for 3
2 4
and 4 in 7C2 ways and from the remaining 5 places
Hence option (b) but correct answer is (d).
select any two for 5 and 6 in 5C2 ways
All forces on sphere pass through its centre except the
Now, the remaining 3 digits can be filled in 3! ways
force of friction exerted by inclined plane. Since net torque Total ways = 9C2 . 7C2 . 5C2 . 3!
on sphere in equilibrium about its centre is zero, the
9! 7! 5!
torque on sphere due to frictional force about its centre = . . .3! = 9.7!
must be zero. Hence frictional force on cylinder is zero. 2!.7! 2!.5! 2!.3!
FT-75
16 x [0, 1]
66. (b) I f (t) dt
0 1 x2
x 2x
x4 2
Consider g(x) f (t) dt g(0) 0 1 + x2 = 2x x=1 | 10a | = 10
0 | 10a | = 10, | 20 2 30 |
LMVT for g in [0, 1] gives, some (0, 1) such that
[ | 10a | ] = 1, 10
g(1) g(0)
g( ) .......... (1) 69. (b) Let y = mx be any line represented by the equation
1 0 ax3 + bx2y + cxy2 + dy3 = 0
Similarly, LMVT in [1, 2] gives, some (1, 2) such ax3 + bx2(mx) + cx (m2x2) + dm3x3 = 0
a + bm + cm2 + dm3 = 0 which is a cubic equation.
g(2) g(1)
that g ( ) .......... (2) It represents three lines out of which two are
2 1 perpendicular hence
Eq. (1) + Eq. (2)
g ( ) g ( ) g(2) g(0) ; but g' (x) = f (x4) . 4x3 a a
m1m2 = –1 and m1m2m3 = m3
zero d d
16 and m3 is the root of the given equation
3 4 3 4 f (t) dt
4 f( ) f( ) 2 3
a a a
0 Hence, a b c d 0
d d d
3u 2 2u 3 1 d2 + bd + ca + a2 = 0
3v2 2v 3 1 0 70. (c) x2 – 2x cos + 1 = 0,
67. (a)
3w 2 2w 3 1 2 cos 4 cos 2 4
x cos i sin
R1 R1 – R2 and R2 R2 – R3 2
Let x = cos + i sin
u2 v2 u3 v3 0 x2n – 2xn cos n + 1
2 2 3 3 = cos 2n + i sin 2n – 2 (cos n + i sin n ) cos n + 1
v w v w 0 0
= cos 2n + 1 – 2 cos2 n + i (sin 2n – 2 sin n cos n )
2 3
w w 1 =0+i0=0
8 A Bx C
u v u2 v2 vu 0 71. (a) Let 2 ........ (i)
(x 2)(x 4) x 2 x2 4
v w v2 w2 vw 0 0 Then, 8= A (x2 + 4) + (Bx + C) (x + 2) ........(ii)
w2 w3 1 Putting x + 2 = 0 i.e. x = – 2 in (ii), we get 8 = 8A A = 1
Putting x = 0 and 1 respectively in (ii), we get 8 = 4A + 2C and
R1 R1 – R2 8 = 5A + 3B + 3C
Solving these equation, we obtain A =1, C = 2, B = – 1
u w (u 2 w 2 ) v(u w) 0 Substituting the values of A, B and C in (i) , we obtain
v w v2 w2 vw 0 0 8 1 x 2
2
w 2
w 3
1 (x 2)(x 4) x 2 x2 4
8 1 x 2
1 u w v 0 dx = dx dx
2 2 (x 2)(x 2 4) (x 2) x2 4
v w v w vw 0 0
w2 w3 1 1 x 1
= dx 2
dx 2 2
dx
(x 2) x 4 x 4
(v2 + w2 + vw) – (v + w) [(v + w) + u] = 0
v2 + w2 + vw = (v + w)2 + u (v + w) 1 1 1 1 x
uv + vw + wu = 0 = log | x + 2 | – dt 2. tan C, (where t = x2 + 4)
2 t 2 2
68. (d) x [–1, 0]
1 x
1 x2 = log | x + 2 | – log t + tan–1 2 + C
x 2x 2
2
x2 + 6x + 1 = 0 1 x
= log | x + 2 | – log (x2 + 4) + tan–1 2 + C
x 2 2 3 | 10a | [| 20 2 30 |] 30 20 2 2
FT-76 The Pattern Target AIEEE
77. (c)
72. (b) n 7iˆ 2jˆ kˆ is normal to plane
1 1 1 1
(Assuming n aiˆ bjˆ ckˆ and using P(8,2,–12)
1 1 2 1 2 3 1 2 3 4
.....,
n
n.AB 0, n.BC 0, n.AC 0) 1 2 1 1
P = (8, 2, – 12) tn 2
1 2 3 .....n n(n 1) n n 1
2
= A (A.B)A (A.A)A B .C = – | A | [A B C] T : ty = x + at2 ......... (1)
zero
Line perpendicular to (1) through
(a, 0)
Now, | A |2 4 9 36 49 tx + y = ta ......... (2)
2
2 3 6 Equation of OP : y x 0 ......... (3)
t
[A B C] 1 1 2
1 2 1 From equations (2) and (3) eliminating t we get locus
= 2 (1 + 4) – 1 (3 – 12) + 1 (–6 – 6) 6R
= 10 + 9 – 12 = 7 79. (a) Box 5B
2 4W
| A | [A B C] 49 7 343
P(E) = P(R R B W or B B R W or W W R B)
3 /4 3 /4 n (E) = 6C2 · 5C1 · 4C1 + 5C2 · 6C1 · 4C1 + 4C2 · 6C1 · 5C1
74. (a) I (sin x cos x) dx x (sin x cos x) dx n (S) = 15C4
0 0 I II 720 4! 48
P (E) =
3 /4 15 14 13 12 91
= (sin x cos x) dx x( cos x sin x) |30 /4 80. (a) ix2 + (1 + i) x + i = 0
0 zero =1
Im ( ) = 0
3 /4
(sin x cos x) dx dy
81. (c) Line touches the curve at (0, b) and also
0 dx x 0
3 /4 dy
=2 (sin x cos x) dx 2 ( 2 1) exists but even if fails to exist tangents line can be
dx
0
drawn.
75. (a) A divides C1C2 externally in the ratio 1 : 3. 82. (b) H
B
1 3 (AK) AB A
cos P O Q
A(–3,0)
C1(–1,0) B C2(3,0) (OA) AK
K
AP AQ 99
(AP) (AQ) = AB (from (1)) B2 : person does not suffer P(B2) =
2 100
2 (AP) (AQ) 99 1
AB P(A/B1) = , P(A/B2) =
(AP AQ) 100 100
83. (b) Truth table has been given below : P(B1 ).P(A / B1 )
P (B1 / A)
P(B1 ).P(A / B1 ) P(B2 ).P(A / B 2 )
p q ~ p p q (p q) ~p ~p q
T T F T F F 1 99
T F F T F F 100 100 99 1
50%
= 1 99 99 1 2 99 2
F T T T T T
100 100 100 100
F F T F F F
84. (b) Clearly, tan [x] = 0 for all x R and period of sin 4 {x} = 1. 88. (a) Since, APB = AQB =
so y = mx + 8 intersect the
2
1 1 1 1 circle whose diameter is AB.
85. (b) 1 1 2 2 Q
2 2 Equation of circle is x2 + y2 = 16
CD < 4 D
1 1 1 1 P A
3 3 .... n n 8
2 2 4 1 m2 4 C
1 m2 B
1 1
Consider r r 2 m ( , 3) ( 3, )
If the line passing throw the point A (–4, 0), B (4, 0)
= r2 ( 2
) r 1 (r 2 r 1)
then APB = AQB = does not formed.
n 2
n(n 1)(2n 1) n(n 1)
= (r 2 r 1) n m ±2
r 1 6 2 89. (d)
(a) Tr (kA) = k (a11 + a22 + a33) = kTr(1)
n n n(n 2 2) (b) Tr (A + B) = a11 + b11 + a22 + b22 + a33 + b33
= [2n2 + 3n + 1 – 3n – 3 + 6] = (2n2 + 4) =
6 6 3 = Tr (1) + Tr (2)
(c) Tr (I3) = 1 + 1 + 1 = 3
86. (d) Let the equation of tangent y mx a 2 m2 b2
(d) Tr (A2) = 2
a11 2
a12 (a11 a 22 a 33 ) 2
Foci (± ae, 0), vertices (± a, 0), C (0, 0)
/2
mae a m 2 2
b 2
mae a m2 2
b 2
90. (a) an (1 sin t) n sin 2t dt
s , s 0
1 m2 1 m2
Let 1 – sin t = u – cos t dt = du
1 1 1
ma a 2 m2 b2 ma a 2 m2 b2 1 1
= 2 u n (1 u) du 2 u n du u n 1du 2
a , a , n 1 n 2
1 m2 1 m2 0 0 0
an 1 1
Hence, n 2
2 2 2 n(n 1) n(n 2)
a m b
c
2 n
1 m an 1 1 1 1 1
lim 2
n 1 n n n 1 2 n n 2
m2 a 2e2 n n
1 1 1 1
ss c2 1 m2 2 =2
e n n 1 n n 2
aa c2 m2a 2 1 1
1 m2 1 1 1 1 1 3
= 2(1) 1 ..... 2
87. (c) A: blood result says positive about the disease 3 2 4 3 5 2 2
1
B1: Person suffers from the disease P(B1) =
100
FT-78 The Pattern Target AIEEE
CH3 25. (b) Hydrated CoCl2. 6H2O is pink coloured and contains
| +– octahedral [Co(H2O)6 ]2+ ions. If this is partially
CH3 — C — Cl + N a O CH2CH3 E2 CH3 — C = CH2 dehydrated by heating, then blue coloured tetrahedral
| |
CH3 CH3 ions [Co(H2O)4]2+ are formed.
[Co(H 2 O) 6 ]2 [Co(H 2 O) 4 ]2 2H 2 O
16. (b) s0 K sp ; s1 K sp / 0.02 M; s2 Ksp / 0.01 M;
pink blue
s3 K sp / 0.05 M
H 3O HCO 3
Obviously s0 s2 s1 s3 26. (b) K
2
As pH = 6.0[H3O]+ =10–6
CO 2 H 2 O
H O OH
H 3O HCO 3
17. (d) H3C C N H3C C N K (H2O is in excess, therefore its
2
O
CO 2 H 2 O
H H O
conc. remains constant)
Rearrangement
18. (b) C6 H5 NH 2 NaNO2 HCl C6 H 5 N NCl HCO 3 6
0 C K 3.8 10
6 = 3.8
OH CO 2 H 3O 10
N 2 Cl +
27. (c) In (NH4)2 [(TiCl6)], Ti4+ (3d0 4s0) has no unpaired
electrons.
OH In K2Cr2O7, Cr6+ (3p6 d0) has no unpaired electrons.
N=N In CoSO4, Co2+ (d7) has unpaired electrons in d-orbitals,
so it is both paramagnetic and coloured.
Red dye In K3[Cu(CN)4], Cu+ (3d10), has no unpaired electron.
19. (c) P PS w / m ; (640–600)/640 = wM/mW Ea 1
28. (b) log K log A (Arrhenius equation)
P W/M 2.303R T
40/640 = 2.175 78/m 39.08 Plot of log K Vs 1/T gives a straight line with slope
m = 2.175 78 640 / 39.08 40 = 69.458 69.60 –Ea/2.303R
20. (c) Multiple bonds formation tendency with carbon and 29. (b) • Li does not form peroxide or superoxide due to it
nitrogen decreases from sulphur to tellurium. small size.
CS2 (S = C = S) is moderately stable, • Solubility of carbonates and biocarbonates
CSe2 (Se = C = Se) decomposes readily whereas, increases on moving down the group.
CTe2 (Te = C = Te) does not exist
• The increasing order of size of hydrated ions of
21. (d) Liquation is the principle based on difference in melting
alkali metals is Li+ > Na+ > K+ > Rb+ > Cs+
points.
• Cesium used in photoelectric cells due to its low
22. (c) In NO2 + odd (unpaired) electron is removed. In
peroxides (O22–) no unpaired electrons are persent as I.E. Hence statements (b) is the only choice correct.
the antibonding pi M.O.’s acquired one more electron 30. (d) Cell reaction Zn Cu Zn Cu
each for pairing. AlO2 containing Al 3+ (2s 2 p6 ) 0.059 0.01
E1 E cell log
configuration and 2 oxides (O2–) ions each of which 2 1.0
does not contain unpaired electron. Superoxide O2–
E1 (E cell 0.059) V
has one unpaired electron in pi antibonding M.O.
23. (a) The two solutions are isotonic hence there will be no 0.059 1.0
movement of H2O. E2 E cell log
2 0.01
24. (b) Nickel salts reacts with dimethyl glyoxime in presence
of NH4OH to give red ppt. of nickel dimethyl glyoxime. E2 (E cell 0.059) V . Thus, E1 > E2.
CH3 – C = NOH
2 + NiCl2 + 2NH4OH 31. (d) Let the scooterist velocity be v. Then
CH3 – C = NOH 1000 + (10 × 100) = v × 100
OH O 2000
100 v = 2000 v 20m / s
CH3 – C = N N = C – CH3 100
Ni + 2NH4Cl + 2H2O 32. (b) We have, F = kx
CH3 – C = N N = C – CH3 where, F, x and k are force, length and constant
O OH respectively.
Nickel dimethylglyoxime (red ppt.)
FT-80 The Pattern Target AIEEE
5 = kx ......(1) or
and 7 = ky ......(2) The rate of cooling decreases with decrease in
Multiplying eq (2) by 2 temperature difference between body and surrounding.
14 = 2ky ......(3)
43. (c) 1 1 2 3C 2C
Subtracting eq (1) from (3),
14 – 5 = 2ky – kx or 9 = k(2y – x) C C C C C C C
Hence, required length = 2y – x
33. (b) Here, mA = 0.5kg ; mB = 1kg 2 1 3
2C 2CC C2 0 C C
2
T
µmAg 44. (b) R increases with increasing temp:
µmAg V = IR
F T
µ(mA + mB)g I 1 1
Slope of graph = = ; Slope of T1 is more i.e R is
Force on block A V R 1
This relation is independent of L. 46. (d) The orbital velocity, if a satellite close to earth is
v rel v0 gR e , While the escape velocity for a body
38. (d) rod point ,
r
vrel represents the velocity of one point w.r.t. other. thrown from the earth’s surface is ve 2 gRe .
3v v v0 gRe 1
= and ‘r’ being the distance between them. Thus, =
r ve 2 gRe 2
2v
=
r or ve 2v0
39. (d) Both are diatomic gases and CP – CV = R for all gases. i.e., if the orbital velocity of a satellite revolving close
40. (c) Here the number of molecules is same. Hence,
to the earth happens to increase to 2 times, the
T T2 200 400 satellite would escape.
Tfinal = 1 = = 300 K
2 2
I0 0 Isin
41. (a) During the time of collision between two elastic bodies, 47. (c) B (sin sin )
elastic potential energy increases as a result of decrease 4 a 2 a
in intermolecular space. Hence, kinetic energy decreases B at any point on Y axis is inversely proportional to a.
during that time. Note that kinetic energy doesn't change a Y
after collision.
42. (b) More the initial temperature more is the rate of cooling.
Hence, T3 > T2 > T1 P O
X
–a/2 I a/2
FT-81
48. (b) Power of source = EI = 240 × 0.7 = 166 55. (b) The process of changing the frequency of a carrier
wave (modulated wave) in accordance with the audio
140 frequency signal (modulating wave) is known as
Efficiency 83.3%
166 frequency modulation (FM).
V0 100 V0 100 0.693
49. (d) iR 5 , iL 10 and 56. (d) 1. 2. R Nt
R 20 XL 10 t1/ 2
Radioactivity at T1 is R1 = N1,
V0 100 Radioactivity at T2 is R2 = N2
iC 5
XC 20 Number of atoms decayed in time
R R2 ( R1 R2 )T
Current, i i 2R (iC iL )2 52 52 5 2 amp. (T1 – T2) = (N1 – N2) or 1
0.693
50. (c) Magnetic induction at O due to coil Y is given by, i.e., (R1 – R2)T
2 I(2 r) 2 57. (b) The figure of merit of a galvanometer is the current
0
BY ......(1) required to produce unit deflection in the galvanometer.
4 3/ 2
(2r)2 d2 Mathematically,
K = I/ , where k is the figure of merit of the
Similarly, the magnetic induction at O due to coil X is galvanometer.
given by 58. (d) In the graph given, slope of curve 2 is greater than the
2 slope of curve 1.
0 2 Ir
BX ......(2) P P
4 2 2 3/ 2
r d/2 V V 2 1
2 1
BY 1 He O2
From eq. (1) & (2)
BX 2 Since, monoatomic diatomic
51. (a) f max = mg, a max = g. Hence, curve 2 corresponds to helium and curve 1
corresponds to oxygen.
2 2
If A is the amplitude a max = A 4 AV 2 g.
1 Q1 Q2
g 59. (b) VA = ;
Therefore, A . 4 0 R 2R
4 2V2
52. (c) Total time taken to travel distance d is : 1 Q2 Q1
VB=
4 0 R 2R
d d n1 n 2 d
d ;
2n 1 2n 2 2n 1 n 2 n eff
2R 2R
3 R R
n 2 3n 1 n eff n1
2
53. (b) On the screen, we have four amplitudes pair wise
A R B
coherent. A1 A 2 A 3 A 4 A12 A 34
However, if A12 and A34 have equal magnitude because
of random phase of A12 and A34, no fringes will be
seen.
nh h 1 Q2 Q1
54. (a) mvr , ; VA– VB = Q1 – Q2 –
2 mv 4 R
0 2 2
nh Work done = Q × V = q × (VA– VB)
Using the two concept we get, mvr (where n = 1)
2
q Q2 Q1
1 h = 4 Q1 – Q2 –
2 r ....(1); 0 R 2 2
mv
h q 1
....(2) 2Q1 Q 2 – 2Q 2 – Q1
mv 4 0 R 2
2 r h mv 1
Divide (2) by (1), 1:1 q Q1 – Q 2 ( 2 – 1)
mv h 1
( 24 0 R)
FT-82 The Pattern Target AIEEE
(a 2 p 2 2abp b2 ) (b 2 p 2 2bcp c2 )
f f
(c 2 p 2 2cdp d2 ) 0
I is divisible by 20n on using statement - 2.
(ap b) 2
(bp c) 2
(cp d) 2
0 65. (b) cos36°cos42°cos78°
= cos36°cos(60° – 18°)cos(60° + 18°)
ap b 0, bp c 0 & cp d 0
5 1
= cos 2 60 sin 2 18
b c d 4
a, b, c and d are in G.P
a b c
2
Also ad = bc 5 1 1 5 1
63. (c) 4 4 4
1
(a) log( a 2 b) log( a 2 b) 2
2 5 1 1 5 1 5 1 2 5
=
1 4 4 4 16
2
= log(a 4b 2 4ab)
2
2 2
5 1 5 1 5 1 5 1 5 1
1 = = 1
= log(12ab 4ab) 16 64 16 4
2
1
log( 2 4 . ab ) 5 1 4 6 2 5 1
2 =
16 4 8
1 66. (c) The given expression is equal to
(4 log 2 log a log b)
2
1 1 1
cos(cos x sin x sin x)
log x log y log z
(b) Let k 1 1 1
b c c a a b cos sin x sin(sin x) x
2 5
log x k (b c), log y k(c a ),
log z k(a b) 1 1
[Using cos x sin x ]
2
x a . yb . z c p k[ a (b c )) b (c a ) c (a b)]
67. (b) – .
k ( 0) 2 2
=p 1
where p is any arbitrary base of the log.
so that cot cot
(c) Given expression 2
log xyz xy log xyz yz log xyz zx
cot cot 1 1
log xyz ( xy . yz . zx) log xyz ( x 2 . y 2 . z 2 ) cot cot cot
x2 y2 – x 1, x 1
68. (a) The hyperbola is 1 . It’s asymptotes are 74. (d) f (x) = | x – 1| =
9 4 x – 1, x 1
2 Consider f (x2) = (f (x))2
y x . Clearly the given line is parallel to
3 If it is true it should be x
2 Put x = 2
y x , hence intersects the hyperbola at exactly
3 LHS = f (22) = |4 – 1| = 3
one point. RHS = ( f (2))2 = 1
69. (b) Let tan 1 , tan 2 be the roots of the equation (a) is not correct
2tan2 – 4tan + 1 = 0. Thus Consider f (x + y) = f (x) + f (y)
tan 1+ tan 2 = 4/2 = 2; tan 1tan 2 = 1/2. Put x = 2, y = 5 we get
Now tan( 1+ 2) = [(tan 1 + tan 2)/ (1 – tan 1tan 2)] f (7) = 6; f (2) + f (5) = 1 + 4 = 5
= 2/[1 – (1/2)] = 4. (b) is not correct
70. (a) arg (z1) = arg (z2) Consider f (| x |) = | f (x) |
Put x = – 5 then f (| –5 |) = f (5) = 4
z1
arg z arg z1 arg z 2 0 | f (– 5) | = | – 5 – 1| = 6
2 (c) is not correct.
71. (a) C1 (1, 0); C2 (0, –2) Hence (d) is the correct alternative.
C1C 2 1 4 5 1 1 1 3
L.H.S. = cos (cos ) cos cos sin
2 2
r1 r2 3 C1C 2 r1 r2
Hence, C2 lies inside C1.
= cos 1 (cos ( / 3 )) /3
72. (d) 3
(a) We have | AB | = | A | | B | so statement 1 is true.
Also for a square matrix of order 3, | kA | k 3 | A | In statement 2, put x sin then /4 /4
because each element of the matrix A is multiplied by k
so sin 1 (2x 1 x 2 ) sin 1 (2sin cos )
and hence in this case we will have k3 common
(b) Since A is invertible, therefore A–1 exists and 76. (a) We have,
det(A 1 )
1 cos 2 cos 2 cos cos sin sin
det(A)
cos sin cos 2 sin 2 cos sin
(c) (A B) 2 (A B)(A B) cos 2 cos sin cos sin sin 2
1 2 1 Since AB = 0, cos( ) 0
4
2 1 3 0 – 5 – 4 0 or
5
1 1 ( ) is an odd multiple of
2
FT-84 The Pattern Target AIEEE
77. (c) Let f image of goh image, fogoh = F (x) = f [goh (x)] x 1/ 3
lim f (x) 1
lim
= f [g ( x 3)] f (cos x 3) x 1/ 3 x 1/ 3 x 1/ 3
82. (b) Squaring both sides we get highest order as 2.
2 83. (c) Differentiating w.r.t.x, we get
F(x)
cos x 3 1 1 1
2f ( x )f (x )
f ( x ) (f (x )) 2
Now x 3 0 and 1 cos x 3 1
1 1
f ' (x) f ( x) x C
2 2
D
fogoh f where C is a constant.
A 2a f (x)
A>–3 goh C 84. (c) I dx ... (i)
0 f (x) f (2a x)
h g
f (2a x)
2a
B I dx ... (ii)
0 f (2a x) f (x)
ABC is equilateral.
1
y x 2 2y x 4
2 A B C 60
i.e. 2y = x + 4 & x + 2y + 4 = 0
sin 2 A sin 2 B sin 2 C
88. (b) Let the point be ( x1, y1).
Therefore y1 = (x1 – 3)2 ...(i)
Now slope of the tangent at ( x1, y1) is (sin 2 60 sin 2 60 sin 2 60 )
2(x1 – 3), but it is equal to 1.
2
7 3 9
Therefore, 2(x1 – 3) 1 x1 3
2 2 4
2
7 1 90. (c) x = tanA, y = tanB, – z = tanC. Then (x + y – z) = –xyz.
y1 –3 .
2 4 tanA + tanB + tanC = tanA tanB tanC
A+B+ C= 2A + 2B = 2 – 2C
7 1
Hence the point is , . tan(2A + 2B) = tan(2 – 2C) = – tan2C
2 4
tan2A + tan2B + tan2C = tan2A.tan2B.tan2C
1 a b 2 tan A 2 tan B 2 tan C
89. (a) Given, in ABC 1 c a 0
1 tan A 1 tan B 1 tan 2 C
2 2
1 b c
2 tan A 2 tan B 2 tan C
1(c2 – ab) – a(c – a) + b(b – c) = 0 . .
1 tan A 1 tan B 1 tan 2 C
2 2
a2 + b2 + c2 – ab – bc – ca = 0
2a2 + 2b2 + 2c2 – 2ab – 2bc – 2ca = 0 Put the value of tanA , tanB, tanC, we get
(a2 + b2 – 2ab) + (b2 + c2 – 2bc) 2x 2y 2z
+ (c2 + a2 – 2ca) = 0 2 2
2 2 2
1 x 1 y 1 z2
(a – b) + (b – c) + (c – a) = 0
Here, sum of squares of three members can be zero if 8xyz
and only if a = b = c
(1 x )(1 y 2 )(1 z 2 )
2
FT-86 The Pattern Target AIEEE
1. (c) The configuration of atom of 4th period with maximum 7. (c) C(graphite) is the stable state of aggregation of carbon.
unpaired electrons is 1s2 2s2 2p6 3s2 3p6 3d5 4s1 . 8. (d) Reaction between diborane and alkene are carried out
Hence its atomic number is 24. in dry ether under an atmosphere of N2 because B2H6
2. (d) The spin quantum number (s) is not the result of and the products are very reactive. The products
solution of schrodinger equation as solved for H-atom. further treated with alkaline H2 O2 to convert into
It was proposed by Goldschmidt and Uren Back.It has alcohols.
alkaline
1 1 B2 H 6 6C2 H 4 B(C2 H4 )3
H 2O 2
values and which signify the direction of reactive
2 2
3CH3CH 2 OH H3BO3
electron on its axis during movement.
3. (a) Aluminium has greater affinity for oxygen and the 9. (b) Liquid crystals on heating first become turbid and then
clear.
reaction is highly exothermic.
10. (c) Ions I and IV are the same (trans), with mirror plane
4. (d) Among alkaline earth metals, barium and radium have
through en groups.
the tendency to form peroxides. 11. (c) By definition, a mixed complex contains more than one
5. (b) The more the reduction potential, the more is the type of ligands.
deposition of metals at cathode. Cation having E° value 12. (b) The concentration of electrolyte, which ionises in water
less than – 0.83 V (reduction potential of H2O) will not shall be more although 80% of it dimerises
deposit from aqueous solution.
6. (d) There is very little difference in acid strength in the 13. (b) Hg2 Cl 2 2 NH 4 OH
series H3 PO 4 , H3 PO 3 and H 3 PO 2 because the (A)
hydrogen in these acids are not all bonded to [H 2 N Hg Cl Hg] NH 4 Cl 2 H 2 O
phosphorus. (Black)
In the above three acids although the number of –OH
groups (ionisable hydrogen increases, yet the acidity HgCl 2 2 NH 4 OH Hg(NH 2 )Cl NH 4Cl H 2 O
does not increase very much. This is due to the fact (B) (White)
CH 3 CH CH CH 3 CH 2 CHCH 2 CH 3 X = OHCH 2 CH 2 NH 2 ,
Major ( B ) Minor ( C ) Ethanolami ne
FT-88 The Pattern Target AIEEE
30. (a) Alkyl or Aryl cyanide react with grignard reagent to 37. (c) When a beaker containing a liquid of density moves
form ketones up with an acceleration a, it will work as a lift moving
C N upward with acceleration a. The effective acceleration
due to gravity in lift = (a + g)
Ether
+C6H5MgBr Pressure of liquid of height h = h (a + g)
38. (d) Apparent wt.
= Real wt.–Upthrust
O
1
C = 12 – (1000 10 6 ) 103 10 12 5 7 N
H 3O+ 2
C6 H5– C=NMgBr
39. (c) As there are two sides of the film created, the total
C6H5 boundary of the film would be 8L. Hence, force of
surface tension will be 8L T.
or C6H5 C O MgBrNH 2
| 40. (a) Both statements are correct and statement-2 is correct
C6 H 5 explanation of statement-1.
31. (a) Time taken in travelling horizontal distance of 200m 41. (b) T1 = 273 K, T2 = 2×273 ( on doubling P at const V, T
with constant velocity of 600m/s is equal to 200/600 = doubles)
1/3 sec. Vertical distance moved under acceleration due T = 273 K, Q = nCvdt, ms = 3J/g/K, nCv = 3
to gravity, with zero initial velocity will be 1
CV = 3
1 1 4
S = ut + at2 = 0 + 10 (1/3)2 = 0.55m. Hence, the
2 2 1
bullet should be aimed 0.55m higher than target. CV = 12 J/mol/K, n' Cv = 12 = 6
2
mv 2 g Q = n'Cvdt = 6×273 = 1638J
32. (d) mg m v or g r 2 or r 2 42. (a) Rate of cooling surface area and for a given mass,
r
33. (b) Using conservation of momentum, surface area of sphere is minimum.
43. (c) Work done = Area of graph ABD;
v m1v1 From above we get,
m1v1+ 0 = m1 1 mv m1v1 mv
3 3 Area = 1/2 × 4 (1 × 105 + 5 × 105) = 2 × 6 × 105 = 12 × 105J
44. (c) Heat flowing per sec. through cylinder of radius R,
(v = 5gl to complete motion along vertical circle)
T1 – T2
2m1v1 2 m1v1 3 m Q1 = K1 ( R2) l
mv v or v1 5g
3 3 m 2 m1
Heat flowing per sec through outer shell of radius 2R,
34. (c) When some portion of the disc is taken out, the mass
concentration of the remaining disc is more on the T1 – T2
Q2 = K2 ( (2R)2 – R2)
opposite side and hence the centre of mass shifts to l
the opposite side of the removed portion.
35. (a) For conservation of momentum, we have, T1 – T2
m1v1 = (m1 + m2) v or v = m1/(m1 + m2) v1 Total Q = Q1 + Q2 = (K1 + 3K2) R2 ....(1)
l
1 1
Now the loss of energy = m v 2 – (m1+ m2)v2
2 1 1 2 2R
Fraction of energy lost R
Cylinder
1 1
m1v12 (m1 m 2 )v 2
2 2 T1 T2
1 l
mv
2 1 1 Let K be the equivalent thermal conductivity of the
= 1 – [(m1+m2) / m1] ( v2/v12) system.
= 1 – [(m1+m2)/m1]×[ m1v12/ [(m1+m2)2]×(1/v12) T1 – T2
= 1 – [ m1/[(m1+m2)] = m2 /(m1+m2) Then Q = K (2R)2 ....(2)
l
36. (c) Inside the sphere, V is constant and is equal to that on
the surface of the sphere. Outside the sphere it comes From eqs. (1) and (2), we have (K1 + 3K2) = 4K
out to be, V = –Gm/r i.e. |V| 1/r. Hence the graph (c) K1 3K 2
is correct. or K =
4
FT-89
0i 0i
Potential of A is equal to potential of C and potential of BTotal
B is equal to potential of D. 4r 4 r
Hence potential difference between AB is equal to 52. (d) K.E. gained by charged particle of charge q when
potential difference between CD. accelerated under a pot. diff. V will be Ek = qV;
Capacitance across AB is equal to capacitance across For a given V, E q.
CD which is C1. For proton, deutron and -particle, the ratio of charges
48. (b) As S2 is open, hence C1 & C3 are in series, also C2 & is 1 : 1 : 2.
C4 are in series combination. 53. (d) At B, the velocity is maximum. Taking vertical downward
3 3 motion of body from A to B, we have, u = 0, a = g, s = H
Q= ×V = × 12 = 9 C; and v = ?
4 4
As v2 = u2 + 2 as ; so v2 = 0 + 2g H or v 2g H
1 3
C'1 3 / 4F 54. (a) A i i' i' A i 30 30 60 0 .
1 3
55. (c) Before the presence of electric field, the free electrons
C1
move randomly in the conductor, so their drift velocity
2 4 4 is zero and therefore, there is no current in the
C'2 F
2 4 3 conductor. In the presence of electric field, each electron
in the conductor experience a force in a direction
opposite to the electric field. Now the free electrons are
C2 accelerated from negative end to the positive end of
the conductor and hence a current starts to flow from
B 12V the conductor.
FT-90 The Pattern Target AIEEE
Imax Imin
2(I1 I2 ) x2 1
log | tan 1 t | C log tan 1 C
57. (a) A maximum frequency deviation of 75 kHz is permitted x
for commercial FM broadcast stations in the 88 to 108
MHz VHF band. x2 1
f (x)
x
13.6Z 2
58. (c) E ; According to question,
n2 64. (a) f (x) Pe 2 x Qe x Rx
f (x) 2Pe 2 x Qe x R
13.6Z2 13.6Z2 13.6Z2
5
47.2
47.2
4 9 36 31 2Pe 2 log 2 Qelog 2 R
Z 2 25 Z 5 8P 2Q R 31 ......... (i)
Also, 0 = P + Q ......... (ii)
59. (c) ne E = m g
log 4 39
19
2 103 & (f (x) Rx)dx
or n 1.6 10 1.8 10 14
10
0 2
0.9 10 2 log 4 39
(Pe2x Qe x ) dx
14 2 0 2
1.8 10 10 .9 10
n or n 5 log 4
19
1.6 10 2 10 3 P 2x 39
e Qe x
60. (a) A convex mirror always forms a virtual image which 2 0 2
can't be cast on a screen. Hence a convex lens is used
P 2 log 4 P 39
to find the focal length of a convex mirror. e Q elog 4 Q
2 2 2
15P 6Q 39 ......... (iii)
61. (d) f (x) (tan x sin x) 2 Solving (i), (ii) and (iii), we get P = 5, Q = – 6, R = 3
tan x & sin x are both continuous & differentiable 65. (c) For ellipse a 3, b 2
at x = 0
area = ab= 6
f ( x) is cont. & diff at x = 0
For circle r = 1 area = (r) .
2
f ( x) 2(tan x sin x)[sec x cos x ] , f (0) 0 Second statement is not necessarily true.
62. (b) Both the curves satisfied by the point (1, y1). 66. (a) The curve of y (x2 + 4a2) = 8a3 is symmetrical about
y-axis and cuts it at A (0, 2a). Tangent at A is parallel to
So, 1 + a – 1 = 6 + b a–b=6 ......(1)
x-axis. x-axis is asymptote. This curve meets x2 = 4ay
dy dy x2 8a 3
Also at x = 1 Where, x4 4a 2 x 2 32a 4 0
dx curve 1 dx curve 2 2
4a x 4a 2
[3x2 + a]x = 1 = [12x]x = 1 a=9 ......(2)
(x 2 4a 2 )(x 2 8a 2 ) 0 x 2a
From (1) & (2), a = 9 and b = 3.
2a 8a 3 2ax2
2 Required area 2 dx dx
(x 1) dx 0 x2 4a 2 0 4a
63. (c) I
2
x 1
(x 4 3x 2 1) tan 1
x
A(0, 2a)
(Dividing Num. and Den. by x2)
(–2a, a) (2a, a)
1
1 dx
x2 dt O
I 2 1
1 1 t 1 tan t
x2 3 2
tan 1
x a2
x x (6 4).
3
FT-91
1 1
Put n = 0 / 2 cot (x 1) tan x tan 1 ( x)
1 1 x 1
/2 tan 1 ( x ) cot 1
( x 1) 84. (a) f {f [f ( x)]} f f f f
1 x 1 x
1
x 1 x 2x 1 0 1 x
1
x=– f (x) is not defined for x = 1; f
1
is not defined for x = 0.
2 1 x
81. (a) Given, k sin + cos 2 = 2k – 7
k sin + 1 – 2 sin2 = 2k – 7 f {f [f(x)]} is discontinuous at x = 0 and 1 i.e., there are two
2sin2 – k sin + (2k – 8) = 0 points of discontinuity.
For the existence of real roots, discriminant 0. 85. (a) f (x) x [x ]
k2 – 4 × 2 (2k – 8) 0 (k – 8)2 0, which is always true.
f ( x 1) x 1 ([ x] 1) x [x]
k 4
Roots of the quadratic equation are , 2, but sin 2. Period of x – [x] is 1
2
f (x) sin(2x [2x ])
k 4
sin but 1 sin 1
2
1 1 1
f x sin 2 x 2 x
k 4 2 2 2
1 1 2 k 4 2 2 k 6
2
sin (2x 1 [2x] 1) sin (2x [2x])
1 cos 3 x (1 cos x ) (1 cos x cos 2 x )
82. (c) Lim Lim
x 0 x sin x cos x x 0 x sin x cos x 1
Period is .
2
x
2 sin 2
2 (1 cos x cos 2 x ) dy y y
Lim 86. (c) Given, cos 2
x 0 x x cos x dx x x
x.2 sin cos
2 2
dy dv
Putting, y = vx so that v x
x dx dx
sin
2 1 cos x cos 2 x 1 3 dv
Lim 3 . We get, v x v cos 2 v
x 0 x x 2 2
2 cos cos x dx
2 2
dv dx dx
2 sec 2 v dv
1 2x cos v x x
83. (c) y sin
1 x2 Integrating, we get
y
dy 1 2(1 x 2 ) 4 x 2 tan v = – ln x + ln c tan ln x ln c
. x
dx 4x 2 2 2
1 x
1
2 2 This passes through 1, ln c 1
1 x
4
2(1 x 2 ) 1 2(1 x 2 )
. e
2
(1 x 2 ) 2 1 x | 1 x 2 | (1 x 2 ) y x tan 1
log
x
If | x | < 1 x2 < 1, then 1–x2 > 0
dy 2 2 x (2 1) 2 x 1 2x
|1 x2 | 1 x2 2 87. (a) y tan 1
= tan
1
1 x dx 1 2 x.2 x 1 1 2 x .2 x 1
If | x | > 1 x2 > 1, then 1 –x2 < 0
1 dy 2x 1
log 2 2 x log 2
dy 2 tan (2 x 1 ) tan 1
(2 x )
|1 x2 | x2 1 dx 1 2 2(x 1)
1 22x
dx 2
1 x
dy 2 3
dy (log 2) 1 log 2
Obviously does not exist for x2 = 1 or | x | = 1 dx x 0 5 5
dx
FT-93
z is | z 0 |2 r