Jee Mains 2009 Question Paper

1
AIEEE2009, BOOKLET CODE(A)

Note: (i) The test is of 3 hours duration.
(ii) The test consists of 90 questions. The maximum marks are 432.
(iii) There are three parts in the question paper. The distribution of marks subjectwise in each part is as under for each correct
response.
Part A Physics (144 marks)
Question No. 1 to 2 and 9 to 30 consists FOUR (4) marks each and Question No. 3
to 8 consists EIGHT (8) marks each for each correct response.
Part B Chemistry (144 marks)
Question No. 31 to 39 and 46 to 60 consists FOUR (4) marks each and Question
No. 40 to 45 consists EIGHT (8) marks each for each correct response.
Part C Mathematics(144 marks)
Question No. 61 to 82 and 89 to 90 consists FOUR (4) marks each and Question
No. 83 to 88 consists EIGHT (8) marks each for each correct response.
(iv) Candidates will be awarded marks as stated above for correct response of each question. 1/4th marks will be deducted for
indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an
item in the answer sheet.
(v) * marked questions are from syllabus of class XI CBSE.
Physics
PART A
1.
This question contains Statement-1 and Statement-2. Of the four choices given after the statements,
choose the one that best describes the two statements.
Statement 1: For a charged particle moving from point P to point Q, the net work done by an
electrostatic field on the particle is independent of the path connecting point P to point Q.
Statement-2: The net work done by a conservative force on an object moving along a closed loop is
zero
(1) Statement-1 is true, Statement-2 is false
(2) Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1.
(3) Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement-1.
(4) Statement-1 is false, Statement-2 is true
Sol:
(2)
Work done by conservative force does not
depend on the path. Electrostatic force is a
conservative force.
2.
The above is a plot of binding energy per nucleon

Eb, against the nuclear mass M; A, B, C, D, E, F
correspond to different nuclei.
Consider four
reactions:
(ii) C A + B +
(i) A + B C +
(iii) D + E F + and
(iv) F D + E +
where is the energy released? In which reactions
is positive?
(1) (i) and (iv)
(2) (i) and (iii)
(3) (ii) and (iv)
(4) (ii) and (iii)
Sol:
(1)
1st reaction is fusion and 4th reaction is fission.
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3.
A p-n junction (D) shown in the figure can act

as a rectifier. An alternating current source
(V) is connected in the circuit.
(1)
(2)
(3)
(4)
Sol:
(3)
Given figure is half wave rectifier
4.
The logic circuit shown below has the input waveforms A and B as shown. Pick out the correct
output waveform.
Sol:
(1)
(2)
(3)
(4)
(1)
A
1
1
0
0
*5.
Truth Table
B
Y
1
1
0
0
1
0
0
0
If x, v and a denote the displacement, the velocity and the acceleration of a particle executing simple
harmonic motion of time period T, then, which of the following does not change with time?
(1) a2T2 + 42v2
(3) aT + 2v
Sol:
aT
x
aT
(4)
v
(2)
(2)
aT 2 xT 42
4 2
=
= 2 T =
= constant.
x
x
T
T
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6.
In an optics experiment, with the position of the object fixed, a student varies the position of a convex
lens and for each position, the screen is adjusted to get a clear image of the object. A graph between
the object distance u and the image distance v, from the lens, is plotted using the same scale for the
two axes. A straight line passing through the origin and making an angle of 45o with the x-axis meets
the experimental curve at P. The coordinates of P will be
f f
2 2
(1) (2f, 2f)
(2) ,
(3) (f, f)
(4) (4f, 4f)
Sol:
(1)
It is possible when object kept at centre of
curvature.
u=v
u = 2f, v = 2f.
*7.
A thin uniform rod of length l and mass m is swinging freely about a horizontal axis passing through
its end. Its maximum angular speed is . Its centre of mass rises to a maximum height of
Sol:
(1)
1 l 2 2
3 g
(2)
1 l
6 g
(3)
1 l 2 2
2 g
(4)
1 l 2 2
6 g
(4)
T.Ei = T.Ef
1 2
I = mgh
2
1 1 2 2
1 l 2 2
ml = mgh h =
2 3
6 g
8.
Let P(r) =
Q
r be the charge density distribution for a solid sphere of radius R and total charge Q.
R 4
for a point p inside the sphere at distance r1 from the centre of the sphere, the magnitude of electric
field is
(1) 0
(3)
Sol:
(2)
Qr12
(4)
4oR 4
Q
4o r12
Qr12
3oR 4
(3)
r1
E4r12 =
r4r 2 dr
Qr12
E=
.
40R 4
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9.
The transition from the state n = 4 to n = 3 in a hydrogen like atom results in ultraviolet radiation.
Infrared radiation will be obtained in the transition from
(1) 2 1
(2) 3 2
(3) 4 2
(4) 5 4
Sol:
(4)
IR corresponds to least value of
2
n1
n22
i.e. from Paschen, Bracket and Pfund series. Thus the transition corresponds to 5 3.
*10.
One kg of a diatomic gas is at a pressure of 8 104 N/m2. The density of the gas is 4 kg/m-3. What is
the energy of the gas due to its thermal motion?
(2) 5 104 J
(1) 3 104 J
4
(3) 6 10 J
(4) 7 104 J
Sol:
(2)
Thermal energy corresponds to internal energy
Mass = 1 kg
density = 8 kg/m3
Volume =
mass
1
= m3
density 8
Pressure = 8 104 N/m2

Internal Energy =
11.
5
P V = 5 104 J
2
This question contains Statement-1 and Statement-2. Of the four choices given after the statements,
choose the one that best describes the two statements.
Statement-1: The temperature dependence of resistance is usually given as R = Ro(1 + t). The
resistance of a wire changes from 100 to 150 when its temperature is increased from 27oC to
227oC. This implies that = 2.5 10 3 / o C .
Statement 2: R = Ri (1 + T) is valid only when the change in the temperature T is small and R =
(R - Ro) << Ro.
(2) Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1.
(3) Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement-1.
Sol:
(1)
Directions: Question numbers 12 and 13 are based on the following paragraph.

A current loop ABCD is held fixed on the plane of the paper as shown
in the figure. The arcs BC (radius = b) and DA (radius = a) of the loop
are joined by two straight wires AB and CD. A steady current I is
flowing in the loop. Angle made by AB and CD at the origin O is 30o.
Another straight thin wire with steady current I1 flowing out of the plane
of the paper is kept at the origin.
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12.
The magnitude of the magnetic field (B) due to loop ABCD at the origin (O) is
(1) zero
(2)
I b a
(3) o
4 ab
Sol:
24ab
oI
(4)
2 (b a ) + ( a + b )
4
3
(2)
Net magnetic field due to loop ABCD at O is
B = BAB + BBC + BCD + BDA
= 0+
13.
o ( b a )
I
I
oI
I
oI
+0 o = o o =
(b a )
4a 6
4b 6 24a 24b
24ab
Due to the presence of the current I1 at the origin

(1) The forces on AB and DC are zero
(2) The forces on AD and BC are zero
oII1
2 (b a ) + ( a + b )
4
3
II
(4) The magnitude of the net force on the loop is given by o 1 ( b a )
24ab
(3) The magnitude of the net force on the loop is given by
Sol:
(2)
The forces on AD and BC are zero because magnetic field due to a straight wire on AD and BC is
parallel to elementary length of the loop.
14.
A mixture of light, consisting of wavelength 590 nm and an unknown wavelength, illuminates Youngs
double slit and gives rise to two overlapping interference patterns on the screen. The central
maximum of both lights coincide. Further, it is observed that the third bright fringe of known light
coincides with the 4th bright fringe of the unknown light. From this data, the wavelength of the
unknown light is
(1) 393.4 nm
(2) 885.0 nm
(3) 442.5 nm
(4) 776.8 nm
Sol:
(3)
31 = 42
2 =
15.
3
3
1770
1 = 590 =
= 442.5 nm
4
4
4
Two points P and Q are maintained at the potentials of 10V and -4V respectively. The work done in
moving 100 electrons from P to Q is
(1) 19 10 17 J
(2) 9.60 10 17 J
(3) 2.24 10 16 J
(4) 2.24 10 16 J
Sol:
(4)
W = QdV = Q(Vq - VP) = -100 (1.6 10-19) ( 4 10)
= + 100 1.6 10-19 14 = +2.24 10-16 J.
16.
The surface of a metal is illuminated with the light of 400 nm. The kinetic energy of the ejected
photoelectrons was found to be 1.68 eV. The work function of the metal is (hc = 1240 eV nm)
(1) 3.09 eV
(2) 1.41 eV
(4) 1.68 eV
(3) 151 eV
Sol:
(2)
1
hc 1240evnm
mv 2 = eVo = 1.68eV h =
=
= 3.1 eV 3.1 eV = Wo + 1.6 eV
2
400nm
Wo = 1.42 eV
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*17.
A particle has an initial velocity 3i + 4j and an acceleration of 0.4i + 0.3j . Its speed after 10 s is
(2) 7 2 units
(4) 8.5 units
(1) 10 units
(3) 7 units
Sol:
(2)
r
r
u = 3i + 4j ; a = 0.4i + 0.3j
r r r
u = u + at
= 3i + 4j + 0.4i + 0.3j 10 = 3i + j + 4i + 3j = 7i + 7j
Speed is
72 + 72 = 7 2 units
*18.
A motor cycle starts from rest and accelerates along a straight path at 2 m/s2. At the starting point of
the motor cycle there is a stationary electric sire. How far has the motor cycle gone when the driver
hears the frequency of the siren at 94% of its value when the motor cycle was at rest? (speed of
sound = 330 ms-1).
(1) 49 m
(2) 98 m
(3) 147 m
(4) 196 m
Sol:
(2)
Motor cycle, u = 0, a = 2 m/s2
Observer is in motion and source is at rest.
330 v O
v vO
330 94
94
330 vO =
n=n
100
100
330
v + vS
94 33 33 6
m/s
v O = 330
=
10
10
v 2 u2 9 33 33 9 1089
s=
98 m.
=
=
2a
100
100
n = n
*19.
Sol:
Consider a rubber ball freely falling from a height h = 4.9 m onto a horizontal elastic plate. Assume
that the duration of collision is negligible and the collision with the plate is totally elastic. Then the
velocity as a function of time the height as function of time will be
(1)
(2)
(3)
(4)
(3)
h=
1 2
gt ,
2
v = - gt and after the collision, v = gt.
(parabolic)
(straight line)
Collision is perfectly elastic then ball reaches to same height again and again with same velocity.
v
+v1
t1
2t2
3t1
-v1
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20.
Sol:
A charge Q is placed at each of the opposite corners of a square. A charge q is placed at each of the
other two corners. If the net electrical force on Q is zero, then the Q/q equals
(1) 2 2
(2) -1
(3) 1
(4)
1
2
(1)
Three forces F41, F42 and f43 acting on Q are shown
Resultant of F41 + F43
= 2 Feach
=
F41
q
1 Qq
2
4o d2
1 QQ
=
4o 2d 2
Q
F43
Resultant on Q becomes zero only when q charges are of

negative nature.
F4,2
F42
dQ Q Q
=
d2
2d2
QQ
2q =
2
Q
Q
q=
or
= 2 2
q
2 2
*21.
Sol:
A long metallic bar is carrying heat from one of its ends to the other end under steady-state. The
variation of temperature along the length x of the bar from its hot end is best described by which of
the following figure.
(1)
(2)
(3)
(4)
(2)
We know that
dQ
d
= KA
dt
dx
In steady state flow of heat
dQ 1
. .dx
dt kA
H - = k x = H - k x
Equation = H - k x represents a straight line.
d =
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22.
A transparent solid cylindrical rod has a refractive index of
2
3
. It is surrounded by air. A light ray is incident at the mid
point of one end of the rod as shown in the figure.

The incident angle for which the light ray grazes along the wall of the rod is
(4) sin1
(2) sin1
(1) sin1
(3) sin1
Sol:
(4)
C
3
SinC =
2
.. (1)
Sin r = sin (90 C) = cosC =
1
2
sin 2
=
sinr 1
2 1
sin =
3 2
1
= sin1
3
*23.
Three sound waves of equal amplitudes have frequencies (v 1), v, (v + 1). They superpose to give
beats. The number of beats produced per second will be
(1) 4
(2) 3
(3) 2
(4) 1
Sol:
(3)
Maximum number of beats = + 1 ( 1) = 2
*24.
The height at which the acceleration due to gravity becomes
g
(where g = the acceleration due to
9
gravity on the surface of the earth) in terms of R, the radius of the earth is
Sol:
(1) 2R
(2)
R
(3)
2
(4)
R
2
2R
(1)
g =
GM
(R + h )
, acceleration due to gravity at height h
g GM
R2
R
= 2 .
= g
2
9 R (R + h )
R +h
R
1
1 R
=
=
R+h 3
9 R + h
3R = R + h 2R = h
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*25.
Two wires are made of the same material and have the same volume. However wire 1 has crosssectional area A and wire-2 has cross-sectional area 3A. If the length of wire 1 increases by x on
applying force F, how much force is needed to stretch wire 2 by the same amount?
(1) F
(2) 4F
(3) 6F
(4) 9F
Sol:
(4)
A1l 1 = A 2 l 2 l 2 =
F1
l1
A
F
x 2 = 2 l 2
3A
x1 =
A1l 1 A l 1 l 1
l
=
=
1 =3
A2
3A
3
l2
.. (i)
..(ii)
Here x1 = x2
F2
F
l 2 = 1 l1
3A
A
l
F2 = 3F1 1 = 3F1 3 = 9F
l2
*26.
In an experiment the angles are required to be measured using an instrument. 29 divisions of the
main scale exactly coincide with the 30 divisions of the vernier scale. If the smallest division of the
main scale is half-a-degree(=0.5o), then the least count of the instrument is
(1) one minute
(2) half minute
(3) one degree
(4) half degree
Sol:
(1)
o
Least count =
27.
An inductor of inductance L = 400 mH and resistors of resistances R1

= 2 and R2 = 2 are connected to a battery of emf 12V as shown in
the figure. The internal resistance of the battery is negligible. The
switch S is closed at t = 0. The potential drop across L as a function
of time is
(1) 6e5t V
(2)
(3) 6 1 e t / 0.2 V
Sol:
value of main scale division

1
1 1
1
MSD =
=
=
= 1 minute
30 2
60
No of divisions on vernier scale 30
12 3t
e V
t
(4) 12e 5t V
(4)
I1 =
F
R1
E=L
12
= 6A
2
dI2
+ R2 I2
dt
I2 = Io 1 e t / tc Io =
E 12
=
= 6A
R2
2
L 400 10 3
=
= 0.2
R
2
I2 = 6 1 e t / 0.2
tc =
Potential drop across L = E R2I2 = 12 2 6 (1 e-bt) = 12 e-5t
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10
Directions: Question numbers 28, 29 and 30 are based on the following paragraph.
Two moles of helium gas are taken over the cycle ABCDA, as shown in the P T diagram.
*28.
Assuming the gas to be ideal the work done on the gas in taking it from A to B is
(1) 200 R
(2) 300 R
(3) 400 R
(4) 500 R
Sol:
(3)
WAB = Q - U = nCpdT nCvdT (at
constant pressure)
= n(Cp Cv)dt
= nRdT = 2 R (500 300) = 400 R
P
2 105 Pa
n = 2, = 1.67
1 105 Pa
300 K
*29.
The work done on the gas in taking it from D to A is

(1) 414 R
(2) + 414 R
(3) 690 R
(4) + 690 R
Sol:
(1)
At constant temperature (isothermal process)
500 K
105
P1
= 2.303 2R 300log
5
P2
2 10
1
= 2.303 600R log
2
WDA = nRT ln
= 0.693 600 R = - 414 R.

*30.
The net work done on the gas in the cycle ABCDA is

(1) Zero
(2) 276 R
(3) 1076 R
(4) 1904 R
Sol:
(2)
Net work done in a cycle = WAB + WBC + WCB + WBA
= 400 R + 2 2.303 500 R ln 2 400R 414 R
= 1000R x ln 2 600R x ln 2 = 400R x ln 2 = 276R
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11
CHEMISTRY
PART B
31.
Knowing that the Chemistry of lanthanoids (Ln) is dominated by its +3 oxidation state, which of the
following statements in incorrect ?
(1) Because of the large size of the Ln (III) ions the bonding in its compounds is predominantly ionic in
character.
(2) The ionic sizes of Ln (III) decrease in general with increasing atomic number.
(3) Ln (III) compounds are generally colourless.
(4) Ln (III) hydroxides are mainly basic in character.
Sol:
(3)
Ln+3 compounds are mostly coloured.
32.
A liquid was mixed with ethanol and a drop of concentrated H2 SO4 was added. A compound with a
fruity smell was formed. The liquid was :
(1) CH3 OH
(2) HCHO
(4) CH3 COOH
(3) CH3 CO CH3
Sol:
(4)
Esterification reaction is involved
H+
CH3 COOH( l ) + C2H5 OH( l )
CH3 COOC2H5( l ) + H2 O( l )
*33.
Arrange the carbanions, ( CH3 )3 C, CCl3 , ( CH3 )2 CH, C6 H5 CH2 , in order of their decreasing stability :
(1) C6 H5 CH2 > CCl3 > ( CH3 )3 C > ( CH3 )2 CH
(2) ( CH3 )2 CH > CCl3 > C6 H5 CH2 > ( CH3 )3 C
(3) CCl3 > C6 H5 CH2 > ( CH3 )2 CH > ( CH3 )3 C
(4)
( CH3 )3 C > ( CH3 )2 CH > C6H5 CH2
Sol:
(3)
2o carbanion is more stable than 3o and Cl is I effect group.
*34.
The alkene that exhibits geometrical isomerism is :

(1) propene
(2) 2-methyl propene
(3) 2-butene
(4) 2- methyl -2- butene
Sol:
(3)
H
CH3
cis
CH3
C=C
C=C
CH3
> CCl3
H3C
Trans
*35.
In which of the following arrangements, the sequence is not strictly according to the property written
against it ?
(1) CO2 < SiO2 < SnO2 < PbO2 : increasing oxidising power
(2) HF< HCl < HBr < HI : increasing acid strength
(3) NH3 < PH3 < AsH3 < SbH3 : increasing basic strength
(4) B < C < O < N : increasing first ionization enthalpy.
Sol:
(3)
Correct basic strength is NH3 > PH3 > AsH3 > BiH3
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12
36.
The major product obtained on interaction of phenol with sodium hydroxide and carbon dioxide is :
(1) benzoic acid
(2) salicylaldehyde
(3) salicylic acid
(4) phthalic acid
Sol:
(3)
Kolbe Schmidt reaction is
OH
ONa
NaOH
OH
OH
COONa
CO2
6atm, 140o C
COOH
H3 O+
Salicylic Acid
37.
Which of the following statements is incorrect regarding physissorptions ?

(1) It occurs because of vander Waals forces.
(2) More easily liquefiable gases are adsorbed readily.
(3) Under high pressure it results into multi molecular layer on adsorbent surface.
(4) Enthalpy of adsorption ( Hadsorption ) is low and positive.
Sol:
(4)
Enthalpy of adsorption regarding physissorption is not positive and it is negative.
38.
Which of the following on heating with aqueous KOH, produces acetaldehyde ?

(1) CH3 COCl
(2) CH3 CH2 Cl
(4) CH3 CHCl2
(3) CH2 Cl CH2 Cl
Sol:
(4)
OH
/
aq.KOH
CH3 CHCl2
CH3 CH
\
OH
*39.
In an atom, an electron is moving with a speed of 600m/s with an accuracy of 0.005%. Certainity
with which the position of the electron can be located is (h = 6.6 10 34 kg m2 s1 ,
mass of electron,
em = 9.1 10 31 kg )
(1) 1.52 10 4 m
(3) 1.92 10 3 m
Sol:
CH3 CHO
H2 O
(2) 5.10 10 3 m
(4) 3.84 10 3 m
(3)
x.m v =
x =
h
4
h
4 mv
0.005
= 0.03
100
6.625 10 34
= 1.92 10 3 m
x =
4 3.14 9.1 10 31 0.03
v = 600
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13
40.
In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is
3
CH3 OH(l ) + O2 (g) CO2 (g) + 2H2 O(l ) At 298K standard Gibbs energies of formation for
2
CH3 OH( l ), H2 O( l ) and CO2 (g) are -166.2, -237.2 and -394.4 kJ mol1 respectively. If standard
enthalpy of combustion of methanol is -726kJ mol1 , efficiency of the fuel cell will be
(1) 80 %
(2) 87%
(3) 90%
(4) 97%
Sol:
(4)
3
O2 (g) CO2 (g) + 2H2 O(l ) H = 726kJ mol1
2
Also Gof CH3 OH(l ) = -166.2 kJ mol-1
CH3 OH(l ) +
Gof H2 O(l ) = -237.2 kJ mol-1
Gof CO2 (l) = -394.4 kJ mol-1

Q G = Gof products Gof reactants.
= -394.4 -2 (237.2) + 166.2
= 702.6 kJ mol-1
G
100
now Efficiency of fuel cell =
H
702.6
=
100
726
= 97%
41.
Two liquids X and Y form an ideal solution. At 300K, vapour pressure of the solution containing 1 mol
of X and 3 mol of Y is 550 mm Hg. At the same temperature, if 1 mol of Y is further added to this
solution, vapour pressure of the solution increases by 10 mm Hg. Vapour pressure (in mmHg) of X
and Y in their pure states will be, respectively :
(1) 200 and 300
(2) 300 and 400
(3) 400 and 600
(4) 500 and 600
Sol:
(3)
PT = PXo x X + PYo x Y
x X = mol fraction of X
x Y = mol fraction of Y
1
o 3
550 = Pxo
+ PY 1 + 3
1+ 3
o
o
P
3P
= X + Y
4
4
550 (4) = PXo + 3PYo .. (1)
Further 1 mol of Y is added and total pressure increases by 10 mm Hg.
1
4
+ PYo
550 + 10 = PXo
1+ 4
1+ 4
o
o
560 (5) = PX + 4PY .(2)
By solving (1) and (2)
We get, PXo = 400 mm Hg

PYo = 600 mm Hg
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42.
The half life period of a first order chemical reaction is 6.93 minutes. The time required for the
completion of 99% of the chemical reaction will be (log 2=0.301) :
(1) 230.3 minutes
(2) 23.03 minutes
(3) 46.06 minutes
(4) 460.6 minutes
Sol:
(3)
Q =
0.6932 0.6932
=
min1
t1/ 2
6.93
Also t =
[ Ao ]
2.303
log
[A]
[ A o ] = initial concentration (amount)

[ A ] = final concentration (amount)
2.303 6.93
100
log
0.6932
1
= 46.06 minutes
t=
43.
o
Given : EFe
= 0.036V,
3+
/ Fe
o
= -0.439V. The value of standard electrode potential for the
EFe
2+
/ Fe
3+
+ e Fe2 + (aq) will be :
change, Fe(aq)
(1) -0.072 V
(3) 0.770 V
Sol:
(2) 0.385 V
(4) -0.270
(3)
Q Fe3 + + 3e Fe; Eo = 0.036V
G1O = nFEo = 3F( 0.036)
= +0.108 F
Also Fe2+ + 2e Fe; Eo = -0.439 V
GO2 = -nF Eo
= -2 F( -0.439)
= 0.878 F
3+
+ e Fe2+ (aq)
To find Eo for Fe(aq)
GO = -nFE o
= -1FE o
o
o
Q G = G1 Go2
Go = 0.108F - 0.878F
-FEo = +0.108F 0.878F
EO = 0.878 - 0.108
= 0.77v
*44.
+
= 0)
On the basis of the following thermochemical data : ( fGo H(aq)
H2 O(l ) H+ (aq) + OH (aq); H = 57.32kJ

1
O2 (g) H2 O(l ) ; H = 286.20kJ
2
The value of enthalpy of formation of OH ion at 25o C is :
(1) -22.88 kJ
(2) -228.88 kJ
(3) +228.88 kJ
(4) -343.52 kJ
H2 (g) +
Sol:
(2)
By adding the two given equations, we have
1
+
+ OH(aq)
; H =-228.88 Kj
H2(g) + O2(g) H(aq)
2
+
Here Hof of H(aq)
=0
Hof of OH = -228.88 kJ
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45.
Copper crystallizes in fcc with a unit cell length of 361 pm. What is the radius of copper atom ?
(1) 108 pm
(2) 127 pm
(3) 157 pm
(4) 181 pm
Sol:
(2)
For FCC,
2a = 4r (the atoms touches each other along the face- diagonal)
2a
=
4
= 127 pm
r=
46.
2 361
4
Which of the following has an optical isomer ?

(1) CO (NH3 )3 Cl
(3) CO (H2 O )4 ( en )
(2) CO ( en )(NH3 )2
3+
2+
(4) CO ( en )2 (NH3 )2
3+
Sol:
(4)
It is an octahedral complex of the type M ( AA )2 X2
Where AA is bidentate ligand.
*47.
Solid Ba (NO3 )2 is gradually dissolved in a 1.0 10 4 M Na2 CO3 solution. At what concentration of
Ba2 + will a precipitate begin to form ?(Ksp for Ba CO3 = 5.1 10 9 ).
(1) 4.1 10 5 M
(2) 5.1 10 5 M
(4) 8.1 10 7 M
(3) 8.1 10 8 M
Sol:
(2)
Ba (NO3 )2 + CaCO3 BaCO3 + 2NaNO3
Here CO32 = [Na2 CO3 ] = 10 4 M
K sp = Ba +2 CO32 5.1 10 9 = Ba2+ 10 4 Ba +2 = 5.1 10 5
At this value, just precipitation starts.
48.
Which one of the following reactions of Xenon compounds is not feasible ?

(1) XeO3 + 6HF Xe F6 + 3H2 O
(2) 3Xe F4 + 6H2 O 2 Xe + XeO3 + 12 HF + 1.5 O2
(3) 2XeF2 + 2H2 O 2Xe + 4HF + O2
(4) XeF6 + RbF Rb(XeF7 ]
Sol:
(1)
Remaining are feasible
*49.
Using MO theory predict which of the following species has the shortest bond length ?
(1) O22 +
(2) O2+
(3) O2
Sol:
(4) O22
(1)
Bond length
1
bond order
no..of bonding e no.of antibonding e

2
Bond orders of O2+ , O2 , O22 and O2+2 are respectively
2.5, 1.5, 1 and 3.
Bond order =
50.
In context with the transition elements, which of the following statements is incorrect ?
(1) In addition to the normal oxidation states, the zero oxidation state is also shown by these elements
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in complexes.
(2) In the highest oxidation states, the transition metal show basic character and form cationic
complexes.
(3) In the highest oxidation states of the first five transition elements (Sc to Mn), all the 4s and 3d
electrons are used for bonding.
(4) Once the d5 configuration is exceeded, the tendency to involve all the 3d electrons in bonding
decreases.
Sol:
(2)
In higher Oxidation states transition elements show acidic nature
*51.
Calculate the wavelength (in nanometer) associated with a proton moving at 1.0 103 ms 1
(Mass of proton = 1.67 10 27 kg and h = 6.63 10 34 Js ) :
(1) 0.032 nm
(2) 0.40 nm
(3) 2.5 nm
(4) 14.0 nm
Sol:
(2)
=
h
6.63 10 34
=
0.40 nm
mv 1.67 1027 103
52.
A binary liquid solution is prepared by mixing n-heptane and ethanol. Which one of the following
statements is correct regarding the behaviour of the solution ?
(1) The solution formed is an ideal solution
(2) The solution is non-ideal, showing +ve deviation from Raoults law.
(3) The solution is non-ideal, showing ve deviation from Raoults law.
(4) n-heptane shows +ve deviation while ethanol shows ve deviation from Raoults law.
Sol:
(2)
The interactions between n heptane and ethanol are weaker than that in pure components.
*53.
The number of stereoisomers possible

CH3 CH = CH CH ( OH) Me is :
for
(1) 3
(3) 4
(2) 2
(4) 6
compound
of
the
molecular
formula
Sol:
(3)
About the double bond, two geometrical isomers are possible and the compound is having one chiral
carbon.
*54.
The IUPAC name of neopentane is

(1) 2-methylbutane
(3) 2-methylpropane
Sol:
(2) 2, 2-dimethylpropane
(4) 2,2-dimethylbutane
(2)
CH3
|
Neopentane is H3 C C CH3
|
CH3
*55.
The set representing the correct order of ionic radius is :

(2) Na+ > Li+ > Mg2+ > Be2+
(1) Li+ > Be2 + > Na+ > Mg2+
(3) Li+ > Na + > Mg2+ > Be2+
Sol:
56.
(4) Mg2+ > Be2+ > Li+ > Na +
(2)
Follow the periodic trends
The two functional groups present in a typical carbohydrate are :
(1) -OH and -COOH
(2) -CHO and -COOH
(3) > C = O and - OH
(4) - OH and -CHO
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Sol:
(3)
Carbohydrates are polyhydroxy carbonyl compounds.
*57.
The bond dissociation energy of B F in BF3 is 646 kJ mol1 whereas that of C-F in CF4 is 515kJ
mol1 . The correct reason for higher B-F bond dissociation energy as compared to that of C- F is :
(1) smaller size of B-atom as compared to that of C- atom
(2) stronger bond between B and F in BF3 as compared to that between C and F in CF4
(3) significant p - p interaction between B and F in BF3 whereas there is no possibility of such
interaction between C and F in CF4 .
(4) lower degree of p - p interaction between B and F in BF3 than that between C and F in CF4 .
Sol:
(3)
option itself is the reason
58.
In Cannizzaro reaction given below

()
: OH
&& ( ) the slowest step is :
2 Ph CHO
Ph CH2 OH + PhCO
2
()
(1) the attack of : OH at the carboxyl group

(2) the transfer of hydride to the carbonyl group
(3) the abstraction of proton from the carboxylic group
(4) the deprotonation of Ph CH2 OH
Sol:
(2)
Hydride transfer is the slowest step.
59.
Which of the following pairs represents linkage isomers ?

(1) Cu (NH3 )4 [Pt Cl4 ] and Pt (NH3 )4 [CuCl4 ]
(2) Pd (P Ph3 )2 (NCS )2 and Pd (P Ph3 )2 ( SCN)2
(3) CO (NH3 )5 NO3 SO4 and CO (NH3 )5 SO4 NO3
(4) Pt Cl2 (NH3 )4 Br2 and Pt Br2 (NH3 )4 Cl2
Sol:
(2)
NCS- is ambidentate ligand and it can be linked through N (or) S
60.
Buna-N synthetic rubber is a copolymer of :

Cl
|
(1) H2 C = CH C = CH2 and H2C = CH CH = CH2
(2) H2 C = CH CH = CH2 and H5 C6 CH = CH2

(3) H2 C = CH CN and H2 C = CH CH = CH2
(4) H2 C = CH CN and H2 C = CH C = CH2
|
CH3
Sol:
(3)
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Mathematics
PART C
a
61.
a +1
Let a, b, c be such that b ( a + c ) 0 . If b b + 1 b 1 + a 1

c c 1 c + 1 ( 1)n + 2 a
value of n is
(1) zero
(3) any odd integer
Sol:
a +1 a 1
b +1
c 1
b 1
c + 1 = 0, then the
( 1)
n +1
( 1)
(2) any even integer

(4) any integer
(3)
a a +1 a 1
a +1 b +1 c 1
a a +1 a 1
a +1 a 1 a
n
n
b b + 1 b 1 + ( 1) a 1 b 1 c + 1 = b b + 1 b 1 + ( 1) b + 1 b 1 b
c c 1 c +1
a
c
c c 1 c +1
c 1 c +1 c
b
a a +1 a 1
a +1 a a 1
a a +1 a 1
a a +1 a 1
n +1
n+ 2
= b b + 1 b 1 + ( 1) b + 1 b b 1 = b b + 1 b 1 + ( 1) b b + 1 b 1
c c 1 c +1
c 1 c c +1
c c 1 c +1
c c 1 c +1
This is equal to zero only if n + 2 is odd i.e. n is odd integer.

62.
If the mean deviation of number 1, 1 + d, 1 + 2d, .. , 1 + 100d from their mean is 255, then the d is
equal to
(1) 10.0
(2) 20.0
(3) 10.1
(4) 20.2
Sol:
(3)
n
a + l)
sum of quantities 2 (
1
Mean ( x ) =
=
= [1 + 1 + 100d] = 1 + 50d
n
2
n
1
1
2 d 50 x 51
xi x 255 = 101[50d + 49d + 48d + .... + d + 0 + d + ...... + 50d] = 101 2
n
255 x 101
d=
= 10.1
50 x 51
M.D. =
*63.
If the roots of the equation bx 2 + cx + a = 0 be imaginary, then for all real values of x, the expression
3b2 x 2 + 6bcx + 2c 2 is
(1) greater than 4ab
(2) less than 4ab
(3) greater than 4ab
(4) less than 4ab
Sol:
(3)
bx 2 + cx + a = 0
Roots are imaginary c 2 4ab < 0 c 2 < 4ab c 2 > 4ab
3b2 x 2 + 6bcx + 2c 2
since 3b2 > 0
Given expression has minimum value
4 3b2 2c 2 36b2 c 2
12b2c 2
Minimum value =
=
= c 2 > 4ab .
2
12b2
4 3b
)(
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*64.
Let A and B denote the statements

A: cos + cos + cos = 0
B: sin + sin + sin = 0
If cos ( ) + cos ( ) + cos ( ) =
3
, then
2
(1) A is true and B is false

(3) both A and B are true
Sol:
(2) A is false and B is true

(4) both A and B are false
(3)
3
2
2 cos ( ) + cos ( ) + cos ( ) + 3 = 0
cos ( ) + cos ( ) + cos ( ) =
2 cos ( ) + cos ( ) + cos ( ) + sin2 + cos2 + sin2 + cos2 + sin2 + cos2 = 0

( sin + sin + sin ) + ( cos + cos + cos ) = 0
2
*65.
The lines p p2 + 1 x y + q = 0 and p2 + 1 x + p2 + 1 y + 2q = 0 are perpendicular to a common line

for
(1) no value of p
(3) exactly two values of p
(2) exactly one value of p

(4) more than two values of p
Sol:
(2)
Lines must be parallel, therefore slopes are equal p p2 + 1 = p2 + 1 p = - 1
66.
If A, B and C are three sets such that A B = A C and A B = A C , then

(1) A = B
(2) A = C
(3) B = C
(4) A B =
Sol:
(3)
67.
Sol:
r r uur
If u, v, w are non-coplanar vectors and p, q are real numbers, then the equality
r
r
uur
r uur
r
uur
r
r
3u pv pw pv w qu 2w qv qu = 0 holds for
(1) exactly one value of (p, q)

(2) exactly two values of (p, q)
(4) all values of (p, q)
(3) more than two but not all values of (p , q)
(1)
r r
pq + 2q2 u v
r r uur
But u v w 0
3p2 pq + 2q2 = 0
( 3p
uur
w = 0
7q2
q
7
q
2p2 + p2 pq + +
= 0 2p2 + p + q2 = 0
4
2
4
2
q
p = 0, q = 0, p =
2
This possible only when p = 0, q = 0 exactly one value of (p, q)
x 2 y 1 z + 2
lies in the plane x + 3y z + = 0 . Then ( , ) equals
=
=
3
2
5
(1) (6, - 17)
(2) ( - 6, 7)
(3) (5, - 15)
(4) ( - 5, 15)
68.
Let the line
Sol:
(2)
Drs of line = ( 3, 5, 2 )
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Drs of normal to the plane = (1, 3, )

Line is perpendicular to normal 3 (1) 5 ( 3 ) + 2 ( ) = 0 3 15 2 = 0 2 = 12 = 6
Also ( 2, 1, 2 ) lies on the plane
2 + 3 + 6 ( 2 ) + = 0 = 7
( , ) = ( 6, 7 )
*69.
From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and
arranged in a row on the shelf so that the dictionary is always in the middle. Then the number of such
arrangements is
(1) less than 500
(2) at least 500 but less than 750
(3) at least 750 but less than 1000
(4) at least 1000
Sol:
(4)
4 novels can be selected from 6 novels in 6 C4 ways. 1 dictionary can be selected from 3 dictionaries
in 3 C1 ways. As the dictionary selected is fixed in the middle, the remaining 4 novels can be arranged
in 4! ways.
The required number of ways of arrangement = 6 C4 x 3 C1 x 4! = 1080
70.
[cot x ] dx , [] denotes the greatest integer function, is equal to

0
(1)
(2) 1
(4)
(3) 1
Sol:
(4)
Let I = [cot x ] dx
(1)
= cot ( x ) dx =
0
[ cot x ] dx
(2)
Adding (1) and (2)
2I =
[cot x ] dx
0
[ cot x ] dx
= =
( 1) dx
0
Q [ x ] + [ x ] = 1 if x Z
= 0 if x Z
= [ x ]0 =
I =
71.
For real x, let f ( x ) = x 3 + 5x + 1 , then

(1) f is one-one but not onto R
(3) f is one-one and onto R
Sol:
(2) f is onto R but not one-one

(4) f is neither one-one nor onto R
(3)
Given f ( x ) = x 3 + 5x + 1
Now f ' ( x ) = 3x 2 + 5 > 0, x R
f(x) is strictly increasing function
It is one-one
Clearly, f(x) is a continuous function and also increasing on R,
f ( x ) = and Lt f ( x ) =
Lt
x
x
f(x) takes every value between and .
Thus, f(x) is onto function.
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72.
In a binomial distribution B n, p = , if the probability of at least one success is greater than or

4
9
, then n is greater than
equal to
10
1
1
(2)
(1)
4
3
4
log10 log10
log10 + log103
(3)
Sol:
log10
9
log103
(4)
log10
4
log103
(1)
n
1 qn
*73.
9
1
1
3

n log 3 10 n
4
10
10
log
log103
4
10
4
If P and Q are the points of intersection of the circles
x 2 + y 2 + 3x + 7y + 2p 5 = 0
and
x + y + 2x + 2y p = 0 , then there is a circle passing through P, Q and (1, 1) for

(1) all values of p
(2) all except one value of p
(3) all except two values of p
(4) exactly one value of p
2
Sol:
(1)
Given circles S = x 2 + y 2 + 3x + 7y + 2p 5 = 0
S' = x 2 + y 2 + 2x + 2y p2 = 0
Equation of required circle is S + S' = 0

As it passes through (1, 1) the value of =
( 7 + 2p )
(6 p )
2
If 7 + 2p = 0, it becomes the second circle

it is true for all values of p
74.
The projections of a vector on the three coordinate axis are 6, - 3, 2 respectively. The direction
cosines of the vector are
6 3 2
(1) 6, 3, 2
(2) , ,
5 5 5
6 3 2
6 3 2
(3) , ,
(4) , ,
7 7 7
7 7 7
Sol:
(3)
Projection of a vector on coordinate axis are x 2 x1, y 2 y1, z2 z1
x 2 x1 = 6, y 2 y1 = 3, z 2 z1 = 2
( x 2 x1 )
+ ( y 2 y1 ) + ( z 2 z1 ) = 36 + 9 + 4 = 7
2
The D.Cs of the vector are
*75.
If Z
(1)
6 3 2
, ,
7 7 7
4
= 2 , then the maximum value of Z is equal to
z
3 +1
(3) 2
Sol:
(2)
5 +1
(4) 2 + 2
(2)
4 4
4 4
Z = Z +
Z = Z +
Z Z
Z Z
4
4
4
Z Z +
Z 2+
Z Z
Z
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Z 2 Z 4 0
( Z (
) ) ( Z (1 5 )) 0 1
5 +1
5 Z 5 +1
*76.
Three distinct points A, B and C are given in the 2 dimensional coordinate plane such that the ratio
of the distance of any one of them from the point (1, 0) to the distance from the point ( - 1, 0) is equal
1
to . Then the circumcentre of the triangle ABC is at the point
3
5
(1) ( 0, 0 )
(2) , 0
4
5
5
(4) , 0
(3) , 0
2
3
Sol:
(2)
P = (1, 0 ) ; Q ( 1, 0 )
Let A = ( x, y )
AP BP CP 1
=
=
=
AQ BQ CQ 3
..(1)
3AP = AQ 9AP2 = AQ2 9 ( x 1) + 9y 2 = ( x + 1) + y 2

2
9x 2 18x + 9 + 9y 2 = x 2 + 2x + 1 + y 2 8x 2 20x + 8y 2 + 8 = 0
5
(2)
x +1= 0
2
A lies on the circle
Similarly B, C are also lies on the same circle
x2 + y2
5
Circumcentre of ABC = Centre of Circle (1) = , 0
4
*77.
The remainder left out when 82n ( 62 )
2n +1
is divided by 9 is
(1) 0
(3) 7
Sol:
(2) 2
(4) 8
(2)
82n ( 62 )
2n +1
= (1 + 63 ) ( 63 1)
= (1 + 63 ) + (1 63 )
n
2n +1
2n +1
= 1 + nc1 63 + nc 2 ( 63 ) + .... + ( 63 )
2
= 2 + 63 n c1 + nc 2 ( 63 ) + .... + ( 63 )
n 1
2n +1)
c1 + (
2n +1)
) + (1
( 2n +1)
c 2 ( 63 ) + .... ( 63 )
c1 63 + (
( 2n )
2n +1)
c 2 ( 63 ) + .... + ( 1)( 63 )
2
( 2n +1)
Reminder is 2
*78.
The ellipse x 2 + 4y 2 = 4 is inscribed in a rectangle aligned with the coordinate axes, which in turn in
inscribed in another ellipse that passes through the point (4, 0). Then the equation of the ellipse is
(1) x 2 + 16y 2 = 16
(2) x 2 + 12y 2 = 16
(3) 4x 2 + 48y 2 = 48
Sol:
(4) 4x 2 + 64y 2 = 48
(2)
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x2 y2
+
= 1 a = 2, b = 1 P = ( 2, 1)
4
1
x2 y2
x2 y2
Required Ellipse is 2 + 2 = 1 2 + 2 = 1
a
b
4
b
(2, 1) lies on it
V
4
1
1
1 3
4
+
= 1 2 = 1 = b2 =
16 b2
4 4
3
b
2
2
2
2
x
y
x
3y
+
= 1
+
= 1 x 2 + 12y 2 = 16
16 4
16
4
3

2 6 10 14
The sum to the infinity of the series 1 + + 2 + 3 + 4 + ...... is
3 3
3
3
(1) 2
(2) 3
(3) 4
(4) 6
x 2 + 4y 2 = 4
*79.
Sol:
P (2, 1)
1
A
V
(4, 0)
(2)
2 6 10 14
+
+
+
+ ....
3 32 33 34
1
1 2
6 10
S = + 2 + 3 + 4 + ....
3
3 3
3
3
Dividing (1) & (2)
1
1 4
4
4
S 1 = 1 + + 2 + 3 + 4 + ....
3 3
3
3
3
Let S = 1 +
(1)
(2)
2
4 4
1 1
2
4 4 1 4 4 3 4 2 6
2
6
S = + 2 1 + + 2 + ...... S = + 2
= + 2 = + = S= S=3
1
3
3 3 3 3
3
3
3
3
3
2
3
3
2
3
3
1
3
80.
The differential equation which represents the family of curves y = c1ec 2 x , where c1 and c 2 are
arbitrary constants is
(1) y ' = y 2
(2) y " = y ' y
(4) yy " = ( y ' )
(3) yy " = y '

Sol:
(4)
y = c1ec 2 x
(1)
y ' = c 2 c1e
c2 x
y ' = c2 y
y " = c2 y '
From (2)
y'
c2 =
y
So, y " =
(2)
( y ')
y
yy " = ( y ' )
81.
One ticket is selected at random from 50 tickets numbered 00, 01, 02, ., 49. Then the probability
that the sum of the digits on the selected ticket is 8, given that the product of these digits is zero,
equals
1
1
(2)
(1)
14
7
1
5
(4)
(3)
50
14
Sol:
(1)
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S = { 00, 01, 02, ., 49 }

Let A be the even that sum of the digits on the selected ticket is 8 then
A = { 08, 17, 26, 35, 44 }
Let B be the event that the product of the digits is zero
B = { 00, 01, 02, 03, ., 09, 10, 20, 30, 40 }
A B = {8}
1
1
= 50 =
Required probability = P ( A / B ) =
14 14
P (B )
50
2x
Let y be an implicit function of x defined by x 2x x cot y 1 = 0 . Then y ' (1) equals
P ( A B)
82.
(1) 1
(3) log 2
Sol:
(2) 1
(4) log 2
(1)
x 2x 2x x cot y 1 = 0
Now x = 1,
(1)
1 2 coty 1 = 0 coty = 0 y =
Now differentiating eq. (1) w.r.t. x

dy
2x 2x (1 + log x ) 2 x x ( c osec 2 y )
+ cot y x x (1 + log x ) = 0
dx

Now at 1,
2
dy
+ 0 = 0
2 (1 + log1) 2 1( 1)
dx 1,
2
dy
dy
2 + 2
=0
= 1
dx 1,
dx 1,
83.
The area of the region bounded by the parabola ( y 2 ) = x 1 , the tangent to the parabola at the
2
point (2, 3) and the x-axis is

(1) 3
(3) 9
Sol:
(2) 6
(4) 12
(3)
Equation
( y 2)
of
tangent
at
(2,
3)
to
2y = x + 4
= x 1 is S1 = 0
x 2y + 4 = 0
Required Area = Area of OCB + Area of
OAPD Area of PCD
3
1
1
= ( 4 x 2 ) + y 2 4y + 5 dy (1 x 2 )
2
2
0
y3
= 4 + 2y 2 + 5y 1 = 4 9 18 + 15 1
3
0
= 28 19 = 9 sq. units
D (0, 3)
P (2, 3)
C (0, 2)
B (-4, 0)
A (1, 2)
A (1, 2)
0
(or)
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Area =
( 2y 4 y
84.
+ 4y 5 dy = y + 6y 5 dy = ( 3 y )
2
( y 3 )3
27
=
dy =
= 9 sq.units
3
3
0
Given P ( x ) = x 4 + ax 3 + bx 2 + cx + d such that x = 0 is the only real root of P' ( x ) = 0 . If P ( 1) < P (1) ,
then in the interval [ 1, 1]
(1) P ( 1) is the minimum and P (1) is the maximum of P
(2) P ( 1) is not minimum but P (1) is the maximum of P
(3) P ( 1) is the minimum and P (1) is not the maximum of P
(4) neither P ( 1) is the minimum nor P (1) is the maximum of P
Sol:
(2)
P ( x ) = x 4 + ax 3 + bx 2 + cx + d
P ' ( x ) = 4x 3 + 3ax 2 + 2bx + c
Q x = 0 is a solution for P' ( x ) = 0 , c = 0
P ( x ) = x 4 + ax 3 + bx 2 + d
(1)
Also, we have P ( 1) < P (1)
1 a + b + d < 1+ a + b + d a > 0
Q P ' ( x ) = 0 , only when x = 0 and P(x) is differentiable in ( - 1, 1), we should have the maximum and
minimum at the points x = - 1, 0 and 1 only
Also, we have P ( 1) < P (1)
Max. of P(x) = Max. { P(0), P(1) } & Min. of P(x) = Min. { P(-1), P(0) }
In the interval [ 0 , 1 ],
P' ( x ) = 4x 3 + 3ax 2 + 2bx = x 4x 2 + 3ax + 2b
Q P ' ( x ) has only one root x = 0, 4x + 3ax + 2b = 0 has no real roots.

2
( 3a ) 32b < 0
2
3a2
<b
32
b>0
Thus, we have a > 0 and b > 0
P' ( x ) = 4x 3 + 3ax 2 + 2bx > 0, x ( 0, 1)
Hence P(x) is increasing in [ 0, 1 ]
Max. of P(x) = P(1)
Similarly, P(x) is decreasing in [-1 , 0]
Therefore Min. P(x) does not occur at x = - 1
85.
Sol:
The shortest distance between the line y x = 1 and the curve x = y 2 is

(1)
3 2
8
(2)
2 3
8
(3)
3 2
5
(4)
3
4
(1)
x y +1= 0
x=y
(1)
1 = 2y
dy
dy
1
= Slope of given line (1)
=
dx
dx 2y
2
1
1
1
1
1
1 1
= 1 y = y = x = = ( x, y ) = ,
2
2
4
2y
2

4 2
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The shortest distance is
1 1
+1
4 2
1+ 1
3
4 2
3 2
8
Directions: Question number 86 to 90 are Assertion Reason type questions. Each of these questions
contains two statements
Statement-1 (Assertion) and Statement-2 (Reason).
Each of these questions also have four alternative choices, only one of which is the correct answer. You have
to select the correct choice
86.
Let f ( x ) = ( x + 1) 1, x 1
2
Statement-1 : The set x : f ( x ) = f 1 ( x ) = {0, 1}

Statement-2 : f is a bijection.
(1) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1
(2) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1
Sol:
(3)
There is no information about co-domain therefore f(x) is not necessarily onto.
87.
Let f ( x ) = x x and g ( x ) = sin x .

Statement-1 : gof is differentiable at x = 0 and its derivative is continuous at that point.
Statement-2 : gof is twice differentiable at x = 0.
Sol:
(3)
f ( x ) = x x and g ( x ) = sin x
sin x 2 ,x < 0
gof ( x ) = sin ( x x ) =
2
,x 0
sin x
2x cos x 2 ,x < 0
( gof ) ' ( x ) =
2
,x 0
2x cos x
Clearly, L ( gof ) ' ( 0 ) = 0 = R ( gof ) ' ( 0 )
gof is differentiable at x = 0 and also its derivative is continuous at x = 0

2cos x 2 + 4x 2 sin x 2 ,x < 0
Now, ( gof ) " ( x ) =
2
2
2
,x 0
2cos x 4x sin x
L ( gof ) " ( 0 ) = 2 and R ( gof ) " ( 0 ) = 2
L ( gof ) " ( 0 ) R ( gof ) " ( 0 )
gof(x) is not twice differentiable at x = 0.

*88.
Statement-1 : The variance of first n even natural numbers is

Statement-2 : The sum of first n natural numbers is
numbers is
n ( n + 1)( 2n + 1)
n ( n + 1)
2
n2 1
4
and the sum of squares of first n natural
6
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Sol:
(4)
Statement-2 is true
Statement-1:
Sum of n even natural numbers = n (n + 1)
n ( n + 1)
Mean x =
= n +1
n
2
1
2
2
2
1
Variance = ( x i ) x = 22 + 42 + ..... + ( 2n ) ( n + 1)
n
n
()
()
1 2 2
4 n ( n + 1)( 2n + 1)
2
2
2 1 + 22 + ..... + n2 ( n + 1) =
( n + 1)
n
n
6
(n + 1) 2 ( 2n + 1) 3 (n + 1) (n + 1) [ 4n + 2 3n 3] (n + 1)(n 1) n2 1
=
=
=
=
3
3
3
3
Statement 1 is false.
=
89.
Statement-1 : ~ ( p ~ q ) is equivalent to p q .
Statement-2 : ~ ( p ~ q) is a tautology.
Sol:
(3)
p
T
T
F
F
90.
pq
q
T
F
T
F
T
F
F
T
~q
F
T
F
T
p ~ q
~ ( p ~ q)
F
T
T
F
T
F
F
T
Let A be a 2 x 2 matrix
Statement-1 : adj ( adj A ) = A
Statement-2 : adj A = A
Sol:
(2)
adj A = A
n 1
= A
adj ( adj A ) = A
n2
2 1
= A
0
A= A A=A
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AIEEE2009, ANSWER KEY

Test Booklet Code-A
Test Booklet Code-B
Test Booklet Code-C
Test Booklet Code-D
PHY
CHE
MAT
CHE
MAT
PHY
MAT
PHY
CHE
CHE
PHY
MAT
1. (2)
31. (3)
61. (3)
1. (1)
31. (2)
61. (2)
1. (4)
31. (4)
61. (4)
1. (1)
31. (4)
61. (4)
2. (1)
32. (4)
62. (3)
2. (4)
32. (1)
62. (2)
2. (4)
32. (2)
62. (3)
2. (1)
32. (3)
62. (4)
3. (3)
33. (3)
63. (3)
3. (1)
33. (2)
63. (1)
3. (4)
33. (3)
63. (4)
3. (1)
33. (4)
63. (1)
4. (1)
34. (3)
64. (3)
4. (3)
34. (3)
64. (3)
4. (3)
34. (4)
64. (1)
4. (1)
34. (4)
64. (1)
5. (2)
35. (3)
65. (2)
5. (3)
35. (2)
65. (1)
5. (2)
35. (2)
65. (3)
5. (4)
35. (1)
65. (2)
6. (1)
36. (3)
66. (3)
6. (1)
36. (2)
66. (1)
6. (1)
36. (3)
66. (4)
6. (2)
36. (3)
66. (3)
7. (4)
37. (4)
67. (1)
7. (1)
37. (1)
67. (2)
7. (4)
37. (1)
67. (3)
7. (1)
37. (3)
67. (2)
8. (3)
38. (4)
68. (2)
8. (1)
38. (4)
68. (4)
8. (4)
38. (4)
68. (4)
8. (1)
38. (4)
68. (4)
9. (4)
39. (3)
69. (4)
9. (2)
39. (1)
69. (1)
9. (4)
39. (2)
69. (4)
9. (1)
39. (3)
69. (4)
10. (2)
40. (4)
70. (4)
10. (1)
40. (3)
70. (4)
10. (3)
40. (2)
70. (4)
10. (4)
40. (1)
70. (2)
11. (1)
41. (3)
71. (3)
11. (2)
41. (4)
71. (4)
11. (1)
41. (1)
71. (3)
11. (4)
41. (4)
71. (3)
12. (2)
42. (3)
72. (1)
12. (4)
42. (1)
72. (1)
12. (3)
42. (4)
72. (4)
12. (4)
42. (4)
72. (3)
13. (2)
43. (3)
73. (1)
13. (3)
43. (2)
73. (3)
13. (4)
43. (2)
73. (1)
13. (4)
43. (2)
73. (1)
14. (3)
44. (2)
74. (3)
14. (3)
44. (2)
74. (4)
14. (3)
44. (1)
74. (4)
14. (1)
44. (4)
74. (4)
15. (4)
45. (2)
75. (2)
15. (2)
45. (3)
75. (3)
15. (2)
45. (4)
75. (4)
15. (3)
45. (2)
75. (4)
16. (2)
46. (4)
76. (2)
16. (2)
46. (2)
76. (2)
16. (4)
46. (3)
76. (3)
16. (3)
46. (2)
76. (1)
17. (2)
47. (2)
77. (2)
17. (2)
47. (1)
77. (1)
17. (3)
47. (1)
77. (4)
17. (1)
47. (2)
77. (2)
18. (2)
48. (1)
78. (2)
18. (1)
48. (4)
78. (1)
18. (3)
48. (2)
78. (4)
18. (1)
48. (2)
78. (1)
19. (3)
49. (1)
79. (2)
19. (1)
49. (2)
79. (2)
19. (4)
49. (3)
79. (3)
19. (1)
49. (1)
79. (1)
20. (1)
50. (2)
80. (4)
20. (2)
50. (1)
80. (1)
20. (1)
50. (3)
80. (2)
20. (4)
50. (1)
80. (3)
21. (2)
51. (2)
81. (1)
21. (2)
51. (2)
81. (4)
21. (2)
51. (3)
81. (2)
21. (4)
51. (3)
81. (1)
22. (4)
52. (2)
82. (1)
22. (2)
52. (2)
82. (3)
22. (3)
52. (4)
82. (3)
22. (1)
52. (3)
82. (4)
23. (3)
53. (3)
83. (3)
23. (2)
53. (4)
83. (3)
23. (3)
53. (1)
83. (3)
23. (2)
53. (1)
83. (4)
24. (1)
54. (2)
84. (2)
24. (1)
54. (1)
84. (1)
24. (3)
54. (3)
84. (4)
24. (4)
54. (3)
84. (1)
25. (4)
55. (2)
85. (1)
25. (1)
55. (2)
85. (4)
25. (4)
55. (1)
85. (1)
25. (2)
55. (4)
85. (1)
26. (1)
56. (3)
86. (3)
26. (1)
56. (1)
86. (1)
26. (1)
56. (2)
86. (1)
26. (4)
56. (4)
86. (3)
27. (4)
57. (3)
87. (3)
27. (3)
57. (1)
87. (3)
27. (2)
57. (1)
87. (2)
27. (2)
57. (1)
87. (4)
28. (3)
58. (2)
88. (4)
28. (2)
58. (4)
88. (4)
28. (3)
58. (3)
88. (3)
28. (1)
58. (2)
88. (3)
29. (1)
59. (2)
89. (3)
29. (2)
59. (4)
89. (2)
29. (2)
59. (3)
89. (4)
29. (2)
59. (3)
89. (1)
30. (2)
60. (3)
90. (2)
30. (2)
60. (3)
90. (2)
30. (2)
60. (2)
90. (1)
30. (4)
60. (2)
90. (1)
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1

AIEEE2009, BOOKLET CODE(A)

Part B Chemistry (144 marks)

Part C Mathematics(144 marks)

The above is a plot of binding energy per nucleon

A p-n junction (D) shown in the figure can act

(1) (2f, 2f)

(4) (4f, 4f)

IR corresponds to least value of

Pressure = 8 104 N/m2

Directions: Question numbers 12 and 13 are based on the following paragraph.

Due to the presence of the current I1 at the origin

(3) The magnitude of the net force on the loop is given by

v = - gt and after the collision, v = gt.

Resultant on Q becomes zero only when q charges are of

In steady state flow of heat

A transparent solid cylindrical rod has a refractive index of

. It is surrounded by air. A light ray is incident at the mid

point of one end of the rod as shown in the figure.

Sin r = sin (90 C) = cosC =

The height at which the acceleration due to gravity becomes

, acceleration due to gravity at height h

An inductor of inductance L = 400 mH and resistors of resistances R1

value of main scale division

Potential drop across L = E R2I2 = 12 2 6 (1 e-bt) = 12 e-5t

The work done on the gas in taking it from D to A is

= 0.693 600 R = - 414 R.

The net work done on the gas in the cycle ABCDA is

(2) ( CH3 )2 CH > CCl3 > C6 H5 CH2 > ( CH3 )3 C

(3) CCl3 > C6 H5 CH2 > ( CH3 )2 CH > ( CH3 )3 C

( CH3 )3 C > ( CH3 )2 CH > C6H5 CH2

The alkene that exhibits geometrical isomerism is :

Which of the following statements is incorrect regarding physissorptions ?

Which of the following on heating with aqueous KOH, produces acetaldehyde ?

Gof H2 O(l ) = -237.2 kJ mol-1

Gof CO2 (l) = -394.4 kJ mol-1

We get, PXo = 400 mm Hg

[ A o ] = initial concentration (amount)

H2 O(l ) H+ (aq) + OH (aq); H = 57.32kJ

Which of the following has an optical isomer ?

Which one of the following reactions of Xenon compounds is not feasible ?

no..of bonding e no.of antibonding e

The number of stereoisomers possible

The IUPAC name of neopentane is

The set representing the correct order of ionic radius is :

(4) Mg2+ > Be2+ > Li+ > Na +

In Cannizzaro reaction given below

(1) the attack of : OH at the carboxyl group

Which of the following pairs represents linkage isomers ?

Buna-N synthetic rubber is a copolymer of :

(2) H2 C = CH CH = CH2 and H5 C6 CH = CH2

Let a, b, c be such that b ( a + c ) 0 . If b b + 1 b 1 + a 1

(2) any even integer

This is equal to zero only if n + 2 is odd i.e. n is odd integer.

Let A and B denote the statements

(1) A is true and B is false

(2) A is false and B is true

2 cos ( ) + cos ( ) + cos ( ) + sin2 + cos2 + sin2 + cos2 + sin2 + cos2 = 0