Sri Chaitanya & Narayana JEE MAIN GTMs 2023-24 Single File

Sri Chaitanya IIT Academy.,India.
 A.P  T.S  KARNATAKA  TAMILNADU  MAHARASTRA  DELHI  RANCHI

A right Choice for the Real Aspirant
ICON Central Office - Madhapur - Hyderabad
SEC: Sr.Super60_Elite, Target & LIIT-BTs JEE-MAIN Date: 09-01-2024
Time: 09.00Am to 12.00Pm GTM-10 Max. Marks: 300
IMPORTANT INSTRUCTION:
1. Immediately fill in the Admission number on this page of the Test Booklet with Blue/Black Ball Point Pen
only.
2. The candidates should not write their Admission Number anywhere (except in the specified space) on the
Test Booklet/ Answer Sheet.
3. The test is of 3 hours duration.
4. The Test Booklet consists of 90 questions. The maximum marks are 300.
5. There are three parts in the question paper 1,2,3 consisting of Physics, Chemistry and Mathematics having
30 questions in each subject and subject having two sections.
(I) Section –I contains 20 multiple choice questions with only one correct option.
Marking scheme: +4 for correct answer, 0 if not attempt and ‐1 in all other cases.
(II) Section‐II contains 10 Numerical Value Type questions. Attempt any 5 questions only, if more than 5
questions attempted, First 5 attempted questions will be considered.
∎ The Answer should be within 0 to 9999. If the Answer is in Decimal then round off to the nearest Integer
value (Example i,e. If answer is above 10 and less than 10.5 round off is 10 and If answer is from 10.5 and
less than 11 round off is 11).
To cancel any attempted question bubble on the question number box.
For example: To cancel attempted question 21. Bubble on 21 as shown below
.
Question Answered for Marking Question Cancelled for Marking
SRI CHAITANYA IIT ACADEMY, INDIA 09‐01‐24_ Sr.Super60_Elite, Target & LIIT-BTs _Jee‐Main_GTM‐10_Q.P
6. Use Blue / Black Point Pen only for writing particulars / marking responses on the Answer Sheet. Use of pencil is
strictly prohibited.
7. No candidate is allowed to carry any textual material, printed or written, bits of papers, mobile phone any electron
device etc, except the Identity Card inside the examination hall.
8. Rough work is to be done on the space provided for this purpose in the Test Booklet only.
9. On completion of the test, the candidate must hand over the Answer Sheet to the invigilator on duty in the Hall.
However, the candidate are allowed to take away this Test Booklet with them.
10. Do not fold of make any stray marks on the Answer Sheet
Name of the Candidate (in Capital): ________________________________________________
Admission Number:
Candidate’s Signature:________________ Invigilator’s Signature: ________________
09‐01‐24_ Sr.Super60_Elite, Target & LIIT-BTs _ Jee‐Main_GTM‐10_Test Syllabus

PHYSICS : TOTAL SYLLABUS
CHEMISTRY : TOTAL SYLLABUS
MATHEMATICS : TOTAL SYLLABUS
Sec: Sr.Super60_Elite, Target & LIIT-BTs Page 2

PHYSICS Max Marks: 100
(SINGLE CORRECT ANSWER TYPE)
This section contains 20 multiple choice questions. Each question has 4 options (1), (2), (3) and (4) for its answer, out of which ONLY ONE option can be
correct.
Marking scheme: +4 for correct answer, 0 if not attempted and –1 in all other cases.
1. A bus is moving with a velocity of 10 m/s on a straight road. A scooterist wishes to overtake
the bus in 100 seconds. If the bus is at a distance of 1 km form the scooterist, at what
velocity should the scooterist chase the bus:
1) 50 m/sec 2) 40 m/sec 3) 30 m/sec 4) 20 m/sec
2. The length of an elastic string is x when the tension is 5N. Its length is y when the tension is
7N. What will be its length, when the tension is 9N?
1) 2 y  x 2) 2 y  x 3) 7 x  5 y 4) 7 x  5 y
3. 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  , of this rod, varies as,   a  bx. What should be the
dimensions of b?
1) M 2 L1T 0 2) M 1L2T 0 3) M 1L3T 1 4) M 1L2T 3

4. 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 circumference of wheel relative to road will be:
1) v A  v, vB  0 2) v A  vB  0 3) v A  0, vB  v 4) v A  0, vB  2v
5. A metallic wire of density d is lying horizontal on the surface of water (surface tension = t).
The maximum length of wire so that it may not sink will be:
2Tg 2 T 2T
1) 2) 3) 4) any length
d dg  dg
6. Two points of a rod move with velocities 3v and v perpendicular to the rod and in the same
direction, separated by a distance r. Then the angular velocity of the rod is:
3v 4v 5v 2v
1) 2) 3) 4)
r r r r
7. For hydrogen gas C p  Cv  a and for oxygen gas C p  Cv  b, so the relation between a
and b is given by:

1) a = 16b 2) 16a = b 3) a = 4b 4) a = b
8. Two bars of thermal conductivities K and 3K and lengths 1 cm and 2 cm, respectively, have
equal cross-sectional area. They are joined length wise as shown in the figure. If the
temperature at the ends of this composite bar is 0 C and 100 C , respectively (see figure),
then the temperature  of the interface is:
100 200
1) 50 C 2) 
3) 60 C 4)
3C 3 C
9. The equivalent capacitance of the network, (with all capacitors having the same capacitance
C.
1)  2) Zero 3) C 
  
3  1 / 2  4) C 
   
3  1 / 2


10. The current I vs voltage V graphs for a given metallic wire at two difference temperatures T1
and T2 are shown in the figure. It is concluded that:
1) T1  T2 2) T1  T2 3) T1  T2 4) T1  2T2
11. In the circuit shown the effective resistance between B and C is:
4 3
1) 3  2) 4  3)  4) 
3 4
12. In the given circuit, the current drawn from the source is:
1) 20 A 2) 10 A 3) 5 A 4) 5 2 A

13. A flat plate P of mass ‘M’ executes SHM in a horizontal plane by sliding over a frictionless
surface with a frequency . A block ‘B’ of mass ‘m’ rests on the plate as shown in figure.
Coefficient of friction between the surface of B and P is . What is the maximum amplitude
of oscillation that the plate block system can have if the block B is not to slip on the plate:
g g g g
1) 2) 3) 4)
4 2 2  2 2 8 2 2 2 2 2
14. A glass slab has the left half of refractive index n1, and the right half of n2  3n1. The
effective refractive index of the whole slab is:
n 3n1 2n1
1) 1 2) 2n1 3) 4)
2 2 3
15. In the arrangement shown L1, L2 are slits and S1, S2 two independent sources on the screen,
interference fringes:
1) Will not be there

2) Will not be there if the intensity of light reaching the screen from S1 and S 2 are equal.
3) If light coming from S1 & S2 have same wavelength we may observe interference fringes
like pattern.
4) We will have only the central fringe.
16. What is the ratio of the circumference of the first Bohr orbit for the electron in the hydrogen
atom to the de Broglie wavelength of electrons having the same velocity as the electron in
the first Bohr orbit of the hydrogen atom?
1)1:1 2)1:2 3)1:4 4) 2:1

17. An electric field is expressed as E  2i  3 j. Find the potential difference VA  VB  between
two points A and B whose position vectors are given by rA  i  2 j and rB  2i  j  3k .
1) 1V 2) 1V 3) 2V 4) 3V
18. P – V plots for two gases during adiabatic processes are shown in the figure. Plots 1 and 2
should correspond respectively to:
1) He AND Ar 2) He AND O2 3) O2 AND N 2 4) O2 AND He

19. Two identical thin rings, each of radius R metres, are coaxially placed at a distance R metres
apart. If Q1 coulomb and Q2 coulomb are respectively, the charges uniformly spread on the
two rings, the work done in moving a charge q from the centre of one ring to that of the
other is:

1) Zero 2)
q  Q1  Q2   2 1 
2 4 0 R
3)
q 2  Q1  Q2 
4)
q  Q1  Q2   2 1 
4 0 R 2 4 0 R
20. A parallel beam of light is incident on the system of two convex lenses of focal lengths
f1  20 cm and f 2  10 cm. What should be the distance between the two lenses so that rays
after refraction from both the lenses pass undeviated?
1) 60 cm 2) 30 cm 3) 90 cm 4) 40 cm
(NUMERICAL VALUE TYPE)
Section-II contains 10 Numerical Value Type questions. Attempt any 5 questions only. First 5 attempted questions will be considered if more than 5
questions attempted. The Answer should be within 0 to 9999. If the Answer is in Decimal then round off to the nearest Integer value (Example i,e. If answer
is above 10 and less than 10.5 round off is 10 and If answer is from 10.5 and less than 11 round off is 11).
Marking scheme: +4 for correct answer, 0 if not attempt and -1 in all other cases.
21. The masses of the blocks A and B are 0.5 kg and 1 kg respectively. These are arranged as
shown in the figure and are connected by a massless string. The coefficient of friction
between all contact surfaces is 0.4. The force (in N) necessary to move the block B with

constant velocity will be: g  10 m / s 2 

22. An artificial satellite is moving in a circular orbit around the earth with a speed equal to half
of the magnitude of escape velocity from the earth. If the satellite is stopped suddenly in its
 
orbit and allowed to fall freely onto the earth, then speed in ms 1 with which it hits the
surface of the earth is 

kgR . Find the value of k. Given g  9.8 ms 2 and R  6.4  106 m 
23. A non-conducting partition divides a container into two equal compartments. One is filled
with helium gas at 200K and the other is filled with oxygen gas at 400K. The number of
molecules in each gas is the same. If the partition is removed to allow the gases to mix, the
final temperature (in K) will be:
24. Two circular coils X and Y, having equal number of turns, carry equal currents in the same
sense and subtend same sense angles 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
BX
to smaller coil X at O as BX then the ratio is:
BY
25. A point charge is placed outside a spherical shell as shown in the figure. Find the electric
potential (in kilo volt) on the surface of shell (Take: q  1 c and R=50cm)

26. In the following circuit, the current through the resistor R   2  is I amperes. The value of I
is_____
a
27. The pressure and volume of an ideal gas are related as P  3
, where ' a ' and ' b ' are
V 
1  
b
constants. The translational kinetic energy due to the thermal motion of the gas sample at
volume V  b, (given that ab  12 KJ ) is n KJ . Find n.
28. 
The potential energy of a particle is determined by the expression U   x 2  y 2 , where  
is a positive constant. The particle begins to move from a point with the co-ordinates  3,3
only under the action of potential fields force. When it reaches the point 1,1 its kinetic
energy is 4 K . Find the value of K .
29. A plane progressive wave is given by x   40 cm  cos  50 t  0.02 y  where y is in cm and t
in s. the particle velocity at y = 25 cm in time t = 1/100s will be 10 n m / s. What is the

value of n.
30. A charge of 5 C is placed at the centre of a square ABCD of side 10 cm. Find the work
done (in  J ) in moving a charge of 1 C from A to B.

CHEMISTRY Max Marks: 100
correct.
31. An aqueous solution of aniline of concentration 0.24M is prepared. What is concentration of
sodium hydroxide is needed in this solution so that anilinium ion concentration remains at
1 x 108 M? ( K b for C6 H 5 NH 3  2.4  10 5 M )
1) 102 M 2) 101 M 3) 103 M 4) 104 M

32. For a cell reaction involving two electron charge, the standard emf of the cell is 0.295V at
250 C . The equilibrium constant of the reaction at 250 C will be:
1) 1 1010 2) 10 3) 29.5 102 4) 2.95  1010
33. Product is :
1) 2)
3) Both (A) and (B) 4) No reaction
34. The standard heat of formation listed for gaseous NH3 is – 11.02 kcal/mol at 298 k. Given
that at 298 k, the constant pressure heat capacities of gaseous N 2 , H 2 and NH3 are
respectively 6.96, 6.89, 8.38 cal/mol. H 773K for the reactions,
1 3
N 2 ( g )  H 2 ( g )  NH 3 ( g )
2 2
1) 10.6kcal mol 1 2) 13.6kcal mol 1
3) 12.4kcal mol 1 4) 11.8kcal mol 1

35. A cobalt amine has the formula Co( NH 3 )6 Cl3 , with AgNO3 solution, one-third of the
chloride is Precipitated. The complex shows cis-trans isomerism. It can have the structure:
1) [Co( NH 3 )6 ]Cl3 , 2) [Co( NH 3 )5 Cl ]Cl2
3) [Co( NH 3 )4 Cl2 ]Cl 4) [Co( NH 3 )5 .H 2O]Cl3
36. The pair in which both species have same magnetic moment is:
1) [Cr ( H 2O ) 6 ]2  ,[CoCl4 ]2  2) [Cr ( H 2 O ) 6 ]2  ,[ Fe( H 2 O )6 ]2 
3) [ Mn ( H 2 O ) 6 ]2  ,[Cr ( H 2 O )6 ]2  4) [CoCl4 ]2  ,[ Fe( H 2 O )6 ]2 
37. Identify the product “Z” in the following series of reactions:
HCN H 3O  HI
C6 H12O6 ( glu cos e)  X 
 Y  Z :

1) hexanoic acid 2) α-methyl caproic acid
3) Heptanoic acid 4) none of these
38. Consider the following diazonium ions:
The order of reactivity toward diazo-coupling with phenol in the presence of dil. NaOH is
1) I < IV < II < III 2) I < III < IV < II
3) III < I < II < IV 4) III < I < IV < III
39. which one is the best method of reducing 3-bromopropanal to 1-bromopropane?
BrCH 2CH 2CHO  BrCH 2CH 2CH 3
1) Wolf-Kishner reduction 2) Clemmensen reduction
3) Either of the two 4) None of the two

40.
1) 2) 3) 4)
41.
Which is true about this reaction?
1) A is meso 2, 3-butan-di-ol formed by syn addition
2) A is meso 1, 2-butan-di-ol formed by anti addition
3) A is racemic mixture of d and  , 1, 2-butan-di-ol formed by anti addition
4) A is racemic mixture of d and  , 1, 2-butan-di-ol formed by syn addition
42. The incorrect statement regarding O( SiH 3 )2 and OCl2 molecule is/are:
1) The strength of back bonding is more in O( SiH 3 )2 molecule than OCl2 molecule
 -Si bond angle in O( SiH ) is greater than Cl - O

2) Si-O  - Cl bond angle in OCl .
3 2 2
3) The nature of back bond in both molecules is (2 p  3d )

4) Hybridisation of central O-atom in both molecules is same.
43. Ammonia forms the complex [Cu ( NH 3 ) 4 ]2  with copper ions in alkaline solution but not
in acidic solution. The reason for this is:
1) In alkaline solution Cu  OH 2 is precipitated which is soluble in excess of alkali
2) Copper hydroxide is amphoteric substance
3) In acidic solution hydration protects Cu 2 ions
4) In acidic solution protons are coordinated with ammonia molecules forming NH 4  ions
44. The hydrolysis constant for ZnCl2 will be:
2 2
K KW KW Kb
1) K h  W 2) K h  3) K h  4) K h 
Kb Kb Kb2 2
KW
Where Kb is effective dissociation constant of base Zn 
45. Consider the following conversions:

(i ) O( g )  e  O  ( g ) , H1 (ii ) F( g )  e  F ( g ) , H 2
(iii ) Cl( g )  e  Cl ( g ) , H 3 (iv) O ( g )  e  O 2( g ) , H 4
That according to given information the incorrect statements is:
1) H 3 is more negative than H1 and H 2
2) H1 is less negative than H 2
3) H1, H 2 and H 3 are negative whereas H 4 is positive
4) H1 and H 3 are negative whereas H 2 and H 4 are positive
46. What is the geometry of the IBr2 ion?

1) Linear
2) Bent shape with bond angle of about 90
47. The formation of PH 4 is difficult compared to NH 4 because:

1) lone pair of phosphorus is optically inert
2) lone pair of phosphorus resides in almost pure p-orbital
3) lone pair of phosphorus resides at sp3 orbital
4) lone pair of phosphorus resides in almost pure s-orbital
48. Which of the following solid salt on heating with solid K 2Cr2O7 and cone. H 2 SO4 orange
red vapours are evolved which turn aqueous NaOH solution yellow?
1) NaBr 2) NaCl 3) NaNO3 4) NaI

49.
Increasing order of stability is
1) I < III < II < IV 2) IV < III < II < I
3) III < IV < II < I 4) II < IV < III < I
50. For the electrochemical cell, M M  X  X , E  M  / M  0.44V and
E  ( X / X  )  0.33V . From this data one can deduce that:

1) M  X  M   X  is the spontaneous reaction.
2) M   X   M  X is the spontaneous reaction.
3) Ecell  0.77V
4) Ecell  0.77V
51. 
 2 NO2 ( g ). To
At 273K and atm, ‘a’ litre of N 2O4 decompose to NO2 as: N 2O4 

what extent has the decomposition (%) proceeded when the original volume is 25% less than
that of existing volume?
52. For a first order reaction, calculate the ratio between the time taken to complete three-fourth
of the reaction between the time taken to complete half of the reaction.
53. Suppose 1017 J of light energy is needed by the interior of the human eye to see an object.
How many photons of green light (  550nm) are needed to generate this minimum
amount of energy?

3 1
54. t1 2 of a reaction, A  Product (Order = ) is represented by t1 2 . The value of 2m
2 (a0 )m
is:
55. The energy of activation of reaction, if its rate double when the temperature is raised from
290 K to 300 K is x kcal then the value of x/3 is _________ (Take R = 2 cal/k/mol)
56. In a constant volume calorimeter, 3.5 g of gas with molar mass 28 g mol 1 was burn in
excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from
298.0 K to 298.45 K due to the combustion process. Given that the heat capacity of the
calorimeter is 2.5 kJ K 1 , the numerical value for the enthalpy of combustion of the gas in
kJ mol 1 is:
57. The pH of an aqueous solution of 0.1 M solution of a weak monoprotic acid which is 1%
ionized is:
58. Find out the % of oxalate ion in given sample of oxalate salt of which 0.3 g is present in
100 mL of solution required 90 mL of N/20 KMnO4 for complete oxidation.
59. The molal freezing point constant of C6 H 6 is 4.90 and its melting point is 5.51o C . A
solution of 0.816 g of compound A when dissolved in 7.5 g of benzene freezes at 1.59o C .

Calculate molar mass of compound A
60. Find the volume of a 0.2 M solution of MnO4 that will react with 50 mL of 0.1 M solution
of C2O42 in acidic medium____

MATHEMATICS Max Marks: 100
correct.
61. The number of integer values of a for which the inequality x 2  ax  a 2  6a  0 is satisfied
for all x  1, 2  is
1) 5 2) 7 3) 8 4) 6
62. The relation R defined on the set A  1,2,3,4,5 is given by R   x : y  : x 2 – y 2  16 then
1) R is reflexive and symmetric 2) R is transitive
3) R is not symmetric 4) R is reflexive and transitive
63. A number is said to be a nice number if it has exactly 4 factors. (Including one and number
itself). Let n  23  32  53  7  112 , then number of factors, which are nice numbers is
1) 36 2) 12 3) 10 4) 9
x n  nx n1  1
64. The value of lim , n 1 is (where [.] denotes greatest integer function)
e 
x  x
1) 1 2) zero 3) n 4) n(n –1)

65.  
The solution of the differential equation 1  xy  x5 y 5 dx  x 2 x 4 y 4  1 dy  0 is given by  
1 1 1 1
xy  x5 y 5 xy  x5 y 5 x 2 y 2  x5 y 5 x 2 y 2  x5 y 5
1) x  ce 2) x  ce
5
3) x  ce 5
4) x  ce 5 5
66. The number of point(s) on the line 3x  4 y  5, which are at a distance of
sec 2  2cosec 2 ,   R, from the point 1,3 is
1) 1 2) 2 3) 3 4) zero
67. The equation of circumcircle of an equilateral triangle is x 2  y 2  2 gx  2 fy  c  0 and one
vertex of the triangle is (1, 1). The equation of incircle of the triangle is:
 
1) 4 x 2  y 2  g 2  f 2
2) 4  x 2
 y   8 gx  8 fy  1  g 1  3g   1  f 1  3 f 
2
3) 4  x 2
 y   8 gx  8 fy  g  f
2 2 2
4) 4  x 2
 y   8 g  f   0
2 2 2

68. If AFB is a focal chord of the parabola y 2  4ax and AF  4, FB  5 (F being the focus of
given parabola) then the latus-rectum of the parabola is equal to
80 9
1) 2) 3) 9 4) 80
9 80
x2 y 2 x2 y2 1
69. If the ellipse  2  1 and the hyperbola   are orthogonal then the value of
16 b 144 81 25
b2
1) 1 2) 5 3) 7 4) 9

70. If cos 2  
3

1 2
a  1 and tan 2  tan 2/3  , then cos2/3   sin 2/3  
2
2/3 1/3
2/3 2 2
1) 2a 2)   3)   4) 2a1/3
a a
71. A and B play a game of tennis. The situation of the game is as follows; if one scores two
consecutive points after a deuce he wins; if loss of a point is followed by win of a point, it is
deuce. The chance of a server to win a point is 2/3. The game is at deuce and A is serving.
Probability that A will win the match is, (serves are changed after each point scored)
1) 3/5 2) 2/5 3) 1/2 4) 4/5
72. If A, B, C are the sets of all positive divisors of 1030 , 2020 and 3010 respectively. Then
 A  B  C  equals
1) 2381 2) 13981 3) 4381 4) 7161
73. If in a moderately asymmetrical distribution, mode and mean of the data are 6 and 9
respectively, then median is
1) 8 2) 7 3) 6 4) 5
74. ' n1 ' men and ' n2 ' women are to be seated in a row so that no two women sit together. If
n1  n2 then total number of ways in which they can be seated, is equal to:
1) n1 Cn2 2) n1 Cn2  n1 ! n2 ! 3) n1 Cn2 1  n1 ! n2 ! 4) n1 1

Cn2  n1 ! n2 !

If coefficient of x y z in  x  y  z  is A, (where A  0 ) then coefficient x 4 y 4 z is

2 3 4 n
75.
nA A
1) 2A 2) 3) 4) nA
2 2
xy
76. Consider the equation  233456 then the number of positive integral solutions of the
x y
equation are
1) 140 2) 819 3) 72 4) 601
77. I 
1  x  dx is equal to
x 1  xe 
x 2
 xe x   1   xe x   1 
1) l n  x 
 x 
c 2) l n  x 
 x 
c
 1  xe   1  xe   1  xe   1  xe 
 xe x   1   xe x   1 
3) l n  x 
 x 
c 4) l n  x 
 x 
c
 1  xe   1  xe   1  xe   1  xe 
  2  1
78. lim  tan  tan  ..  tan  
n 
 3n 3n 3n
3 2 3 3
1) ln2 2) ln2 3) l n3 4)
   
            
79. Let a, b, c be three non zero vectors such that a  b  c  0 . Then  b  a  b  c  c  a  0,
where  is equal to
1) 1 2) 2 3) –1 4) –2
 
80. If a  xi   x  1 j  k and b   x  1 i  j  ak always make an acute angle with each other
for every value of x  R , then
1) a   ,2  2) a   2,   3) a   ,1 4) a  1,  

81. Let ABC be a triangle. A be the point (1, 2), y  x is the perpendicular bisector of AB and
x  2 y  1  0 is the angle bisector of angle C. If the equation of BC is given by
ax  by  5  0, then the value of a  b is:
ba bc
82. If a, b, c are in H.P., then the value of  is equal to
ba bc
83. The tangents drawn from the origin to the circle x 2  y 2  2rx – 2hy  h 2  0 are
perpendicular then the value of r 2 – h2 is ___
84. If minimum distance between the curve y 2  4 x and x 2  y 2  12 x  31  0 is equal to  ,
then  2 is equal to
85. If the area of the region consisting of points (x, y) satisfying | x  y | 2 and x 2  y 2  2 is
equal to  k1  k2  sq.unit , then value of  k1  k2  is where k1 , k2  N .
 1
 if i  j
86. A square matrix An is defined as An   aij  , where aij   i 2  j 2 , then the value
nn '
 0 if i  j

1 5 
of   trace of  An1   is equal to
10  n1 
87. Four people sit around a circular table, and each person will roll a normal six sided die once.
The probability that no two people sitting next to each other will roll the same number is
N N 
. Then the value of   is (where [.] denotes the greatest integer function)
1296 10 

  
88. 
In a quadrilateral ABCD, AC is the bisector of the AB  AD which is  2
3
,
      

 
p
15 AC  3 AB  5 AD , if cos BA CD  (where p and q are co-prime and a  b
2 
q
 
denotes the angle between a and b ) then p  q is equal to.
 
89. Let u be a vector on rectangular coordinate system with sloping 60 . Suppose that u  i is

  
geometric mean of u and u  2i where i is the unit vector along x-axis then u has the
a  b where a, b  N , then the value  a  b    a  b  is

3 3
value equal to
90. Consider a system of equations x 4  y 4  z 4  1, x 4   y 4  z 4    2 and
x 4  y 4   z 4   , where  is a real parameter. If the number of real solutions of the system

of equations is n, then n is equal to


only.
.
Admission Number:

correct.
1. A uniform metre scale of length 1m is balanced on a fixed semi–circular cylinder of
radius 30 cm as shown in figure. One end of the scale is slightly depressed and released.
The time period (in seconds) of the resulting simple harmonic motion is
(Take g  10ms 2 )
Cylinder
  
1)  2) 3) 4)
2 3 4
2. The combination of gates shown in figure yields.
1) NAND gate 2) OR gate 3) NOT gate 4) XOR gate

3. The orbital speed of the electron in the ground state of hydrogen is . What will be its
orbital speed when it is excited to the energy state – 3.4 eV?
  
1) 2 2) 3) 4)
2 4 8
4. A bullet is fired normally on an immovable wooden plank. It loses 25% of its momentum
in penetrating a thickness of 3.5cm. The total thickness penetrated by the bullet is
1) 8cm 2) 10cm 3) 12 cm 4) 14cm

5. A cylinder of radius R made of a material of thermal conductivity k1 , is surrounded by a
cylindrical shell of inner radius R and outer radius 2R made of a material of thermal
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
k1k2 k  3k2 3k1  k2

1) k1  k2 2) 3) 1 4)
k1  k2 4 4
6. A toroidal solenoid has 3000 turns and a mean radius of 10cm. It has a soft iron core of
relative permeability 2000. What is the magnitude of the magnetic field in the core when
a current of 1A is passed through the solenoid?
1) 0.012 T 2) 0.12 T 3) 1.2 T 4) 12T
7. If the source of light used in a Young’s double slit experiment is changed from red to
violet, then
1) the consecutive fringe lines will come closer
2) the central bright fringe will become a dark fringe
3) the fringes will become brighter
4) the intensity of minima will increase
8. If the velocity-time graph has the shape AMB, what would be the shape of the
corresponding acceleration-time graph?
Velocity(v)
A B
Time(f)
a a a a
l l l l
1) 2) 3) 4)

9. A charge q is placed at one corner of a cube as shown in figure. The flux of electrostatic
field E through the shaded area is
Z
q
Y
X
q q q q
1) 2) 3) 4)
48 0 4 0 8 0 24 0
10. Match List-I with List-II
List-I List-II
P) Rectifier 1. Used either for stepping up or stepping down
the AC voltage
Q) Stabilizer 2. Used to convert AC voltage into DC voltage
R) Transformer 3. Used to remove any ripple in the rectified
output voltage
S) Filter 4. Used for constant output voltage even when
the input voltage or load current change
Choose the correct answer from the options given below.
1) P-2,Q-1,R-3,S-4 2) P-2,Q-4,R-1,S-3
3) P-2,Q-1,R-4,S-3 4) P-3,Q-4,R-1,S-2
11. An unpolarised light beam is incident on the polariser of a polarisation experiment and
the intensity of light beam emerging from the analyser is measured as 100 lumens. Now,
if the analyser is rotated around the horizontal axis (direction of light) by 30 in clockwise
direction, the intensity of emerging light will be….. lumens.
1) 133 2) 75 3) 50 4) 0
12. Match List I with List II.
List I List II
   
(a) C  A B 0 (i) C

A 
B
   
(b) AC  B 0 (ii) C 
B

A
   
(c) B  AC 0 (iii) C

A 
B
   
(d) A  B  C (iv)
A

C 
B
Choose the correct answer from the options given below:
1)  a    iv  ,  b    i  ,  c    iii  ,  d    ii 
2)  a    iv  ,  b    iii  ,  c    i  ,  d    ii 
3)  a    iii  ,  b    ii  ,  c    iv  ,  d    i 
4)  a    i  ,  b    iv  ,  c    ii  ,  d    iii 
13. Given below are two Statements. One is labelled as
Assertion A and the other is labelled as Reason R.
Assertion A: Two identical balls A and B thrown with same speed ‘u’ at two different
angles with horizontal attained the same range R. If A and B reached the maximum
height h1 and h2 respectively, then R  4 h1h2
 u 2 sin 2    u 2 cos 2  
Reason R: Product of said heights, h1h2    . 
 2g   2g 
  
Choose the CORRECT answer:

1) Both A and R are true and R is the correct explanation of A
2) Both A and R are true but R is NOT the correct explanation of A
3) A is true but R is false
4) A is false but R is true
14. Given below are two Statements:
Statement-I: When N moles of an ideal gas undergoes adiabatic change from state
NR T2  T1  Cp
 P1,V1,T1  to state  P2 ,V2 ,T2  , the work done isW  , where   and
1  Cv
R= universal gas constant,
Statement-II: In the above case. When work is done on the gas the temperature of the
gas would rise.
1) Both Statement-I and Statement-II are true
2) Both Statement-I and Statement-II are false
3) Statement-I is true but Statement-II is false
4) Statement-I is false but Statement-II is true

15. Given below are two statements:
Statement I: An electric dipole is placed at the centre of a hollow sphere. The flux of
electric field through the sphere is zero but the electric field is not zero anywhere in the
sphere.
Statement II: If R is the radius of a solid metallic sphere and Q be the total charge on it.
The electric field at any point on the spherical surface of radius r (< R) is zero but the
electric flux passing through this closed spherical surface of radius r is not zero.
In the light of the above statements, choose the correct answer from the options given
below:

1) Both Statement I and Statement II are true
2) Statement I is true but Statement II is false
3) Both Statement I and Statement II are false
4) Statement I is false but Statement II is true

16. In a meter bridge experiment, the circuit diagram and the corresponding observation table
are shown in figure.
R X
Resistance
box
Unknown
G resistance
E K
S.No R  I(cm)

1. 1000 60
2. 100 13
3. 10 1.5
4. 1 1.0
Which of the readings is inconsistent?
1) 3 2) 2 3) 1 4) 4
17. An   particle, a proton and an electron have the same kinetic energy. Which one of the
following is correct in case of their De-Broglie wavelength?
1) a   p  e 2) a   p  e 3) a   p  e 4) a   p  e
18.  
A charge particle is moving in a uniform magnetic field 2iˆ  3 ˆj T . If it has an
acceleration of  aiˆ  4 ˆj  m / s 2 , then the value of a will be
1) 3 2) 6 3) 12 4) 2
19. Two concentric circular loops of radii r1  30cm and r2  50cm are placed in X-Y plane as
shown in the figure. A current I = 7A is flowing through them in the directions as shown
in figure. The net magnetic moment of this system of two circular loops is approximately:
50 cm
I
30 cm I
7ˆ 2 7ˆ 2 ˆ 2 ˆ 2
1) kAm 2)  kAm 3) 7kAm 4) 7kAm
2 2
20. A tuning fork of frequency 480 Hz is used in an experiment for measuring speed of sound
(v) in air by resonance tube method. Resonance is observed to occur at two successive
lengths of the air column, l1  30cm and l2  70cm . Then, v is equal to
1) 332ms 1 2) 384ms 1 3) 379ms 1 4) 338ms 1
21. A particle executes SHM of amplitude 25 cm and time period 12s. What is the minimum
time required for the particle to move between two points located at 12.5 cm on either
side of the mean position? (i.e. separation between points is 25 cm)
[Report your answer in seconds]
22. Photoelectric emission from a metal begins at a frequency of 6  1014 Hz . The emitted
electrons are fully stopped by a retarding potential of 3.3V. Find the wavelength (in nm)
of the incident radiation. Take h  6.6  1034 Js , e  1.6  1019 C
23. A smooth thick uniform coin has mass M = 250g and radius R = 6 cm and is initially at
rest. A rod held horizontal at a height h above centre C hits the ball. The ball begins to
roll without slipping. Find the value of h. [Assume that the impulse imparted by the stick
is horizontal]
[Give your answer in cm]

F
h
C
3
24. A body is projected vertically up with a velocity equal to of the escape velocity from the
4
surface of the earth. The maximum height it reaches is: (Radius of the earth = R) [Height is
measured from surface of earth]
aR
If your answer is (Report the value of a + b (a and b are coprime positive integers)
b
25. An observer can see through a small hole on the side of a jar (radius 15 cm) at a point at
height of 15 cm from the bottom (see figure). The hole is at a height of 45 cm. When the
jar is filled with a liquid up to a height of 30cm the same observer can see the edge at the
bottom of the jar. If the refractive index of the liquid is N/100, where N is an integer, the
value of N is _____.
45 cm
15 cm
15 cm
3
26. Four identical rectangular plates with length, l  2 m and breadth, b  m are arranged as
2
X 0
shown in figure. The equivalent capacitance between A and C is in SI units. The
d
value of X is _____. (Round off to the Nearest Integer)
A B C D
d d d
27. Two travelling waves of equal amplitudes and equal frequencies move in opposite
directions along a string. They interfere to produce a stationary wave whose equation is
 2 t 
given by y  10cos  x  sin   cm , x is in meter and t in second. The amplitude of the
 T 
4
particle at x  m will be _____cm.
3
28. The diagram shows part of a standard vernier having least count 0.1mm
0 10
3cm 4cm
If the correct reading is “n” cm. Then the value of “100n” is____
29. A container filled with liquid of density 103 kg / m3 is kept on horizontal surface. An
orifice of area 10cm2 is made in the wall of container at a distance 100 cm below the free
surface of liquid. The thrust on the container at the instant when height of liquid is 50 cm
above the orifice is ________ (in Newton) (Assuming container is at rest and area of
orifice is very small than the area of the container) Take g  10m / s 2
4 1
30. One mole of an ideal gas whose adiabatic exponent   undergoes process P  200 
3 V
then change in internal energy of gas when volume changes from 2m3 to 4m3 is (in
Joules).

correct.
31. The phase diagram for the pure solvent (solid lines) and the solution (non volatile-non
Electrolyte solute) (dashed lines) are recorded below:
1 atm
P L
The quantity indicated by L in the figure is

1) P 2) T f 3) K f m 4) Kb .m
32. Which of the following graph is correct representation between atomic number (Z) and
magnetic moment of d-block elements?[Outer electronic configuration :  n  1 d xns1or 2 ]
Magnetic
Magnetic
moment
moment
1) Z 2) Z
Magnetic
Magnetic
moment
moment
3) Z 4) Z

33. The following mechanism has been proposed for the exothermic catalysed complex
reaction:
slow k k

A  B  IAB 
1  AB  I 
2 P A
fast
If k1 is much smaller than k2 , the most suitable qualitative plot of potential energy (P.E.)
versus reaction co-ordinate (R.C.) for the above reaction.
P.E.
P.E.
AB+I A+P
AB+I
A+B IAB
A+B
IAB A+P
1) R.C. 2) R.C.
P.E.
P.E.
A+B A+B
IAB AB+I IAB
A+P AB+I
A+P
3) R.C. 4) R.C.
34. Which has maximum dipole moment?
Cl Cl
Cl Cl Cl Cl
Cl Cl
1) Cl 2) 3) Cl 4) Cl

35. Geological conditions are sometimes so extreme that quantities neglected in normal
laboratory experiments take on an overriding importance. For example, consider the
formation of diamond under geo-physically typical conditions. The density of graphite is
2.4 g / cm3 and that of diamond is 3.6 g / cm3 at a certain temperature and 500 k bar. By
how much does Utrans differs from Htrans for the graphite to diamond transition?
1) 83.33 kJ/mol 2) 0.83 kJ/mol
3) 8.33  107 kJ/mol 4) 83.33 J/mol

36. The correct increasing bond angle among BF3, PF3, and ClF3 follows the order.
1) BF3  PF3  ClF3 2) PF3  BF3  ClF3
3) ClF3  PF3  BF3 4) BF3  PF3  ClF3
37. Statement-1: Trityl amine (Triphenylmethylamine) is a tertiary amine.
Statement-2: N in Trityl amine is attached with no hydrogen.

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
3) Statement-1 is True, Statement-2 is True
4) Statement-1 is False, Statement-2 is False
38. An organic compound A upon reacting with NH3 gives B. On heating, B gives C. C in
presence of KOH reacts with Br2 to give CH3CH 2 NH 2 . A is
1) CH 3CH 2COOH 2) CH 3COOH
CH3  CH  COOH
CH 3
3) CH 3CH 2CH 2COOH 4)

39. Which is the correct reaction coordinate diagram for the following solvolysis reaction?
CH3 CH3 OH
H 2O
Br  OH + CH3

G G
1) Reaction Coordinate 2) Reaction Coordinate
G G
Reaction Coordinate Reaction Coordinate

3) 4)
40. The labelled O18 will be in _____.

O O
C O H C O CH3
HCl
CH 3  O 18 
 H   H 2O
Catalyst
Methyl benzoate
1) H 2O 2) Methyl benzoate
3) Both (A) and (B) 4) Benzoic acid

41. The approach to the following equilibrium was observed kinetically from both directions.
PtCl42  H 2O  Pt  H 2O  Cl3  Cl 
At 25 C , it was found that
d  PtCl42 
 
dt
   
  3.9  105 s 1  PtCl 2   2.1  103 L mol 1s 1  Pt H O Cl   Cl  
 4  
 2  3  
The value of K eq (equilibrium constant) for the complexation of the fourth Cl  by Pt (II)
is
1) 53.8 mol L1 2) 0.018 mol L1 3) 53.8 L mol 1 4) 0.018 L mol 1
42. The electric charge for electrode deposition of one equivalent of the substance is
1) one ampere per second 2) 965000 C per second
3) one ampere per hour 4) charge on 1 mole of electrons
43.
Column I Column II
(A) Glycosidic Linkage (p) Globular protein
(B) Maltase (q) Connects two monosaccharide units
(C) Peptide bond (r) Monomeric unit of nucleic acids
(D) Nucleotide (s) Connects two amino acid units
1) A-s ; B-r ; C-q ; D-p 2) A-r ; B-s ; C-p ; D-q
3) A-q ; B-p ; C-s ; D-r 4) A-p ; B-q ; C-r ; D-s
44. Statement-1: p-Methoxybenzaldehyde undergoes nucleophilic addition more readily
than p-nitrobenzaldehyde.
Statement-2: The carbon atom of the aldehyde group is more electrophilic in
p-nitrobenzaldehyde than in p-Methoxybenzaldehyde.
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
3) Statement-1 is True; Statement-2 is False
4) Statement-1 is False; Statement-2 is True
45. An aromatic compound ‘X’ with molecular formula C9 H10O gives the following
chemical tests:
(i) forms 2, 4-DNP derivative
(ii) reduces Tollens’ reagent
(iii) undergoes Cannizzaro reaction and
(iv) on vigorous oxidation, 1, 2-benzenedicarboxylic acid is obtained X is

CHO
COCH3
1) CH3 2) C 2H5
CHO
CHO
C2 H5
3) 4) C2 H5
46. Iron powder is added to 1.0 M solution of CdCl2 at 298 K. The reaction occurring is
Cd 2  aq   Fe  s   Cd  s   Fe2  aq  . If the standard potential of a cell producing this
reaction is 0.037 V, the concentrations of Cd 2 and Fe2 ions in the above reaction at
equilibrium respectively will be (Hint:- 101.3  20 )

1) 0.05 M, 0.95 M 2) 0.95 M, 0.05 M
3) 0.40 M, 0.60M 4) 0.60 M, 0.40 M
47. Column-I Column-II
A) B p) +1.25 V (Eo for M3+/M)
B) Al q) 303 K(m.p.)
C) Tl r) -1.66 V (Eo for M3+/M)
D) Ga s) 801 kJ mol-1 (IE1)
1) A-s; B-r; C-p; D-q 2) A-p; B-s; C-q; D-r
3) A-q; B-p; C-r; D-s 4) A-r; B-q; C-s; D-p
48. Statement 1: In the brown ring test of NO2 or NO3 , the FeSO4 solution must be freshly
prepared.
Statement 2: On exposure to sunlight, Fe2 is converted into Fe3 which does not give
the brown ring test.
1) If both Statement-1and Statement-2 are True and Statement-2 is the correct
explanation for Statement-1
2) If both Statement-1 and Statement-2 are True but Statement-2 is NOT the correct
explanation for Statement-1
3) Statement-1 is True but Statement-2 is False
4) Statement-1 is False but Statement-2 is True
49. For electron gain enthalpies of the elements denoted as  eg H , the incorrect option is:
1)  eg H (Cl )   eg H ( F ) 2)  eg H ( Se)   eg H ( S )
3)  eg H ( I )   eg H ( At ) 4)  eg H (Te)   eg H ( Po )
50. Column-I Column-II
A) Sn p) Highest density amongst group 14 elements
B) Pb q) Lowest m.p. amongst group 14 elements
C) Si r) Lowest IE1 amongst group 14 elements
D) Ge s) Highest electrical resistivity
1) A-s; B-r; C-p; D-q 2) A-p; B-s; C-q; D-r
3) A-q,r; B-p; C-s; D-s 4) A-r; B-q; C-s; D-p
51. An artificial fruit beverage contains 30.0g of tartaric acid  H 2C4 H 4O6  and 18.8 g of its
salt, potassium hydrogen tartrate per litre. What is the pH of the beverage? For tartaric
acid, Ka1  5.0 104 , Ka2  4 109  log2  0.3 .

(Atomic weight of C = 12, H = 1, O = 16, K = 39)
52. Amongst the following, the maximum number of water soluble vitamins are:
A, B1, B2 , B6 , B12 , C , D, E and K
53. How many of the following elements are lanthanoids?
Cs, Ra, Sn, Sm, Pb, Er , Se, Gd
54. How many of the following ethers cannot be prepared by Williamson’s synthesis?
CH 3OCH 2CH 3 , C6 H 5OCH 3 ,  C6 H5 2 O ,
 CH 3 3 COCH 3, C6 H5CH 2OCH 3 ,  C2 H 5  2 O ,
C6 H5CH 2OC6 H5 ,  CH3  CH 2OCH 2CH3,  CH 3 3 C  O  C  CH 3 3
55. Find out number of compounds which are more stabilize in ionic structure, from
following.
O O O
, , , ,
56. If x and y are total number of electrons which are present in non-axial and axial set of d-
x
orbitals respectively in Ni cation of  Ni  DMG 2  , then calculate value of  
 y
57. An aqueous solution contains 3% and 1.8% by mass, Urea and glucose respectively.
What is the freezing point of solution in nearest integer? (Kf =1.86o C/m)
58. In a sample of excited hydrogen atoms electrons make transition from n  2 to n  1 .
Emitted photons strike on a metal of work function   4.2eV . Calculate the wavelength
o
(in A ) associated with ejected electrons having maximum kinetic energy.  174.9  13.2 
59. HCHO dis-proportionates to HCOO and CH3OH in the presence of OH 
(Cannizzaro’s reaction).
2 HCHO  OH   HCOO   CH 3OH
E
If the equivalent weight of HCHO is E, then the value of is
10
60. The final value (in L) of one mole of an ideal gas initially at 27 C and 8.21 atm pressure,
if it absorbs 420 cal of heat during a reversible isothermal expansion is  ln 2  0.7 

correct.
61. Let the position vectors of the points A, B, C and D be
           
5 i  5 j  2 k , i  2 j  3 k , 2 i   j  4 k and  i  5 j  6 k . Let the set
    3
2
S={   R : the points A, B, C and D are coplanar}. Then is equal to:
S
1) 37 2) 13 3) 61 4) 41
2
62. Let A   aij  , where aij  0 for all i, j and A2  I . Let ‘a’ be the sum of all diagonal
22
elements of A and b  A . Then 3a3  5b 4 is equal to:
1) 5 2) 4 3) 3 4) 7
 x 3 
    
63. Let A   x  R :  x  3   x  4  1 , B   x  R : 3   3r 
x  3 x
 3  , where [t] denotes
 10 
  r 1  
 
greatest integer function. Then,
1) A  B 2) A  B   3) A  B 4) B  A
64. The number of triplets  x, y, z  , where x , y , z are distinct non negative integers satisfying
x  y  z  15 , is:
1) 136 2) 114 3) 80 4) 92
65. Let sets A and B have 5 elements each. Let mean of the elements in sets A and B be 5
and 8 respectively and the variance of the elements in sets A and B be 12 and 20
respectively. A new set C of 10 elements is formed by subtracting 3 from each element of
A and adding 2 to each element of B. Then the sum of the mean and variance of the
elements of C is_________.
1) 36 2) 40 3) 32 4) 38
66. A circle touching the x-axis at (3, 0) and making an intercept of length 8 units on the
positive y-axis passes through the point:
1) (3, 10) 2) (2, 3) 3) (1, 5) 4) (3, 5)

1 1
67. If the sum   k  2 k k k 2

a b
where a, b  N then a  b equals to
k 1
1) 24 2) 20 3) 16 4) 22
68. For the system of linear equations  x  y  z  1, x   y  z  1, x  y   z   , which one
of the following statements is NOT correct?
1) It has infinitely many solutions if   2 and   1
2) It has no solution if   2 and   1
3
3) x  y  z  if   2 and   1
4
4) It has infinitely many solutions if   1 and   1
69. Let P  x0 , y0  be the point on the hyperbola 3x 2  4 y 2  36, which is nearest to the line
2 y  3x  1 . Then 2  y0  x0  is equal to:
1) 9 2) -3 3) 3 4) -9
70. A pair of fair dice is thrown independently three times. The probability of getting a score of
exactly 9 twice is
8 1 8 8
1) 2) 3) 4)
243 729 9 729
71. Let R be a relation defined on the set of real numbers by aRb  1  ab  0 . Then R is:
1) Equivalence relation 2) Transitive relation
3) Symmetric relation 4) Anti-Symmetric relation
1
72. Consider the function f  x   xe x  x where x  R  0 , then which of the following
xe
statement is CORRECT?
1) f(x) attains its local maxima at x  x0 , where x0   0,1
2) f(x) attains its local minima at x  1
3) f(x) attains its local minima at x  x0 , where x0  1,  
4) f(x) attains its local minima at x  x0 , where x0   0,1
 m  sec x  tanx 2  n   c then  1  1  is equal

2sec x tan x 11
73. If  dx   sec x  tan x   
 
 sec x  tan x 10 m n
to (where ‘c’ is constant of integration)

1) 1 2) -1 3) 2 4) -2
, and g  x    f   x   f  x   .
1
74. Let f :  0,1  R be a function defined by f  x  
1  e x
Consider two statements
(I) g is a decreasing function in (0,1)
(II) g is one-one in (0,1) Then,

1) only (II) is true 2) both (I) and (II) are true
3) only (I) is true 4) neither (I) nor (II) is true
75. Let A, B, C be 3  3 matrices such that A is Symmetric and B and C are Skew-symmetric.
Consider the statements.
(S1) A31B52  B52 A31 is Symmetric
(S2) A52C 31  C 31A52 is Symmetric Then,

1) only S2 is true 2) both S1 and S2 are false
3) both S1 and S2 are true 4) only S1 is true
If the coefficients of three consecutive terms in the expansion of 1  x   n  N  are in
n
76.
the ratio 1:5:20, then the coefficient of the fourth term is
1) 5481 2) 3654 3) 2436 4) 1817
77. The area of the region  x, y  : x2  y  8  x2 , y  7 is (in sq. units)

1) 24 2) 21 3) 20 4) 18
78. Statement 1: If 2cos  sin   1 0      then the value of 7cos  6sin  is 2
Statement 2: if cos 2  sin   1 ,0     , then sin   cos6  0

2 2
1) both Statement-1 and Statement-2 is true
2) both Statement-1 and Statement-2 is false
3) Statement-1 is true, Statement-2 is false
4) Statement-1 is false, Statement-2 is true

6k
79. Statement 1: 
k 1  3  2  3
k k k 1
 2 k 1 
2
  k 3   k  1 n
n
Statement 2: 3 3 for any natural number n
k 1
80. Statement-1: 1125  1225 when divided by 23 leaves the remainder zero.
Statement-2: a n  bn is always divisible by  a  b  , a, b, n  N


4 2
 cos x 2023
   sin x 2023   cos x 2023
81. The value of dx 
0
  is equal to
sin 6 x 2
 
82. The absolute value of lim
x 0 log cos 2 x 2  x 
e 

83. Let f : R  R and g : R  R be respectively given by f ( x)  1  x and g ( x)  x3  1.

 min  f ( x ), g ( x ) , if x  1
Define h : R  R by h( x )   number of points at which h(x) is
 max  f ( x ), g ( x ) , if x  1
k
not differentiable is k then is
3
x  2 x 1
84. Let f : 1,    2  B, f ( x )  4 is an Onto function then sum of all
2  x
integers in set B, is equal to
85. If f ( x )  x 2  x3  x  0  and g ( x) is the inverse of f ( x) , then the value of  5 g '(2)  3 is
x2 y 2 x2 y 2
86. e1 & e2 are eccentricities of   1 and   1 respectively. If  e1, e2  lies on
18 4 9 4
k
15 x 2  3 y 2  k , then value of 
2
x2
87. Let f ( x)  1  x  ln 1  x   x  , x  1 and f '  x   0 x  0,  , then value of  
4
is______ (where [.] denotes greatest integer function)
1
88. Suppose f is a function satisfying f  x  y   f ( x)  f ( y ) for all x, y  N and f 1  . If
5
m
f ( n) 1
 n(n  1)(n  2)  12 , then m is equal to___________.
n 1
11

9 1 
11
9  1 
If the co-efficient of x in   x3  in   x 
3 
89. and the co-efficient of x are
  x  
  x 
equal, then   is equal to_________.
2
90. Let the mean and variance of 8 numbers x , y , 10, 12, 6,12, 4, 8 be 9 and 9.25
respectively. If x  y , then 3x  2 y is equal to____________.


only.
(II) Section-II contains 10 Numerical Value Type questions. Attempt any 5 questions only, if more than 5
.
SRI CHAITANYA IIT ACADEMY, INDIA 05-01-2024_ Sr.S60_Elite, Target & LIIT-BTs _Jee-Main_GTM-08_Q.P
Admission Number:
05-01-2024_Sr.Super60_Elite, Target & LIIT-BTs _ Jee-Main_GTM-08_Test Syllabus

correct.
1. A bag of sand of mass 9.8 kg is suspended by a rope. A bullet of mass 200g travelling with a
speed of 10ms 1 gets embedded in it, then loss of kinetic energy will be
1)4.9 J 2) 9.8J 3)14.7J 4) 19.6J
2. The length of metallic wire is l1 when tension in it is T1 . It is l2 when the tension is T2 . The
original length of the wire will be
T l T l l l T l T l T l T l
1) 1 1 2 2 2) 1 2 3) 2 1 1 2 4) 2 1 1 2
T2  T1 2 T1  T2 T2  T1
3. Two point charges Q each are placed at a distance d apart. A third point charge q is placed at
a distance x from midpoint on the perpendicular bisector. The value of x at which charge q
will experience the maximum Coulomb’s force is:
d d d
1) x  d 2) x  3) x  4) x 
2 2 2 2
4. When the switch S, in the circuit shown, is closed then the valued of current I will be:
i1 C i2
20V 10V
A 2 4 B
i
2
V 0
1)3A 2) 5A 3) 4A 4) 2A

5. A small square loop of side ‘a’ and one turn is placed inside a larger square loop of side b
and one turn (b >>a). The two loops are coplanar with their centres coinciding. If a current I
is passed in the square loop of side ‘b’, then the coefficient of mutual inductance between
the two loops is:
 a2 0 8 2 0 b2 0 8 2
1) 0 8 2 2) 3) 8 2 4)
4 b 4 a 4 a 4 b
6. An isolated and charged spherical soap bubble has a radius ‘r’ and the pressure inside is
atmospheric. If ‘T’ is the surface tension of soap solution, then charge on drop is:
2rT 2rT
1) 2 2) 8 r 2rT  0 3) 8 r rT  0 4) 8 r
0 0
7. For the arrangement shown in the figure, let a and T be the acceleration of the blocks and
tension in the string respectively. The string and the pulley are frictionless and massless.
Which of the graphs show the correct relationship between a and T for the system in which
sum of the two masses m1 and m2 is constant.
m1  m2
m2
m1
T T
1) a2 2) a2

T T
1 1
3) a2 4) a2
8. The dimensions of  b4 (  = Stefan’s constant and b= Wien’s constant ) are:
1)  M 0 L0T 0  2)  ML4T 3  3)  ML2T 3  4)  MLT 2 

       
9. Assertion:-A body becomes massless at the centre of earth.
 d
Reason:- Acceleration due to gravity at a depth ‘d’ from the surface of earth is g 1   .
 R
1)Assertion is correct, reason is correct; reason is a correct explanation for assertion.
2)Assertion is correct, reason is correct; reason is not a correct explanation for assertion
3)Assertion is correct, reason is incorrect
4)Assertion is incorrect, reason is correct.
10. Statement I - A bar magnet can exert torque on itself due to it’s own field.
Statement II - One element of a current carrying wire can exert a force on another element of
same wire.
1)Statement I is true; Statement II is true
2)Statement I is false; Statement II is false
3)Statement I is true; Statement II is false
4)Statement I is false; Statement II is true
11. Assertion : A flask contains argon and chlorine in the ratio of 2 : 1 by mass at a temperature
of 27 C . The ratio of average kinetic energy per molecule is 1 : 1

Reason : The average kinetic energy of a molecule is proportional to the absolute
temperature of gas and independent of pressure, volume or the nature of ideal gas.
1)Assertion is correct, reason is correct; reason is a correct explanation for assertion.
2)Assertion is correct, reason is correct; reason is not a correct explanation for assertion
3) Assertion is correct, reason is incorrect
4) Assertion is incorrect, reason is correct.
12. Assertion : A variable frequency ac source is connected to a capacitor. If the frequency of ac
source decreases the displacement current also decreases.
Reason : The capacitive reactance increases with decrease in frequency of ac.
1) Assertion is correct, reason is correct; reason is a correct explanation for assertion.
2) Assertion is correct, reason is correct; reason is not a correct explanation for assertion
13. Assertion : The conductivity of a semi conductor increases with increase in temperature.
Reason : In a semi conductor, the number density of carriers and relaxation time increases
with increase in temperature.
1) Assertion is correct, reason is correct; reason is a correct explanation for assertion.
2) Assertion is correct, reason is correct; reason is not a correct explanation for assertion
14. Un polarised light is incident on a plane glass surface of refractive index 1.5 . What should
be the angle of incidence so that the reflected and refracted rays are perpendicular to each
other?
1) sin 1 1.5 2) cos1 1.5 3) tan 1 1.5 4) sin 1  0.667 
15. A projectile is thrown with a velocity of 10 2 m / s at an angle of 45 with horizontal. The

interval between the moments when speed is 125m / s is g  10m / s 2 
1)1.0 s 2) 1.5 s 3) 2.0 s 4) 0.5 s
16. A particle leaves the origin at t = 0 and moves in the positive x-axis direction. Velocity of
 t
the particle at any instant is given by v  u 1   . If u  10ms 1 and t '  5s , find the x-
 t' 
coordinate of the particle at an instant of 10 s.
v  ms 1 
t'
t s
1)-10m 2) 0 3) 10 m 4) 20 m
P2
17. During an experiment, an ideal gas is found to obey a condition  cons tan t ,

   density of the gas  . The gas is initially at temperature T, pressure P and density  . The
gas expands such that density changes to  /2. Then,
1) The pressure of the gas changes to 2P
T
2) The temperature of the gas changes to
2
3) The temperature of the gas changes to 2T
4) The graph of the above process on P-T diagram is hyperbola.
18. Equal masses of three liquids A, B and C have temperatures 10 C , 25 C and 40 C
respectively. If A and B are mixed, the mixture has a temperature of 15 C . If B and C are
mixed, the mixture has temperature of 30 C . If A and C are mixed, the mixture will have a
temperature of
1) 16 C 2) 20 C 3) 25 C 4) 29 C

19. A body executes S.H.M. with an amplitude A. Its energy is half kinetic and half potential
when the displacement is
1)A/3 2) A/2 3) A / 2 4) A / 2 2
20. A resistor of 200 and a capacitor of 15.0 F are connected in series to a 220 V, 50Hz ac
source. Calculate the current in the circuit.
1)0.755 A 2) 0.925 A 3) 1.225 A 4) 1.565 A
21. Two bodies, a ring and a solid cylinder of same material are rolling down without slipping
on inclined plane. The radii of the bodies are same. The ratio of velocity of the centre of
x
mass at the bottom of the inclined plane of the ring to that of the cylinder is . then, the
2
value of x is---------.
22. A diatomic gas    1.4  does 400 J of work when it is expanded isobarically. The heat given
to the gas in the process is --------J
23. An organ pipe 40cm long is open at both ends. The speed of sound in air is 360ms 1 . The
frequency of the second harmonic is ----------Hz.
24. The same size images are formed by a convex lens when the object is placed at 20 cm or at
10 cm from the lens. The focal length of convex lens is---------------cm.
25. A nucleus disintegrates into two smaller parts, which have their velocities in the ratio 3 : 2.
1
x
The ratio of their nuclear sizes will be   3 . The value of ‘x’ is :
3

26. A slab of material of dielectric constant K = 5 has the same area as the plates of a parallel
3
plate capacitor but has a thickness d , [Where d is the separation between the plates] is
4
xc0
inserted between the plates, then the capacitance of the capacitor becomes ( c0 =
2
capacitance without di-electric). Find x ?
27. A storage battery of emf 8.0 V and internal resistance 0.5 is being charged by a 136 V dc
supply using a series resistor of 15.5 . Find the terminal voltage of the battery during
charging?
28. Two long and parallel straight wires A and B carrying currents of 8.0 A and 5.0 A in the
same direction are separated by a distance of 4.0 cm. The magnitude of force on a 10 cm
section of wire A is x  105 N . Find x?
29. The work functions of two metals A and B are 4 eV and 2eV respectively. The ratio of
slopes of the graph drawn between maximum kinetic energy of emitted electron and
frequency of incident radiation for two metals is…………
30. An electron is in an excited state in a hydrogen atom. It has a total energy -3.4 eV. Its de
Broglie wavelength is x  1010 m . Find x ? [Planck’s constant = 6.64  1034 Js , mass of
electron = 9  10 31 kg ].

correct.
31. The number of electrons present in 3d orbital of Cu  is

1) 20 2) 10 3) 16 4) 24
32. What is the mass ratio of ethylene glycol ( C2 H 6O2 , molar mass = 62 g/mol) required for
making 500 g of 0.25 molal aqueous solution and 250 mL of 0.25 molar aqueous solution?
1) 1 : 1 2) 3 : 1 3) 2 : 1 4) 1 : 2
33. In the cumene to phenol preparation in presence of air, the intermediate is
1) 2) 3) 4)
34.
List-1 List-II
Reaction Reagents
(A) Hoffmann Degradation (1) Conc.KOH, 
(B) Clemenson reduction (II) CHCl3 , NaOH / H 3O 
(C) Cannizaro reaction (III) Br2 , NaOH
(D) Reimer-Tiemann reaction (IV) Zn  Hg / HCl
1)  A  III ,  B   IV ,  C   II ,  D   I
2)  A  II ,  B   IV ,  C   I ,  D   III
3)  A  III ,  B   IV ,  C   I ,  D   II
4)  A  II ,  B   I ,  C   III ,  D   IV

35. The correct order in aqueous medium of basic strength in case of methyl substituted amines
is:
1) Me2 NH  MeNH 2  Me3 N  NH 3
2) Me2 NH  Me3 N  MeNH 2  NH 3
3) NH 3  Me3 N  MeNH 2  Me2 NH
4) Me3 N  Me2 NH  MeNH 2  NH 3
36. What is the number of unpaired electron(s) in the highest occupied molecular orbital of the
following species : N 2 , N 2  , O2 , O2 ?

1) 0, 1, 2, 1 2) 2, 1, 2, 1 3) 0, 1, 0, 1 4) 2, 1, 0, 1
37. The bond dissociation energy is highest for
1) Cl2 2) I 2 3) Br2 4) F2
38. The primary and secondary valencies of cobalt respectively in Co  NH 3 5 Cl  Cl2 are:
1) 3 and 5 2) 2 and 6 3) 2 and 8 4) 3 and 6

39. During the borax bead test with CuSO4 a blue green colour of the bead was observed in
oxidizing flame due to the formation of
1) Cu3B2 2) Cu 3) Cu  BO2 2 4) CuO
40. Which of the following conformations will be the most stable?
1) 2) 3) 4)

41. The rate constant of a reaction is equal to rate of reaction:
1) When concentrations of reactants do not change with time.
2) When concentrations of all reactants and products are equal.
3) At time t=0.
4) When concentrations of all reactants are unity.
42. For the redox change;
Zn( s )  Cu 2 (1M )  Zn 2  (0.1M )  Cu ( s )
In a cell E o cell is 1.10 volt. Ecell for the cell would be
1) 1.13 V 2) 0.82 V 3) 2.14 V 4) 180 V
43. In a mixture A and B components show negative deviation as
1) Vmix   ve
2) Vmix  ve
3) A  B interaction is weaker than A  A and B  B interaction.
4) None of the above.
44. Calculate the value of H o / kJ for the following reaction using the listed thermochemical
equations:
2C ( s)  H 2 ( g )  C2 H 2 ( g )
2C2 H 2 ( g )  5O2 ( g )  4CO2 ( g )  2 H 2O(l ) H 0 / kJ  2600 kJ
C ( s)  O2 ( g )  CO2 ( g ) H 0 / kJ  390 kJ
2 H 2 ( g )  O2 ( g )  2 H 2O(l ) H 0 / kJ  572 kJ
1) +184 2) +214 3) +202 4) +234
45. The reaction below can be used as a laboratory method for preparing small quantities
of Cl2  g  . If a 49 g sample that is 96 % K 2Cr2O7 by mass is allowed to react
with 325 mL of HCl( aq ) with a density of 1.15 g/ml and 30.1% HCl by mass, how many
grams of Cl2  g  are produced?
K 2Cr2O7  HCl  CrCl3  Cl2  H 2O  KCl

(Atomic weights: K = 39, Cr = 52, Cl = 35.5)
1) 68.16g 2) 45.64g 3) 34.08g 4) 53g
46. Column-I & Column-II contain data on Schrondinger Wave-Mechanical model, where
symbols have their usual meanings. Match the columns.
Column-I Column-II
(P) 4s
(A)
(Q) 5 p x
(B)
(C)  (  ,  )=K (independent of  &  ) (R) 3s
(D) At least one angular node is present (S) 6 d xy
1) A-P; B-PQS; C-PR; D-QS 2) A-PQS;B-P; C-QS; D-PR

3) A-PQS;B-QS; C-P; D-PR 4) A-QS;B-PQS; C-PR; D-P
47. Titration curves for 0.1M solutions of three weak acids HA1,HA2 and HA3 with ionization
constants K1,K2 and K3 respectively with a strong base are plotted as shown in the figure.
Which of the following option is false?
1) K 2   K1  K3  / 2 2) pK1  pK3 3) K1  K 2 4) K 2  K3

48. Match the column for the substrate, reagent and product.
Column – I Column – II
OH
ph
A) ph MnO2
 
OH P) Dicarbonyl compound
distilation
B)  CH 3CO2 2 Ca   Q) Me2CO
C) HO OH HSO4
 
R) MeCHO
4 Pd / BaSO
D) CH3COCl  S) Product gives test iodoform test
H2
1) A -P, B-Q,S, C-S, D- R 2) B-R,S,A-Q,C-R,D-P,Q

3) A-R,B-Q,P,C-P,D-S 4) B-R, A-R,S,C-Q,D-P
49. Match the following
A. Glucose – Fructose P. Anomers
B. Fructose – Mannose Q. Diastereomers
C. Glucose – Mannose R. Functional isomers
D.  -D-Glucopyranose S. Lobry-De-Bruyn-Alberada-Van Ekenstein
 -D-Glucopyranose Transformation
1)A-R,S; B-R,S ; C-Q,S; D-P,Q; 2)A-Q,S ; B-P,Q ; C-R,S ; D-P,Q;
3)A-P,S ; B-P,R ; C-R,S ; D-R,S; 4)A-R,S ; B-S,Q; C-R,S;D-P,R;

50. Predict the product of the following reaction
1) 2)
3) 4)
NUMERICAL VALUE TYPE)
51. If a pure enantiomer of 1-bromo-2-butene is treated with Br2/CCl4 in presence of FeBr3, how
many different products (including stereoisomers) would result?
52.
Ni
 H 2 

 product
1mole
1mole
If the number of moles of benzene which remain after complete reaction is x then 100x = ?
53. How many non axial d orbitals are involved in the hybridization of CrO2Cl2 ?
54. Calculate Z(effective) for the f-electron in Ce3

55. The number of bridged oxygen atoms present in compound B formed from the following
reactions is
673 K
Pb  NO3 2  A  PbO  O2
D im erise
A 
B
56. How many of the following orders/statements are correct.
a) IP1 of nitrogen is not less than that of oxygen
b) IP1 of Pb is less than that of Sn
c) Electronegativity order B  Tl  In  Ga  Al
d) IP2 order B  C  Be
e) Electron gain enthalpy order Ne  Ar  Kr  Xe  He  Rn
f) Electron affinity of Be+4 is 217.6 ev/ion nearly.
g) Electronegativities of K & Rb are same
h) Order of sizes Tl  In  Al  Ga  B
i) Official IUPAC name of element with atomic number 105 is Unnilquadium.
2
57. With respect to the complex ion  Mn  OH 2 6  how many of the following statements are
true?
I. It is diamagnetic.
II. It is a low spin complex.
III. The metal ion is a d 5 ion.

IV. The ligands are weak field ligands.
V. It is octahedral.
58. A real gas is subjected to an adiabatic process from (4 bar, 40 L) to (6 bar, 25L) against a
constant pressure in a signal step. Calculate the enthalpy change for the process (in bar litre).

59. If 4 A current produced in fuel cell, use for the deposition of Ag  in 1L AgNO3 aq  solution
for 241.25 sec using Pt electrode.
Calculate the pOH of solution after electrolysis at 25 C .

60. Identify number of reactions that would give salicylic acid as major product.

correct.
50 n
61. Let  X i   Yi  T , where each X i contains 10 elements and each Yi contains 5 elements.
i 1 i 1
If each element of the set T is an element of exactly 20 of sets X i ’s and exactly 6 of sets Yi
’s, then n is equal to
1) 15 2) 50 3) 45 4) 30
62. For t  R & f be a continuous function. Let
1 cos2 t 1 cos 2 t
I
I1   x f ( x(2  x )) dx & I 2   f ( x(2  x))dx . Then 1
2 I2
sin t sin 2 t
1 2
1) 1 2) 3 3) 4)
3 5
 x  x 
63.  2 xy y   y 2 xy 
The general solution of the differential equation x  y .e  e  dy  y  e  y .e  dx is
   
   
x y x y
1) e xy  e y  c xy
2) e  e x  c 3) e xy  e y  c
4) e xy
ex c
64. If from each of the three boxes containing 3 white & 1 black, 2 white & 2 black, 1 white & 3
black balls, one ball is drawn at random, then the probability that 2 white & 1 black ball will
be drawn is
13 1 1 3
1) 2) 3) 4)
32 4 32 16

65. Consider the points A   0,1 & B   2,0  , P be a point on the line 4 x  3 y  9  0 Co-
ordinates of the point ‘P’ such that PA  PB is maximum is
 12 17   84 13   6 17   24 17 

1)  ,  2)  ,  3)  ,  4)  , 
 5 5  5 5  5 5  5 5
66. If one of the diameters of the circle x 2  y 2  2 x  6 y  6  0 is a chord to the circle with
centre (2,1), then the radius of the circle is
1) 4 2) 3 3 3) 3 4) 5
67. Two families with three members each and one family with four members are to be seated in
a row. In how many ways can they be seated so that the same family members are not
separated?
3 2 3
1) 2!3!4! 2)  3! .  4! 3)  3! . 4! 4) 3! 4!
2
68. If f : R  R, f  f  x     f  x   then f ( f ( f ( f ( x)))) is equal to
2 2
  
1) f x 4 2)  f  x  
4
  
3) f x 2  
4) f x 4
69. If both roots of quadratic equation ax 2  x  c  a  0 are imaginary and c > -1, then
1) 3a  2  4c 2) 3a  2  4c 3) c  a 4) None of these
70. Let z1, z2 and z3 be complex numbers satisfying z  i 3  1 and 3 z1  i 3  2 z2  2 z3 then
z1  z2 is
1
1) 2) 2 3) 2 4) 2 2
2

3 3 1
71. 
Statement –I: The equation sin 1 x    cos 1 x   a 3  0 has a solution for all a 
32
.
2
    9 2
Statement –II: For any x   1,1,sin 1
x  cos 1
x  , and 0   sin 1 x    .
2  4 16
1) Both statement I and II are true.
2) Both statement I and II are false.
3) Statement I is true and statement II is false
4) Statement I is false sand statement II is true
50 1
72. Statement –I: The largest term in the expansion of  4  3x  when x  is the 8th term
4
n   n  1 a 
Statement –II: Greatest term in the expansion of  x  a  is rth term where r   
 xa 
x
when [y] represents greatest integer and 0
a ,

4) Both statement I and II are true
 2 1 1
73. Statement 1: As A   0 1 1  satisfies the equation x3  5 x 2  6 x  2  0 , therefore A is
1 1 2
 
invertible.
Statement II: If a square matrix A satisfies the equation
a0 x n  a1x n 1  .........an 1x  an  0, and an , a0  0,

Then A is invertible

74. 
Statement I: The sum of algebraic distances from the points n , n 2 , where
n=1,2,3,………….2023 to a variable straight line is zero and if such a line always passes
through the point  ,   then     1366200
Statement II: If sum of algebraic distances from the points  xi , yi  , where i  1,2,3........n to a
 n n 
variable straight line is zero then that the line passes through the point   xi ,  yi 
 
 i 1 i 1 
75. Consider of the following statements is/are true?
f ( c  h )  f (c  h )
Statement I: If f is differentiable at x=c, then Lim exists and equals
h 0 2h
f ' (c ).
 2 1
 x sin 2 , x0
Statement II: Let g ( x)   x , then g ' exists and g ' is continuous, everywhere
 0, x0

on R
1) Both I and II are true 2) I is true, II is false
3) I is false, II is true 4) both I and II are false

76. The mean of five observations is 4 and their variance is 5.2. If three of these observations are
1, 2 and 6, then the other two observations can be
1) 2 and 9 2) 3 and 8 3) 4 and 7 4) 5 and 6
    
77. If the vectors AB  3 i  4 k and AC  5 i  2 j  4 k are the sides of a triangle ABC, then the
length of the median through A is
1) 18 2) 72 3) 33 4) 45
x  2 y 1 z
78. The vertices B and C of a triangle ABC lie on the line,   such that BC=5
3 0 4
units. Then the area (in sq. units) of this triangle, given that the point A(1,-1,2),is :
1) 5 17 2) 2 34 3) 6 4) 34
79. The area of the region bounded by the parabola ( y  2)2  ( x  1), the line touches to it at the
point whose ordinate is 3 and the x-axis is:
1) 9 2) 10 3) 4 4) 6
80. The area (in sq. units) of an equilateral triangle inscribed in the parabola y 2  8 x , with one
of its vertices on the vertex of this parabola , is :
1) 64 3 2) 256 3 3) 192 3 4) 128 3
81. Let an1  2  an , n  1,2,3........ & a1  3then Lim an is _____________

n 
2
82. Number of points where f ( x ) | x 2  2 x  3 | .e|9 x 12 x  4| is not differentiable is/are
_______
n n
2
83. If x1, x2 ,.....xn are n observations such that  xi  400 and  xi  80 then the least value
i 1 i 1
of n is

3 2
84. Let a , b , c be three unit vectors such that a .b  b.c  a .c  and ( a , c)  , then the value
2 3
 
of 4 a . b  b.c  a .c is __________
85. A and B are two non-empty sets and A is proper subset of B. If n( A)  4, then minimum
possible value of n( AB) is __________ (where  denotes symmetric difference of set A
and B)
86. Let a,b,c be any real numbers. Suppose that there are real numbers x, y, z not all zero such
that x  cy  bz, y  az  cx and z  bx  ay then a 2  b2  c 2  2abc is equal to
87. The value of “ a ” for which the equation x 3  2ax  2  0 and x 4  2 ax 2  1  0 have a
common root is “  ” then | 8 | is
88. If M and m are the maximum and minimum distances of the point P(2,6) from the ellipse
( x  2) 2 ( y  1) 2
  1, then the value of M-2m is ___________
9 16
89. If a rectangular hyperbola  x  1 y  2   4 cuts a circle x 2  y 2  2 gx  2 fy  c  0 at
points  3,4  ,  5,3 ,  2,6  and  1,0  then the value of g  f is ____________
2
 1 1   1
90. If tan15    tan195  2a then the value of  a   is
tan 75 tan105  a


SRI CHAITANYA IIT ACADEMY, INDIA 07‐01‐24_ Sr.Super60_Elite, Target & LIIT-BTs _Jee‐Main_GTM‐09_KEY &SOL’S

KEY SHEET
PHYSICS
1) 3 2) 2 3) 2 4) 1 5) 3
6) 4 7) 1 8) 1 9) 4 10) 2
11) 2 12) 2 13) 1 14) 1 15) 2
16) 4 17) 2 18) 2 19) 2 20) 2
21) 2 22) 214 23) 3 24) 16 25) 158
26) 2 27) 5 28) 274 29) 10 30) 1200
CHEMISTRY
31) 4 32) 4 33) 2 34) 2 35) 1
36) 3 37) 4 38) 1 39) 2 40) 2
41) 3 42) 4 43) 3 44) 4 45) 3
46) 1 47) 1 48) 1 49) 2 50) 3
51) 3 52) 5 53) 3 54) 2 55) 3
56) 3 57) 1 58) 5 59) 3 60) 6
MATHEMATICS
61) 3 62) 1 63) 1 64) 2 65) 4
66) 1 67) 1 68) 1 69) 3 70) 1
71) 3 72) 4 73) 3 74) 1 75) 1
76) 2 77) 3 78) 4 79) 1 80) 3
81) 1 82) 12 83) 0 84) 6 85) 4
86) 8 87) 2 88) 10 89) 1 90) 25

SOLUTIONS
PHYSICS
1. The magnitude of the restoring torque = force  perpendicular distance
 mg  AB
 mg  R sin 


B
A
mg
R 
Since  is small, sin    . Here  is expressed is radian. The equation of motion of

d 2
the scale is I   mgR
2
dt
d 2   mgR 
Or I 
2  I 
dt 
I mL2
  .Now I  Hence
mgR 12
L
T
3gR
Using the values L  1m, g  10ms 2 and R = 0.3m, we get T   / 3 second.
2. If the inputs A and B are inverted and then applied to a NAND gate, the output of the
NAND gate will be the same as that of an OR gate. The input of the last NAND gate is
A.A and B.B.
Hence the correct choice is (b)

3. Energy state –3.4 eV corresponds to a level n given by 13.6eV / n 2  3.4eV which

1
gives n = 2. Now, orbital speed 0 . Hence the orbital speed in the excited state is
n
 / 2.
4. Let u cms 1 be the speed of the bullet. Since the mass of the bullet remains unchanged,
its speed becomes   cms 1 after it penetrates a distance x = 3.5 cm. The
3u
4
retardation a due to the resistance of the wooden is given be u 2   2  2ax
2
 3u 
Or u 2     2a  3.5
 4 
u2
Which gives a  cms 2 . The bullet will come to rest when its velocity  '  0. If x '
16
is the thickness penetrated by the bullet, then u 2   '2  2ax '
u2 u2
Or x'  . But a  cms 2 .
2a 16
u 2  16
Therefore x '   8cm
2u 2
Hence the correct choice is (a)
5. Let the length of the cylinder be l and let its ends be maintained at temperatures 1
and  2 . Area of the cross-section of the inner cylinder   R 2 . Area of cross – section
of outer cylinder    2 R 2   R 2  3 R 2
Rate of flow of heat across inner cylinder is
Q2 
 
k2 3 R 2 1   2 
(i)
l
Rate of flow of heat across the outer shell is
Q2 
 
k2 3 R 2 1   2 
(ii)
l

Let the effective thermal conductivity of the compound cylinder be k. The rate of flow
of heat across the compound cylinder is
Q
 
k 4 R 2 1   2 
(iii)
l
Now Q  Q1  Q2 (iv)
Using (i), (ii) and (iii) in (iv), we get 4k  k1  3k2
k  3k2
Or k 1
4
Hence the correct choice is (c)
6. The magnetic field in the core is given by

B   nI
When  is the permeability of soft iron and n is the number of turns per unit length of
the solenoid.
Now
 3000 3000
r  and n  
0 2 r 2  0.1
 B  r 0nI
 2000  4  107 
3000
 1  12T
2  0.1
Hence the correct choice is (d)
7. Fringe width decreases with decrease in wave length.
8. Slope of velocity-time graph gives acceleration.
9. Net flux passing through a closed surface depends upon net charge enclosed by the
surface.
10. Rectifier converts ac into dc.
11. Application of Law of Malus.
12. We get it using triangle law of vector addition.

13. Projectile on horizontal ground.
14. There is no heat exchange in adiabatic process.
15. Electric field is zero inside a uniformly charged spherical shell.
16. If one value in not close to remaining three, that is the in-consisted one.
17. De-Broglie wavelength is inversely proportional to square root of mass for given
kinetic energy.
18. Magnetic force is perpendicular to velocity.
19. Magnetic moment is a vector quantity, it should be added as vectors.
20. L 2  L1   / 2
21.
-25 cm -12.5 cm Zero +12.5 cm +25 cm
A D 0 C B
Let the displacement of the particle in SHM be given by
x  t   A sin t    ….(i)
2 2
Where A  25cm and    rads 1
T 3
Let us supposed that at time t = 0, the particle is at extreme position B. Setting x = A

and t = 0 in Eq.
(i) We have A  A sin 
Giving    / 2
Putting    / 2 in Eq. we get
x  t   A cos t  ii 
Now let us say that the particle reaches point C at t  t1 and point D at t  t2 . At
C, the displacement x  t1   12.5cm and at D, it is x  t2   12.5cm . So from (ii)
we have 12.5  25cos t1

And 12.5  25cos t2
Or cos t1  0.5 or t1   / 3
2
And cos t2  0.5 or t2 
3
2  
Hence   t2  t1    
3 3 3
22. eV0  h  v  v0 
   
 1.6  1019  3.3  6.6  1034  v  6  1014 
 v  1.4  1015 Hz
c 3  108
    2.14  107 m  214nm
v 1.4  1015
23. The horizontal force F imparts a linear impulse
I   Fdt  change in linear momentum
 I  M   0  MR (i)
Where  is the velocity of the centre of mass of the coin.
Since it rolls without slipping,   R , where  is the angular velocity. The torque
due to F imparts an angular impulse
J  Ih  change in angular momentum
 I  0  I
1   1 2
Or Ih   MR 2    I  MR  …(ii)
2   2 
Dividing (ii) by (i)

1
MR 2 R
h 2   3.0cm
MR 2

2GM
24. Escape velocitye .
R
3 3 2GM
Velocity of projection  e  .
4 4 R
The total energy of the body when it is projected is, Ei  KE  PE
1 GmM
 m 2 
2 R
1 9 2GM GmM
 m  
2 16 R R
9 GmM GmM 7 GMm

  
16 R R 16 R
Let h be the maximum height attained by the body.
The distance of the body from the centre of the earth is r = R + h. At this height, the
GmM GmM GmM
total energy of the body is, E f  KE  PE  0   
r r  R  h
From the principle of conservation of energy, Ei  E f , i.e.
7 GmM GmM
 
16 R  R  h
9R
Or 7 (R + h) = 16 R or 7h = 9R or h 
7
25. Light bends away from normal going out of water.
26. Two bodies connected by a wire are at same potential.
27. Amplitude is maximum distance from its mean position.
28. VS = 4, MS = 2.7  MS  VS  LC  2.7  0.04  2.74
29. Thrust   a.v 2   a.2 gh  103  103  2  10  50  102  10N
30. RT  200V  1
 R 
U  n  
  1
 
T f  Ti  1
4
R

T f  Ti    
 3R T f  Ti  3  200 V f  Vi 
1
3
 1200J

CHEMISTRY
31. L represent Tb which is equal to Kb .m
32. Magnetic moment  n  n  2  BM
n : Number of unpaired e 
As atomic number increases in d-block element number of unpaired e  first increases
upto middle then decreases.
slow
33. 
A  B   IAB; So Ea f  is high and Ea b  is low. k1  k2 ; So, Ea for this step is
fast
very high and next step is low and overall reaction is exothermic.
1
34. 

35. H  U  P.V  P VD  VG 
N  12 12 
 500  103  105     106 m3
M 2  3.6 2.4 
 8.33  104 J/mol
36. Bond angle of BF3,PF3,ClF3 are 1200,<109.281,<900 respectively.
37. Trityl amine structure is:
NH
38. CH 3CH 2COOH 
3  CH CH COONH
3 2 4
Pr opanoic acid  A  Amm.Pr opanoate  B 
 KOH  Br2
 CH 3CH 2CONH 2  CH 3CH 2 NH 2
  H 2O  Pr opanamide C 
( Hofmannbromanide
reaction)
39.
Me Me Me
Me
H 
Br 
  H O

2 
o
H   OH
 H
I1 I2 Product
nT .S  nInt  1
40.
O O
18
PhCOH HO18 CH3 PhCO CH3
41. K eq for the reaction in backward direction
Kb 2.1  103 L mol 1s 1

   53.846 L mol 1
 5  1
Kf 3.9  10 s
n
42. equivalents deposited = n Faradays.
43. Polysaccharides – Glycosidic
Maltase – Globular Protein
Peptide bond – Amide linkage in Protein
Nucleotide – Monomer of Nucleic acid
44. Electron with drawing group contain Aldehydes attracts Nucleophile to greater extent.
45.  2,4  DNP test positive means Aldehyde (or) Ketone is present in compound.
 Tollen’s & Cannizaro Positive means Aromatic Aldehyde.
 1,2  Dicarboxylic acids means ortho Isomer.
46. We have Cd 2  aq   Fe  s   Cd  s   Fe 2  aq 
1.0 M  x x
 2 
 RT  Fe 
Hence E  E  ln
2 F Cd 2 
 
At equilibrium E = 0. Hence
 Fe2   E 
ln  
RT x
EH o  i.e.  anti log  
2 F Cd 2  1.0 M  x  RT / 2 F 
   
x  0.037V  1.254 18
or  anti log    10  18 or x  M  0.95M
1.0M  x  0.0295V  19
Thus,  Fe2   0.95M and Cd 2  aq   0.05M
 
47. The first ionization enthalpy of B is 801 kJ mol-1,Eo for Al3+/Al is -1.66 V, Eo for Tl3+/Tl+
is +1.25 V and the m.p. of Ga is 303 K.
Fe2 
 Fe 3
air
48.
hv
Fe 3 Can’t give Brown ring with “NO” Gas released from NO2
49.

50. Sn has the lowest m.p. and the lowest IE1 and Pb has the highest density among carbon
family members and both Si and Ge have high electrical resistivity.
 HC H O  
K a1  4 4 6 O
51. pH  P  log
 H 2C4 H 4O6 O
18.8 / 188
 3.3  log 3
30 / 150
52. B Complex and Vitamin C are soluble.
53. Sm, Er and Gd are three elements which are lanthanoids.
54. Williamson’s synthesis requires that the alkyl halide should be 1 and alkoxide may be
1 ,2 or 3 . Thus, the two ethers which cannot be prepared by Williamson’s synthesis are:
 C6 H5 2 O,  CH3 3 C  O  C  CH3 3
55. Ionic forms give stable Aromatic compounds
Ni 2  aq  DMG   Ni  DMG     2H 
3 AqNH
56.  2
green Rosy Re d volu min ous ppt .
Hyd . of Ni 2  :dsp 2
Distribution of electrons in the d-orbitals of Ni
2
 
3d 8 in  Ni  DMG 2  : d
x2  y 2
d xy
d
z2
d yz d zx
Hence x  6, y  2
n n   3  1.8  1000 
57. T f  k f  1 2  100   1.86  60 180   1.172 T f  1.172 C
 W   95.2 
 
58. Ein  10.2eV   4.2eV
150 o o
KEmax  10.2  4.2  6eV   e  A  5A
6
30
59. E  30
 2 2 
 
22
 V 
60. q  U  w  0   nRT ln 2 
 V1 
V2
Or, 420  1  2  300  ln  V2  6 L
 1  0.082  300 
 
 8.21 

MATHEMATICS
61. A, B, C, D are coplanar
4 3 3  2

  AB AC AD   0  7   5 4  2  0
6 0 6  2
 6 6  12     5  3  2     6  2   20  4  21  0
 6  6  12  2 2  15  13    6  2   4  1  0
 
 12 2  42  18  8 2  22  6  0
 4 2  20  24  0   2  5  6  0
 2
  3  2 0
 3
    3
2
Now  36  25  61
S
2
62. A2  I  A  1  A  1  b
 
Let A  
  
    
A2   I
      
 2       1 0 
     2    1
       2  0 1 
      0      0  a
      0
   2  0

63. A   x   :  x  3   x  4  1 ,
2 x   7  1
2  x   6
 x  3  x  2 ...(A)
 x 3 
    3 x 
B   x  R : 3x   3r  3 ,
 10 
  r 1  
 
x 3
  
3x   3r   33 x
 10 
 r 1 
x 3
 1 
32 x 3  101   33 x
 1 
 10 
9
x 3
 1  35 x  3
 36 2 x  335 x
 6  2 x  3  5x
 3  3x
 x  1 .......(B)
64. x  y  z  15
Total no. Solution  1531C2  136...(1)
Let x  y  z
2 x  z  15  z  15  2t
 t  0,1,2,.....7  5
 7 solutions
there are 21 solutions in which exactly

Two of x,y,z are equal ........(2)
There is one solution in which x  y  z ......(3)
Required answer  136  21  1  114 .
65. A  a1, a2 , a3 , a4 , a5
B  b1, b2 , b3 , b4 , b5
5 5
Given,  ai  25, bi  40
i 1 i 1
2 2
5  5  5
2 
5 
 ai2 
 i 1 
ai  bi   bi 
i 1
  12, i 1   i 1   20
5 5  5 5
   
   
5 5
 ai2  185,  bi2  420
i 1 i 1
Now, C  C1, C2 ,........C10
  ai 15  bi 10 
 Mean of C, C 
10
C  1050  6 .
10
66. Equation of circle is
 x  32   y  5 2  25
 k  2 k k k  2
67. Tk 
k  k  2 k 2  k  2
2
 k 2 k k k  2

2k  k  2 
  1  1 
1
2  k k  2 

T1   1  1 
1
2  1 3 
T2   1  1 
1
2  2 4 
T3   1  1  and so on
1
2  3 5 
 as k   sum
1 2 1 2
 1  1  
1 1
 
2  2  2 2 8 16  8
68. For   2,   0
69. 3 x 2  4 y 2  36
dy 3 x 3 x0
 
dx 4 y 4 y0
Point  x0 , y0  on curve.
3 x0 3

4 y0 2
x0  2 y0
3 x02  4 y02  36
y0  3 / 2 x0  2 y0
 6 3   6 3 
 , , , 
 2 2  2 2
Calculating  distance from both co-ordinate
 6 3 
 ,  is nearest to line.
 2 2
 
2
3 1 8 8
70. Probability  C2 .  .  
 9   9  243

71. If  a, b   R
Then  b, a   R
Hence it is a symmetric relation.
1
72. f  x   xe x  where x  0
xe x
f ' x  
  x  e x 
e3 x  x  1 x  e x
 xe x 
2
f '  x   0  x  1; e  x   x (no solution)
e x  x (exactly one solution)
At x  1, local maxima.
At x  x0   0,1 local minima.
2sec x tan x
73.  dx
 sec x  tan x 10
Let sec x  tan x  t
 sec x  tan x  t
1
2sec x  t  1t
 1
 2sec x tan x  dx  1   dt
 t2 
 1
1 2  dt
I   t10 
t

1 1
I   c
9 11
9t 11t
11  1 2 1
I   sec x  tan x     sec x  tan x     c
 9 11 
1  ex
74. g  x 
1  ex
75. A31B52  B52 A31 is skew Symmetric
A52C 31  C 31A52 is Symmetric

n n
Cr Cr 1
76. 5 4
n n
Cr 1 Cr
n  r 1  5 n  5r  4 ......(2)
r
n  6r  1 .......(1)
 n  29r  5
Coeff of 4th term  29C3
 3654
77. y  x2 y  8  x2 y7
x2  8  x2
x2  4
x  2

  1
   
2
2 1.7   8  2 x dx   2 x 2 dx
2
 
 1  0
  3 2   x3 
1
 2 7   8 x     2 
2 x
  3    3 
  1   0
  16   2  1
 2 7  16     8     2  
  3  3  3
 32 22   10  2
 2 7     2 7   
 3 3  3 3
60
  20 .
3
78.
2cos  sin   1  4cos 2   1  sin 2   2sin 

4  4sin 2   1  sin 2   2sin 
5sin 2   2sin   3  0
1 
cos 2  sin    
2 10
k
2
 k   
6 1 3
79. 
k 1  3  2  3
k k k 1
 
 2  3 k 1   2     2  k 1 
k 1 k
2
1      1    
  3    3  

80. a n  bn is divisible by a  b only if n is odd


81. I 
4 2
 cos x 
2023
.dx ………………(i)
  sin x 2023   cos x 2023
0
a a
As  f ( x)   f (a  x)
0 0
 /2
4  sin x 
2023
   cos x 2023   sin x 2023

I dx ………………..(ii)
0
Adding (i) & (ii)


2
4
2I 
  dx
0
 x 0 /2
4
2I 

4  
2I  .
  2 
I  1.
82.
lim
   12
2sin 6 x 2
 2 x2  x 
x 0 2
83.

 x3  1;   x  0

h( x)   x3  1; 0  x  1
 3
 x  1; 1  x  
 Number of points at which h(x) is not differentiable is 0.
1  
2
x 1 1 x 1
84. f ( x)  4 4
2 x 2 x
4. 1 x 1  4 1 x  2
 2 x 1 x 1

4. x 11  4 x2
 2 x 1 x 1
If x  1,2  , then y   2,4
If x   2,   , then y   2,0 
Sum  3  4  1  6 .
85. Given f ( x)  x 2  x3
As we know that, if g is the inverse of f, then

y  f ( x)  x  g ( y )
dy
Now,  2 x  3x 2
dx
dx 1
 g '( y )  
dy 2 x  3 x 2
When y=2, then x=1
 dx  1 1
Now,  g '(2)     
 dy  x 1 x  3 5
y 2
Hence, the value of  5 g '(2)  3
1 3
 4.
86. 
For ellipse 4  18 1  e12 
7 7
e12  e1 
9 3

For hyperbola : 4  9 e22  1 e2 
13
3

Put  e1, e2  in15 x 2  3 y 2  k k  16
x2
87. f ( x)  1  x  n 1  x   x 
4
1 3
lim f ( x)  0  1  
x 1 4 4
And f (0)  0
x
f '( x)  n 1  x  
2
f '(0)  0, f '(2)  0, f '(3)  0
    2,3
n
88. f n 
5
1 m 1 1 1 1  5
    
5 n 1 (n  1)(n  2) 12  2 m  2  12
11
11 
5
 3 1 
9
89. The co-efficient of x in   x  is C6
  x  6
11
9  1  11 
6
the co-efficient of x in   x  3  is  C5
  x  5

 2  1
x  y  52
90.  9  x  y  20
8
For variance
x-9, y-9, 3, 3, 1, -5, -1, -3
x0

 x  9    y  9   54
2 2
 9.25
8
 x  9 2  11  x 2  20
x  7 or 13  y  13,7
3 x  2 y  3  13  2  7  25 .

only.
.
Admission Number:


correct.
1. A spherical insulator of radius R is charged uniformly with a charge Q throughout its
volume and contains a point charge Q/4 located at its centre. Which of the following graphs
best represents qualitatively, the variation of electric field intensity E with distance r from
center?
1) 2)
3) 4)
2. Figure shows the graph of stopping potential versus the frequency of radiation incident on a
photosensitive metal. Threshold frequency of the metal is
1)
V2 v2  V1v1  2)
V2 v1  V1v2  3)
V2 v1  V1v2  e 4)
V2 v1  V1v2 
V2  V1 V2  V1 v2  v1 V2  V1
3. Two identical balls P and Q are projected with same speeds in a vertical plane from same
point O making projection angles with horizontal 30º and 60º respectively and they fall
directly on a plane AB at points P ' and Q ' respectively. Which of the following statements
is true about distances as given in options ?
1) AP '  AQ '
2) AP '  AQ '
3) AP '  AQ '
4) As these are complimentary projection angles, then AP '  AQ '.
4. The average mass of a nucleon is taken as one amu. Any element necessarily consists of an
integral number of nucleons; still the atomic mass of an element (on amu scale) is not an
integer, primarily because
1) the mass of a proton is slightly less than that of a neutron
2) of the pressure of extra nuclear electrons
3) of the mutual conversion of proton into neutron and vice-versa
4) of the existence of isotopes, with different relative abundances.
5. Force–displacement  F  x  graphs for four different particles moving along a straight line
are shown. If W1 , W 2 , W 3 a n d W 4 are the works done corresponding to figures a, b, c and d
respectively, then  x0  x1  x0  x2  x1  x3  x2 
a) b)

c) d)
1) W 3  W 2  W1  W 4 2) W 3  W 2  W 4  W1 3) W 2  W 3  W 4  W1 4) W 2  W 3  W1  W 4
6. Assertion: When a sphere is rolls freely on a horizontal table, it slows down and eventually
stops.
Reason: When the sphere rolls on the table, both the sphere and the surface deform near the
contact. As a result, the normal force does not pass through the centre and provide an
angular deceleration.
1) Both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
2) Both Assertion and Reason are correct, but Reason is not the correct explanation of
Assertion.
3) Assertion is correct but Reason is incorrect.
4) both the Assertion and Reason are incorrect.
7.
  
A particle is moving in an elliptical orbit as shown in figure. If p, L and r denote the linear
momentum, angular momentum and position vector of the particle (from focus O)
  
respectively, when the particle is point at A, then the direction of   p  L is along
1) –ve x-axis 2) +ve x-axis 3) +ve y-axis 4) –ve y-axis

8. Six stars of equal mass m are moving about the centre of mass of the system such that they
are always on the vertices of a regular hexagon of side length a. Their common time period
of revolution around center will be
1) 4
a3
2) 2 4 3a 3
3) 4
3a3
4) 2

Gm 5 3  4 
Gm 
Gm 5 3  4  Gm 4 3a 3
9. A body is projected vertically upwards from the surface of the earth with a velocity
sufficient to just carry it to infinity. The time taken by it to reach a height of three times the
radius of the earth is
(acceleration due to gravity  9.8 m s  2 and radius of the earth = 6400 km)
1) 44.44 min 2) 22.22 min 3) 18.76 min 4) 37.52 min
10. Assertion : In a pressure cooker, the water is brought to boil. The cooker is then removed
from the stove. Now on removing the lid of the pressure cooker, the water starts boiling
again.
Reason: The impurities in water bring down its boiling point.
1) Both Assertion and Reason are correct and the Reason is a correct explanation of the
Assertion.
2) Both Assertion and Reason are correct but Reason is not a correct explanation of the
Assertion.
3) The Assertion is correct but Reason is incorrect.
4) Both the Assertion and Reason are incorrect.
11. A uniform rod is suspended horizontally from its mid-point. A piece of metal whose weight
is w is suspended at a distance l from the midpoint. Another weight W 1 is suspended on the
other side at a distance l1 from the mid-point to bring the rod to a horizontal position. When
w is completely immersed in water, w1 needs to be kept at a distance l2 from the mid-point to
get the rod back into horizontal position. The specific gravity of the metal piece is
w wl1 l1 l1
1) 2) 3) 4)
w1 wl  w1l2 l1  l2 l2

12. Figure shows a capillary rise H. If the air is blown through the horizontal tube in the
direction as shown, then rise in capillary tube will become
1) = H 2) > H 3) < H 4) zero

13. An ideal gas has molar heat capacity CV at constant volume. The gas undergoes a process as
 
T  T0 1  V 2 where T and V are temperature and volume respectively, T 0 and  are
positive constants. The molar heat capacity C of the gas is given as C  CV  Rf V  , where
f V  is a function of volume. The expression for f V  is
V 2 1  V 2
1) 2) 3) V 2 1  V 2  4) 1
1  V 2 2V 2 2 V 2
 1  V 2 
14. The displacement of particle executing SHM is described by x  t   Acos t    . If at t  0,
the position of the particle is 1 cm and its initial velocity is  cm s  1 , then its amplitude and
initial phase angle respectively are
1) 2 cm, 75° 2) 2 cm, 45° 3) 2 cm, 45 4) 2 cm,30

15. Assertion: The fundamental frequency of an open organ pipe increases as the temperature is
increased.
Reason: As the temperature increases, the velocity of sound increases more rapidly than
length of the pipe.

Assertion.
4) Both the Assertion and Reason are incorrect.
16. A neutral conducting solid sphere of radius R has two spherical cavities of radii a and b as
shown in the figure. Centre to centre distance between two cavities is c. q a and qb charges
are placed at the centers of cavities respectively. The force between qa and qb is
1 qa qb 1  1 1
1) 2) qa qb  2  2  3) zero 4) insufficient data
40 c2 4 0 a b 
17. A rod of length l having charge q uniformly distributed moves towards right with constant
speed v. At t  0, it enters in an imaginary cube of edge l / 2. Sketch the variation of electric
flux through the cube with respect to time.

1) 2) 3) 4)
18. What is equivalent capacitance of circuit between points A and B?
1) 2
3
F 2) 4
3
F 3) Infinite 4) 1  3   F
19. In a metallic conductor, under the effect of applied electric field, the free electrons of the
conductor
1) drift from higher potential to lower potential.
2) move in the curved paths from lower potential to higher potential
3) move with the uniform velocity throughout from lower potential to higher potential
4) move in the straight line path in the same direction as applied electric field.
20. Statement – I : More the value of magnetic flux linked with a coil, more will be the induced
e.m.f developed in the coil.
Statement – II : Lenz’s law is direct consequence of law of conservation of energy.
1) Statement I is false but statement II is true
2) Statement I is true but statement II is false
3) Both statements I and II are false
4) Both statements I and II are true
21. Two wedges each of mass 600 g are placed next to each other on a rough horizontal surface.
The coefficient of static friction between the wedges and the surface is 0.4. A cube of mass
‘M’ is balanced on the wedges as shown in the figure. If there is no friction between the
cube and wedges, the largest mass ‘M’ of the cube that can be balanced without motion of
n
the wedges is kg , find n.
10
22. A string of area of cross-section 4mm2 and length 0.5 m is connected with a small rigid body
of mass 2kg. The body is rotated in a vertical circular path of radius 0.5m. The body acquires
a speed of 5m / s at the bottom of the circular path. Strain produced in the string when the
body is at the bottom of the circle is ..........  10  4.
(Use Young's modulus 1011 N / m 2 and g  10 m / s 2 )
23. In an experiment to verify Newton's law of cooling, a graph is plotted between the
temperature difference  T  of the water, surrounding and time as shown in figure. The
initial temperature of water is taken as 80ºC. The value of t2 as mentioned in the graph will
be_______  ln 1.5   0.405, ln  3   1.099 
24. Two long parallel wires carrying currents 8A and 15A in opposite directions are placed at a
distance of 7cm from each other. A point P is at equidistant from both the wires such that the

lines joining the point P to the wires are perpendicular to each other. The magnitude of
magnetic field at P is _______ 10  7 T .
(Given : 2  1.4 )
25. A domain in a ferromagnetic substance is in the form of a cube of length 1 m . If it contains

8  1010 atoms and each atomic dipole has a dipole moment of 9  10  24 Am 2 , then the
magnetization of the domain is ______  10 4 A.m 1 .
26. The alternating current in a circuit is described by the graph shown in figure. If r.m.s current
n
from this graph (in A) is , find n.
10
27. An infinitely long thin wire carrying a uniform linear static charge density  is placed along
the z-axis (figure). The wire is set into motion along its length with a uniform velocity

V  V kˆ. The magnitude of poynting vector S at the curved surface of imaginary long
W
cylinder with its axis at the wire = _________  10 6
m2
Given    C / m,V  4 mm / s, a  2m
 

28. Calculate the current (in A) produced in a germanium plate of thickness 0.25 mm and area
1 cm 2 when a potential difference of 1.5V is applied across the thickness. The hole density in
germanium is 2  1019 m  3 and the mobilities of electrons and holes are 3 5 0 c m 2V 1
s 1 and
1200 cm 2V 1
s 1 respectively.
29. A ball of mass 0.45 kg which is initially at rest is hit by a bat. The bat remains in contact with
the ball for 3  10 3 s. During this time period, the force on the ball by the bat varies with time
t(in s) as    
F  t      10 6 t    10 9 t 2  N where  and  are constants. The ball's speed,
immediately as it loses contact with the bat is 20 m/s. If the relation between  and  is
  2   n, find n.
30. The velocity displacement (v-s) graph shown represents the motion of a particle moving
along a straight line. The graph is a circle of radius 2 m and centre is at (2, 0) m.
If the value of acceleration for this particle at a point 2  2 m , 2 ms 1  is z ms2, find z.

correct.
31. In the preparation of methanol by catalytic hydrogenation of carbon monoxide catalyst used
is
1) Zn + Cr 2) Ni 3) ZnO - Cr2O3 4) VO
2 5
32. In the given reaction sequence, the major product B is
1) 2) 3) 4)
33. Consider the following compounds
The compounds which will give iodoform test are

1) (ii) and (iii) only 2) (iii) and (iv) only
3) (i), (ii) and (iv) only 4) (ii), (iii) and (iv) only
34. Hinsberg’s reagent is
1) C6H5SO3H 2) C6H5OH 3) C6H5SO2Cl 4) C6H5SO2Na
35. Match List-I with list-II
List-I List-II
(Name of Vitamins) (Deficiency diseases)
(a) Thiamine (i) Convulsions
(b) Pyridoxine (ii) Bleeding gums
(c) Riboflavin (iii) loss of appetite
(d) Ascorbic acid (iv) Burning sensation of the skin
Choose the correct answer from the options given below
1) (a)-(iii), (b) –(i), (c)-(iv),(d)-(ii) 2) (a)-(i), (b)-(ii), (c)-(iii), (d)(iv)
3) (a)-(iii), (b)-(ii), (c)-(i), (d)-(iv) 4) (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)
36. Select the incorrect statement(s) among the followings
A) The unit of rate constant for second order reaction is mol–1Ls–1
B) Inversion of cane sugar is a pseudo first order reaction
C) Order of a reaction can only be a whole number
D) The exponents of concentration terms in rate law expressions are always equal to the
stoichiometric coefficients of the reactions
1) (A) and (B) only 2) (B) and (D) only
3) (C) and (D) only 4) (A) and (C) only
37. The following graph is given, between total energy and distance between the two nuclei for
species H 2 , H 2 , He 2 & He 2 . Which of the following statements is/are correct?
1) H2 is less stable than He 2

2) Bond dissociation energy of H2 is more than bond dissociation energy of He 2 .
3) Since bond orders of He 2 and H2 are equal both will have equal bond dissociation
energy.
4) Bond length of H2 is less than bond length of H2 .

1
38. For Arrhenius equation the graph between lnk vs is straight line. The slope & intercept of
T
the graph respectively are
-2.303Ea -R
1) , 2.303 log A 2) , ln A
R Ea
-Ea -Ea
3) ln A, 4) and ln A
2.303R R
39. Select the incorrect statement(s) among the following
a) If D0 > P , it becomes more kinetically favourable for the fourth electron to occupy t 2g
4
orbital with t 42g e 0g configuration for d ions
b) Nature of bonding in co-ordination compound cannot be explain by MOT
c) Solvated isomerism is a special class of co-ordination isomerism
1) b and c only 2) b only 3) a and c only 4) a, b and c
40. Which among the following ions has highest number of unpaired electrons?
1) Er(III) 2) Lu(III) 3) Ho(III) 4) Gd(III)
41. Complete the following fill in the blanks by given options.
A) For the same metal, the same ligands and metal ligand distance, D0 = ......Dt
B) Anhydrous CuSO4 is ……… and CuSO4 . 5H2O is …………. in colour
C) In tetrahedral crystal field …………. configurations are rarely observed
3+
D) Ruby is Al2O3, containing about 0.5-1% Cr ions, which are randomly distributed in
position normally occupied by ………
4
1) (A) , (B) white, blue, (C) low spin, (D) A l 3+
9
4 2-
2) (A) , (B) white, black, (C) high spin, (D) O
9
9
3) (A) , (B) white, blue, (C) low spin, (D) A l 3+
4
9 2-
4) (A) , (B) white, black, (C) low spin,(D) O
4

42. Which of the following alkenes does not give acetaldehyde as one of the products on
reductive ozonolysis?.
1) 2) 3) 4)
43. Statement-I : The single N-N bond is weaker than the single P-P bond.
Statement-II : The catenation tendency of N is weaker than P.
In the light of the above statements, choose the correct answer from the options given below.
1) Statement-I is incorrect but statement-II is correct
2) Both statement-I and statement-II are correct
3) Both statement-I and Statement-II are incorrect
4) Statement-I is correct but statement-II is incorrect
44. Correct order of Lattice enthalpy is
1) NaCl > MgCl2 > AlCl3 2) MgCl2 > NaCl > AlCl3
3) AlCl3 > MgCl2 > NaCl 4) AlCl3 > NaCl > MgCl2
45. The major product of the following reaction is
CH 3
NBS, benzoyl peroxide CH 3CH 2ONa

CCl4 CH 3CH 2OH
Br

CH 2 Br CH 3
CH 2OCH 2CH 3 CH 2OCH 2CH 3
Br
OCH 2CH 3 OCH 2CH 3 OCH 2CH 3

1) 2) Br 3) 4)
46. In S4O62- , the oxidation numbers of four Sulphur atoms are
1) +4, 0, 0 and +6 2) +4, -2, -2 and +6
3) +5, 0, 0 and +5 4) +5, -2, -2 and +5
47. Consider the following statements:
A) Halogens have maximum negative electron gain enthalpy in the corresponding periods
B) Bond dissociation enthalpy of F2 is greater than Br2
C) Negative electron gain enthalpy of fluorine is less than that of chlorine.
The incorrect statement(s) is/are
1) A and B only 2) B only 3) B and C only 4) C only
48. Correct order of first ionization enthalpy of the given element is
1) Al > Ga > In > Tl 2) Tl > Ga > Al > In
3) Ga > Tl > In > Al 4) Tl > In > Ga > Al
49. Select the correct statement among the following
1) Standard enthalpy of formation of H2O (l) is zero
2) Naphthalene sublimes slower than solid CO2

3) Enthalpy of vapourization of acetone is more than that of water
4) If water freezes then amount of heat is given off to the surrounding will be equal to
enthalpy of sublimation of water

50. The correct order of freezing point for 10 g aqueous solution each of urea, glucose and sucrose
is
1) Tf (glu cos e) > Tf (urea ) > Tf (sucrose) 2) Tf (sucrose) > Tf (glu cos e) > Tf (urea )
3) Tf (urea ) > Tf (glu cos e) > Tf (sucrose) 4) Tf (sucrose) > Tf (urea ) > Tf (glu cos e)
51. Number of copper (atomic weight 64) atoms present in a coin made up of bell metal is
3.0115 × 1022. The weight of the coin in grams is ………..
52. Cu 2  2e   Cu ; log Cu 2  vs Ered graph is of the type as shown in figure where
OA = 0.34 V then electrode potential of the half cell of Cu | Cu 2 (0.1 M) will be x, the value
of 10  x is  
2.303RT
 0.06 
 F 
53. 0.2 M, 100 ml of CH3 COOH is mixed with 0.1 M, 100 ml of NaOH. The pH of resulting
solution is (pKa ( CH3 COOH) = 4.76)
54. If 5 mol PCl5 on decomposition gives 2 mol PCl3 as per given equilibrium
 PCl 3(g ) + Cl 2(g )
PCl5(g ) 
Then % degree of dissociation of PCl5 will be

55. Consider the following statements about SN2 mechanism
The number of correct statements among the following is
I) Follows second order kinetics
II) Nucleophilic substitution reaction
III) Racemic mixture is obtained
IV) Order of reactivity of alkylhalides is primary halides > secondary halides > tertiary
halides
V) Reaction occurs via carbocation formation
56. Among P4 (s), S8 (s), Cl2 (g) and NO2 (g) , how many species can undergo disproportionation
in the alkaline medium?
57. Out of following groups, number of groups exerting +R effect, is
–CN,–OH, –CHO,–OCOR,– NH2 ,–NHCOR
58. How many of the following compounds are more reactive than benzene for EAS reaction?
CH 3 NH 2 CN OMe
OH CHO COOH
, and
, , , ,
(a) (b) (c) (d) (e) (f) (g)
59. The (E) emf of Pt / H 2 1 atm  / H 2 SO4 is 0.295 V, then the pH of the acid is
60. Number of degenerate orbitals present in third shell of L i 2 + ion is

correct.
61. Statement -1: The sum of the series 1  1  2  4   4  6  9   9 12 16  ....   361  380  400 is
8000
 k   k 1   n , for any natural number n .

n
3 3 3
Statement – 2:
k 1
1) Statement -1 is false, Statement – 2 is true.

2) Statement -1 is true, statement -2 is true; Statement- 2 is a correct explanation for
statement-1.
3) Statement-1 is true, Statement- 2 is true; Statement -2 is not a correct explanation for
Statement -1.
4) Statement -1 is true, statement -2 is false
62. If a line having negative slope passing through the point (8,2) meets the X-axis at A and Y-
axis at B. Then
Statement – 1 : The minimum value of OA  OB is 18
Statement – 2 : The minimum value of D OAB is 32
1) Statement -1 is false, Statement – 2 is true.
2) Statement -1 is true, statement -2 is true; Statement- 2 is a correct explanation for
statement-1.
3) Statement-1 is true, Statement- 2 is true; Statement -2 is not a correct explanation for
Statement -1.
4) Statement -1 is true, statement -2 is false
63. In a shop there are five types of ice –creams available. A child buys six ice-creams.
10
Statement-1: The number of different ways the child can buy the six ice –creams is C5
Statement-2 : The number of different ways the child can buy the six ice-creams is equal to
the number of different ways of arranging 6A’s and 4B’s is a row
1) Statement – 1 is false, Statement -2 is true
2) Statement -1 is true, Statement -2 is true; Statement-2 is a correct explanation for
Statement -1
3) Statement -1 is true, Statement-2 is true; Statement -2 is not a correct explain for
Statement – 1
4) Statement – 1 is true, Statement -2 is false
n2 1
64. Statement – 1:The variance of first n odd natural numbers is
3
Statement – 2: The sum of the first n odd natural number is n2 and the sum of square of
n  4 n  1
2
first n odd natural numbers is

3
1) Statement is true, Statement 2 is false
2) Statement 1 is true, Statement 2 is true ; Statement 2 is not a correct explanation for
Statement 1
3) Statement 1 is false, Statement 2 is true
4) Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement
1
65. Assertion (A): If (-1, 3, 2) and (5, 3, 2) are respectively the orthocenter and circumcentre of
a triangle, then (3, 3, 2) is it’s centroid.
Reason (R): Centroid of the triangle divides the line segment joing the orthocenter and the
circumcentre in the ratio 1 : 2
Assertion.
4) Assertion is incorrect but Reason is correct
1
66. The domiain of the function f  x  is (when [ ] is G.I.F)
 x2  3 x 18
1)  , 4   7,   2)  , 3  [7, )
3)  , 4   6,  4)  ,4    7,  
67. The value of 54  4 cos2 90  1 4 cos2 270  1  4 cos 2 810  1 4 cos 2 2430  1 is
1) 54 2) 18 3) 27 4) 36
68. For two non-zero complex number z1 and z2 , if Re(z1 z2)= 0 and Re (z1+z2) = 0 , then which
of the following are possible?
A) Im ( Z1 ) > 0 and Im( Z2) > 0 B) Im( Z1 ) < 0 and Im( Z2) > 0
C) Im ( Z1 ) > 0 and Im( Z2) < 0 D) Im ( Z1 ) < 0 and Im( Z2) < 0
Choose the correct answer from the option given below;
1) B and D 2) B and C 3) A and B 4) A and C

    
x2  4 x2 4
69. Let S   x : x  R 3 2  3 2  10 . Then sum of the values of n  S  is equal to
 
1) 4 2) 0 3) 6 4) 2
70. If the letters of the word MATHS are permuted and all possible words so formed are
arranged as in a dictionary with serial numbers, then the serial number of the word THAMS
is
1) 103 2) 104 3) 101 4) 102
71. The point (4,1) undergoes the following three transformations successively.
i) Reflection about the line y  x
ii) Translation through a distance 2 units along the positive direction of x  axis
iii) Rotation through an angle  / 4 about the origin in the counter clockwise direction.
Then the final position of the point is given by coordinates
 1 7   1 7 
1)  ,  2)   2, 7 2  3)   ,  4)  2, 7 2 
 2 2  2 2
72. The minimum length of the chord of the circle x 2  y 2  2x  2y  7  0 which is passing
through (1, 0) is :
1) 2 2) 4 3) 2 2 4) 5
x 2 y2 x 2 y2
73. If e1 and e2 are the eccentricities of the ellipse + =1 and the hyperbola - =1
18 4 9 4
respectively and (e1, e2 ) is a point on the ellipse 15x + 3y2 = k then ‘k’ is
2
1) 14 2) 15 3) 16 4) 17
74. Let R1 and R2 be relations on the set {1,2,.....50} such that
R1   p , p  : p is a prime and n  0 is an int eger and
n
 
R2  p, pn : p isa prime and n  0 or 1 
Then, the number of elements in R1  R2 is___________.
1) 8 2) 7 3) 9 4) 10
 /x  / x2 dy
75. If y  1    , then is
1  1  1  1  1  1  dx
                 
x  x  x  x  x  x 
 
     
y    
1) y     2)    
  x   x   x  x  1  1   1  
x x x 
      
     y  
3) y     4)  x  x  x 
1 1 1 x  1  1   1  
    
x x x  x x x 
é
tr ê(AB - BA) ú
3ù
76. Let A and B be 3´ 3 matrices, then ë û=

det (AB - BA)
1 1
1) 3 2) 9 3) 4)
3 9
2 1 
77. Let ‘f’ is a function , continuous on [0,1] such that f  x   5 x  0,1 and f  x  x   ,1
x 2 
1
then the smallest ' a ' for which  f  x dx  a holds for all ' f '
0
is
5  5  5
1) 5 2)  2ln 2 3) 2  ln   4) 2  2ln  
2 2
  2
 
78. The differenctial equation whose general solution is given y  C1 cos  x  C2   C3e xC  C5 sin x, 4
where C1 , C 2 , ......C5 are constants is.

d4y d2y d 3 y d 2 y dy
1) 4  2  y  0 2) 3  2   y  0
dx dx dx dx dx
d y d 2 y dy
3
d 3 y d 2 y dy
3) 3  2   y  0 4) 3  2   0
dx dx dx dx dx dx
If  ,  ,  ,   R satisfy 
  1     1     1    1
2 2 2 2
79. 4
    
If biquadratic equation a0 x  a1x  a2 x  a3x  a4  0 has the roots
24 3
 1   1   1   1 
     1 ,      1 ,      1 ,      1 . Then the value of a2 / a0 is:
1) 4 2) –4 3) 6 4) none of these
80. It is given that complex numbers z1 and z 2 satisfy z1  2 and z2 =3. If the included angled
of their corresponding vectors is 600, then z1  z2 can be expressed as

n
, where ‘n’ is
z1  z2 7
natural number then n =
1) 126 2) 119 3) 133 4) 19
81. The number of elements in the set , n  N , 20  n  100 and 3n  3 is a multiple of 7 is
82. The total number of ordered pairs (x, y) which satisfies
y =| sinx |, y = cos 1  cos x  where  2  x  
83. Let (x,y,z) be poitns with integer coordinate satisfying the system of homogeneous
equations;
3x  y  z  0
 3x  z  0
3 x  2 y  z  0
Then the number of such poitns for which x  y  z  90 is
2 2 2
9 9
  x  5  9 and  x  5
2
84. If i i  45 , then the standard deviation of the 9 items x1, x2 ,...., x9 is
i 1 i 1
85. If  ,  are two distinct real roots of the equation ax 3  x  1  a  0 ,  a  1,0 , none of which
1 a x3  x2  a al  k   
is equal to unity, then the value of is . Find the value of kl .
lim
x  1/  
 e1ax 1  x 1 
 ,x  0
86. Let f  x    xe ax
. Where a is a positive constant. The interval in which f 1  x  is
 x  ax  x ,x  0
2 3
k a
increasing is  ,  . Then k l is equal to.
a l 
 3 x  2  x 2  6 1  x 2  x 2  dx
  
87. If f(x) = 
   ; x  0,1 and f(0) = 0 then the value of
  f (2-1/6 ) is
3
1 x 2
88. Consider y  x2 and f  x where f  x , is a differentiable function satisfying

f  x  1  f  z 1  f  x  z  x, z  R and f  0  0 ; f 1  0  4 . If area bounded by curve y  x2
3 
and y  f  x  is  , find the value of    .
 16 
89. For a unique value of  &  , the system of equations given by
x y z 6
x  2 y  3 z  14
2x  5y   z  
Has infinitely many solutions, then    is equal to

4
90. X and Y are two weak students in mathematics and their chances of solving a problem
correctly are 1/ 8 and 1/12 respectively. They are given a question and they obtain the same
answer. If the probability of common mistake is 1 , then probability that the answer was
1001
correct is  /b ( a and b are co-primes). Then a  b 


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

SOLUTIONS
PHYSICS
KQ KQ 5 KQ
1. E 3
r  2 at r  R and E  at r  R
R 4r 4 r2
dE KQ KQ R
At r  R, E is minimum when 0 3  0  r
dr R 8r 2
E   at r  0 and E  0 at r  
These variations are best represented by option (1)
2. Ve  h  h0
V1e  h1  h0 ______ 1
V2e  h2  h0 ______  2 
1  V    
1 1 0
 2  0 V1  1  0 V2
 2 V    2 2 0
V2v1  V1v2
 0 
V2  V1
3. Projection angle of ball P = 30° and projection angle of ball Q = 60°
Q
P R
O
Q' P'
For complimentary angle ranges, OR is same for P and Q as O and R are on same
horizontal plane.
From figure we can say that AP '  AQ '
4. Conceptual
5. From the graph,
Work done = Area under F – X curve.
Also, area below X-axis is negative & area above X-axis is positive.
From the given graph, it is clearly visible that,
A3  A2  A1  A4  W3  W2  W1  W4 W  A; A  area
6. When a sphere rolls on the table it slows down and eventually stop because in this case,
normal force acting at every point is different which provides necessary torque about
centre of mass to cause angular deceleration.
7.
From figure, Linear momentum towards +ve y-axis  pjˆ  .


From right hand thumb rule, Angular momentum  L  is towards +ve z-axis Lkˆ  
Now,

 
  
  p  L   pjˆ   Lkˆ  pL ˆj  kˆ  pLiˆ 

From properties of cross product of 2 vectors, the direction of  is along +ve x-axis.
8.
Gm1m2
We know that, F 
r2
From the geometry of hexagon
Gm 2 Gm2 Gm2 F
F1  F5   F  let  F2  F4   
 3a 
2
a2 3a 2 3
Gm 2 Gm 2 F
F3   
 2a 
2
4a 2 4
Resultant gravitation Force,
F
F '  F12  F52  2 F1 F5 cos120  F22  F42  2 F2 F4 cos 60 
4
F F 5 1   5 1  Gm 2
F     F  4  2
3 4 4 3  3 a
Since, Centripetal Force = Gravitational Force
Gm 2  5 1  Gm  5 1  4 3a 3
m 2 a       2
a 2  4  a 3  4 
 
T
3 3 Gm 5 3  4
9. If at a distance r from the centre of the earth, the body has velocity v,
1 2  GMm  1 2  GMm 
mv      mve    
2  r  2  R 
2GM  R   GM 
v 2  ve2    1 Ve  2 gR , g  2 
R r   R 
Rh
2 gR 2
t
R  dr 2g 1
v  2 gR  2 gR   1  v   R  dt   r1/2 .dr
2
r  r dt r 0 R 2g R
2 1 2 1  h
3/2

  R  h   R 3/ 2   
3/ 2
t   R 3/2 1    1
3 R 2g   3 R 2g
 R  
1 2 R  h   1 2 R  3 R  
3/2 3/2
1 2R
t 1    1  t   1    1 Given h  3R   t 7
3 g  R   3 g  R   3 g
1 2  6400  103
Putting all values, we get t
3 9.8
7 R  6400 10  3
80
t  14.2  7 s  2666.65 s  44.44 min
3
10. In a pressure cooker the water is brought to boil, if temperature is 100°C. When the lid of
cooker is opened, pressure is lowered so that boiling point decreases and water boils
again. Generally impurities increase the boiling point.
11.

In equilibrium condition in air

wl  w1l1 ......  i 
Now, in water
W
W  FB  l  W1l2 (where FB = buoyant force  )

W
Hence,  W  l  W1l2
  
 1 l  W   1  l2 1 l2 l
W 1   l  W . .l2 W1  l 1     1  2
  l1  l1     l1  l1 l1  l2
12.
According to Bernoulli equation,

1 2
P1   v   gh  constant ...........  i 
2
If velocity of air will increases the pressure at point (A) must be decreases to maintain
equation (i).
According to question, air will start to blown through the horizontal tube so pressure at
point (A) will decrease.
 The height of liquid rise in capillary tube will increases.
13. Given,
C  CV  Rf V  .......  i    
T  T0 1  V 2  T0  T0 V 2
dT
dT  0  T0 .2V .dV dT  2T0VdV   2T0V 
dV
According to the first law of thermodynamics
dQ  dU  dW nCdT  nCV dT  PdV
P dV
C  CV  ...........  ii 
n dT
 .T  T0 1  V 2  
P R R
 PV  nRT
n V V
 dT  P
Putting this value   and in eqn  ii  we get.
 dV  n
1 1  V 2 
C  CV  T0 1  V 2   
R
C  CV  R  2 
V 2T0V  2V 
1  V 2
Compare with equation (i), we get, f V  
2V 2
14. Displacement equation is given by,
x  t   A cos  t    .......  i 
At t  0, x  1 cm  1  A cos   0     1  A cos  .......  ii 
Differentiating x  t  with respect to t, we get-
V   A sin t   
At t  0, v   cm / sec     A sin   1   A sin  ........  iii 
Squaring and adding equation (ii) and (iii), we get-
A2 cos 2   A2 sin 2   1  1  A  2
1 A cos 
Dividing equation (ii) from (iii), we get-   cot   1
1  A sin 
 
cot   cot       135 or  45,   45
 4
15. The fundamental frequency of an open organ pipe increases as the temperature is
V
increased because the fundamental frequency of an open organ pipe is n  as
2l
temperature increases the velocity of sound increases more rapidly than length of the
pipe.
16. Electrostatic force between two point charges is independent of any other medium. Net
force on one charge changes due to presence of other medium.
17. At time t , length of rod inside cube = vt
l q
At, t  0 to t   Charge inside cube qin  vt
2v l
qin qv
Then,    t    t
 0  0l
qv
Graph will be straight line of slope
 0l
l q
At t  0,   0  t  ,  
2v 2 0
l l q
t  to t  , half of rod inside cube,  
2v v 2 0
l 3l
t  to t  ,
v 2v
charge inside the cube
q l  3q qv
qin  q   vt     t
l 2 2 l
 3q qv 
  t  3q qv
Then,    2 l    t
0 2 0  0l
Hence, the graph will be straight line of negative slope and positive intercept.
18. Since a middle arm which connect the upper short the circuit. So, it will neglected.

The effective circuit is shown in figure.

The capacitance of upper series,
1 1 1 1 1 2
    ......  C  C
Ceq 1 3 9 27 3
2 4
C AB  2C  2   F C AB   F
3 3
19. Path of electron between two consecutive collisions with lattice ions under the influence
of electric field is curved.
20. Conceptual
21. Given,   0.4, m  600 gm  0.6 kg
Mg
2 N cos 45  Mg  2 N  Mg  N  f L    N cos   mg 
2
 Mg 1  M 
fL      mg    g   m  ...............  i 
 2 2   2 
Mg Mg
If wedge is balanced, F  N sin   sin 45  ...............  ii 
2 2
 Mg Mg Mg  Mg 2 m 2  0.4  0.6
From equation (i) & (ii),   mg   mg   M  
2 2 2 2 1    1  0.4
M  0.8 kg
22.
2  5
2
mv 2
2 mg  20 
F mv 0.5 strain  30  10 5
Strain  Force  F   mg  strain  R strain  6
AY R AY 4 10  1011
23.

T T
dT dT
  K T  T0     K  dt
dt Ti
T  T0 Ti
T  T0
 exp   Kt 
Ti  T0
40
 exp  K  6 
60
  0.405 
l n  2 / 3   K  6  K   0.068
6
Again Newton's Law of cooling,
20
 exp   Kt2 
60
l n 1 / 3   0.068  t2
t2  16.15min  16 min
24. Given, i1  8 A, i2  15 A
Distance between two wire = 7 cm
0 8  15
B1   and B2  0 
2 3.5 2 2 3.5 2
So, Bnet  B12  B22
0 1
Bnet   82  152
2 3.5 2
 1
Bnet  0  17
2 3.5 2
Bnet  68  10 7 T
Hence, the magnetic field at P is 68  10 7 T .
 
3
25. Volume of cube  domain   106  1018 m3
Number of atoms  N   8  1010
Dipole moment  M   9  10 24 Am 2 for each atom
The maximum possible dipole moment M max is obtained for case when all atomic moment
at perfectly aligned.
Hence, M max  NM  8  1010  9  10 24  72  10 14 Am 2
M max 72 1014
 Magnetization   18
 72 104 A / m
Volume 10
26. I rms  rms current
12  2 2 5
   1.58 A  1.6 A
2 2

 ˆ
27. Given E  j
2 0 a
i  v
B  0 iˆ  0 iˆ I  V 
2 a 2 a
1 1   ˆj  
 S   E  B    0 Viˆ 
0 0  2 0 a 2 a 
V2
 V 2
2 
 j  iˆ   2 2 kˆ
4  0 a
2
4  0 a
28. The conductivity of a pure semiconductor is given by   eni  e   h 
Here, ni  2  1019 m 3 ,  e  0.35 m 2V 1s 1
and  h  0.12 m 2V 1s 1
 
   1.6  1019 C 2  1019 m 3 
0.35  0.12 m 2V 1s 1
 1.504  1.
1 l
 The electrical resistance of the semiconductor piece is R 
 A
25  10 5 m
  1.66 .
 
1.504  1m 1 1 10 4 m 2 
 The current through the semiconductor is, from Ohm’s law.
V 1.5V
i or, i   0.902 A.
R 1.66 
29. mv  mu   F  t  dt
T
  
0.45  20  0.45  0      10 t   109 t 2  dt 
0
T2 T3
9    106    109 
2 3
6
9 10 27 109
9   106    109 
2 3
9
9  9    2  2
2
 x  a   y  b  r2
2 2
30.
 s  2  v  0  22   s  2   v 2  4
2 2 2
Differentiating with respect to t,

ds dv
2  s  2  2v 0
dt dt
 ds dv 
 s  2  a  0  v  and a  
 dt dt 
At s   2  2  m,  2  2   2  a  0
So, a  2 m / s 2

CHEMISTRY
31. Reference NCERT (XII) -Page 344
32. Reference NCERT (XII) Page No.384
35. Reference NCERT(XII) -Page 426
36. Reference NCERT (XII) Page No 102, 103
37. Conceptual
38. Reference NCERT Page No 114
39. Reference NCERT (XII) Page No. -258
40. NCERT – Page 236
41. Reference NCERT Page No. – 259
42. Reference NCERT. XI Page No.391
43. Reference NCERT (XII) Page No. 173
44. AlCl3 > MgCl2 > NaCl
CH 3 CH 2 Br CH 2OCH 2CH 3
NBS, benzoyl peroxide CH 3CH 2ONa

CCl4 heat allylic bromination CH 3CH 2OH
45. Br Br Br
46. Reference NCERT (XI) Page No.: 273

47. Reference NCERT XII Page198
49. Reference NCERT (XI) Page No. -171,172,173
51. Weight of 6.023 × 1023 Cu atoms = 64 g
Weight of 3.0115 × 1022 Cu atoms = 3.2 g
80 g of copper is present in 100 g of bell metal.
100
Therefore, weight of bell metal coin, in grams = 3.2  4
80
0 .0 6
52. E C u  2 / C u  E c0 u  2 / C u  lo g  C u  2 
2
0.06 0 .0 6 0.06
 0.34  log 10 1   0 .3 4   E Cu / Cu 2  0.34  = -0.31
2 2 2
53. CH 3COOH + NaOH  CH 3COONa + H 2 O
20 m.eq. 10 m.eq. 0
10 m.eq. 0 10
pH = pKa = 4.76
54.  PCl3 + Cl2
PCl5(g) 
5 0 0
5– a a a
2
3 2 2 %a = ´100 = 40%
5
56. Reference NCERT Page No.: 272
57. Reference NCERT(XI) Page 354
58. a, b, e, g
59. EOP  0.059 pH 0.295  0.059 pH pH  5.0
60. Reference NCERT(XI) Page 60
MATHEMATICS
2 3
61. nth term Tn = (n -1) + (n -1) n + n 2 = n 3 -(n -1)
n
3
Sn = å k 3 - (k - 1) = 8000
k=1
3
 n = 8000  n = 20
62.
Let m be the slope a line

 y  2  m  x  8  my  y  8m  2
x y
 OA  OB  10  8  m  
 2   10  2 8.2  18
  1
8
 2 8  m   2 m
m
63. The given situation in statement 1 is equivalent to find the non negative integral solutions
of the
equation x1  x2  x3  x4  x5  6 . Which is coeff, of x 6 in the expansion of
1  x  x  x3  ....   coeff .of x 6 in 1  x 
2 5 5
5.6 2
= coeff. of x 6 1  5x  x ...........
2!
5.6.7.8.9.10 10! 10
 =  C6
6! 6!4!
 Statement 1 is wrong.
Number of ways of arranging 6A’s and 4B’s in a row
10 10
  C6 which is same as the number of ways the child can buy six icecreams.
6!4!
Statement 2 is true
64. Statement 2 : sum of the squares of first ‘n’ odd natural numbers is not equal to
n
( 4n 2 + 1)
. So statement -2 is false
3
65. 3G  2S  O
66.  x   3   x   6   0
sin 3
67. 4cos2   1  4 1  sin 2    1  3  4sin 2   ............(i)
sin 
So, given expression cane be written as
270 sin 810 sin 2790
 54   
sin 90 sin 270 sin 2430
 By(i)

0
sin 729
=  54   54
sin 90
68. Let z1  x1  iy 1 and z2  x2  iy2 since, given Re  z1 z2   x1 x2  y1 y2  0 So, x1 and x2 are of
opposite sign and y1 and y2 are of opposite sign
    1
x2  4 x2  4
69. Let 3 2  t, 3 2 
t
1
t   10  t 2  10t  1  0  t  5  2 6
t
 
x2  4
 3 2  5  2 6,5  2 6
 x 2  4  2, 2 or x 2  6 ,2  x  2,  6
There are total 4 solutions
70. A______________________=4!
H______________________=4!
M_____________________=4!
T A___________________=3!
T H A M S =1
Total = 4  4! 1 3! 1  103
71. Reflection about the line y=x, changes the point (4,1) to (1,4).
On translation of (1,4) through a distance of 2 units along positive direction of x  axis ,
the point becomes 1  2, 4  , i.e, (3,4)
On rotation about origin through an angle  / 4 the point P Also OP=5= OP1 and
3 4 
cos   ,sin   Now, x  OP1 cos    
5 5 4 
     3 4  1
 5  cos cos   sin sin    5   
 4 4  5 2 5 2  2
       3 4  7  1 7 
y  OP1 sin      5  sin cos   cos sin    5     P1    , 
4   4 4  5 2 5 2  2  2 2
72. Length of chord = 2 32  5  4

4 7 4 13
73. e1 = 1 - = ; e2 = 1 + =
18 3 9 3
 (e1, e2 ) lies on 15x 2 + 3y2 = k  15e12 + 3e 22 = k  k = 16
74. Given a set {1,2,....50}
Possible choices of P are
2,3,5,7,11,13,17,23,29,31,37,41,43 and 47. So , we can calcualte no, of elemetns in R1 as
 2, 2  ,  2, 2  .... 2, 2 
0 1 5
 3,3  ,.....  3,3 

0 3
 5,5  ,.... 5,5 

0 2
 7, 7  ,.... 7, 7 
0 2
11,11  ,..... 11,11 

0 1
Every number of P n should lie in the given set 1, 2,....50

Total number of relations are
1, 2,....50
Similarly for relation R2 ,
 2, 2  ,  2, 2  ..... 47, 47  ,  47, 47 
0 1 0 1

n  R2   2  14  28
Required difference
n  R1   n  R2   36  28  8
1  
2
x x x
75. y  
1  1  1   1  1  1 
                  
x  x  x   x  x  x 
1 1
x2 x /2 x3
  
1  1  1  1  1  1  1  1 
                      
x  x  x  x  x  x  x  x 
1  1  1 
log y  3ln x  ln      ln      ln    
x  x  x 
1 1 1
1 1 3 x x x2 2 2
y    
y x 1   1   1 
          
x  x  x 
 1 1 1 
y  
y1   3  x  x  x 
x 1  1  1 
            
 x  x  x 
y    
y1     
x  1/ x   1/ x   1/ x  y 
76. Let AB - BA = X 3´3 Use Cayley – Hamilton theorem.

X3 - C1X 2 + C2 X - C3I3 = O3 where C1 = tr X = tr (AB - BA) = 0
C3 = det (X ), taking trace of the above equation, we have,
tr (X 3 )
tr (X ) - 3det (X ) = 0 
3
=3
det (X )
2  2 
77.  5  x ,1
x  5 
1 2/ 5 1  2  1 2
  f  x  dx   f  x  dx   f  x  dx  5   0 +  dx
0 0 2/ 5
 5  2/ 5 x
1
 f  x  dx  2  2  ln x 
1
 1/ 5
0
 5
 a  2  2 ln  
 2 
78. y   C1 cos C2  cos x   C5  C1 sin C2  sin x  C3eC e  x 4
y  A cos x  B sin x  Ce x

 C1 , C2 , C3 , C4 are arbitrary constants.
  1,   1,   1,   1 as   1     1     1    1  0
2 2 2 2
79.
 The roots of given equation is equal to 1.
a
 S2  2  6
a0
z2 3 i  /3
80.  e
z2 2
3   
1  ei /3  2  3cos   3i sin
z1  z3 3 3
 2  
3
z1  z2 1  ei /3  
2  3cos  3i sin
2 3 3
2
7 3 3
2
   
2  2  49  27 133
 
1 3 3
2
1  27 7
   
2  2 
81. Since n   20,100
3n  7  3 for n  20 , n  25,......97
82.
83. Given ststem of equations 3x  y  z  0 3x  z  0 And 3x  z  0

Let x  p, where p is an integer, then y  0 and z  3 p
But x 2  y 2  z 2  90  p 2  9 p 2  90
 p 2  9  p  0, 1, 2, 3
i.e. p can take 7 different values,
 Number of points (x,y,z) are 7
84. Let xi  5 = yi
9 9
  yi  9 and y 2
i  45
i 1 i 1
So, required standard deviation is

2
 9 9

 y   yi 2
i  45  9 
2
  i 1   i 1      2
9  9  9 9
 
 
 1  a  2 
 a  x  1 x  1   x 2      x  1
   
85.  lim a  lim a
x
1

1   x  x
1
1   x 

 lim1 a
1    x  1     x   a    
x
a
1   x  
 axeax  e ax ;x  0
86. f  x  
1  2ax  3 x ; x  0
2
Clearly, f 1  x  continuous at x  0
a 2 xeex  2aeax ;x  0
 f 1  x   ; f 1  x  increasing if  ax  2  ae ax  0 and 2a  6 x  0
 2a  6 x ;x  0
3 2  6 2  2  x2  x
 x  2  x  1  x 2  x  dx
3
x  2  x2 3
   = 2
87.  3
1 x 2  3
1  x2
dx
1
 21/6  dx  2 6 x  C
88. f  x   f  z   f  x  z  and f  0   0 and f 1  0  =4  f  x  =4x
So, area bounded =   0  4x  x 2  dx

4
1 1 1 1 1 16
89. D  1 2 3   8    8 D3  1 2 14    36  0    36
2 5  2 5 
90. E1  be the event of both getting the correct answer
E2  both getting wrong answers
E  both obtaining same answer.
1 1 1  1  1  77
p  E1    , P  E2   1   1   
8 12 96  8   12  96

1
E  E  1.
1  E1  96 13
P    1; P    P   
 E1   E2  1001  E  1. 1  1 . 77 14
96 1001 96

SRI CHAITANYA IIT ACADEMY, INDIA 05-01-2024_ Sr.S60_Elite, Target & LIIT-BTs _Jee-Main_GTM-08_KEY &SOL’S

SEC: Sr.Super60_NUCLEUS & ALL_BT JEE-MAIN Date: 05-01-2024
KEY SHEET
PHYSICS
1) 2 2) 4 3) 4 4) 2 5) 1
6) 2 7) 2 8) 2 9) 4 10) 4
11) 1 12) 1 13) 3 14) 3 15) 1
16) 2 17) 4 18) 1 19) 3 20) 1
21) 3 22) 1400 23) 900 24) 15 25) 2
26) 5 27) 12 28) 2 29) 1 30) 7
CHEMISTRY
31) 2 32) 3 33) 4 34) 3 35) 1
36) 1 37) 1 38) 4 39) 3 40) 1
41) 4 42) 1 43) 2 44) 4 45) 3
46) 1 47) 1 48) 1 49) 1 50) 3
51) 4 52) 75 53) 3 54) 4 55) 0
56) 6 57) 3 58) 80 59) 12 60) 4
MATHEMATICS
61) 4 62) 1 63) 1 64) 1 65) 4
66) 3 67) 2 68) 1 69) 2 70) 1
71) 4 72) 1 73) 1 74) 3 75) 2
76) 3 77) 3 78) 4 79) 1 80) 3
81) 2 82) 2 83) 16 84) 2 85) 1
86) 1 87) 17 88) 7 89) 8 90) 16

SOLUTIONS
PHYSICS
1. As no external force is acting on system so, Pi  Pf
0.2 10  10  v  v  0.2m / sec
1 1 2
Loss in K .E.    0.2   102   10  0.2 
2 2
1
  10   0.2  10  0.2  9.8 J
2
2. T l T  K l 
T1 l1  l0 T l T l
  or , T1l2  T2l0  T2l1  T2l0  l0  2 1 1 2
T2 l2  l0 T2  T1
3. We have, from the given figure
F F

q
Q d d Q
2 2
KQq x 2 KQqx
Fnet  2 F cos  Fnet  2
2
. 1 / 2 Fnet  3/ 2
d  d2   2 d2 
x2  2
4 x  4  x  4 
   
dF
For maximum Fnet  net  0
dx
5 / 2 3/ 2
3 d2   d2 
 x    x2   .2 x   x 2   0
2 4   4 
5/ 2
 d2   2 d2  2 d2 2 d2 d
  x2    3 x  x 2
   0  2 x   x  x
 4   4 4 8 2 2
4. Let voltage at C  xV
From kirchhoff’s current law,
i1 C i2
20V 10V
A B
2 i 4
2
V 0

KCL : i1  i2  i
20  x 10  x x  0 V X 10
   x  10  i     5 A
2 4 2 R R 2
5. Magnetic field at centre due to each side
b
a
I
0 I

b
 sin 45  sin 45 
4
2
Field due to all 4 sides
0 I
B
4 b
 sin 45  sin 45  4
2
 
 0 I   I
B  2sin 45  4  2 2 0
b
 4   b
 2 
Flux through small square will be
0 I
  2 2  a2
 b
Coefficient of mutual induction, Mib  a is given as
 2 20 a 2 0 a2
M   8 2
I b 4 b
4T
6. Inside pressure must be greater than outside pressure in bubble. This excess pressure
r
is provided by charge on bubble.
Pa
Pa
4T  2 4T Q2  Q 
  ...  
r 2 0 r 16 2 r 4  2 0  4 r 2 
Q  8 r 2rT  0
7. Let m1  m2  m  cons tan t and m1  x  m2  m  x
x  m  x 2x  m  x  m 2
a
m
gT
m
g T 
2g
 g  a2 
w
8. SI unit of   2 4
mk
SI unit of b  mk
w
 b4  2 4
 m 4 k 4  mL4T 3
mk
9. Mass of the body will not be zero.
10. The current element can exert force on another element.
11. Kinetic energy of gas molecule depends only on absolute temperature.
V
12. i  V  2 fc 
xc
13. With increase in temperature, the number density increases but relaxation time decreases.
14.   tan ic  1.5  tan ic
ic  tan 1 1.5 
15. Here, u  10 2m / s,   45 , v  125m / s
At any instant t, velocity is given by
2 2
v  vx2  v y2   u cos     u sin   gt 
2 2
 125  10 2 1/ 2   10 2  1/ 2  10t 
 125  100  t 2  2t  2   t 2  2t  0.75  0
 t1  0.5s and t2  1.5s  t2  t2  1.0s
10
16. At t  10, v  10 1    10ms 1
 5
Displacement from t = 0 to 10
= Area (from t = 0 to 5) – Area from (t = 5 to 10)= 0.
v  ms 1 
10
10
0 t s
5
10
Since, particle started from origin, its x-coordinate at t = 10 s is zero

17. P 2 /   const.
As  change to  /2 , P changes to P / 2
 (a) is wrong
PV Pm c m cm 1
T    
nR n R n R nR 
As  change to  /2 , T changes to 2T
(b) is correct and (c) is wrong.
nRT nRT nRT P 2
P   
V m m c
 PT  const.
Hence, graph on PT diagram is hyperbola.

 (d) is correct
18. Let S A , SB and SC be the specific heats of liquids
A, B and C respectively.
Then, mS A 15  10   mS B  25  15
 S A  2SB
mS B  30  25   mSC  40  30 
 S B  2 SC
Let the temperature on mixing A and C be T.
Then, mS A T  10   mSC  40  T 
2S B  T  10    S B / 2  40  T 
 4 T  10   40  T T  16 C
19. K  U and K  U  E  U  E / 2
1 2 11 2
kx   kA   kA / 2
2 2 2 
20. Given
R  200, C  15.0 F  15.0  106 F
V  220V , v  50 Hz
a) In order to calculate the current, we need the impedance of the circuit. It is
2
Z  R 2  X C2  R 2   2 vC 
2 2
 200    2  3.14  50 15.0 10 6 F 
2 2
 200    212.3 
 291.67
Therefore, the current in the circuit is
V 220V
I   0.755 A
Z 291.5
21. Velocity at the lowest point of inclined plane
2 gh
V
K2
1 2
R
2
K   K2  1
 2  1&  2 
 R  Ring  R  Solid cylinder 2
VR 2 gh 3 1 3
/   
2 gh 1 2 2 2
VC  1
11 2
x  3
22. Work done = PV
 400  PV  400  nRT  PV  nRT at cons tan t pressure 
Now, Q  nC p T
R  1.4
n T  400   400   1400 J
 1  1 0.4

V V 360
23. f     900 Hz
 L 40
100
f
24. m ; As, m1  m2
f u
f f
  f  15cm
f  10 f  20
v 3
25. Given, 1 
v2 2
Apply conservation of momentum
m1 2
m1v1  m2 v2  
m2 3
Since, Nuclear mass density is constant
1   2
3 1
m1 m2 r  m r  2 3
  1   1  1  
4 3 4 3  r2  m2 r2  3 
 r1  r2
3 3
So, x = 2
26.
0 A 0 A
C 
 1 3d  1 
d  t 1   d  1  
 K 4  5
0 A 0 A
 
3d  4  2d 5 0 A
d   
4 5 5 2d  5c0
2
 128
i  net   8A
27. Rnet 16
V  ir V  8  8  05 V  12V
28.
0i1i2l 2 107  8  5 101
F 
2 d 4 102  20  106
F  2  105 N
29. Slope of kmax and frequency of incident radiation is equal to Planck’s constant.
30. K  3.4  1.6  1019 J
h

2mk
6.64  1034

2  9 1031  3.4  1.6 1019
 7  1010 m

CHEMISTRY
31. Ten d-electrons
W1 1000 W2 1000
32. 0.25   0.25   W1  2W2
62 500 62 250
33.
34. Reactions Reagent used

(A) Hoffmann degradation Br2 / NaOH
(B) Clemenson reduction Zn  Hg / HCl
(C ) Cannizaro reaction conc.KOH / 
(D) Reimer-Tiemann reaction CHCl3 , NaOH / H 3 O 
35. In aqueous medium basic strength is dependent on electron density on nitrogen as well as
solvation of cation formed after accepting H  . After considering all these factors overall
basic strength order is Me2 NH  MeNH 2  Me3 N  NH3
36. Conceptual
37. Bond energy of F2 less than Cl2 due to lone pair
Lone pair repulsions.
Bond energy order Cl2  Br2  F2  I 2
38. Co( NH 3 )5 Cl  Cl2
Oxidation number of Co is +3.
So primary valency is 3.
It is an octahedral complex so secondary valency 6
Or Co-ordination number 6.
39. Blue green colour is due to formation of Cu(BO2 )2

CuSO4   CuO  SO3
CuO  B2 O3  Cu ( BO2 )2
40.
Van der waal and torsional strain. Hence it must be most stable.
41.
1   A 1   B m n
aA  bB  P , Rate    K  A  B 
a t b t
if  A  1 M and  B   1 M , Rate  K
0 0.059 Cu 2 
42. (A) E  E cell  log
2  Zn 2 

0 0.0591  Zn 2 
Ecell  E cell  log
2 Cu 2  
Since
0.059 0.1
 1.10  log  1.13V
2 1
43. (b) For a non-ideal solution with negative deviation,

 Hmix=-ve
 Vmix=-ve
44.
2 C2 H 2 ( g )  5 O2 ( g )  4 CO2 ( g )  2 H 2O( I ) (i )
H 0 / kJ  2600 kJ
C (s )  O2 ( g )  CO2 ( g ) H 0 / kJ  390 kJ (ii )
2 H 2 ( g )  O2 ( g )  2 H 2O ( I ) H 0 / kJ  572 kJ (iii )
 r H for 2C ( s )  H 2 ( g )  C2 H 2 can be calculated by
 r H   (1300)  2(390)   ( 572 / 2)
 r H  234
45. K2Cr2O7 +14HCl→ 2CrCl3+3Cl2+7H2O+2KCl
1 112.5
nCr2O72−= 49  0.96   0.16 wHCl=325×1.15×0.301=112.5 g nHCl=  3.08
294 36.5
Limiting reagent is dichromate.wCl2=31×0.16×71=35.01 g
46. According to probability diagrams only for S orbital, wave function will have maximum
at ZERO
47. From graph, K2=(K1+K3)/2
48. Conceptual
49. Conceptual
50. Good leaving group.
51. Conceptual
52. Conceptual
53. Like K2Cr2O7, this also has d3s hybridization, it has dxy, dyz, dzx. All are non axial.
54. Slater's rule, Z=58, electronic configuration of ion is [Xe] 5f1
55. B = N 2O 4
56. a, c, d, f, g, h are correct
57. statement III, IV, V are true.
58. q  0, U  w , H  U  PV
4  241.25
59. Moles of H    0.01
96500
pH = 2
60. Conceptual

MATHEMATICS
50 n
 X i  Yi  T  n  X i   10, n Yi 

i 1 i 1
61. 50 n
500 5n
So,  X i  500,  Yi  5n    n  30
i 1 i 1 20 6
62.
b b
 f ( x ) dx   f ( a  b  x) dx
a a
2
1 cos t 1 cos 2 t
I1   xf ( x(2  x )) dx  2  f ( x(2  x ))  I1
sin 2 t sin 2 t
1 cos 2 t
I1
 2 I1  2  f ( x(2  x )) dx  2 I1  2 I 2  1
sin 2 t
I2
63.
x
 2 xy   xy 
x  y .e  e y
 dy  y  e  y 2 .e xy  dx
   
   
x
 y 2 ( xdy  ydx )e xy  e y ( ydx  xdy )

x
xy x/ y ydx  xdy x
 e d ( xy )  e 2
 e y .d  
y  y
x
c
 e xy  e y
64. you can have W,W,B or W, B,W or B,W,W
Required probability= 3 . 2 . 3  3 . 2 . 1  1 . 2 . 1 
9

3

1

13
4 4 4 4 4 4 4 4 4 32 32 32 32
65. AB  PA  PB
So PA  PB is maximum when P is collinear with AB.
Equation of AB  y  1  1  x  0   x  2y  2  i
2
& solve with 4x  3 y  9  0   ii 
 24 17 
(i) & (ii) P  , 
 5 5
66. AB  4
M is the midpoint
2 2
AM  2 MA2  CM 2   CA  4  5   CA
CA  3

3
67. Number of arrangement   3! 3! 4!  3!   3! 4!
2
68. Replacing f  x  by x, we f  x   x2  
 f ( f ( f ( f ( x))))  x16 also f  x4   x16
69. Let f  x   ax2  x  c  a f 1  c  1  0 c  1
Given expression is positive for every x  R
  1 a 1
So, f    0    c  a  0  4c  3a  2  0  4c  2  3a .
2 4 2
70. A, B, C lie on the circle with centre  0, 3  and radius 1 unit
3 AO AO
DO  AD 
4 4
AO  BO  CO  1 using phythagoras theorem
7
BDO BD 
4
using phythagoras theorem in BAD
1
AB  Z1  Z2 
2
71.
  
sin 1 x    , 
 2 2
3  1  
   sin x   
4  4 4
2
  9
0   sin 1 x     2 ........(i)
 4  16
Statement II is true
(sin 1 x)3  (cos 1 x )3  a 3
 (sin 1 x  cos 1 x)  (sin 1 x  cos 1 x) 2  3sin 1 x cos 1 x   a 3
2
  3sin 1 x cos 1 x  2a 2
4
2
 1   2 2
  sin x    (8a  1) 
 4 12 16
2
   2
  sin 1 x    (32a  1)
 4 48
Putting this value in equation (i)
 0  32a  1  27
1 7
a
32 8
Statement 1is false

 3
 51 
 51 3x  4 8
72. r   
 4  3x   4  3 
 4
73. A satisfies the equation x 3  5 x 2  6 x  2  0
Since a0 An  a1 An1 ....  an1 A  an I  0 as an  0
So if we multiply both sides by A  1 , we get the inverse of matrix
 2023 2023 2 
n n    2024  4047    1012,1365188
74.  ,     n 1 , n 1   1012,   
 2023 2023   6 
 
f  c  h  f  c  f  c  f  c  h
75. lim
h0 h
f  c  h  f  c  f  c  h  f  c 
lim  lim
h0 h h0 h
f  c  f  c
' '
 2 f '  c. f is differentiable

(2) is false
76. Let the two unknown items be x and y, then
1 2  6  x  y
Mean  4  4
5
x  y  11
And variance = 5.2
12  22  62  x2  y2 2
  mean  5.2
5
2
41 x2  y2  5 5.2   4 
 
41 x2  y2 106
x2  y2  65 ………….(ii)
Solving Equations (i) and (ii) for x and y, we get
x  4, y  7 orx  7, y  4
77. We have,
AB  BC  CA  0  BC  AC  AB
Let M be mid point of BC
 
Now, BM  AC  AB  BM  BC 
2 2  
Also, we have
AC  AB AB  AC   
 AB   AM   4 i  j 4 k  AM  33
2 2
78. Let a point D on BC= (3 , 2,1, 4 )
A 1, 1, 2
B
D
x  2 y 1 z
C  
3 0 4
   
AD  (3  3) i  2 j  (4  2) k
   
 AD  BC , AD . BC  0
 (3  3)  3  2(0)  (4  2)4  0
17
 
25
 1 68 
Hence, D   ,1, 
 25 25 
  1 
2
2  68 
2
| AD |   1  (2)    2 
 25   25 
(25)2  4(25) 2  (18)2 3400 2 34
  
25 25 5
1  
Area of triangle   | BC |  | AD |
2
1 2 34
  5  34  BC  5
2 5
79. Since, ordinate is 3
So, y  3  x  2
Po int of contact is (2,3)
Given equation of parabola ( y  2)2  ( x  1)
Diff .w.r.t.x,
2( y  2) y1  1
1
 y1 
2( y  2)
1 y3 1
 y1 (2,3)     x  2y  4  0
2 x2 2
Equation of tan gent to the given parabola is ,
x  2y  4  0
Re quired area of the bounded region
3
   ( y  2) 2  1  (2 y  4) dy
0
 9 sq. units

80.
A  2t 2 , 4t 
O 30
 0, 0  30
B  2t 2 , 4t 
Let A  (2t 2 , 4t ) and B  (2t 2 , 4t )
For equilateral triangle (AOM  300 )
4t 1 4t
tan 300  2   2 t2 3
2t 3 2t
1
Area  .8(2 3).2.24  192 3.
2
81.
lim an  lim an 1  L
n  n 
lim an 1  lim 2  an
n  n 
 L  2  L  L  2& L  1
L  1, an is a sequence of positive nol numbers.
82.
2
12 x 1|
Given, f ( x ) | x 2  2 x  3 | .e|9 x
2
f ( x ) | ( x  3)( x  1) | .e (3 x  2)
2
 ( x  3)( x  1).e(3 x  2) ; x  (3, )

(3 x  2)2
f ( x )  ( x  3)( x  1).e ; x   1, 3
 2
 ( x  3)( x  1).e(3 x 2) ; x(, 1)

Hence at x  1,3 f ( x)  0
Clearly , non  differentiable at x  1 & x  3.
1 2
83.  2  0,  xi 2  x  0
n

3 2 3 1
84. a.b  b.c  c.a  (a.b  b.c  c.a)  2c.a   2 cos  1 
2 3 2 2
85.
A  B and A  B
A  B   and B  A  
n( AB )  n( A  B )  ( B  A)  n( B  A)  1
86.

The equation system x  cy  bz  0
cx  y  ax  0
bx  ay  z  0
1 c b
have nontrivial solutionif c 1 a  0  0
b a 1
Expand by Ri ,1(1  a 2 )  c(c  ab)  b(ac  b)  0
1  a 2  c 2  abc  abc  b 2  0
a 2  b 2  c 2  2abc  1
87.
Let a be a common root then  3  2a  2  0 and  4  2a 2  1  0
i.e. 4  2a 2  2  0 ..............(i)
4 2
  2a  1  0 ...............(ii )
1
From (i) and (ii) 2  1  0   
2
3
1 1 1 1 17
So,    2a  2  0   a  2  0  a  2    .
2 2 8 8 8
88. Centre is (2,1). Length of major axis=4

Equation of major axis is x=2
Ends of major axis A(2,5), A1 (2,-3)
P(2,6) is outside ellipse
M=PA+A A1 =9
m=PA=1
 M  2m  7
89. If a circle a rectangular hyperbola then arithmetic mean of point of intersections is the
midpoint of centre of hyperbola and circle, g  f  8
90. Given,
1 1
tan15  
 
 tan195  2a ----------(i)
tan 75 tan105
We know that,
tan15  2  3
1
 cot 75  2  3
tan 75
1

 cot105   cot 75  3  2
tan105
tan195  tan15  2  3


Time: 03.00Pm to 06.00Pm GTM-06 Max. Marks: 300
KEY SHEET
PHYSICS
1) 3 2) 2 3) 3 4) 1 5) 1
6) 1 7) 2 8) 2 9) 4 10) 1
11) 3 12) 2 13) 3 14) 2 15) 2
16) 3 17) 4 18) 2 19) 2 20) 2
21) 6 22) 161 23) 34 24) 3 25) 48
26) 4 27) 70 28) 12 29) 5 30) 6
CHEMISTRY
31) 2 32) 4 33) 3 34) 3 35) 4
36) 3 37) 1 38) 2 39) 2 40) 2
41) 2 42) 3 43) 4 44) 3 45) 2
46) 4 47) 2 48) 2 49) 1 50) 4
51) 282 52) 40 53) 1 54) 3 55) 2
56) 1000 57) 718 58) 548 59) 54 60) 1
MATHEMATICS
61) 1 62) 2 63) 3 64) 4 65) 4
66) 3 67) 4 68) 2 69) 3 70) 2
71) 3 72) 4 73) 3 74) 3 75) 2
76) 1 77) 4 78) 4 79) 2 80) 1
81) 384 82) 27 83) 5 84) 432 85) 13
86) 405 87) 36 88) 14 89) 8 90) 122

SOLUTIONS
PHYSICS
1. 3
Sol: XL 10 ; XC  4 ; R  6
R
Power factor cos  
Z
X L  Xc
Z  R2   X L  Xc   36  10  4
2 2
 36  6  6 2
R 6 1
cos   
Z 6 2 2
2. 2
V0
Sol: From the graph V  x  V0
X0
Differentiating w.r.t, fine
dV  V0   dx   V0 
a     V 
dt  x0   dt   x0 
2
 V   V   V0  V0 2
 a   0   0 x  V0   a    x 
 x0   x 0   x0  x
So, the graph is a st.line with negative intercept
3. 3
3RT 8RT Vrms 3
Sol: Vrms  and Vavg  
M M Vavg 8
4. 1
mv 2m  KE   P2 
Sol: Radius of a particle in the magnetic field is r   KE 
Bq Bq  2 x 
2 1 kp 2 Kp 4
rp ; r  2:1 then rp
 
m p k p q
,    
r 1 m k  q p 4 k 1 K 1
5. 2
mv2 k
Sol: F 
du k
 2   2
dr r r r
1
 v parabola
r
6. 1
Sol: Since all element of wire are same distance from axis, moment of Inertia is I  Mr 2

2
ML2
is radius of semicircle I  M    I  2
L L
Where r
   
7. 2
Sol: Magnetic energy in a solenoid is given by
1 2  
Rt

U  LI where I  I 0  1  e L
 in R – L circuit
2  
2 2
1 2  Rt
   Rt
  1 2
U  LI0 1 e L  U  U0 1  e L  U0  LI0 
2      2 
2  Rt
25U0   RT
 1
 U0 1  e L   1 e L
100   2
Rt L
Applying logarithm on both sides ln 2  or t  ln 2
L R
8. 2
Sol: Before collision, u1  u;u2  0 , A her collision v1 v;v2  v
If the collision is elastic, then we can write u1 u2  v2 v1 u  2v ___(1)
Applying principle of conservation of momentum m1u   m2  m1  v ____  2
m2
From (1) & (2) we get 3
m1
9. 4
Sol: Equation of adiabatic process is PV   constant
dp  dV
PV 1dv V dp  0 Or  is take firmed charge in pressure
p V
10. 1  
Sol: EM wave is propagating in y-direction. E and B vectors are perpendicular to direction
of propagation of wave. Hence neither of them have a component along y-axis. Both
vectors have to be in x-z plane. Ex , Bz or Ez , Bx is the correct answer
11. 3
Sol: Assuming SAB a triangle
B
r
dA V(dt)

A
r
SUN
1
dA  Vdt  r
2
L dA L
Areal velocity momentum L  mvr  vr   
m dt 2m
12. 2 and 3
 
Sol: y  3cos   2t   y  3cos  2 t   / 4 It is in SHM
4 
2 
T  
2 
Or
y  sin2 x 1 1
  cos 2  wt 
2 2
13. Key : 3
Sol : E  V  B
E  3108  2 108 kˆ
E  6 ˆj
As E & B remains perpendicular
14. Key: 2
Sol: T  2 l / g
2 T l g
 100   100   100
T l g
So error in calculation of g in error in T = 2% & error in length = 1%
So total = 3%
15. 2
Sol:
mg 60 f
600
Translational Equilibrium
mg sin   T  f ____(1)
Rotational Equilibrium (about CM)
T  R  f  R  T  f ____(2)
mg sin 60 3
f   mg  0.433mg
2 4
But maximum frictional fore is
fmax  mg cos60 = 0.2 m g
Applied force in greater than maximum static frictional force so friction = 0.2 mg
1
 mg
5
16. Key : 3
dv
Sol : P  maV p  m v
dt
dv
P  ma v pdt  mdvv
. .
dx
P dx  m v2.dv p dt  m v.dv

3 2
v mv
P.x  m pt 
3 2
xv3 t  v2 x  t 3/2
17. Key : 4
Sol :
vr
vs 30 120
vr 1
sin 300    10  vr  5
vs 2
Take angle with river vertical side as 300
18. Key : 2
Sol :
l
R v  0.1v
A
l  0.1cm
I  0.01A
d  0.01mm
 R l r
 100   100   100  2  100
 R l r
19. 2
Sol: Floating means weightlessness
g1  g  R2 cos2  (   00 at aquifer)
 given g 1  0  g  r 2    g  T  2 R  84 min
R g
20. Key : 2
Sol : Momentum conservation L  I 
Mr2  Mr2' 2mr2'
21. Key : 6
2
Sol: Time taken to travel from half amplitude is T / 12 so  1/ 6
12
22. Key : 161
0 A
Sol : C 
d t t / k

12
8.85  10  100
C  161 pf
5
10  5 
10
    6  10  10 m
23. 34
4
Sol: Volume of sphere, V   R3
3
V 3 r V 3 r V
  100   100  100  34
V r V r V
24. 3
Sol: Applying Bernouli’s theorem, at ‘A’ and ‘B’
1 mg 1
PA   gh  PB  v2 P0    gh  P0   v 2
2 A 2
Simplifying, we get v  3m / s
25. 48
12 x
Sol: Applying Wheatstone bridge principle, We can write 
6 72  x
Simplifying, we get x  48 cm
26. 4
Sol: If length of each resistance is ‘ ’ then R1  6 and R2 3 . Equivalent resistance in parallel
combination is, R
R1 R2
R
 6  3   2
R1  R2  6    3 
So, effective resistivity is 2 cm
27. Key : 70
Sol :
50  20 170
Here Req  10  
70 7
So current through 10 resistance  7A
 vR  70v
28. 12
Sol: Let q1 1C and q2 1C
Force on 1C charge due to infinite charges can be written as
 
 1 1 1   1  4
F  kq1q2 1    2  .... F  kq1q2     kq1q2 
  2   4   8 1
2 2
 1   3
 4
4
F   9  10 9  1 1  10 6   F  12  10 3 N
3
29. 5
Sol: Between ‘A’ and ‘B’, horizontal component of momentum remains same. Vertical
0
component of momentum charges from mu sin45 to  mu sin 45 0
So, charge in momentum is
p  2mu sin 450   2 5 103  5 2   12 p  5102 Kg  m / s

30. 6
Sol: Let compression in the spring be ‘e’. K.E. of ball gets converted into spring P.E. at its
maximum compression
1 2 1 2 1 1
 4 10   10  e 2  e  2m
2
mv  ke 
2 2 2 2
Length, l  x  e  x  l  e x  8  2  6m
CHEMISTRY
31. Key : 2
Sol :
Br
PPh3
Ni
Ph3 P Br
Trans – [ NiBr2 (PPh3 )2 ]
NO2
NH 2
NO2
CO
NO2 NH 3
NH 3
Meridonial - [CO(NH3 )3 (NO2 )3
32. (4)
Sol
It is most basic because there is no amine inversion.
33. 3)
Sol
34. (3)
Sol
In NaCN; carbon is more nucleophilic atom. Whereas in AgCN; Ag – C has covalent bond.
35. (4)
Sol
36. (3)
Sol
37. (1)
Sol
38. (2)
Sol
39. Key : 2
Sol : no of Angular nodes = l  0
no of radial nodes = n  l  1  n  0  1  2
n3
So orbital is 3S
40. (2)
Sol 3H C l  3H 
41. (2)
Sol : A) BCl3  Even Electron molecule B) NO  Odd Electron molecule

SF6  Expanded octet molecule H2SO4  Expanded octet.
C) SF6  Even Electron molecule D) BCl3  Even Electron molecule
H2SO4  Expanded octet. NO  Odd Electron molecule
S  12 e  in outer orbit.
42. (3)
Sol : N2  g   3H 2  g   2 NH3  g 
W2  20g 5g
20 5
n
28 2
Stoichiometric Amount:
20 / 28 20 5/ 2 5
N2   H2  
1 28 3 6
N2 is the Limiting Reagent.
20
n  NH3   2  n  N2   2 
28
=1.42
43. Key: 4
Sol: Na2 S   Na2 Fe  CN 5 NO   Na4  Fe  CN 5 NOS 
Sodium nitro prusside Violet coloured complex
When Na2 S reacts with sodium nitro prusside forming violet coloured complex
44. (3)
Sol IE :N a < A 1 < M g < S i
Option (C), matches the condition.
i.e IE  AI   577 KJmol 1
45. Key: 2
Sol:
C C
 NaI

( i ) NaOH
( ii ) dil HNO3

CH 2  I CH 2  OH
AgNO3
AgI
yellow ppt

46. Key:4
H H
Sol: P range of strong acid weak base reaction is form 4 to 7 and methyl range P range is
H
3.2 to 4.5 hence in strong acid weak base reaction methyl orange can be used. P range of
H
weak acid strong base reaction is form 7 to 11 and P range of phenolphthalein is 8.7 to
10.5. Hence phenolphthalein can be used. Statement I is correct II is false.
47. Key: 2
Sol: The velocity of electron is an orbit can be calculated by using the formula.
2.18106  z
v m / sec
n
 ‘v’ is directly proportional to ‘z’ and is inversely proportional to
‘n’
‘n’  principal quantum number.
 statement I false statement II true.
48. (2)
Sol
49. (1)
3 2 4 2
Sol A) Sc 0 , Zn 10 B) Ti 0 , Cu 9
3d 3d 3d 3d
2 3 2 2
C) V 3, Ti D) Zn 10 , Mn 5
3d 3d 1 3d 3d
50. (4)
7 4
 2 2 
Sol 8 MnO4  3S2O3  H2O  8 MnO2  6SO4  2OH
Change in oxidation state of Mn is from +7 to +4 which is 3.
51. 282
Key K sp  S 2
S  Ksp  8 1028  2 2 1014
 2 .8 2  10  1 4
 282  10  16
=282
52. 40
Sol r  K  x   y 0   K  x 
Using I & II
3
4  10  L 
3    L  0.2
2  10  0.1 
Using I & III
M 103 0.4 M 8
 M 8   40
2 103 0.1 L 0.2
53. 1
Sol In Tetrapeptide,
No. of Amino Acids = 4
No. of Peptide bonds = 3
Ans. =4-3 .
54. 3
Br Br
| |
Sol CH3  C  CH  2Br2  CH3  C  CH
| |
Br Br
1 1
   360  0.27  0.30375
160 2
 3.0375  10  1
55. 2
3
Sol Fe  CN6 
C N  is strong field ligand
Fe  3 3 d 5  t 25 g e g0 
CFSE  5  0.4 0   2.00

56. 1000

Sol:  129m1
A
KCl solution 1:
74.5 ppm, R1 100
KCl solution 2:
149ppm, R2  50
 W1 
ppm1 m1  Mo V 
Here,   
ppm2 M 2  V W2 
 m0 

1000
k1 
1 M1 k M 50 1
  1 2  2   1, 000  10 3
 2 k  1000 k2 M 1 100 2
2
M2
57. Key: 718


 BE   EA
1
Sol: U  H f   SE  IE 
 2 
 1 
 436.7   89.2  419.0   243.0   348.6   717.8 kj / mol
 2 
58. 548
Sol: Heisenberg’s uncertainty principle
h h
x  p x   2a0  mvx   minimum 
4 4
h 1 1 6.63  10 34
 vx    
4 2 a 0 m 4  3.14  2  52.9  10 12  9.1  10 31
 548273ms1  548.273km s 1  548kms 1
59. 54
HNO3 NaOH
Sol: 600 mL  0.2 M 400 mL  0.1M
 120 m mol  40 m mol
HNO3  NaOH  NaNO3  H2O
bef . 120 40
aft 80 0 40m mol
 r H  40mmol   57  103 
J
mol
J
 40  103 mol  57  103  2280J
mol
m ST  2280
2280 22800
T   10 3   103  542.86 103
4.2 42
T  54.286 102 K T  54.2861020C
T  54.286
60. 1
Sol P  KH  x
mol of O2
0.920  46.82  103 bar
bar mol of H 2O
mol of O2
0.920  46.82  103 
1000
18
0.920  46.82  nO2 18
0.920
  nO2
46.82 18
 1.09  10 3  nO2 mmolof O2 1

MATHEMATICS
61. 1
3 3
3 3
4
48 1  2x  4
Sol:-  3  (2x)
2 2
dx  48  sin 1  
2  3 3 2
3 2
4
4
  2 3 3 
1 2 3 2

 24  sin 1     sin 
 3 4 
 3 4   
  3  1 
 24  sin 1    sin 1   
  2  2 
    
 24     24    2
 3 4  12 
62. Key: 2
cos 1 1  x 2  sin 1 1  x   cos 1 1  x 2 
Sol : RHL  lim  lim
x 0 x 1  x 2  2 x 0 x
 1
 lim  2 x  (L’Hospital Rule)
2 x 0 1  1  x 
2 2
x 1 
  lim   lim 
x 0
2 x2  x4 x 0
2  x2 2
LHL  lim
 2

cos 1 1  1  x  sin 1   x 


lim
sin 1 x


lim
sin 1 x
1  x   1  x  1  x  1  x   1 2 x0 x  2 x
3 2 2
x 0 2 x 0 
 1 
    As LHL  RHL so f  x  is not continuous at x  0 .
22 4
Therefore no such  exists
63. 3
Sol:- f (x)  a satisfies f (x )  f ( y)  f (x )  f (y)
x
f(1)  a1  3  a f(x)  ax  3x
n
 f (x)  3279
k 1
3  3 2  3 3  ......  3 n  3279
3(3n 1)
 3279 3 n  1  2186  3 n  2187  3 7  n  7
2
64. 4
1 0 0 0 0
0 1 0 0 0
 
Sol:- A  0 0 1 0 0
 
0 0 0 1 0
 0 0 0 0 1 
Number of such matrices = 5! = 120
65. 4
 1
2
  1
Sol:- 3  x    2   2  x   5 0
 x  x
1 3(y2  2)  2y  5  0
Put x  y
x
3y2  2y 1  0 3y2  2y 1  2 (3y 1)2  2
 1
3y 1   2 3 x    1  2  0
 x
3(x2  1)  x(1 2)  0 3x2  (1  2)x  3  0
3x2  (1 2)x  3  0 3x2  ( 2 1)x  3  0
D1  (1  2)2 4(3)(3)  0 , D2  ( 2  1)2  4(3)(3)  0
No, of real roots = 0
66. 3
22x
Sol:- f(x)  ,x R
22x  2
4x 4x 41x
f(x)  x f (x)  f (1 x)  x  1x
4 2 4 2 4 2
4x
4 4 2
x
 x   x 1  f (x )  f (1  x )  1
4  2 4  2 4 4  2
x
Now
 1   2   2022 
f f  ...  f 
 2023  2023  2023
 1   2022   2   2021  1011   1012 
f  f   f   f    ...  f  f
 2023  2023  2023  2023  2023  2023
= 1 + 1 + 1 + .... 1011 times = 1011
67. 4
Sol:- Lt ([x  5]  [2x  2])  0
xa
 [a  5]  [2a  2]  0
 [a  5]  [2a  2]
Here a  [  7.5,  6.5)
68. 2
Sol: Total number of 6 digited numbers from0, 1, 2, 3, 4, 5, 6
= 6  6  5  4  3  2 n  S   6  6! 0  1  2  3  4  5  6  21
Therefore 6 digited number divisible by 3 can be formed by using the digits
1,2,3,4,5,6 in 6! ways by using the digits 0,1,2,4,5,6 in 5  5! ways and
6! 2  5  5! 6  10 4
by using the digits 0,1,2,3,4,5 in 5  5! ways P  E    
6  6! 36 9
69. 3
Sol:- Given a1,a2 ,a3,a4 ,a5 ,a6 are in A.P
Let T1  a1  a
Common difference =d
Also given a1  a3  10
 a  a  2d  10  2a  2d  10.........(1)
Mean of six numbers is 19
2
a 1  a 2  a 3  a 4  a 5  a 6 19
 
6 2
6  19
 a  (a  d)  (a  2d)  (a  3d)  (a  4d)  (a  5d) 
2
 6a  15d  3  19
 2a  5d  19.........(2)
Eq. (1) & (2) 3d = 9  d  3
Also 2a  2(3)  10  2a  4  a  2
 six numbers are 2, 5, 8, 11, 14, 17
Wkt variance   2

 xi 2
 2
n
2
2 2  52  82  112  14 2  17 2  19 
  
6  2
2
4  25  64  121  196  289  19 
  
6  2
699 361 1398  1083 315 105
    
6 4 12 12 4
105 105
2   8 2  8   2  105  210
4 4
70.
Key: 2
x  1 x  1   x  1 2
f  x   3  
Sol: x 1  x  1
2
 x  1
2
2  3 1  2
     4x  2
 x  1  x  1 x  1   x  12  x  1
 1  
f ' ( x )  0 x   ,  1    ,    1 
2  
71. 3
Sol:- Let A  (0, y1 )
y1  4........(1)
And (1)y1  ......(2)
From (1) & (2)
y1  2, A = (0, 2),   2
AB  3x  2y  4, AC  2x  y  2  0
Let C   h,2h  2
 3  2h  1  1 
m1  m2 1     1 h   C    ,1
 2   h 1 2  2 
a  1  m  1  3  m = 1, -3
Equation of tangent is y  mx   
m 2 2m
 1 
C   ,1
2 
 a 2a   3 
P   2 ,    ,3
m m   2 
CP  2 2
72. 4  
 
Sol:- Given 1 parallel to  1  t
   
2 perpendicular to  2   0
  
  1 2

Now
 do
 dot product
  with  on both sides
   
   1   2 (2   0)
   
(4iˆ  3jˆ  5k)(i
ˆ ˆ  2jˆ  4k)
ˆ  (t)   0 (1  t)
1
4  6  20  t(16  9  25)  0  10  t(50)  t 
5
 1 ˆ ˆ
1  ˆ
(4i  3 j  5k)
5
  
From   1 2
    1 
2    1  (i  2j  4k)   (4i  3j  5k)
 5 
 1
 
  (9iˆ  13ˆj  15k)ˆ 5  (iˆ  ˆj  k)
2
ˆ  9  13 15 52  (iˆ  ˆj  k)
ˆ 7
2
5
73. 3
Sol:- Given digits are 3, 5, 6, 7, 8
i) Four digits numbers starts with 7, 8
7, 8
2  4 P3  48
ii) Five digit numbers
= 5P4
= 120
= 120 + 48 = 168
74. 3
Sol:- x  2 y  3z  3
4  3y  4z  4
8x  4 y   z  9   has infinitely solution ( ,  ) 

1 2 3
72
  4 3 4  0  5  72  0   
5
8 4 
1 2 3
21
3  4 3 4  0  5  21  0   
5
8 4 9
75. 2
Sol:- Let f(x)  x  bx  cx  d
3 2
f 1(x)  3x2  2bx  c

f 11(x)  6x  2b
f111(x)  6
Given, f(x)  x  x f (x)  xf (2)  f (3)
3 2 1 11 111
x3  bx2  cx  d  x3  x2 (3  2b  c)  x(12  2b)  6

By comparing coefficients
- b = 3 + 2b + c c = 12 + 2b
-3 = 3b + c ......(1) 2b – c = -12 .......(2)
Eq. (1) & (2)
3b  c  3
2b  c  12
----------------
5b = -15
b = -3, c = 6, d = -6
f(x)  x3  3x2  6x  6
From option 2
2f(0) – f(1) + f(3) = f(2)
2(-6) – (-2) + 12 = 2
76. 1
3
 2 2 
1  sin  i cos
 9 9 
Sol:- E  
2 2 
 1  sin  i cos 
9 9 
3
   2    2     
1  cos  2  9   i sin  2  9  cos  4  9  
 
1  cos    2   i2sin      cos      
      
4 9 4 9 4 9  
           
 2 cos  4  9  cos  4  9   i sin  4  9   
  
           
 2 cos  4  9  cos  4  9   i sin  4  9   
  
   3          
  cis 3      3     
   4 9     4 9  

5 5
 cis  i sin
6 6
1
  ( 3  i)
2
77. 4
Sol:- 1   ,  2  2  , P(x1,y1) be the mid point of the chord AB
3 3
(x  4) 2  (y  5) 2  3
Then  cos 
2 6 2
 r1  3
Similarly r3  1

Given r12  r22  r32  r2  2  
4
78. 4
Put y = vx and variables separable then 6[y(e)]  2e
2 2
Sol:-
79. 2
Sol:- | A| 2 3
80. Key: 1
Sol: g  x  1  n  f  x  1   n  xf  x    nx  n  f  x    nx  g ( x)
1 1
 g  x  1  g  x   nx g   x  1  g   x    g   x  1  g   x    2 …..(i)
x x
1
Putting x  1,2,3,4 in (i), we get  g   2   g  1  1 g   3  g   2   
4
1 1
g   4   g   3   g   5  g   4   
9 16
1 1 1 205 205
Adding. g   5  g  1  1      g   5   g  1 
4 9 16 144 144
81. 384
Sol:- Given lines are
x  6 y  6 3 6 x   y  2 6 3 2 6
  and  
2 3 4 3 4 5
 
a   6i  6 j  6k c   i  2 6 j  2 6k
 
b  2i  3j  4k d  3i  4 j  5k
 
a  c  ( 6  )i  6 j  3 6k
 6   6 3 6
   
[a  c b d]  2 3 4  ( 6 )(15 16)  6(10 12)  3 6(8  9)
3 4 5
 6  2 6  3 6   4 6
i j k
b  d  2 3 4  i (15  16)  j(10  12)  k(8  9)   i  2 j  k | b  d | 1 4 1  6
3 4 5
| [a  c b d] |
Shortest distance is ‘6’  6 |   4 6 | 6 6
|bd|
  4 6  6 6   10 6,   2 6
82. 27
x
Sol:- f (x)   f (t) 1  (log e f (t)) 2 dt  e
0
By Leibnitz condition
f 1 (x)  f (x) 1  (log e f (x)) 2 (1)  0  0 f 1 (x)  f (x) 1  (log e f (x)) 2  0
1 f 1 (x)
f 1 (x )   f (x ) 1  (log e f (x )) 2  1
1  (log e f (x)) 2 f (x)
Integrate on both sides
1 f 1 (x)
 1  (loge f (x))2 f (x)
dx   dx loge f(x)  t
f 1 (x) 1
Differentiate on both sides w.r.t to x
f (x)
dx  dt  1 t2
dt   dx
sin1(t)  x  c sin1(loge f(x))  x  c..........(1)


Put x = 0  sin1 (loge f (0))  0  C  f (0)  e sin 1 (1)  c  c 
2
From (1)
 log f (x)  sin    x loge f(x)  cosx
sin 1 (log e f (x))   x   
2 
e
2
    3
Put x log e f    cos 
6  6 6 2
   3
6loge f    6    3 3
 6  2
2
  
S.O.B  6 log e f     (3 3 )  27
2
83. 5
13  23  33  ..... up to n terms 9
Sol:- 
1.3  2.5  3.5  .... up to n terms 5
n2 (n  1)2
13  23  33  ...... up to n terms 
4
n(n  1)(2n  1) n(n  1)
1.3  2.5  3.5  .....up to n terms  
3 2
t n  n(2n 1)  2n2  n
2(n(n  1)(2n1) n(n  1)
Sn   n   n 2   n  
6 2

n (n  1)
2 2
n(n  1) n2  n
4 9 4 9 42 9
   
n(n  1)(2n  1) n(n  1) 5 2n  1 1 5 4n  2  3 5
 
3 2 3 2 6
n2  n 3 9
  5n 2  5n  24n  30
2 4n  5 5
5n 2  19n  30  0 5n 2  25n  6n  30  0
5n (n  5)  6(n  5)  0
(5n  6)(n  5)  0  n  5
84. 432
Sol:- Urn Red Black
A 4 6
B 5 5
C  4
Let E be the event to set red ball from ‘C’
P(C)  P(E / C)
P(C / E) 
P(A)P(E / A)  P(B)P(E / B)  P(C)P(E / C)
1 

Given 0.4  3 4
1 4 5  
   
3 10 10   4 
49   
   
10 10   4   4
76  144  100  24  144    6
Given parabola y  x y  6x
2 2
B
a
600 a/2
300
A D
0
a/2
a 60
C
BD a
sin 30 0   BD 
a 2
AD 3
cos300   AD  a
a 2
 3 a
B   a,  lies on parabola y2  6x
 2 2
a2  3a 
 6   a  12 3
4  2 
a 2  144  3  a 2  432
85. 13
Sol:- Let A  {a , b, c, d}
Given R  {(a, b), (b, c), (b, d)}
We can made R as equivalence
We have to add order pairs (a,a), (b,b), (c,c), (d,d), (b,a), (c, b), (d, b), (a, c), (c, a), (c, d),
(d, c)
(a, d), (d, a)
Then R = {(a,a), (b,b), (c,c), (d,d), (b,a), (c, b), (d, b), (a, c), (c, a), (c, d), (d, c), (a, d),
(d, a)}
Then R is equivalence
Minimum number of ordered pairs = 13
86. 405
n
Sol:- Given binomial expansion is  x  2 

3
 x 
Sum of the coefficient of first 3 terms
n (n  1)
n
C0  nC1(3)  nC2  32  372  1  3n   9  372
1 2
 9n 2  15n  750  0
 (n  10)(9n  75)  0  n  10
2
Now T21  10 C 2 x 8  
3
2
 10 C 2  9  x 4
 x 
 3
 Coefficient x 4 in  x  2   10C2  9  405
 x 
87. 36
Sol:- Given curves are

y2  2y  x.............(1)
And y = -x .........(2)
From (1) & (2)  y  2y  y
2
 y2  3y  0  y  0,y  3
Point of intersections are A (0, 0) B (-3, 3)
3
Required area A   [2y  y 2  (y)]dy
0
y
y=3
(-3,3)
x
(0,0) y=0
88. Key :14

b 2 1 1 1
Sol : a 2  and   6 Solving, we get a  or a  [rejected]
16 b a 12 4

1 1
 b   72  a  b   72     14
1 1
If a 
12 9  12 9 
89. 8

a  î  2jˆ  kˆ ,  c  t(  2i  7 j  2  k )
Sol:- ac  bc  ac  7
 (a  b)  c  0  t(2 14  22 )  7
a  b,c are parallel t(22 12)  7

 c  t(a  b)  c  t(  2i  7 j  2  k ) , b  3iˆ  5jˆ  k,
43 43
 bc   t(  6  35  2  2 )  
2 2
 7  43
 2  (41  2 )  
2
(287 142 )  43(2  6)
2  12 2
287  14  2  43  2  258  29  2  29  2  1   1
Now | a  b||310  | | 7  1 | 8
2
90. 122
Sol:- 2x + y = 0 ------(1)
x – y = 0 –----(2)
x + py = 21a -------(3)
solving (1) & (2)  A (1, -2)
A(1,-2)
=3
0=
x- y
+y
2x
G (2,a)
BH2
(s, -2s) x+py = 21a C (t+3, t)
centriod of triangle ABC is
 4  s  t 2  2s  t 
 ,   (2,a)  s  t  2............(4)
3 3
-2s + t = 3a + 2..............(5)
Solving (4) 7 95) we get  s   a, t  2  a
B(  a , 2a ); C(a  5, a  2)
 B, C lies on x + py = 21a
  a  2ap  21a  P  11
and pa  2p  20a  5
If p = 11  27  9a  a  3
 B(  3, 6), C(8, 5) , (BC)2  (8  3)2  (5  6)2
Distance (BC)  122
2


SEC: Sr.Super60_Elite, Target & LIIT‐BTs JEE-MAIN Date: 29-12-2023
only.
.
SRI CHAITANYA IIT ACADEMY, INDIA 29‐12‐23_ Sr.Super60_Elite, Target & LIIT‐BTs _Jee‐Main_GTM‐05_Q.P
Admission Number:
29‐12‐23_Sr.Super60_Elite, Target & LIIT‐BTs_Jee‐Main_GTM‐05_Test Syllabus

PHYSICS : Theory Based
CHEMISTRY : Theory Based
MATHEMATICS : Theory Based
Sec: Sr.Super60_Elite, Target & LIIT‐BTs Page 2

correct.
1. A rod is supported horizontally by means of two strings of equal length as shown in figure.
If one of the string is cut. Then tension in other string at the same instant will:
Mg
1) remain unaffected 2) increase

3) decreases 4) become equal to weight of the rod
2. Curie temperature is the temperature above which:
1) a ferromagnetic material becomes paramagnetic.
2) a paramagnetic material becomes diamagnetic.
3) a ferromagnetic material becomes diamagnetic.
4) a paramagnetic material becomes ferromagnetic.
3.  2 2

Potential is varying with x and y as V  2 x  y . The corresponding field pattern is:
y y
x x
1) 2)
y y
x x
3) 4)
4. Two electrons are moving with the same speed V. One electron enters a region of uniform
electric field while the other enters a region of uniform magnetic field. After sometime if the
de Broglie wavelength of the two are 1 and 2 , then (select the best alternative)
1) 1  2 2) 1  2
3) 1  2 or 1  2 4) 1  2 or 1  2 or 1  2
5. A rod PQ is connected to the capacitor plates. The rod is placed in a magnetic field (B)
directed downward perpendicular to the plane of the paper. If the rod is pulled out of
magnetic field with velocity v as shown in figure.
 B   
   

P
M
    
V
N
 Q    
1) plate M will be positively charged
2) plate N will be positively charged
3) both plates will be similarly charged
4) no charge will be collected on plates
6. In hydrogen like atoms the ratio of difference of energies E4 n  E2 n and E2 n  En varies with
atomic number z and principle quantum number n as
z2 z4 z
1) 2 2) 4 3) 4) none of these
n n n
7. Consider two identical iron spheres A and B, A lies on a thermally insulating plate, whilst B
hangs from an insulating thread as shown. Equal amounts of heat are given to the two
spheres. Then

1) Temperature of A is greater than that of B

2) Temperature of B is greater than that of A
3) A and B will have equal temperature
4) None of the above
8. A stream of particles   particles encounters a stationary unknown nucleus in a cloud
chamber. After a collision, the two nuclei are scattered in mutually perpendicular directions.
If some kinetic energy is lost in the collision, the unknown nucleus must be
1) A proton 2) a nucleus lighter than an   particle
3) an   particle 4) a nucleus heavier than an   particle
9. Assertion(A): A capacitor is connected to an ideal battery. A dielectric slab is slowly
inserted into the gap between the plates of capacitor. Work required in the process is
positive.
Reason (R): Force due to capacitor will try to pull dielectric into the capacitor. To slowly
insert external force needs to be outward i.e. opposite to displacement.
1) Assertion is false but Reason is true.
2) Both Assertion and Reason are true but Reason is NOT the correct explanation of
Assertion
3) Both Assertion and Reason are true and Reason is the correct explanation of Assertion
4) Assertion is true but Reason is false.
10. Two concentric conducting shells can be connected by a switch ‘s’ as shown. Inner shell has
charge Q.
Statement-1: Charge on inner sphere flows from inner shell to outer shell when switch is
closed.
Statement-2: Charge will not flow if Q is negative as charge can not flow from low to high
potential.
'S'
1) Statement-1 is incorrect and statement-II is correct

2) Statement-1 is correct and Statement-II is incorrect
3) Both Statement-1 Statement-II are incorrect
4) Both Statement-1 and Statement-II are correct
11. Two concentric conducting spheres are as shown. A point charge (Q) is at common center.
Inner sphere is a hollow shell having charge 2Q and outer sphere is a grounded thick hollow
sphere. Which of the following is correct graph of potential as a function of distance from
centre?
2Q
R Q
3R
2R
V V
r r
R 3R R 3R
1) 2)
V
V
r r
R 2R R 2R 3R
3) 4)
12. Consider the following four statements.
A. A paramagnetic substance displays greater magnetization when cooled.
B. Diamagnetism is independent of temperature.
C. If a solenoid uses Iron for its core instead of air then field will be greater.
D. Magnetic field lines are closed loops
Which statements are true?
1) Only A and B 2) A, B and C 3) A, B and D 4) All four
13. ‘Parsec’ is the unit of-
1) Time 2) Distance 3) Frequency 4) Angular acceleration
14. Fusion reaction takes place at high temperature because
1) atoms get ionized at high temperature
2) kinetic energy is high enough to overcome the coulomb repulsion between nuclei
3) molecules break up at high temperature
4) nuclei break up at high temperature
15. The equation has Statement-I and Statement-2. Of the four choices given after the
statements, choose the one that best describes the two statements.
Statement-1: Very large size telescopes are reflecting telescopes.
Statement-2: It is easier to provide mechanical support to large size mirrors than large size
lenses
1) Statement-1 is true and Statement-2 is false
2) Statement-1 is false and Statement-2 is true
3) Statement-1 and Statement-2 are true and Statement-2 is correct explanation for
Statement-1
4) Statement-1 and Statement-2 are true and Statement-2 is not the correct explanation for
Statement-1

16. Assertion: Earth has an atmosphere but the moon does not.
Reason: Moon is small in comparison to earth
1) Assertion and Reason are true but Reason is not correct explanation for Assertion
2) Assertion and Reason are false
3) Assertion is true and Reason is false
4) Assertion and reason are true but Reason is correct explanation for Assertion
17. Assertion: Two conducting rods of same material and same lengths are joined end to end as
shown in figure. Heat current H is flowing through them as shown. Temperature difference
across rod-1 is more than the temperature difference across rod-2.
H
1 2
Reason: Thermal resistance of rod-1 is less compared to rod-2.

1) If both Assertion and Reason are true and Reason is the correct explanation of Assertion.
2) If both Assertion and Reason are true but Reason is not the correct explanation of
Assertion
3) If Assertion is true but Reason is false
4) If both Assertion and Reason are false
18. Assertion: If a wire is stretched, only half of the work done in stretching the wire remains
stored as elastic potential energy.
1
Reason: Potential energy stored in the wire is  stress    strain 
2
1) If both Assertion and Reason are true and Reason is the correct explanation of Assertion.
2) If both Assertion and Reason are true but Reason is not the correct explanation of
Assertion.
3) If Assertion is true but Reason is false.
4) If Assertion is false but Reason is true.
19. Two soap bubbles coalesce to form a single large drop.
Match the following columns.
Column I Column II
A. Surface energy in the process will p. Increase
B. Temperature of the drop will q. Decrease
C. Pressure inside the soap bubble will r. Remain same
1) Aq, B  p, C q 2) Aq, B r, C  p
3) A  p, B r, C q 4) A q, B  p, C r
20. Assertion: In the circuit shown in figure, battery is ideal. If a resistance R0 is connected in
parallel with R, power across R will increase.
Reason: Current drawn from the battery will increase.
E
1) If both Assertion and Reason are true and Reason is the correct explanation of Assertion
2) If both Assertion and Reason are true but Reason is not correct explanation of Assertion
3) If Assertion is true but Reason is false
4) If Assertion is false but Reason is true
Section-II contains 10 Numerical Value Type questions. Attempt any 5 questions only. First 5 attempted questions will be considered if more than 5 questions attempted. The
Answer should be within 0 to 9999. If the Answer is in Decimal then round off to the nearest Integer value (Example i,e. If answer is above 10 and less than 10.5 round off is
10 and If answer is from 10.5 and less than 11 round off is 11).
Note:
From: Question number’s 21 to 30 rules for Answer marking
If your answer is options 2,4 Then you have to fill the OMR sheet
as ‘24’
If your answer is options 1,3,4 Then you have to fill the OMR sheet
as ‘134’
And if your answer is options 1,2,3,4 then you have to fill the OMR sheet as
‘1234’

21. A block of mass 1 kg moves under the influence of external forces on a rough horizontal
surface. At some instant, it has a speed of 1 m/s due east and an acceleration of 1 m/s2 due
north. The force of friction acting on it is F
1) F acts due west
2) F acts due south
3) F acts in the south west direction
4) The magnitude of F cannot be found the given data
22. A particle is taken from point A to point B under the influence of a force field. Now it is
taken back from B to A and it is observed that the work done in taking the particle from A to
B is not equal to the work done in taking it from B to A. If Wnc and Wc is the work done by
non conservative forces and conservative forces present in the system respectively, U is
the change in potential energy,  k is the change is kinetic energy, then
1) Wnc U  k 2) Wc U 3) Wnc Wc  k 4) Wnc U  k
23. If electron of the hydrogen atom is replaced by another particle of same charge but of double
the mass, then:
1) Bohr radius will increase
2) Ionisation energy of the atom will be doubled
3) Speed of the new particle in a given state will be lesser than the electron’s speed in same
orbit
4) Gap between energy levels will now be doubled
24. An object is floating in a liquid, kept in a container. The container is placed in a lift. Choose
the correct option(s)
1) Buoyant force increases as lift accelerates up
2) Buoyant force decreases as lift accelerates up
3) Buoyant force remains constant as lift accelerates
4) The fraction of solid submerged into liquid does not change
25. Charges Q1 and Q2 lie inside and outside respectively of a closed Gaussian surface S. Let E
be the field at any point on S and  be the flux of E over S:

1) If Q1 changes, both E and  will change 2) If Q 2 changes E will change
3) If Q1  0 and Q2  0 then E  0 but   0 4) If Q1  0 and Q2  0 then E  0 but   0
26. The reactance of a circuit is zero. It is possible that the circuit contains:
1) An inductor and a capacitor 2) An inductor but no capacitor
3) A capacitor but no inductor 4) Neither an inductor nor a capacitor
27. For a body executing SHM with amplitude A, time period T, maximum velocity vmax and
T
phase constant zero. Which of the following statements are correct for 0  t  (y is
4
displacement from mean position particle is at mean position at t = 0)?
1) At y   A 2  , v   v max 2  2) For v   v max 2  , y   A 2 
3) For t  T 8 , y   A 2  4) For y   A 2  , t  T 8
28. An astronaut leaves his spaceship for some experiment and floats freely in space at rest
relative to his spaceship. His friend in the spaceship ignites a rocket installed on the
spaceship for a very short duration. Which of the following observations can you make
relative to a reference frame moving together with the astronaut outside the spaceship?
1) The spaceship has lesser kinetic energy than ejected gases
2) The spaceship and the ejected gases have equal kinetic energies
3) The spaceship has greater kinetic energy than the ejected gases
4) Magnitudes of momenta of spaceship and ejected gases are equal
29. In an X-ray tube, electrons emitted from a filament (cathode) carrying current I hit a target
(anode) at a distance d from the cathode. The target is kept at a potential V higher than the
cathode resulting in emission of continuous and characteristic X-rays. If the filament current

I
I is decreased to , the potential difference V is increased to 2V, and the separation
2
d
distance d is reduced to , then
2
1) the cut-off wavelength will reduce to half, and the wavelengths of the characteristic X-
rays will remain the same
2) the cut-off wavelength as well as the wavelengths of the characteristic X-rays will remain
the same
3) the cut-off wavelength will reduce to half, and the intensities of all the X-rays will
decrease
4) the cut-off wavelength will become two times lager, and the intensity of all the X-rays
will decrease
30. The breakdown in a reverse biased p-n junction is more likely to occur due to
1) large velocity of the minority charge carriers if the doping concentration is small
2) large velocity of the minority charge carriers if the doping concentration is large
3) strong electric field in a depletion region if the doping concentration is small
4) strong electric field in the depletion region if the doping concentration is large

correct.
31. For the process : H2O l  (1 atm, 373.15K) H2O  g  (1 atm, 373.15K), the correct set of
thermodynamic parameter is
1)  G  0;  S   ve 2)  G  0;  S   ve
3)  G   ve;  S  0 4)  G   ve ;  S   ve
32. Assertion:- When aniline is subjected to nitration by conc. HNO3 &H2SO4 meta nitro aniline
is formed in considerable amount.
Reason: - NH2 is o/p directing but ring deactivating group.
1) Assertion is True, Reason is true: Reason is the correct explanation for Assertion.
2) Assertion is true Reason is true: Reason is not the correct explanation for Assertion.
3) Assertion is True, Reason is False.
4) Assertion is False, Reason is true.
+
NH 2 OH H LiAlH 4
O (A) (B) (C) ; The product 'C' is
33.
OH
N N N O N OH
1) H 2) 3) H 4) H H
34. The equilibrium constant for ionization constant of acetic acid in an aqueous solution of
concentration ‘C’ is given by
c  c2 c  c2 c  c2 c  c2
1) K  2) K  3) K  4) K 
 c   (    c )   c   (    c )
35. Which one of the following is correct (Major product)
NMe2 NMe2 OCH 3 OCH 3
Na / liq NH 3 Na / liq NH 3
1) 2)
COOH COOH NMe2 NMe2
Na / liq NH 3 Na / liq NH 3
3) 4)
36. Which of the following is/are aromatic in character?
(I) (II) (III) (IV)

1) II only 2) I, III and IV 3) II ,III and IV 4) I, II, III and IV
37. About CH 2 F2 the correct statement is.
1) The order of bond angles is HCH  HCF  FCF
2) The shape of the molecule is tetrahedral
3) All bond angles are equal.
4) The order of bond angles is HCH  HCF  FCF
38. The major product formed in the reaction given below is
O
1.NaOH,Br2
NH
2.H3O+
O
NH2 Br
COOH CONH 2
1) 2)
O
COOH
O
CONH 2
3) 4) O
39. Assertion (A): Components of a mixture of red and blue inks can be separated by
distributing the components between stationary and mobile phases in paper chromatography.
Reason (R): The coloured components of inks migrate at different rates because paper
selectively retains different components according to the difference in their partition
between the two phases.
1) Both A and R are correct and R is the correct explanation of A.
2) Both A and R are correct but R is not the correct explanation of A.
3) Both A and R are not correct.
4) A is not correct but R is correct
40. A metal ‘M’ shows the following observable changes in the sequence of reactions. Identify
the metal. M
M 
dil . H 2 SO4
 colourless solution 
aq NaOH

White ppt 
excess NaOH
colourless solution 
H2S
 White ppt
1) Bi 2) Pb 3) As 4) Zn
41. The rate of hydrolysis of boron halides will be in the order of
1) BF3  BCl3  BBr3  BI3 2) BCl3  BF3  BBr3  BI3
3) BI3  BBr3  BCl3  BF3 4) BI3  BCl3  BF3  BBr3
42. The equilibrium:
P4  g   6Cl2  4PCl3  g  is obtained by mixing equal moles of P4 and Cl2 in an evacuated
vessel. Then at equilibrium:
1)  Cl2    PCl3  2)  Cl2    P4  3)  P4    Cl2  4)  PCl3    P4 
43. Biuret test is not given by :
1) proteins 2) urea
3) polypeptide 4) carbohydrates
44. Which of the following complex has highest C  O bond energy?
1) Ni  CO  
 4
2)  Co  CO  
 6
3
3) Mn  CO  
 6

4)  Fe  CO 5 
45. Which of the following diazonium coupling reactions is not feasible?
N2Cl + OH N N OH
1)
H3C N2Cl + OCH3 H3C N N OCH3
2)
O 2N N 2 Cl + OCH 3 O 2N N N OCH 3
NO 2
3) NO 2
NO 2 H 3C NO 2 H 3C
O 2N N 2 Cl + CH3 O 2N N N CH3
NO 2 H 3C
4) NO 2 H 3C

46. Select the law that corresponds to data shown for the following reaction
A  B  Products
Exp [A] [B] Initial rate
1 0.012 0.035 0.1
2 0.024 0.070 0.8
3 0.024 0.035 0.1
4 0.012 0.070 0.8
1) Rate  k  B  2) Rate  k  B  3) Rate  k  A B  4)Rate  k  A  B 
3 4 3 2 2
47. In the following

NO3  As2 S3  H 2O  AsO43  NO  SO42  H 
Equivalent weight of As2S3 (Molecular weight M) is
3M M M M
1) 2) 3) 4)
28 4 24 28
48. Which of the following is incorrect?
1) PCl5 exists as PCl4 

PCl6  in solid state
 
2) PBr5 exists as  PBr4   in solid state

 
 PBr6
3) Liquid IF5 selfionises to give IF4  and IF6 

 
4) ICl can undergo ionization as I  and ICl2 

 
49. In the given reaction, the structure of X is:

H
N

Br2
 X 
H
Br
N
Br
N
1) Br 2)
H N
N
Br
3) Br 4) Br
50. Which failed to form Osazone with excess of C6H5NHNH2 in strongly alkaline medium
1) D-glucose 2) D-Mannose 3) 4) All of these

  
51. In the reaction: CN  H  NO3  CO2  NO  H2O (unbalanced ), if the stoichiometry of
, H  & NO3 are x, y & z respectively,. Find  y  z /x ?


CN 
52. How many of the following reactions give haloform as one of the products?
Cl  3
i H O
 i  H 2O /  
 ii  I 2 / NaOH 
(i) 
 ii  I 2 / NaOH 
(ii)
Cl
O
O O
(iii) 
I 2 / NaOH
 (iv)
OCH 3 
I 2 / NaOH

O
O
 i  Hg  OAC 
 
2 2 4 / H O, NaBH
(v)   2
ii I / NaOH , 

53. CH3  CH  CH  O  CH2  CH3 
dil acid
 X  Y (Y gives positive haloform test)
The number of possible isomers for X excluding stereo isomers and including X.
0 0 0
54. If E1 , E2 and E3 are standard oxidation potentials for Fe Fe 2 , Fe 2 Fe3 and Fe Fe3 , then
E 20  2 E10
E 
0
3 . The value of n is……….
n
55. Among the following, the total number of reactions/ processes in which the entropy
increases are:
a. 2NaHCO3  Na2CO3  s  CO2  g   H 2O  g 
b. A liquid cyrstallises into a solid.
c. Temperature of crystalline solid is raised from zero K to 100 K.
d. Hard boiling of an egg.
e. Intermixing of gases at constant temperature.
f. Boiling of water
g. Desalination of water.
h. NH3  g ,10atm  NH3  g ,1atm
56.
CH 2
KMnO4
H 2C CH 2 A  gas   B

1 mole  LiAlH 4
H C
D

If X is no of moles of CO2 and Y is no. of hydrogens that can participate in hyperconjugation
in D ,then (X + Y) is……

57. Xenon trioxide in strongly alkaline solution forms a product A which slowly
disproportionate into B and xenon gas. The oxidation number of xenon in A is
58. Total number of moles of P-H bonds is in the product(s) formed when one mole of white
phosphorous completely react with NaOH solution.
59. How many of the following undergo decarboxylation on heating

O O
HOOC
O
HOOC O
COOH
COOH
COOH
COOH
COOH
COOH
COOH  H 2C  2
COOH
60. How many of the following compounds are more reactive than chlorobenzene towards
nitration?
CH 3
F NO2 O O O CH 3 OH NMe2
CN CHO OMe
Br

correct.
61. S1 : function  : 5, 81  R defined by   x    x  5 2  x  812  x , takes the value 43 for
some x  5, 81
S2 : If a function y  g  x is defined on 5, 81 , then for any k   g  5  , g  81  there is some
point c  5, 81 such that g  c   k
1) S1 is true and S2 is true for every function g
2) S1 is true and S2 is not true for every function g
3) S1 is not true and S2 is true for very function g
4) S1 is not true and S2 is not true for every function g
62. Which of the following is always true
1) One root of the equation ax  bx  c  0,  a, b  R & c  R  Q & a  0 in the form of
2
p  q then other root is p  q  p  z, q  z  
2) Exactly one of the root of the equation ax 2  bx  c  0,  a  0  lies in the given interval
 k1 , k2  if   k1 .  k2   0
3) If a,b,c, d,e,  are positive real numbers and d   e2  0. The equations
d e 
ax 2  2bx  c  0,  a  0  and dx 2  2ex    0  d  0  have a common root, then  
a b c
4) a,b,c are in A.P and G.P then (a, b, c) can be (0, 0, 0)
63. In a parallelogram as shown in the figure  a  b 
D u1  bx  ay  ab
C
u3  ax  by  ab
u4  ax  by  2ab
A u2  bx  ay  2ab B
Equation of the diagonal AC is
1 4  u2u3  0
1) uu 2) u1  u2  u3  u4  0
1 2  u3u4  0
3) uu 1 3  u2u4  0
4) uu
  01, then
n
64. If 9  80  I   where I, n are integers and
1) I is an odd integer 2) I is an even integer
3)  I    I     2  
n
4) 12   2  9  80
65. Let S1 : A function f always has a local maximum between any two local minima. Then f
must be differentiable
S2 : If a function is defined on  a , b  and continuous on  a, b , then it takes its extreme
values on  a, b.
S3 : Every continuous and bounded function on  ,   takes its extreme values.
The number of statements among the S1, S2 , S3 which are always true for every function
1) 0 2) 1 3) 2 4) 3
66. Graph of y    x  is given below
 0, 3
2 x
1 3
1
1
Then graph of y  is best represented by
  x
y y
12 2 13 2
0 x x
1 3 0 1 3
1 1
1) 2)

y y
2 x y 1
0 1 3
1 13
0 x
1 2 3
3) 4)
67. Which of the following is true
1) If   x  is continuous at x  a then at x  a limit need not exist for ƒ 𝑥
2) For every continuous functions   x  , g  x  on  0,  satisfying   x   g  x  for all x  0
and both lim x and limg  x exist finitely
x x
3) If both functions   x  and g  x  have limits at x  a, then Lt    g  x      Lt   g  x   

x a  x a 
[ ] denotes G.I.F
4) for every continuous monotonic functions f  x  , g  x on  , Then their sum
  x   g  x  is monotonic on 
68. Let   x   x 2  2ax  b, g  x   cx 2  2dx  1 be quadratic expressions whose graph is as
shown in the figure  a  0, cd  0 
y
y  g  x
y    x
A
B
A' O
B'
Here it is given that AA'  BB ' and OA '  OB .
Absolute values of sum of roots of equations   x   0 and g  x   0 are p and q respectively
then p  q 
2d 2d
1) 2 a  2) 2  a  d  3) 1  b 4) 2 a 
c c
69. f(x) and g(x) both defined R  R are two non-constant continuous functions then
Statement – 1: If g (x) is periodic then f(g(x)) is also periodic
Statement – 2: If g(x) and f(x) are both aperiodic then f(g (x)) is aperiodic.
Statement – 3: If g(x) is periodic with fundamental period T then f(g(x)) is not periodic
Statement – 4: If g(x) is periodic with fundamental period T then f(g(x)) is also periodic with
period T
Which of the following must be truth value of above statements in that order.
1) TFFT 2) TFTT 3) FFTT 4) TTTT
1
Let function y    x  satisfies the differential equation x 2
dy
70.  y e  x  0  and
2 x
dx
lim f  x   1 . Identify the incorrect statement?
x 0 
1) Range of   x is  0, 1   1  2)   x is bounded
2
3) lim   x 1
e 1
x
4) 0 f  x  dx  0 f  x dx
1
dx 1
71. STATEMENT–1: The inequality 0 1  x n  1 
n
is true for n  N .
1
x n1  x n  x 2n1
STATEMENT–2:  dx  0,  n  N .
0
1  x n
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
3) Statement-1 is True, Statement-2 is False
4) Statement-1 is False, Statement-2 is False

72. f(x) and g(x) are functions defined on R  R
Statement 1: If f(x) is continuous and g(x) is discontinuous at x = a then their product is
continuous at x = a if f(a) = 0
Statement 2: If f(x) is continuous and g(x) is discontinuous at x = a and if g(a) = 0 then their
product is continuous at x = a
Statement 3: If f(x) is differentiable and g(x) is not derivable at x = a then their product is
derivable at x = a if g(x) is continuous and f(a) = 0
Statement 4: If f(x) is differentiable and g(x) is not derivable at x = a then their product is
derivable at x = a if g(x) is continuous and g(a) = 0
1) FFFF 2) FFTF 3) TFFF 4) TFTF
73. Which of the following is true?
1) Number of equivalence relations on vowels in English language is 50.
2) Brother among humans is symmetric relation
3) RoS and SoR are symmetric only when R  S (for relations R & S)
4) RoS and SoR are symmetric if RoS  SoR (for relations R & S)
74. Statement – 1 : If  : 1,2,3,4,5  1,2,3,4,5 then the number of onto functions such that
  i   i and  1  2 is 11
Statement – 2 : If n(n >1) things are arranged in row, the number of ways in which they can
be de- arranged so that no one of them occupies its original place is
 1 1 n 1 
n ! 1    ......   1 
 1! 2! n! 
Statement–3:Number of ways of distribution of 12 identical balls into 3 identical boxes is 19
Statement–4: Number of ways of distribution of n identical objects among r persons, each
nr 1
one of whom can receive any number of objects is Cr 1
1) FTFT 2) FFTF 3) TTTT 4) TFTF
75. A bag contains four tickets having numbers 112, 121, 211, 222 written on them. Denote by
Ai  i  1,2,3 the event that the ith digit from left of the number on a randomly drawn ticket is
1.
1) A1, A2, A3 are independent events
2) P  Ai A j   P  Ai  P ( A j ) for all i j
3) P  Ai  A j   P  Ai  P( Aj ) for all 1  i  j  3
4) P  Ai  A j   P  Ai  P ( A j ) for all 1  i  j  3
STATEMENT – 1 : If the circles x  y  2gx  2 fy  0 and x  y  2g x  2 f y  0 are

2 2 2 2 1 1
76.
concentric g 1   g1 and they represent same circle
STATEMENT- 2: Let C be any circle with centre (0, 2) . There can be atmost 2 points with
rational coordinates on C.
STATEMENT- 3: Circle x 2  y 2 – 6 x – 4 y  9  0 bisects the circumference of the circle
x 2  y 2  8 x  6 y  23  0
STATEMENT- 4: If centre of first circle lie on the second circle then it bisects
circumference of second circle
1) TTTT 2) FTTF 3) TFTF 4) TTTF
77. STATEMENT -1: Through the vertex ‘O’ of the parabola y2  4x chords OP & OQ are
drawn at right angles to one another. For all positions of the line through P and Q cuts the
axis of parabola at a fixed point
STATEMENT-2: The focus of the parabola y2  4ax is of the form  x1, 0
1) Both Statements are true and Statement-2 is the correct explanation of Statement-1
2) Both Statements are true but Statement-2 is not the correct explanation of Statement-1

78. For square matrices A and B of same order which of the following is not true. (with real
entries)
(O is null matrix and I is Identity matrix of same order as A and B)
1) If AB = O and A is not null matrix then B is singular matrix
2) If A is skew symmetric then adjoint of A can be skew symmetric.
3) If AB = O then  det A 2   det B 2  0
4) If AB = I then BA =I
79. Identify the correct statement.
1) If system of n simultaneous linear equations has a unique solution, then coefficient matrix
is singular
2) If system of n simultaneous linear equations has a unique solution, then coefficient matrix
is non singular
3) If A1 exists,  adj . A 1 may not exist
cos x  sin x 0
 
4) If F  x    sin x cos x 0 , then F  x  .F  y   F  x  y 
 0 0 0

80. Which of the following conclusion(s) hold(s) true for a non-zero vector a ?
         
1) .
ab  .
ac  b  c 2)  b  a  c  b  c
a
       
3) a.b  a.c and a  b  a  c  b  c 4) a  b  a  b  a .b  0

81. Let V1 be the variances of 2024 observations which are in A.P with first term 2024 and
common difference is 2024 and Let V2 be the variances of 2024 observations which are in
A.P with first term 2023 and common difference 2024 then V1 :V2  m: n where the greatest
2 digit value of m  n is m, nN

82. Area of mid point triangle of a triangle whose vertices are complex numbers z1, z2, z3 is
z1 z1 1
i
z2 z2 1 then k 
k
z3 z3 1
     
83. Given a, b and c are three vectors such that b and c are unit like parallel vectors and a  4.
  
If  c  2b then the sum of all possible values of  is equal to ______
a
84. The points P and Q are 1, 1, 1 and  2,1,1 The point of intersection of lines containing the
 
D.r.s of PR 1, 2,1 and QS 1, 4, 2  is A, if AB is perpendicular to PR and QS and
 AB 
2
 32, and possible position of B are B1, B2 then  B1 B 2 2 
2x
85. For x  1, sin 1  k  2 tan 1 x then 3  k 
1 x 2
3
d 2x 2  d y  dx 
2
86. l 2  k  2    0 Then number of possible ordered pair  l, k  is (for twice
dy  dx   dy 
dx d 2 x
differentiable invertible function y    x  , & are finite values.)  l , k  Z and l  3 .
dy dy2
2
87. If I   x  sin 2  sin x   cos 2  cos x   dx , then  I   ___ , where . denotes the greatest integer
0
function
1 ac
88. sin12 0.sin 48 0.sin 54 0  ,  b  tan18 tan 78, c  tan54 tan 48  then the value of is
a b
89. A point P(x, y) moves in such a way that [x + y + 1] = [x] (where [.] greatest integer
function) and x  (0, 4). Then the area representing all the possible positions of P equals
90. A triangle is formed by the points A  0, 0  , B  3, 0  and C  3, 4  . A and C are foci of ellipse
7
and B lies on the ellipse. If area of ellipse is p  p  N  . then the value of p is
2


only.
.
Admission Number:


correct.
1. In a series LCR circuit, the inductive reactance  X L  is 10 and the capacitive reactance
 X C  is 4 . The resistance (R) in the circuit is 6 . The power factor of the circuit is:
3 1 1 1
1) 2) 3) 4)
2 2 2 2 2
2. The velocity- displacement graph of a particle is shown in the figure.
The acceleration – displacement graph of the same particle is represented by:
1) 2)
3) 4)
3. Consider a sample of oxygen behaving like an ideal gas. At 300 K, the ratio of root mean
square (rms) velocity to the average velocity of gas molecule would be:
( Molecular weight of oxygen is 32 g / mol; R  8.3JK  1 mol 1 )
8 3 3 8
1) 2) 3) 4)
3 3 8 3
4. A proton and an  - particle, having kinetic energies K p and K respectively, enter into a
magnetic field at right angles. The ratio of the radii of trajectory of proton to that of -
particle is 2 : 1. The ratio of K p : K is:
1) 4 : 1 2) 8 : 1 3) 1 : 4 4) 1 : 8
C
5. A particle of mass m moves in a circular orbit under the central potential field,U  r    ,
r
where C is a positive constant. The correct radius–velocity graph of the particle’s motion is:
1) 2) 3) 4)
6. Consider a uniform wire of mass M and length L. It is bent into a semicircle. Its moment of
inertia about a line perpendicular to the plane of the wire passing through the centre is:
ML2 1 ML2 1 ML2 2 ML2
1) 2) 3) 4)
2 4 2 2 2 5 2
7. The time taken for the magnetic energy to reach 25 % of its maximum value, when a
solenoid of resistance R, inductance L is connected to a battery is:
L L L
1) In 5 2) In 2 3) In 10 4) Infinite
R R R
8. An object of mass m1 collides with another object of mass m2 , which is at rest. After the
collision the objects move with equal speeds in opposite direction. The ratio of the masses
m2 : m1 is
1) 1 : 2 2) 3 : 1 3) 1 : 1 4) 2 : 1
9. For an adiabatic expansion of an ideal gas, the fractional change in its pressure is equal to
(where  is the ratio of specific heats):
dV 1 dV V dV
1) 2)  3)  4) 
V  V dV V

10. A plane electromagnetic wave propagating along y – direction can have the following pair of
 
  
electric field E and magnetic field B components.
1) Ex , Bz or Ez , Bx 2) Ex , By or E y , Bx 3) E y , B y or Ez , Bz 4) E y , Bx or Ex , By
11. The angular momentum of a planet of mass M moving around the sun in an elliptical orbit is

L . The magnitude of the areal velocity of the planet is:
L 2L L 4L
1) 2) 3) 4)
M M 2M M

12. The function of time representing a simple harmonic motion with a period of is :

1) sin t   cos t  2) 3 cos    2 t 

4 
3) sin 2  t  4) cos t   cos  2t   cos  3t 
13. A plane electromagnetic wave of frequency 100 MHz is travelling in vaccum along the x-

 . (Where, k is unit
direction . At a particular point in shape and time B  2.0  108 kT

vector along z-direction) What is E at this point ? (speed of light c  3 108 m / s )
1) 6.0 k V / m 2) 0.6 k V / m 3) 6.0 j V / m 4) 0.6 j V / m
l
14. The time period of a simple pendulum is given by T  2 . The measured value of the
g
length of pendulum is 10cm known to a 1 mm accurancy. The time for 200 oscillations of
the pendulum is found to be 100 second using a clock of 1 s resolution. The percentage
accuracy in the determination of ‘g’ using this pendulum is ‘x’. The value of ‘x’ to the
nearest integer is
1)2% 2) 3% 3) 5% 4) 4%

15. A solid cylinder of mass m is wrapped with an inextensible light string and, is placed on a
rough inclined plane as shown in the figure. The frictional force action between the cylinder
and the inclined plane is:
[The coefficient of static friction,  ' is 0.4]

S
mg 7
1) 0 2) 3) 5mg 4) mg
5 2
16. A constant power delivering machine has towed a box, which was initially at rest, along a
horizontal straight line. The distance moved by the in time ‘t’ is proportional to:
1) t 2) t 2/3 3) t3/2 4) t1/2
17. A person is swimming with a speed of 10 m/s at an angle of 1200 with the flow and reaches
to a point directly opposite on the other side of the river. The speed of the flow is ‘x’ m/s .
The value of ‘x’ to the nearest integer is __
1) 1 2) 4 3) 8 4) 5
18. In the experiment of Ohm’s law, a potential difference of 5.0V is applied across the end of a
conductor of length 10.0 cm and diameter of 5.00mm. The measured current in the
conductor is 200 A. The maximum permissible percentage error in the resistivity of the
conductor is :
1)8.4 2)3.9 3)3.0 4)7.5

19. If the angular velocity of earth’s spin is increased such that the bodies at the equator start
floating, the duration of the day would be approximately:
[ Take g  10 ms 2 , the radius of earth, R  6400  10 3 m , Take   3 .1 4 ]
1) 60 minutes 2) 84 minutes
3) does not change 4) 1200 minutes
20. A thin circular ring of mass M and radius is rotating about is axis with an angular speed  .
Two particles having mass m each are now attached at diametrically opposite points. The
angular speed of the ring will become.
M M M  2m M  2m
1)  2)  3)  4) 
M m M  2m M  2m M
21. A particle performs simple harmonic motion with a period of 2 second . The time taken by
the particle to cover a displacement equal to half of its amplitude from the mean position is
1
s. The value of “a” to the nearest integer is
a
22. A parallel plate capacitor has plate area 100 m 2 and plate separation of 10m. The space
between the plates is filled upto a thickness 5 m with a material of dielectric constant of 10.
The resultant capacitance of the system is ‘x’ pF The value of  0  8.85  10  12 F .m  1
23. The radius of a sphere is measured to be  7.50  0.85 cm . Suppose the percentage error in
its volume is x. The value of x, to the nearest x, is_____
24. Consider a water tank as shown in the figure. It’s cross – sectional area is 0.4m2 . The tank
has an opening B near the bottom whose cross-section area is 1 cm 2 . A load of 24 kg is
applied on the water at the top when the height of the water level is 40 cm above the bottom.
The value of  , to the nearest integer, is______[Take value of g to be 10 ms 2 ]

25. Consider a 72 cm long wire AB as shown in the figure. The galvanometer jockey is placed at
P on AB at a distance x cm from A. The galvanometer shows zero deflection.
The value of x, to the nearest integer, is __________

26. Two wires of same length and thickness having specific resistances 6 cm and 3 cm
respectively are connected in parallel. The effective resistivity is  cm. The value of  , to
the nearest integer, is______
27. The voltage across the 10 resistor in the given circuit is x volt.

28. An infinite number of point charges, each carrying 1  C charge, are placed along the y –axis
at y 1m, 2m,4m,8m,....... . The total force on a 1 C point charge, placed at the origin, is
x 103 N . The value of x, to the nearest integer, is ______
1
[ Take  9  10 9 Nm 2 / C 2 ]
4 0
29. The projectile motion of a particle of mass 5 g is shown in the figure.
The initial velocity of the particle is 5 2 ms  1 and the air resistance is assumed to be
negligible. The magnitude of the change in momentum between the points A and B is
x  10  2 kgms  1. The value of x, to the nearest integer , is______
30. A ball of mass 4 kg, moving with a velocity of 10 m s  1 , collides with a spring of length 8m
and force constant 100 Nm  1. The length of the compressed spring is x m. The value of x,
to the nearest integer, is________

correct.
31. The correct structures of trans -  NiBr2  PPh3 2  and meridonial - Co  NH3 3  NO2 3  ,
respectively, are :
1)
NO2
NH 3
NO2
Br
PPh3 CO
Ni NH 3
NO2
Ph3 P Br NH 3
2) and
3)
4)
32. Which among the following is the strongest Bronsted base?
1) 2) 3) 4)
33. Which among the following pairs of the structures will give different products on
ozonolysis? (Consider the double bonds is the structures are rigid and not delocalized.)
1) 2)
3) 4)
34.
Considering the above reactions, the compound ‘A’ and compound ‘B’ respectively are:
1) 2)
3) 4)

35.
Consider the above reaction sequence, the Product ‘C’ is:
1) 2)
3) 4)
36.
Consider the above reaction, the compound ‘A’ is:

1) 2)
3) 4)
37.
Which among the following represent reagent ‘A’?
1) 2) 3) 4)
38. Consider the following reaction sequence:
The product ‘B’ is:

1) 2)
3) 4)
39. A certain orbital has no angular nodes and two radial nodes. The orbital is:
1) 2p 2) 3s 3) 3p 4) 2s
40. A compound ‘X’ is acidic and it is soluble in NaOH solution, but insoluble in NaHCO3
solution. Compound ‘X’ also gives violet colour with neutral FeCl3 solution. The compound
‘X’ is:
1) 2) 3) 4)
41. Which of the following pair of molecules contain odd electron molecule and an expanded
octet molecule?
1) BCl3 and SF6 2) NO and H2SO4 3) SF6 and H2SO4 4) BCl3 and NO
42. N 2  g   3H 2  g   2 NH 3 g 
20 g 5g
Consider the above reaction, the limiting reagent of the reaction and number of moles of
NH3 formed respectively are:
1) H2,1.42 moles 2) H2,0.71moles
3) N2,1.42 moles 4) N2,0.71moles

43. Which one of the following complexes is violet in colour?
4 4
1)  Fe  SCN 6  2)  Fe  C N 6 
4
3) Fe4  Fe  CN 6  .H 2O 4)  Fe  CN 5 NOS 
3
44. The first ionization enthalpy of Na, Mg and Si, respectively, are:496, 737 and 786 kJ mol1.

The first ionization enthalpy kJ mol 1 of Al is: 
1) 487 2) 768 3) 577 4) 856
45. Among the following compounds I-IV, which one forms a yellow preceipitate on reacting
sequentially with (i) NaOH (ii) dil.HNO3 (iii) AgNO3 ?
C
C Br C
CH 3
H 3C CH 3 CH 2 I
I II III IV
1) I 2) IV 3) II 4) III
Statement I: In the titration between strong acid and weak base methyl orange is suitable as
an indicator.
Statement II: For titration of acetic acid with NaOH phenolphthalein is not a suitable
indicator.
In the light of the above statements, choose the most appropriate answer from the options
given below:
2) Both statement I and Statement II are false
3) Both statement I and Statement II are True
Statement I: According to Bohr’s model of an atom, qualitatively the magnitude of velocity
of electron increases with decrease in positive charges on the nucleus as there is no strong
hold on the electron by the nucleus.
Statement II: According to Bohr’s model of an atom, qualitatively the magnitude of velocity
of electron increases with decrease in principal quantum number. In the light of the above
statements, choose the most appropriate answer from the options given below:
3) Both Statement I and Statement II are false
4) Both Statement I and Statement II are true
48. Number of lone pairs of electrons in the central atom of SCl2,O3,ClF3 and SF6 , respectively,
are:
1) 0, 1, 2 and 2 2) 2, 1, 2 and 0 3) 1, 2, 2 and 0 4) 2, 1, 0 and 2
49. In following pairs, the one in which both transition metal ions are colourless is:
1) Sc3 , Zn 2  2) Ti 4  , Cu 2  3) V 2  , Ti3 4) Zn 2  , Mn 2 
50. In neutral or faintly alkaline medium, KMnO4 being a powerful oxidant can oxidize,
thiosulphate almost quantitatively, to sulphate. In this reaction overall change in oxidation
state of manganese will be:
1) 5 2) 1 3) 0 4) 3
51. If the solubility product of PbS is 8 1028 , then the solubility of PbS in pure water at 298 K
1
is x  10 16 mol L . The value of x is __________. (Nearest Integer)
[Given: 2  1.41 ]
52. The reaction between X and Y is first order with respect to X and zero order with respect to
Y.
X  Y  Initialrate
Experiment
mol L1 mol L1 molL1 min 1
I 0.1 0.1 2  103

II L 0.2 4  103
III 0.4 0.4 M 103
IV 0.1 0.2 2  103
Examine the data of table and calculate ratio of numerical values of M and L. (Nearest
Integer)
53. In a linear tetrapeptide (Constituted with different amino acids), (number of amino acids) –
(number of peptide bonds) is ___________.
54. In bromination of Propyne, with Bromine 1, 1, 2, 2-tetrabromopropane is obtained in 27%
yield. The amount of 1, 1, 2, 2-tetrabromopropane obtained from 1 g of Bromine in this
reaction is _______ 101 g. (Nearest integer)

(Molar Mass: Bromine =80 g/mol)
3
55.  Fe  CN   should be an inner orbital complex. Ignoring the pairing energy, the value of
 6
crystal field stabilization energy for this complex is ( – ) __________  0 . (Nearest integer)
1
56. Resistance of a conductivity cell (cell constant 129 m ) filled with 74.5 ppm solution of
KCl is 100  (labelled as solution 1). When the same cell is filled with KCl solution of 149
ppm, the resistance is 50  (labelled as solution 2). The ratio of molar conductivity of
solution 1 and solution 2 is i.e. 1  x  10  3 . The value of x is ___________. (Nearest
2
integer)
Given, molar mass of KCl is 74.5 g mol1.
57. The Born-Haber cycle for KC is evaluated with the following data:
 f H  for KC    436.7 kjmol  1 ;  sub H  for K  89.2 kJ mol  1 ;


 ionization H  for K  419.0 kj mol  1  electrongain H for C   g   348.6 kj mol 1
 bond H  for C  2  243.0 kj mol  1
The magnitude of lattice enthalpy of KC in kj mol 1 is ______. (Nearest integer)

58. The minimum uncertainty in the speed of an electron in an one dimensional region of length
2a0 (Where a0  Bohr radius 52.9 pm) is __________ km s 1 .
(Given: Mass of electron = 9.1  1031 kg, Planck’s constant h  6.63  1034 Js)
59. When 600 mL of 0.2 M HNO3 is mixed with 400 mL of 0.1 M NaOH solution in a flask,
the rise in temperature of the flask is __________  10  2 C .
(Enthalpy of neutralisation = 57 kJ mol1 and Specific heat of water = 4.2 JK  1g 1 )

(Neglect heat capacity of flask)
60. If O2 gas is bubbled through water at 303 K, the number of millimoles of O2 gas that
dissolve in 1 litre of water is __________. (Nearest Integer)
(Given: Henry’s Law constant for O2 at 303 K is 46.82 k bar and partial pressure of
O2  0.920 bar)
(Assume solubility of O2 in water is too small, nearly negligible)

correct.
3 3
4
48
61.  2
dx is equal to
3 2 9  4x
4
  
1) 2 2) 3) 4)
2 3 6
 cos 1 1   x2 sin 1 1   x
  ,x  0

62. Let   R be such that the function f  x    3

 
x  x
 , x0
is Continuous at x  0, where  x  x   x , x is the greatest integer less than or
equal to x. Then :
 
1)   0 2) no such  exists 3)   4)  
4 2
63. Let f(x) be a function such that f (x  y)  f (x)  f ( y) for all x, y  N . If f (1)  3 and
n
 f (k )  3279 then the value of n is
k 1
1) 6 2) 8 3) 7 4) 9
64. The number of square matrices of order 5 with entries from the set {0, 1}, such that the sum
of all the elements in each row is 1 and the sum of all the elements in each column is also 1,
is
1) 125 2) 150 3) 225 4) 120
2  1   1
65. The number of real solutions of the equation 3 x  2   2  x    5  0 is
 x   x
1) 4 2) 2 3) 3 4) 0
22 x
66. If f ( x)  , x  R , then f  1   f  2   ...  f  2022  is equal to
22 x  2  2023   2023   2023 
1) 2011 2) 2010 3) 1011 4) 1010
67. The set of all values of a for which lim  x  5   2 x  2  0 , where [ ] denotes the
xa
greatest integer less than or equal to  is equal to

1) [-7.5, -6.5] 2) (-7.5, -6.5] 3) [-7.5, -6.5) 4) (-7.5, -6.5)
68. Let A denote the event that a 6-digit integer formed by 0, 1, 2, 3, 4, 5, 6 without repetitions,
be divisible by 3. Then probability of event A is equal to:
11 4 9 3
1) 2) 3) 4)
27 9 56 7
69. Let the six numbers a1, a2, a3, a4, a5, a6 be in A.P and a1  a3 10. If the mean of these six
19
numbers is and their variance is  2 , then 8 2 is equal to
2
1) 105 2) 200 3) 210 4) 220
70. Let f be a real valued function, defined on R  1, 1 and given by
x 1 2
f  x   3log e  . Then in which of the following intervals, function f  x  is
x  1 x 1
increasing?
  1  1
  
1)  ,   1 2)  ,  1    ,    1 
 2  2  
3)  1, 
1
4)  ,    1, 1
 2
71. The equations of the sides AB and AC of a triangle ABC are ( 1)x  y  4 and
x  (1 )y    0 respectively. Its vertex A is on the y-axis and its orthocentre is (1, 2).
The length of the tangent from the point C to the part of the parabola y 2  6 x in the first
quadrant is

1) 2 2) 6 3) 2 2 4) 4
    
72. Let   4i  3 j  5k and   i  2 j  4k . Let 1 be parallel to  and  2 be perpendicular
ˆ ˆ ˆ ˆ ˆ ˆ
    
to  . If   1  2 , then the value of 5 2 .(iˆ  ˆj  kˆ) is
1)11 2) 6 3) 9 4) 7
73. The number of integers greater than 7000 that can be formed using the digits 3, 5, 6, 7, 8
without repetition is
1) 48 2) 120 3) 168 4) 220
74. If the system of equations
x  2y  3z  3
4x  3y  4z  4
8x  4y  z  9  
Has infinitely many solutions, then the ordered pair (, ) is equal to
1)   72 ,  21  2)   72 , 21  3)  72 ,  21  4)  72 , 21 
 5 5   5 5   5 5   5 5 
75. If f ( x )  x 3  x 2 f '(1)  xf ''(2)  f '''(3), x  R , then
1) 3 f (1)  f (2)  f (3) 2) 2 f (0)  f (1)  f (3)  f (2)

3) f (3)  f (2)  f (1) 4) f (1)  f (2)  f (3)  f (0)
3
 2 2 
 1  sin 9  i cos 9 
76. The value of   is
2 2
 1  sin  i cos 
 9 9 
1 1 1 1
1)  ( 3  i) 2) ( 3  i) 3) (1  i 3) 4)  (1  i 3)
2 2 2 2

77. The locus of the mid point of the chords of the circle C1 : ( x  4) 2  ( y  5) 2  4 which
 2
subtend an angle i at the centre of the circle C1, is a circle of radius ri . If 1  , 3 
3 3
and r12  r22  r32 , then 2 is equal to
  3 
1) 2) 3) 4)
4 6 4 2
78. Let y = y(x) be the solution of the differential equation ( x 2  3 y 2 ) dx  3 xy dy  0 , y(1) =1.
Then 6 y 2 ( e ) is equal to
3 2
1) 3e2 2) e2 3) e 4) 2e2
2
79. Let A be a 3  3 matrix such that | adj ( adj ( adjA)) | 12 4 .Then | A1adjA | is equal to
1) 12 2) 2 3 3) 1 4) 6
80. Let f :S  S where S   0,  be a twice differentiable function such that
f  x  1  xf  x  .
If g : S R be defined as g  x   loge f  x  , then the value of g "  5   g " 1 is equal to :
205 197 187

1) 2) 3) 4) 1
144 144 144
81. If the shortest distance between the lines x  6  y  6  z  6 and

2 3 4
x y2 6 z2 6

  is 6, then the square of sum of all possible values of  is
3 4 5

82. Let f be a differentiable function defined on 0,   such that f(x) > 0 and
 2
x 2
2      
f ( x )   f (t ) 1  (log e f (t )) dt  e, x   0,  .Then  6 log e f  6   is equal to _
0
 2   
13  23  33  ....up to n terms 9
83. If  , then the value of n is
1.3  2.5  3.7  ....up to n terms 5
84. Three urns A, B and C contain 4 red, 6 black, 5 red, 5 black, and  red, 4 black balls
respectively. One of the urns is selected at random and a ball is drawn. If the ball drawn is
red and the probability that it is drawn from urn C is 0.4, then the square of the length of the
side of the largest equilateral triangle, inscribed in the parabola y 2   x with one vertex at
the vertex of the parabola is__
85. The minimum number of elements that must be added to the relation
R {(a,b),(b, c),(b, d)} on the set {a,b,c, d} so that it is an equivalence relation, is _____
86. Let the sum of the coefficient of the first three terms in the expansion of
n
 3  4
 x  2  , x  0, n  N , be 376. Then the coefficient of x is____
 x 
87. If the area of the region bounded by the curves y 2  2 y   x , x  y  0 is A, then 8A is equal
to ____
1 1 1
88. Let , a and b be in G.P. and , ,6 be in A.P., where a , b  0 . Then 72  a  b  is equal to
16 a b
           
89. Let a  iˆ  2 ˆj   kˆ, b  3iˆ  5 ˆj   k , a  c  7, 2b  c  43  0, a  c  b  c . Then | a  b | is
equal to__
90. The equations of the sides AB, BC and CA of a triangle ABC are:
2x  y  0, x  py  21a, (a  0) and x  y  3 respectively. Let P(2, a) be the centroid of
ABC . Then ( BC ) 2 is equal to


only.
(II) Section-II contains 10 Numerical Value Type questions. Attempt any 5 questions only, if more than 5
For example: To cancel attempted question 21. Bubble on 21 as shown belo w
.
SRI CHAITANYA IIT ACADEMY, INDIA 27-12-23_ Sr.Super60_Elite, Target & LIIT-BTs _Jee-Main_GTM-04_Q.P
Admission Number:
27-12-23_Sr.Super60_Elite, Target & LIIT-BTs _ Jee-Main_GTM-04_Test Syllabus


correct.
1. The velocity – displacement graph describing the motion of a bicycle is shown in the

v ms 1 
50
10
O
200 400 x  m
The acceleration –displacement graph of the bicycle’s motion is best described by.

a ms 2  
a ms 2 
18 18
2 2
O
200 400 x  m  O
200 400 x  m 
1) 2)

a ms 2 

a ms 2

18
18
2
2
O
O
200 400 x  m 200 400 x  m 
3) 4)
2. In order to determine the young’s Modulus of a wire of radius 0.2 cm (measured using a
scale of least count=0.001 cm) and length 1m (measured using a scale of least count=1mm),
a weight of mass 1kg(measured using a scale of least count =1 g)was hanged to get the
elongation of 0.5 cm(measured using scale of least count 0.001 cm).What will be the
fractional error in the value of Young ‘s Modulus determined by this experiment?
1) 0.9% 2) 0.14% 3)1.4% 4)9%
3. For changing the capacitance of a given parallel plate capacitor, a dielectric material of
dielectric constant K is used, which has the same area as the plates of the capacitor. The
3
thickness of the dielectric slab is d, where‘d’ is the separation between the plates of
4
parallel plate capacitor. The new capacitance  C ' in terms of original capacitance  C0  is
given by the following relation:
4 3 K 4K 4 K
1) C '  C0 2) C '  C0 3) C '  C0 4) C '  C0
3 K 4K K 3 3
4. An RC circuit as shown in the figure is driven by a AC source generating a square wave.
The output wave pattern monitored by CRO would look close to:
1) 2)
3) 4)
5. A block of 200 g mass moves with a uniform speed in a horizontal circular groove, with
vertical side walls of radius 20cm. If the block takes 40 s to complete one round, the normal
force by the side walls of the groove is:
1) 9.859  104 N 2) 6.28  103 N
3) 0.0314N 4) 9.859  102 N
6. The stopping potential in the context of photoelectric effect depends on the following
property of incident electromagnetic radiation:
1) Amplitude 2) Intensity 3) Frequency 4) Phase
7. A conducting wire of length ‘ l ’, area of cross-section A and electric resistivity  is
connected between the terminals of a battery. A potential difference V is developed its ends,
causing an electric current. If the length of the wire of the same material is doubled and the
area of cross-section is halved, the resultant current would be:
VA 1 VA 3 VA 1 l
1) 4 2) 3) 4)
l 4 l 4 l 4 VA
8. Four equal masses, m each are placed at the corners of a square of length ( l ) as shown in the
figure. The moment of inertia of the system about an axis passing through A and parallel to
DB would be:
D l C
m m
l
l
m m
A l B
1) 3ml 2 2) ml 2 3) 2ml 2 4) 3ml 2

9. The maximum and minimum distances of a comet from the Sun are 1.6  1012 m and
8.0  1010 m respectively. If the speed of the comet at the nearest point is 6  104 ms 1 , the
speed at the farthest point is :
1) 6.0  103 m / s 2) 1.5  103 m / s 3) 3.0  103 m / s 4) 4.5  103 m / s
10. In thermodynamics, heat and work are:
1) Path Functions 2) Extensive thermodynamics state variables
3) Point functions 4) Intensive thermodynamics state variables
11. The volume V of an enclosure contains a mixture of three gases, 16 g of oxygen, 28 g of
nitrogen and 44 g of carbon dioxide at absolute temperature T. Consider R as universal gas
constant. The pressure of the mixture of gases is:
4RT 88RT 3RT 5 RT
1) 2) 3) 4)
V V V 2 V
  
12. A charge Q is moved dl distance in the magnetic field B . Find the value of work done by B .
1) -1 2) 0 3) Infinite 4) 1
13. One main scale division of a vernier calipers is ‘a’ cm and nth division of the vernier scale
coincide with (n-1)th division of the main scale. The least count of the calipers in mm is:
10na  n 1 10a 10a
1) 2)  a 3) 4)
 n  1  10n   n  1 n
14. For an electromagnetic wave travelling in free space, the relation between average energy
densities due to electric (Ue) and magnetic (Um) fields is:
1)U e  U m 2)U e  U m 3)U e  U m 4)U e  U m
15. A block of mass m slides along a floor while a force of magnitude F is applied to it an angle
 as shown in figure. The coefficient of kinetic friction is  K . Then, the block’s acceleration
‘a’ is given by: (g is acceleration due to gravity)

SRI CHAITANYA IIT ACADEMY, INDIA

27-12-23_ Sr.Super60_Elite, Target & LIIT-BTs _Jee-Main_GTM-04_Q.P
F  F  F  F 
1) cos   K  g  sin   2)  cos   K  g  sin  
m  m  m  m 
F  F  F  F 
3) cos   K  g  sin   4) cos   K  g  sin  
m  m  m  m 
16. A conducting bar of length L is free to slide on two parallel conducting rails as shown in the
figure


R1  R2
B
Two resistors R1 and R2 are connected across the ends of the rails. There is a uniform

magnetic field B pointing into the page. An external agent pulls the bar to the left at a
constant speed .
The correct statement about the directions of induced currents I1 and I 2 flowing through R1
and R2 respectively is:
1) I1 is in clockwise direction and I 2 is in anticlockwise direction
2) I1 is in anticlockwise direction and I 2 is in clockwise direction
3) Both I1 and I 2 are in anticlockwise direction
4) Both I1 and I 2 are in clockwise direction

17. Time period of a simple pendulum is T inside a lift when the lift is stationary. If the lift
moves upwards with an acceleration g/2, the time period of pendulum will be:
3 T 2
1) T 2) 3) 3T 4) T
2 3 3
18. A plane electromagnetic wave of frequency 500 MHz is travelling in vacuum along y –

direction. At a particular point in space and time, B  8.0  108 zT
ˆ . The value of electric
field at this point is: (Speed of light  3  108 ms 1 ) xˆ , yˆ , zˆ are unit vectors along x, y and z
directions.
1) 24 xˆ V/m 2) 2.6 xˆ V/m 3) 2.6 yˆ V/m 4) 24 xˆ V/m
  
19. A body of mass 2kg moves under a force of (2 i + 3 j +5 k ) N. It starts from rest and was at
the origin initially. After 4s, its new coordinates are (8,b,20). The value of b is__________
(Round off to the Nearest Integer)
1) 12 2) 10 3) 9 4) 5
20. The pressure acting on a submarine is 3  105 Pa at a certain depth. If the depth is doubled,
the percentage increase in the pressure acting on the submarine would be:
(Assume that atmosphere pressure is 1  105 Pa; density of water is 103 kg m  3 , g = 10 ms 2 )
5 200 3 200
1) % 2) % 3) % 4) %
200 5 200 3
V
21. The resistance R  , where V  (50  2)V and I  (20  0.2) A . The percentage error in R is ‘x’
I
%.The value of ‘x’ to the nearest integer is……………...

22. Consider a frame that is made up of two thin mass less rods AB and AC as shown in the

figure. A Vertical force P of magnitude 100N is applied at point A of the frame.

Suppose the force is P resolved parallel to the arms AB and AC of the frame.
The magnitude of the resolved component along the arm AC is xN.
The value of x, to the nearest integer, is……………….
[Given : sin(350 )  0.573 , cos(350 )  0.819
sin(1100 )  0.939 , cos(1100 )  0.342]
23. The first three spectral lines of H-atom in the Balmer series are given 1 , 2 , 3 considering
 1 
the Bohr atomic model, the wave lengths of first and third spectral lines   are related by a
 3 
factor of approximately ' x ' 101 .The value of x, to the nearest integer, is………………...
24. Consider a 20 kg uniform circular disk of radius 0.2 m. It is pin supported at its center and is
at rest initially. The disk is acted upon by a constant force F=20 N through a mass less string
wrapped around its periphery as shown in the figure.

F= 20 N
Suppose the disk makes n number of revolutions to attain an angular speed of 50 rad s -1.
The value of n, to the nearest integer, is……………. [Given : In one complete revolution,
the disk rotates by 6.28 rad]
25. A fringe width of 6 mm was produced for two slits separated by 1 mm apart. The screen is
placed 10 m away. The wavelength of light used is ‘x’ nm. The value of ‘x’ to the nearest
integer is……………..

26. A force F  4iˆ  3 ˆj  4kˆ is applied on intersection point of x=2 plane and x-axis. The
magnitude of torque of this force about a point (2, 3, 4) is________ (Round off to the
Nearest Integer)
27. In the figure given, the electric current flowing through the 5 k resistor is ‘x’ mA.
3k
5k 3k
3k
21V ,1k
The value of x to the nearest integer is ……………..

28. The value of power dissipated across the zener diode Vz  15V  connected in the circuit as
shown in the figure is X 101 watt.
Rs  35
22V Vz =15V ~ RL  90
The value of X, to the nearest integer, is…………...

29. A sinusoidal voltage of peak value 250V is applied to a series LCR circuit, in which R  8 ,
L=24 mH and C= 60 F . The value of power dissipated at resonant condition is ‘x’ kW.
The value of x to the nearest integer is……………...
30. A ball of mass 10 kg moving with a velocity 10 3ms1 along X-axis, hits another ball of mass
20kg which is rest. After collision, the first ball comes to rest and the second one
disintegrates into two equal pieces. One of the pieces starts moving along Y-axis at a speed
of 10 m/s. The second piece starts moving at a speed of 20 m/s at an angle  (degree) with
respect to the X-axis. The configuration of pieces after collision is shown in the figure.
The value of  to the nearest integer is………………………..
After ollision
Y
X-Axis


correct.
31. Which of the following reaction does not involve Hoffmann bromamide degradation?
O
CH 2 - C -CH3 CH 2  NH 2
i)Br2 ,NaOH/H +

ii)NH3 /Δ
iii)LiAlH 4 /H 2 O
1)
O
NH 2
Cl

i)NH3 ,NaOH
ii)Br2 ,NaOH

2)
O
CH 2  NH 2
CH2 -C-NH 2

Br2 , NaOH

3)
CN NH 2
KOH , H 2O
Br2 , NaOH
4)
32. The exact volume of 1 M NaOH solution required to neutralize 50 mL 1M H 3 PO3 solution and
100 ml 2M H 3 PO2 respectively are
1) 100 ml , 50 ml 2) 50 ml , 50 ml 3) 100 ml , 200 ml 4) 100 ml, 100 ml

33. Arrange the following metal complex/compounds in the increasing order of spin only
magnetic moment. Presume all the three, high spin system.
(Atomic numbers Ce=58, Gd=64 and Eu=63)
a)  NH 4 2 Ce  NO3 6  b) Gd  NO3 3 c) Eu  NO3 3 answer is
1)a<c<b 2)c<a<b 3)a<b<c 4)b<a<c

34. Given below are two statements: one is labeled as Assertion A and the other is labeled as
Reason R:
Assertion A: Size of Bk 3+ ion is less than Np3+ ion.
Reason R: the above is a consequence of the lanthanoid contraction.
In the light of the above statements, choose the correct answer from the option given below:
2) Both A and R are true but R is not the correct explanation of A
35. Identify the elements X and Y using the ionization energy values given below:
Ionization energy (kJ/mol)
1st 2nd
X 495 4563
Y 731 1450
1)X=Na;Y=Mg 2)X=F;Y=Mg 3)X=Mg;Y=F 4)X=Mg;Y=Na
36. An unsaturated hydrocarbon X on ozonolysis gives A. compound A when warmed with
ammonical silver nitrate forms a bright silver mirror along the sides of the test tube. The
unsaturated hydrocarbon X is:
CH 3  C  C  CH 3
CH 3 CH 3
1) CH3  C  C  CH3 2)
CH 3
CH 3  C
3) 4) HC  C  CH 2  CH 3
Statement I: The E  value for Ce 4+ /Ce3+ is +1.74V .
Satement II: Ce is more stable in Ce 4  state than Ce 3  state.
In the light of the above statement, choose the most appropriate answer from the option
given below:

1) Both statement I and statement II are correct
2) Both statement I and statement II are incorrect
3) Statement I is correct but statement II is incorrect
4) Statement I is incorrect but statement II is correct
38. Which among the following pairs of vitamins is stored in our body relatively for longer
duration?
1) Vitamin A and Vitamin D 2) Ascorbic acid and Vitamin D
3) Thiamine and Ascorbic acid 4) Thiamine and Vitamin A
39. Match List –I with List-II
List-I List-II
(a) Lassaigne’s test (i) Carbon
(b)Cu(II) oxide (ii) Sulphur
(c)Silver nitrate (iii) N,S,P and halogen
(d) The sodium fusion extract (iv) Halogen specifically
gives black precipitate with acetic
acid and lead acetate
The correct match is:
1) a  iii; b  i; c  iv, d  ii 2) a  i; b  iv; c  iii, d  ii
3) a  i; b  ii; c  iv, d  iii 4) a  iii; b  i; c  ii, d  iv
40.
H3C OH CH3 Cl
20%H3PO 4  CH3 3 CO K 

"A' "B"
358k (Major Product) (Major Product)
The products “A” and “B” formed in above reaction
CH 3 CH 2 CH 3 CH 3
A B A B
1) 2)
CH 2 CH 2 CH 2 CH 3
A B A B
3) 4)
41.
OH
NH2
NaNO 2 ,HCl "A"

"X"
273-278K
Major Product
In the above chemical reaction, intermediate “X” and reagent/condition “a” are:
N 2 Cl  NO2
X
; A  H 2O /  X
; A  H 2O / NaOH
1) 2)
NO2 N 2 Cl 
X
; A  H 2O /  X
; A  H 2O / NaOH
3) 4)
42. A group 15 element which is a metal and forms a hydride with strongest reducing power
among group 15 hydrides. The element is:
1) As 2) Sb 3) Bi 4) P

43. In chromatography technique the purification of compound is independent of:
1) Solubility of the compound 2) Length of the column or TLC plate
3) Mobility or flow of solvent system 4) Physical state of the pure compound
44. Given below are two statements: one is labeled as assertion A and the other is labelled as
Reason R:
Assertion A: the H  O  H bond angle in water molecule is 104.5
Reason R: the lone pair-lone pair repulsion of electrons is higher than the bond pair –bond
pair repulsion. In the light of the above statements, choose the correct answer from the
option given below:
2) Both A and R are true, and R is the correct explanation of A
4) Both A and R are true, but R is not the correct explanation of A
45. Assertion A: Enol form of acetone CH 3COCH 3  exists in <0.1% quantity. However, the
enol form of acetyl acetone CH 3COCH 2OCCH 3  exits in approximately 15 % quantity.
Reason R: Enol form of acetyl acetone is stabilized by intramolecular hydrogen bonding
Which is not possible in enol form of acetone?
Choose the correct statement:
2) Both A and R are true but R is not the correct explanation of A
46. Which of the following is Lindlar catalyst?
1) Sodium and Liquid NH 3
2) Zinc chloride and HCl
3) Partially deactivated palladised charcoal
4) Cold dilute solution of KMnO4
47.
o
o
DIBAL-H,Toluene,-780C "P"
ii) H 3O
+
 Major product 
The product “P” in the above reaction is :

O

O  C H
CHO
1) 2)
OH
CHO COOH
3) 4)
48. Among the following, the aromatic compounds are:
CH2
A) B) C) D)
Choose the correct answer from the following option:
1) (B) and (C) only 2) (A) and (B) only
3) (A) (B) and (C) only 4) (B) (C) and (D) only
49. A central atom in a molecule has two lone pairs of electrons and form three single bonds.
The shape of this molecule is:
1) T-shaped 2) see-saw
3) trigonal pyramidal 4) planar triangular
50. Which of the following compound CANNOT act as a Lewis base?
1) SF4 2) PCl5 3) ClF3 4) NF3

51. Then decomposition of formic acid on gold surface follows first order kinetics. If the rate
constant at 300 K is 1.0 103 S 1 and the activaton energy Ea  11.488 KJ mol 1 , the rate constant
at 200K is ……….. 105 s 1 . (Round off to the Nearest Integer). (Given: R  8.314 J mol 1 K 1 )
52. 2MnO4 +bC2O42- +cH+  xMn 2+ +yCO2 +zH2O
If the above equation is balanced with integer coefficients, the value of c is…….
(Round off to the Nearest Integer).
53. AB2 is 10% dissociated in water to A 2+ and B the boiling point of a 10.0 molal aqueous
solution of AB2 is ………..  C . (Round off to the Nearest Integer).
[Given: Molal elevation constant of water K b  0.5 K kg mol 1 boiling point of pure water
=100  C
54. When light of wavelength 248nm falls on a metal of threshold energy 3.0 eV, the de-Broglie
wavelength of emitted electrons is …….. A . (Round off to the Nearest Integer).
[Use: 3  1.73, h  6.63  10 34 ] Js
31
[ me  9.110 kg; c  3.0 108 ms1;1eV  1.6 1019 J ]
For the reaction A( g )  B ( g ) at 495 K  r G  9.478kJ mol 1 .

55.
If we start the reaction in a closed container at 495K with 22 mill moles of A, the
amount of B in the equilibrium mixture is……… mill moles (Round off to the Nearest
Integer)
[ R  8.31J mol 1 K 1 ; ln10  2.303 ]
56. The equivalent of ethylene diamine required to replace the neutral ligands from the
coordination sphere of the trans-complex of CoCl3 4 NH 3 is ………( Round off to the nearest
Integer).
57. A 6.50 molal solution of KOH (aq). Has a density of 1.89g cm-3. The molarity of the solution
is….. mol dm-3. (Round off to the nearest Integer). [Atomic masses: K:39.0u; o:16.0u; H:1.0
u]
58. Two salts A2 X and MX have the same value of solubility product of 4.0  1012 . The ratio of
S  A2 X 
their molar solubilities i.e. =…(Round off to the nearest integer).
S ( MX )
59. Complete combustion of 750g of an organic compound provides 420 g of CO 2 and 210 g of
H2O. Then percentage composition of carbon and hydrogen in organic compound is 15.3
and……. Respectively. (Round off to nearest Integer).
60.  6

0.01 moles of a weak acid HA Ka  2.0 10 is dissolved in 1.0 L of 0.1 M HCl solution.
The degree of dissociation of HA is ______ 10 5 (Round off to the Nearest Integer).
[Neglect volume change on adding HA. Assume degree of dissociation <<1]

correct.
61. Let [x] denote greatest integer less than or equal to x. If for n  N ,
 3n   3 n 1 
 2  2 
3n    
1  x  x    a x
3 n
j
j
, then a 2j 4  a2 j 1 is equal to :
j 0 j 0 j 0
n1
1) n 2) 2 3) 2 4) 1
  1
62. If for x   0,  ,log10 sin x  log10 cos x  1and log10  sin x  cos x    log10 n  1 , n  0 ,
 2 2
then the value of n is equal to:
1) 16 2) 9 3) 20 4) 12
k
 6 r

63. Let Sk   tan 1  2 r 1 2 r 1  . Then lim Sk is equal to :
r 1 2 3  k 
 3  3
1) 2) tan 1   3) cot 1   4) tan 1  3
2 2  2
64. Let the position vectors of two points P and Q be 3iˆ  ˆj  2kˆ and iˆ  2 ˆj  4kˆ , respectively.
Let R and S be two points such that the direction ratios of lines PR and QS are (4, -1, 2) and
(-2, 1, -2), respectively. Let lines PR and QS intersect at T. If vector TA is perpendicular to
  
both PR and QS and the length of vector TA is 5 units, then the modulus of a position
vector of A is:
1) 171 2) 5 3) 482 4) 227
65. In a group of 100 persons 75 speak English and 40 speak Hindi. Each person speaks at
least one of the two languages. If the number of persons, who speak only English is α and
the number of persons who speak only Hindi is β , then the eccentricity of the
ellipse 25 β 2 x 2  α 2 y 2   α 2β 2 is
119 3 15 129 117
1) 2) 3) 4)
12 12 12 12
66. The least value of z where ‘z’ is complex number which satisfies the in-equality
  z  3 z  1 
exp  log e2   log 5 7  9i , i  1, is equal to:

  z  1 

2
1) 5 2) 8 3) 2 4) 3
67. Given that the inverse trigonometric function take principal value only. Then the number
of real value of x which satisfy sin 1 

3x  1  4 x 
  sin    sin x is equal to:
1
 5   5 
1) 2 2) 3 3) 0 4) 1
       
68. Let a  iˆ  2 ˆj  3kˆ and b  2iˆ  3 ˆj  5kˆ . If r  a  b  r , r .( iˆ  2 ˆj  k )  3 and
    2
r .(2 i  5 j   k )  1,   R , Then the value of   r is equal to :
1) 11 2) 9 3) 15 4) 13
 
1 1 60
69. If n is the number of irrational terms in the expansion of 3  5 4 8
, then (n – 1) is
divisible by :
1) 26 2) 7 3) 8 4) 30
70. Let the functions f : R  R and g : R  R be defined as:
 x  2, x0  x3 x 1
f  x   2 And g  x   
 x , x0 3x  2, x  1
Then, the number of points in R where (fog) (x) is NOT differentiable is equal to:
1) 0 2) 2 3) 1 4) 3
71. A pack of cards has one card missing. Two cards are drawn randomly and are found to be
spades. The probability that the missing card is not a spade, is :
3 52 22 39
1) 2) 3) 4)
4 867 425 50
 z  11 
72. Let a complex number z , z  1, satisfy log 1    2 . Then, the largest value of z is
2  z  1
 2

 
equal to _________
1) 8 2) 7 3) 5 4) 6

 i i  8 x  8 
73. Let A    , i  1 . Then, the system of linear equations A 
 y   64  has:
 i i     
1) Infinitely many solutions 2) Exactly two solutions
3) No solution 4) A unique solution
74. Let a vector  iˆ   ˆj be obtained by rotating the vector 3iˆ  ˆj by an angle 450 about the
origin in counterclockwise direction in the first quadrant. Then the area of triangle having
vertices  ,   ,  0,   and (0, 0) is equal to :
1 1
1) 2) 1 3) 2 2 4)
2 2
sin 2 x 1  cos 2 x cos 2 x

75. The maximum value of f  x   1  sin 2 x cos 2 x cos 2 x , x  R is :
sin 2 x cos 2 x sin 2 x
3
1) 5 2) 7 3) 4) 5
4
dy  
76. If y = y(x) is the solution of the differential equation, 2 y tan x  sin x, y    0 , then
dx 3
the maximum value of the function y(x) over R is equal to :
1 15 1
1) 8 2) 3)  4)
8 4 2
77. 
The number of elements in the set x  R :  x  3 x  4  6 is equal to: 
1) 3 2) 4 3) 1 4) 2
78. The range of a  R for which the function
x x
f  x    4a  3  x  log e 5  2  a  7  cot   sin 2   , x  2n , n  N has critical points, is:
2 2
 4 
1)  3,1 2) [1, ) 3) (,  1] 4)   ,2
 3 

dy
79. Let C1 be the curve obtained by the solution of differential equation 2 xy  y 2  x 2 , x  0.
dx
2 xy dy
Let the curve C2 be the solution of 2 2
 . If both the curves pass through (1, 1), then
x y dx
the area enclosed by the curves C1 and C2 is equal to :
 
1)  1 2) 1 3) 1 4)   1
4 2
80. Consider three observations a, b and c such that b = a + c. If the standard deviation of
a + 2, b + 2, c + 2 is d, then which of the following is true ?
2

1) b  3 a  c  d
2 2 2
 2) b 2  a 2  c 2  3d 2
3) b
2
 3 a 2
 c 2   9d 2 2 2

4) b  3 a  c  9d
2 2

81. Consider an arithmetic series and a geometric series having four initial terms from the
set {11, 8, 21, 16, 26, 32, 4}. If the last terms of these series are the maximum possible four
digit numbers, then the number of common terms in these two series is equal to _________
dy
82. Let the curve y = y(x) be the solution of the differential equation,  2  x  1 . If the
dx
4 8
numerical value of area bounded by the curve y = y(x) and x – axis is , then the value of
3
y (1) is equal to _______
83. Let f : R  R be a continuous function such that f  x   f  x  1  2 , for all x  R . If
8 3
I1   f  x  dx and I 2   f  x  dx , then the value of I 1  2 I 2 is equal to _______
0 1
 30 20 56  2 7 2 
  1  i 3
84. Let P   90 140 112  and A   1  1  where   , and I3 be the
  2
120 60 14   0    1

identity matrix of order 3. If the determinant of the matrix  P 1 AP  I 3  is  2 , then the
2
value of  is equal to ___________

85. The total number of 3  3 matrices A having entries from the set {0, 1, 2, 3} such that the
T
sum of all the diagonal entries of AA is 9, is equal to _________
ae x  b cos x  ce x
86. If lim  2 , then a + b + c is equal to _____________
x 0 x sin x
87. For real numbers  , ,  and  , if
 x2 1 
x  1  tan 1 
2
 x  dx   log  tan 1  x  1     tan 1    x  1    tan 1  x  1   C

 2  2
 2
 4 2 1  x  1 
2 e 
x 
  x  
x 

 x  3x  1 tan  x      
 
where C is an arbitrary constant, then the value of 10        is equal to _____
88. Let ABCD be a square of side of unit length. Let a circle C1 centered at A with unit radius is
drawn. Another circle C 2 which touches C1 and the lines AD and AB are tangent to it, is
also drawn. Let a tangent line from the point C to the circle C 2 meet the side AB at E. If the
length of EB is α+ 3β , where α,β are integers, then α+β is equal to _________
z+i
89. Let z and w be two complex numbers such that w=zz-2z+2, =1 and Re (w) has
z-3i
minimum value. Then, the minimum value of n  N for which wn is real, is equal to
_______
90. If the variance of the frequency distribution
xi 2 3 4 5 6 7 8
Frequency fi 3 6 16  9 5 6
is 3, then  is equal to ______.


SRI CHAITANYA IIT ACADEMY, INDIA 27‐12‐23_ Sr.Super60_Elite, Target & LIIT‐BTs _Jee‐Main_GTM‐04_KEY &SOL’S

KEY SHEET
PHYSICS
1) 2 2) 3 3) 3 4) 4 5) 1
6) 3 7) 2 8) 4 9) 3 10) 1
11) 4 12) 2 13) 4 14) 2 15) 1
16) 1 17) 4 18) 1 19) 1 20) 4
21) 5 22) 82 23) 15 24) 20 25) 600
26) 20 27) 3 28) 5 29) 4 30) 30
CHEMISTRY
31) 1 32) 3 33) 1 34) 3 35) 1
36) 4 37) 3 38) 1 39) 1 40) 1
41) 1 42) 3 43) 4 44) 2 45) 4
46) 3 47) 3 48) 1 49) 1 50) 2
51) 10 52) 16 53) 106 54) 9 55) 20
56) 2 57) 9 58) 50 59) 3 60) 2
MATHEMATICS
61) 4 62) 4 63) 3 64) 1 65) 1
66) 4 67) 2 68) 3 69) 1 70) 3
71) 4 72) 2 73) 3 74) 4 75) 4
76) 2 77) 4 78) 4 79) 3 80) 3
81) 3 82) 2 83) 16 84) 36 85) 766
86) 4 87) 6 88) 1 89) 4 90) 5

SOLUTIONS
PHYSICS
1. Key: 2
dv
Sol: Slope of velocity – time graph is acceleration. Similarly, V =a
dx
dv
Where  0  a  0 . So 2nd or 4th option is correct
dx
dv
is constant & V is increasing linearly . ‘a’ also increases linearly .
dx
So only 2nd option is correct.
2. Key: 3
mgL dY dm dL dr de
Sol: Y  , %(  2  )  100
 r 2e Y m L r e
[103 103 1(102 )  2(103)]100 1.4%
3. Key: 3
Sol:
3d / 4
d /4
0 A
C
3d d

4k 4
4. Key: 4
Sol: During charging, V=V0 1-e -t/  
Voltage increases exponentially with time.
During discharging, V=V0 e -t/
Voltage decreases exponentially with time.
2
200 20  2 
5. Key: 1, N  mr 2    
1000 100  40 
6. Key: 3
Sol: hv=W0 +Kmax =W0 +ev0  Frequency decides stopping potential.
7. Key: 2
V ρ  2l  V
Sol: i  , R '= =4R, i ' =
R A/2 4R
8. Key: 4
Sol:
1
D C
B
A
2 2
 l   l 
I1 =2m   , I 2 =I1 +4m  
 2  2
9. Key: 3
Sol: Labout sun =consant
V1r1  V2 r2
 6  104 8  1010   V2 1.6  1012 

10. Key: 1
Sol: Heat and work are path functions.
Internal energy, entropy are state functions.
11. Key: 4
Sol:
P=PO 2 +PN 2 +PCO 2
1 RT RT RT 5 RT
= + + =
2 V V V 2 V
12. Key: 2
  
Sol: F  qV  B
 
P  F V  0 W  0
13. Key: 4
Sol: 1MSD= a cm
1MSD a 10a
LC= = cm  mm
number of vsd n n
14. Key: 2
Sol: <Magnetic field energy density >=<Electric Field energy density >
15. Key: 1
F cos    k  mg-Fsinθ 
Sol: a
m

F sin + N
Fcos
mg
16. Key: 1
Sol: Induced current always opposes change in flux
17. Key: 4
Sol: S
l l 2
T1  2   2  T
g eff 3g / 2 3
18. Key: 1
Sol:
E
C=
B
E=CB=24v/m
Direction of  E×B  is direction of wave.
19. Key: 12

 F 2iˆ  3 ˆj  5 kˆ    1
Sol: a  r f  ri  ut  at 2
m 2 2
1  2iˆ  3 ˆj  5kˆ  2

 
r f  0iˆ  0 ˆj  0kˆ   
2  2
  4 


r f  8iˆ  12 ˆj  20 kˆ  b  12
20. Key: 4
P gh 200
Sol: Po +gh=3×105 Pa gh  2 105 Pa, 100  100  %
P  Po  gh 3
21. Key: 5
dR dV dI 2 0.2
Sol:  100   100   100   100   100  4 1
R V I 50 20
22. Key: 82
Sol:
700 A
A 350
B C
700
100N P

Component of ‘P’ along AC  Pcos35  100 0.819  81.9N  82N

23. Key: 15
1 1 1  5  1 1 1   21 
Sol:  R     R     (1)  R     R     (2)
 4 9  36  3  4 25   100 
21 36
(2) / (1)    1.5
100 5
24. Key: 20
Sol:
2
20  0.2 
c  20  0.2   I       10rad / S 2
2
2500
2  o 2  2,502  0  2 10     125  n (2)
20
125
n  19.9  20
6.28
D  10 
25. Key: 600,   6mm   ,   600 n m
d 1mm
26. Key: 20
  
  r  F ; As,       
Sol: r  (2 i  3 j  4 k )  2 i  3 j  4 k
iˆ ˆj k
 
 0 3 4   iˆ 12  12   ˆj  0  16   kˆ  0  12 
4 3 4
 
  16 ˆj  12 kˆ   16 2   12 2  256  144  20
21V
27. Key: 3, Reff  7k , i 
=3mA
7k 
15 1
28. Key: 5, Current through 90Ω= = A
90 6
Voltage across 35Ω=22-15=7v
1 1 1 1
Current through Zener diode=  A , Power =Vi =15    0.5W
5 6 30  30 
250 250 1
29. Key: 4, At resonance XL =XC power = Vrmsirms    3.906 kW
2 2 8
30. Key: 300
Sol: No net Force is along Y-axis, we can conserve momentum along Y-axis
 Pi    Pf 
y y
0   5  10  ˆj  20sin   - j 
sin   1 / 2
CHEMISTRY
31. Key: 1
Sol:
O
O
CH 2 -C- CH 3 NH 2
CH 2 -C- Br C

1) Br2 / nAOH
2) H 

NH 3


O
NH 2

LiAlH 4
H 2O

32. Key: 3
Sol: 2NaOH  H3PO3  Na2HPO3  2H2O
M b  1M M a  1M
M aVa M bVb 1 50 1 x
Vb  x Va  50 mL  
na nb 1 2
nb  2 na  1
x = 100 ml
NaOH  H3PO2  NaH2PO2  H2O
M b  1M M a  2M
μaVa μbVb 2 100 1 x
Vb  x Va  100  
na nb 1 1
nb  1 na  1
x  Vb  200mL
33. Key: 1
Sol: a)In  NH 4 2 Ce  NO3 6 
Ce oxidation state is +4
EC of Ce :[Xe]6s 5d 4f0, therefore n=0, hence magnetic moment =0
+4 0 0
EC of Gd+3:[Xe]6s05d04f7, therefore n=7, hence magnetic moment =7.93

EC of Eu+3:[Xe]6s05d04f6, therefore n=6, hence magnetic moment =6.92
34. Key : 3
Sol: Size decreases from left to right in 5f series due to actinide contraction.
35. Key: 1
Sol: For X IP1<<<IP2 there fore second electron is in (n-1) orbit – hence x is Na
For Y IP1<IP2 there fore second electron is in (n) orbit– hence Y is Mg
36. Key: 4
Sol: 1) CH3  C  C  CH3  
ozonolysis
CH 3  C  0

ammonical . AgNO3
 No reaction
H 3C  C  0
H 3C CH 3 CH 3
C =0   ammonical . AgNO
C =C 
ozonolysis
 2 3
No Reaction
2) H 3C CH 3 H 3C
CH 3
=O + O =
CH 3
C= 
Ozonalysis

3)
H 3C H 3C
H C  0
HO
H C  C  CH  CH 3  
Ozonalysis
 
2 AgNO3
H 3C  C  0
NH 3 l  C =0 + Ag (mirror)
C =0
4) H 3C

37. Key : 3
Sol: Statement I is correct but statement II is incorrect.
Reason: Ce is more stable in Ce3+ state than Ce4+ stable due to higher SRP value.
Ce4+ is a good oxidizing agent and converted to Ce3+
38. Key: 1
Sol: Sol: Vitamin ‘A’ and vitamin ‘D’ are stored in our body for longer duration, they are fat
soluble vitamin.
39. Key: 1
Sol: a) Lassaigne’s test is used to identify the elements N,S,P and halogens
b) Cu  II  O is used to identify ‘Carbon’
c) AgNO3 is used to identify halogen specifically
 
d) 2 Na   s 2  Na2 S   H 2 S   pbs  (black )
CH COOH 3
pb CH3COO 2
40. Key: 1
Sol:
CH 3
H2C OH
Saytzeff's product 
20%H 3 PO 3
358k
CH 2
H2C Cl
 CH3 3 CO K   Hoffmann's Product 
41. Key : 1
Sol:
NH 2 N 2  Cl OH

NaNO 2
HCl
 
H 2O


42. Key: 3
Sol: As we go down the group M-H bond strength decreases, and hence reducing power
increases.
43. Key: 4
Sol: Purification process in chromatography doesn’t depend on physical state of pure
compound.
44. Key: 2
Sol: Due to lp-lp repulsion bond angle decreases
45. Key: 4
Sol: Enol form of acetyl acetone is stailised by intermolecular H-bonding hence enol
content increases
46. Key: 3
Sol: Lindlar’s catalyst consists of palladised charcoal poisoned by lead, sulfur or quinoline.
47. Key: 3
Sol:
o CHO
OH
o
CH 2
0

i)DIBAL-H,TOLUENE,-78 C
ii)H O+

3
48. Key: 1
Sol: B&C have cyclic conjugation and they satisfy huckel’s rule  6  e   . So they are
aromatic.
49. Key: 1
Sol:
AB3E2  type T-shaped
B
E
A B
E
B
50. Key: 2
Sol:
( PCl5 has no lone pairs on central atom to donate)
Cl
Cl
Cl P
Cl
Cl No lone pairs
F
F
S
F
F One lone pair
F F
F One lone pair
F
Cl F
F Two lone pairs

51. Key: 10
Sol:
T1  200, T2  300
K1  ? K 2  10 3
 10 3  11.488  103  1 1 
log     
 
 K1  2.303  8.314  200 300 
 10 3  11488  100   103   11488 
log    ,log  
 K1  19.147  200  300   K1   11488 
   
log10 3  log K1  1, 3  1  log K1
log K1  4, K1  10 4 , K1  10  10 5
52. Key: 16
1 2 + 2+
Sol: 2Mn O 4  5C 2 O 4  16H  2Mn +10CO 2  8H 2 O, C =16
53. Key: 106
Tb  1.2  0.5  10  6
Sol: Given α=0.1,n=3 α= i  1  i=1.2
n 1 B.Pof solution  100  6  1060 C
54. Key: 9
hc
Sol: Energy of incident radiation =
λ
6 .6 3  1 0  3 4  3  1 0 8
2 4 8  1 0 9
0 .0 8 0 2  1 0  1 7
8 .0 2 × 1 0 -1 9 J
 w  w o r k f u n c t i o n = 3 e v = 3  1 .6 0 2  1 0  1 9 J
 4 .8 0 6 × 1 0 -1 9 J
K .E o f p h o t o e - = Î i n c -w
= 8 .0 2 × 1 0 -1 9 - 4 .8 0 6  1 0  1 9
 3 .2 1 4  1 0  1 9 J
h
d e b r o g li w a v e le n g t h λ =
2m K E
6 .6 3  1 0  3 4
λ=
2  9 .1  1 0  3 1  3 .2 1 4  1 0  1 9
6 .6 3  1 0  3 4 6 .6 3
λ=   1 0 9
5 8 .4 9  1 0 50 7 .6 4
 0 .8 6  1 0  9 m
0
λ = 8 .6 A
55. Key: 20
Sol:

Given  A   22
G   9.478
9.478  103  2.303  8.314  495log

 B
A
log
 B

9.478  103
 A  2.303  8.314  495
log
 B  9478
 A  9478
log
 B  1
A
 B  10
A
x   A   10  x   B
  22  x   10  20
  22  x   2
x  20
56. Key: 2
Sol:
NH 3
H3 N
Cl
Co Cl
Cl
H3 N
NH 3
NH 2 NH 2
CH 2  CH 2
Ethylene diamine is a bidentate ligand . So 2 equivalents of Ethylene
diamine replace 4 NH3 ligands.
57. Key: 9
Sol:

1000×M
m=
1000×d  -  M×  GMW solute 
1000×M
6.5=
1000×1.89  -  M×56 
6.5 1890-56M  =1000M
1890-56M=153.8M
209.8M=1890
M=9
58. Key: 50
Sol:
For AX2
4s3  4  1012  S1  3 1012
S1  104 M
For MX
S22  4  1012
S2  2  106
S1 104 100
 6
  50
S 2 2  10 2
59. Key: 3
Sol: Mass of H2Oformed=210g
We know that 18g of H2Oproduced by 2g of H2
210g of H2O____________ ?
=23.33g H2
23.33
Mass % of H 2  100  3
750
60. Key: 2
 H    A 
  
Sol: ka 
 HA
0.1  A 
6
2  10 
0.01
 A  2  10 7  c

 
2 107 102   2105 =2

MATHEMATICS
61. Key: 4
Sol: x  1 a 0  a 1  a 2  a 3  ............  1
x = -1 a 0  a 1  a 2  a 3  .........   1
 a 0  a 2  .... ....  1 & a1  a 3  ......  0
62. Key: 4
1 n
 sin x  cos x 
2
Sol: sin x.cos x  
10 10
n 1 n  6
1  sin 2 x  1  n    10 12
10 5 10  5
63.
Key: 3
Sol:
 3r    3  r 1  3  r 
       
1   2  2
K r 1
S K   tan 1  2
2 r 1
   tan r 1

r 1  3   3 3
r

1     1    .  
 2   2 2 
3
r 1
3
r
  3 3
  tan  1 1
 tan     tan1    cot 1
2 2 2  2 2
64. Key: 1
Sol:
d.r of PR (4, -1, 2)
d.r of GS (-2, 1, -2)

d.r of TA= (0, 4, 2)

Pt. on PR  3  4 ,  1   , 2  2 
Pt. on GS 1  2 , 2   ,  4  2 
Equating x and y co ordinates
3 4 1 2
2  1 ….. (1)
   3 …..(2)
  2
T(11, -3, 6)
 11,  3  4 , 6  2  co-ordination of A
TA2 162  42  5
1
202  5  
2

OA 11,  1, 7  or (11, -5, 5)

 OA  171
n  A  B  100
65.  75  40  n  A  B  100  n  A  B  15
α  60,β  25
 
25 β 2 x 2  α 2 y 2  α 2β 2
25  (25) 2 x 2  (60) 2 y 2   (25) 2 (60) 2
 25 25 25x 2  144y2   (25)2 (60)2

25x 2 144y2
 1
(60)2 (60)2
x2 y 2
 1
144 25
144  25 119
e 
144 12
66. Key: 4
 z  3  z 1
Sol: 2  z 1  2 3 
 z  3 z  1  3  z 2  2 z  3  3 z  3
z 1
2
 z  z 60  z  3  z  2   0
67. Key: 2
3x 16 x 2 4 x 9 x2
Sol: Taking sine on both sides 1  1 x
5 25 5 25
 3 x 25  16 x 2  25 x  4 x 25  9 x 2

 x  0 or 3 25  16 x 2  25  4 25  9 x 2
 9  25  16 x2   625  200 25  9 x2  16  25  9 x2 
 200 25  9 x 2  800  25  9 x 2  4
 x2 1  x   1 and x0  Total number of solutions  3
68. Key: 3
        
Sol: r  a  b  r  0  r  (a  b )  0  r  (a  b)  (3i  j  2k)
  (3i  j  2 k ).( i  2 j  k )  3  3  2  2  3    1
Also  (3i  j  2 k ).(2 i  5 j   k )   1  (6  5  2 )   1,   1   1
69. Key : 1
3  51/8 
1/4 60
Sol:
60  r
60
Cr 3 4
.5r /8
For rational terms r = 0, 8, 16, 24, …….. 56
i.e. 8 terms
Irrational terms 61 – 8 = 53 = n
n – 1 = 52
70. Key: 3
Sol:
 x3  3 ; x0

fog   x 6 ; 0  x 1

 3 x  2  ;
2
x 1
At x = 0 discontinuous hence not differentiable
At x = 1 cont. and differentiable
71.
Key: 4
Sol: E  Lost card not spade
A  both cards drawn are spade
3 1
P E 
4
P EC 
4
 
13 12
 A C  A C
P    51 2 P  C   52 2
E C2 E  C2
 A
PEP 
E E 39
P   
 A  P  E  P  A   P E C .P  A  50
 
E
   E C 
72. Key: 2

z  11 1
Sol: 
 z  1
2
2
 z  7  z  3   0
3  z  7
73. Key: 3
 i i   i i   2 2 
Sol: A2     
 i i   i i   2 2 
 8 8  128 128
A4    ; A8   
 8 8   128 128 
128  x  y   8
| No solution
128   x  y   64
74. Key: 4
Sol:
i   j
2
75o 3i  j
45 o
30o
3 1
  2cos75o 
2
3 1
  2sin750 
2
1 1  3 1  1
Area   .   
2 2 2  2
75. Key: 4
2 1  cos 2 x cos 2 x
Sol: C1 C1  C2 f  x  2 cos 2 x cos 2 x ,
1 cos 2 x sin 2 x

0 1 0
R1  R1  R2 f  x   2 cos 2 x cos 2 x ,
1 cos 2 x sin 2 x
f  x   2sin 2 x  cos 2 x f  x max  5
76.
Key: 2
Sol:
I .F  e 
2 tan xdx
2
 elogsec x  sec2 x
sin x
y.sec 2 x   dx
Solution cos 2 x
y.sec2 x  sec x  c
0×4=2+c  c = -2
y = cos x-2cos 2 x
Max. Will occur for
1 1 1 1 1
cos x    y  2 
2  2 4 4 16 8
77. Key: 4
Sol: (1) x   4  x  3 x  4  6
x  1,  6  x  6
(2) x   4, 0 
 x  3 x  4   6  0 No root
(3) x0
 x  4 x  3  6
 1  D 1  D  One root (one is positive)
x ,
2 2
78.
Key: 4
Sol:
f '  x   4a  3   a  7  cos x  0
Should have solution
3  4a
So   1, 1
7a
3  4a
 1
7a
3  4a  7  a
0
7a

2a
0 a(, 2] [7, )
7a
3  4a
1
7a
3  4a  7  a
0
7a
3a  4  4 
0 a   , 7 
a7  3 
 4 
Take intersection  a   , 2
 3 
79. key: 3
dy y2  x2
Sol:  Put y  vx
dx 2xy
vx
dv v 2 x 2  x 2 v 2  1 dv v  1  2v
  x 
2 2

 v2  1  
dx 2 vx 2 2v dx 2v 2v
2v

v2  1
dv  
dx
x
n  v 2
 1   nx  nc  v 2
 1 
c
x
y2 c
 2
 1   x 2  y 2  cx If pass through (1, 1)  x2  y 2  2 x  0
x x
dx x2  y2
Similarly for second differential equation 
dy 2xy
Equation of curve is x 2  y 2  2 y  0
Now the required area is   1    12  1  1  1   2     1  sq. units
4 2  2 
y
1,1
 0,1
O C 1,0
x
80.
Key : 3
Sol:
x1  a  2 , x 2  b  2 , x 3  c  2 n=3
abc
x 2
3

 x  x 
2
S.P = d = i
3
2
 a bc
 a  3 

3
27d 2   2a  b  c
2
27 d 2  6 a 2  6 ab
 6 a2  b2  c2  ac  2a2  2c2  2ac
9d 2  a2  c2  ac  2a2  2c2  2ac
b2  a2  c2  2ac
Add b  9d  3 a  c
2 2
 2 2

81. Key: 3
Sol: +A.P 11, 16, 21, 26, 31, 36, ……
G.P 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192
Common terms 16, 256, 4096
82. Key: 2
Sol:  dy   2 x 1 dx
y   x  1  C
2
C should be negative
1  C  x  1
2
C 0
x  1  C
-1
  x  1
3
1  c
  x  1  Cx |11 C

2
 C  dx 1  C 
 3 C
1  c
    
3 3
 C C   C 3/2  C 3/2 
    2  C 
3/2
=  C 2 C   
3 3  3 3 
 2  4 8 4 4 8 4.23/2
   2   C     C  
3/2 3/2

 3  3 3 3 3
–C=2  C = –2 y 1  2 2  2 =2
83. Key: 16
Sol: f  x   f  x  1  2 f  x  1  f  x  2  2  f  x  2  f  x 
2 3
I1  4 f  x  dx I 2  4  f  x  dx x  t 1  dx  dt
0 1
2 2 2
I2 
2
 f  t  1 dt  2 f  x  1 dt I
0
1  2 I 2  4  f  x   f  x  1 dx
0
= 16
84. Key: 36
P AP  I    P 1 AP  P 1P    P 1  A  I  P   P1 A  I P
1 2 2 2 2 2 2 2
Sol:  A I
2
2
2
1 7 2 1 7
 1   1 1  0   6 
0   0  
R2  R1
      6    2    2  6   2    6 
2 2 2
  36
85. Key: 766
a b c  a d g   a 2  b 2  c 2 ad  be  cf ag  bh  ci 
     
Sol: A   d e f   A   b e h 
T
AAT   da  eb  fc d 2  e 2  f 2 dg  eh  fi 
 g h i   c f i   ga  hb  ic gd  he  if g 2  h 2  i 2 

Given, a  b  c  d  e  f  g  h  i  9
2 2 2 2 2 2 2 2 2
a, b, c, d , e, f , g , h, i 0,1,2,3
(i) One of them is 3, remaining are 0s
9 ways
(ii) 2 of them are 2s, one of them is 1 and remaining 0s
9C2 .7 C1  252
ways
(iii) One of them is 2, five of them are 1s and remaining 0s
9C1.8C5  504
Ways
(iv) All are 1s
1 way
Total = 9 + 252 + 504 + 1 = 766 ways
86. Key: 4
Sol:
 x2   x2   x2 
a 1  x    b 1    c 1  x  
2  2  2 
lim  
2
 2
x 0 x
x2
a  b  c  x a  c  a  b  c
lim 2 2
x0 x2
abc
a  b  c  0 and a  c  0 and 2  abc4
2
87. Key: 6
 1
x 2 1  2  dx
Sol : I  
 x2 1 dx  4
dx
I   x  
dx
,
4 2 1  x  1 
 x  3x  1 tan  x 
2
 x  3x  1
2
2 2 1  1  x  1 
x  x  3  2  tan 
2
 x 2 2
 x  3 
1

   x   x   x2 
 1 
 1  2  dx  1  1
I1    x  , Take tan 1  x     x  1  tan  1  2  dx  sec2  d
 1
2
 1  1  x x  x 
  x    1 tan  x  
 x   x
 1  1
1 1  2   1  2 
sec  d 
2
  1   x   x dx
  log   log  tan 1  x    I 2  
1  tan  
2
  x  2 1
x2  3  2
x
 1  1  1  1
 1  2  dx  1  2  dx 1 1  x   x 
1 1 1
I2   
x 
 
x   tan 1  x  tan 1 x
  
2  1 2
2  1 2
2 5  5  2  1 
 x  5  x   1    
 x  x
 1  1
 1  1  1 1 1
 x  x  1 1  x  x 
 log  tan  x     tan    tan  
  x  2 5  5  2  1 
   
 1 2 
 1  1  1 1   x  1 
1  x2  1 
 log  tan  x     tan 1  5   tan 1  
  x  2 5  x  2  x 
 
 
1 1 1  1 1
  1,   ,  ,  10        10 1     10  1  5  6
2 5 5 2  10 2 
88. Key: 1
Sol:
 0,1 1, 1
t, t 
1, 0 
 0, 0 
 t  2t 1  0  t  2  1
2
1  2t  t
Now &
y  1  m x  1  mx  y 1 m  0
P=r

  
2 1 m  
2 1 1 m

1  m2
  2 1

m  2 2   2 2  2 1    m  1 2  1
1 m 2 2
1 m

 2 m  1  2m  m  1
2 2
  m  4m1 0
2
 m  4  12 2 3
2
y  1  2  3  x  1 y=0
1 1 3 1
  x  1 x 1 =
2 3 2 3 32
 3 1   2 3  3  2  3  3 1
1
 EB = 1   3 1 
= 2 3  
= 2 1  1
89. Key: 4
Sol: z  xiy and z  i  z  3i  y  1
w  zz  2 z  2  x2  y 2  2 x  2iy  2
  x2  2x  3   2 i
Re  w   x 2  2 x  3   x  1  2
2
Re(w) is minimum  x + 1 and Re(w) = 2


i
So, w  2  2i  w  2e 4
 w4  24 ei
 Minimum value of ‘n’ = 4
90. Key: 5
Sol:

x
 fx i i
f i
6  18  64  5  54  35  48 225  5
  5
45   45  
 f i xi2 12  54  256  25  324  245  384 1275  25
 
 fi 45   45  
 f i xi2
( ) 2   ( x )2
 fi
1275  25
3  25
45  
28  45     1275  25  1260  28  1275  25
3  15    5

SRI CHAITANYA IIT ACADEMY, INDIA 24‐12‐23_Sr.Super60_Elite,Target & LIIT‐BTs_Jee‐Main_GTM‐03_KEY &SOL’S

KEY SHEET
PHYSICS
1) 4 2) 1 3) 2 4) 4 5) 3
6) 2 7) 2 8) 1 9) 1 10) 3
11) 1 12) 4 13) 3 14) 4 15) 2
16) 4 17) 4 18) 3 19) 3 20) 2
21) 21 22) 400 23) 3 24) 14 25) 32
26) 22 27) 8 28) 63 29) 3 30) 69
CHEMISTRY
31) 3 32) 1 33) 3 34) 2 35) 3
36) 3 37) 3 38) 1 39) 3 40) 2
41) 4 42) 2 43) 1 44) 2 45) 1
46) 3 47) 4 48) 1 49) 3 50) 3
51) 2 52) 7 53) 3 54) 2 55) 56
56) 10 57) 222 58) 3 59) 24 60) 3
MATHEMATICS
61) 3 62) 3 63) 4 64) 2 65) 3
66) 4 67) 4 68) 3 69) 2 70) 2
71) 4 72) 4 73) 1 74) 2 75) 1
76) 2 77) 1 78) 3 79) 1 80) 1
81) 15 82) 16 83) 22 84) 24 85) 4
86) 13 87) 64 88) 288 89) 150 90) 76

SOLUTIONS
PHYSICS
1. The volume of liquid flowing through both the tubes i.e., rate of liquid is same i.e., Q =
P r 4
constant. Q [By Poiseuille equation]
8 L
 P1r14  P2r24 P r4 P r4
i.e.,   11  2 2  P2  4 P1 and l2  l1 / 4
8l1 8l2 l1 l2
P1r14 4 P1r24 4 r1
4
  r2  r2  r1 / 2
l1 l1 / 4 16
2. When P  N junction is formed an electric field is generated from N-side to P-side due to
which barrier potential arise & majority charge carrier cannot flow through the junction
due to barrier potential so current is zero unless we apply forward bias voltage.
 f f 60
3. M  0    0    20  240
 fe fe 5
4. The entropy change of the body in the two cases is same as entropy is a state function.
5. One side of mirror is opaque and another side is reflecting this is not in case of lens
hence, it is easier to provide mechanical support to large size mirrors than large size
lenses. Reflecting telescopes are based on the same principle except that the formation of
images takes place by reflection instead of refraction.
6.
Partially
polarized
glass
r
900
air
 B1 B1
Plane
polarized
Unpolarized
By Brewster’s law
tan  B1   g ……….. (i)
From (i) & (ii), we get
 
tan  B1  cot  B 2   B1    B2   B2    B1
2 2
So, (I) is true.
By Brewster’s law
1
tan  B 2   cot  B 2   g
g
  B 2  cot 1   g 
So, (II) is false

Unpolarized
Plane
polarized
qB 2
900 glass
air
Partially
polarized
7. For secondary minima,

n
b sin   n  sin  
b
Distance of nth secondary minima x  D sin 
x 2
Or sin 1  1  sin 1  n4
D b
4 x2 x2  x1 4 2 2
sin  2     
b D D b b b
2
Width of central maxima  D  3 cm
b
8. Diamagnetic substances tend to move from stronger to weaker part of the field.
Magnetic susceptibility of diamagnetic substance is negative
  q
9. By Gauss law of Electrostatics  E.dA  in
0
   dB
By Faraday’s law  E.dl 
dt 

By Gauss law in magnetism  B.dA  0
  d E
By Ampere – Maxwell law  .dl  0ic  0 0
B
dt
10. A. In instrinsic semiconductor, Fermi level is between the two bands.
B. In n-type semiconductor, Fermi level is close to conduction band.
C. In p-type semiconductor, Fermi level is near to valance band.
D. In metal, Fermi level inside the conduction band.
11. It forms erect and real image
12.   tan i p  tan 54.740  1.414  2
sin  A   m  / 2
From    2
sin A / 2
 600    0
600   m
sin  m   2 sin 60 1
 2 
1
 450
 2  2 2 2 2
 
 m  900  600  300
d sin 300 d
13. From d sin   n 

1 2
For first secondary maximum

3 3d 3 3
d sin  '   2n  1 
 sin  '  :  '  sin 1  
2 2 4 4 4
1
14. According to Curie law,  
T
1
 The graph between  and will be a straight line .
T
I 1
15. From T  2 ,T 
MB M
Joining pole to pole means sum position, then
T2 M1  M 2 3M  2M
   5 T2  5  T1  5  5s
T1 M1  M 2 3M  2M
16. As 1   m  r  m  r  1  1000  1  999
H  n i  600  0.5  300 I   m H  999  300  2.997  105 A / m
17. In a paramagnetic substances, intrinsic magnetic moment is not zero. Further, in the
absence of external magnetic field, spin exchange interaction is present.
18. If half wave rectifier the output voltage across C is the max. voltage V0  Vrms 2
 200  2  283V
19. Here, output of NAND gate I , y1  A.B
Output of NAND gate II , y2  A. A.B
Output of NAND gate III , y3  B. A.B
Output of NAND gate IV , y  y2 . y3
A B y1  A.B y2  A. A.B y3  B. A.B y y2 . y3
1 0 1 0 1 1
1 1 0 1 1 0
0 0 1 1 1 0
20. The energy flowing per second per unit area in electromagnetic wave is
 1   EB sin 900
S  EB 
0 0
EB EB 2
The rate at which energy flows at a face of area A  SA  A r
0 0
 

As the propagation of electromagnetic wave is along the direction of E  B , i.e., along
z  axis, so the energy flow is through faces parallel to x  y plane and zero to others.
21.

P
3P0
B C
P0 D
A
V0 2V0 V
From the figure,

Work, W  2 P0V0
Heat given, Qin  WAB  WBC  n.CV TAB  nCP TBC
3R n5R
TC  TB   Cv  and CP  
3R 5R
 n TB  TA  
2 2  2 2 
3 5
  PBVB  PAVA    PCVC  PBVB 
2 2
3 5
  3P0V0  P0V0    6 P0V0  3P0V0 
2 2
15 21
 3P0V0  P0V0  P0V0
2 2
W 2 P0V0 4 400
Efficiency,      %  .
Qin 21 P V 21 21
2 0 0
22. He is short sighted and he is to use a concave lens of f  100 cm
x2
  400
25
23. Here, f0  1.5 cm, fe  2.5 cm
Distance between objective and eye piece  0  f e  25cm
1 1 1
 0  25  fe  25  2.5  22.5 cm Using  
0 u0 f0
1 1 1 1 1 1 21 1.5  22.5
      u0   1.6 cm
22.5 u0 1.5 u0 1.5 22.5 1.5  22.5 21
 D 22.5 25
For relaxed eye, magnification, M  0     140.6
u0 f e 1.6 2.5
2 D
24. Width of central maximum  y
a
2 ' D
New width, y ' 
a'
y'  'D' a 700  D  a 14

   
y a'  D a  400 D 4
2
 n  14
1 1
25. Intensity of polarized light transmitted from first polarizer. I1  I   64  32W m2 .
2 2
26. If C is torque per unit angular twist of the wire, then for a twist  ,
  C  MH sin 
In the 1st case, 1  5400  450  4950 ,1  450
In the 2nd case, 2  3600  450  3150 ,
 2  450
  
C 4950  M1H sin 450 ....... i 
 
And C 3150  M 2 H sin 450 .... ii 
Dividing (i) by (ii), we get
M1 495 11 22
  
M 2 315 7 14
I 4.8  103
27. J   4800 A / m 2
A

4  10  3
25  10
 5

J 4800
Drift velocity, d  

ne 1022  1.6  1019

 3.0 ms 1
l 6  102
Time taken t    0.02s  400t  8
d 3.0
28. Here, R  5  103 ,Vi  220V ;
Zener voltage, VZ  50V
V 50
Load current, I L  Z 
RL 20  103
 2.5  103 A
Current through R ,
220  50
I  34  103 A  I z  31.5mA and 2 I z  63 mA
3
5  10
29. Intensity of e.m. wave is
P 1
I  av  0 E02C
4 r 2 2
 
E0  3000
V
m

30. Sound waves, cathode rays (i.e., a stream of electrons) and cosmic rays (i.e., energetic
protons coming from outer space) are not electromagnetic waves.

CHEMISTRY
31. nucleus
32. Lead storage battery consists of lead anode and a grid of lead packed with lead oxide
 PbO2  as cathode, a 38% solution of H 2 SO4 is used as an electrolyte. On charging the
battery the reaction is reversed and PbSO4  s  on anode and cathode is converted into
Pb and PbO2 respectively.
33. As per Einstein’s equation of photoelectric
1 2 hv hc
Effect hv  hv0  K .E.  mv  hv  hv0  
2  0
1/2 1/2
2 2hc  1 1   2hc  1 1    2hc  0    
v    ; v        
m   0   m   0    m   0 
34. The two possible isomers for the given octahedral complex are  M  NH 3 5 SO4  Cl and
 M  NH 3  Cl  SO4 . They respectively give chloride ion (indicated by precipitation with
 5 
AgNO3 ) and sulphate ion (indicated by precipitation with BaCl2 ). Hence, the type of
isomerism exhibited by the complex is ionization isomerism.
35. If statement I is TRUE and statement II is FALSE
36. At equivalence point, pH is 7 and pH increases with addition of NaOH . So correct
graphs is (c).
37. Conceptual
38. Henry’s law is m  K .P; where, m  mass of gas absorbed by given volume of the
solvent.
P  pressure of gas;  log m  log K  log P
39. hv  hv0  KE
Invertase
40. C12 H 22O11  H 2O 
 C3H12O6  C6 H12O6
Sucrose Clu cos e 
 
Fructose
Invert sugar
Zymase
C6 H12O6 
 2C2 H 5OH  2CO2
Glu cos e
Invertase enzyme catalyses the hydrolysis of sucrose and give mixture of glucose and
fructose which is also
known as invert sugar. While zymase enzyme catal7yses the fermentation of glucose into
ethanol and CO2 .
41. Na2 B4O7  7 H 2O 
 2 NaOH  4 H 3 BO3
Strong base Weak acid

Alkaline

Na2 B4O7  7 H 2O  2 NaBO2  B2O3

Glassy bead

Na2 B4O7  H 2 SO4  5H 2O4  4 H 3BO3
White crystals 
42. First oxide 2nd oxide
M  50% M  40%
O  50% O  60%
Formula  MO2 formula = ?
Let the atomic weight of metal  x
1  x  100
% of m is MO2   50
x  32
x  32
EF of 2nd oxide.
60 3.75
0  3.75 3
16 1.25
40 1.25
M  1.25 1
32 1.25
Formula of 2nd oxide  MO3 .
43. Valence electron and positively charged metal ions
44. Formation of a bright
Yellow precipitate layer of ammonium phosphomolybdate
45. Both statements are correct
46. Law of conservation of mass.
1  1 1 
47. For maximum energy, n1  1 and n2    RH Z 2   
  n2 n2 
 1 2
Since RH is a constant and transition remains the same
2
1 He Z H
2 1 1
Z ;   . Hence, He   91.2  22.8 nm
 H Z He 4
2 4
48. The shortest wavelength transition in the Balmer series corresponds to the transition
n  2  n   . Hence, n1  2, n2  
 1 1 

v  RH  
 n2 n2 
 1 2
 
 109677cm 1 
 1
2 2 

1 
2
 
 27419.25 cm 1
49. NaCN and Na2 S are decomposed by HNO3

50. Conceptual
51. Moles of A2 B  Moles of AB3  0.15
w w
  2a  b  a  3b  a  2b .
2a  b a  3b
52. Gd  64    Xe 4 f 7 5d16s 2  Gd 2   Xe 4 f 7 5d16s0
So, the number of 4 f electrons in the ground state electronic configuration of Gd 2  is 7.
53. H 2O2  KI  I 2
54. PCO2  K H 
PCo2 2.533125  105

X 
KH 1.67  108
X  0.00152
500
x H 2O   27.78 moles of H 2O
18
Amount of CO2 is 500 m soda water  0.042  44 .
 1.848g .
1.4  N1V1
55. %N 
weight of organic compound
56.  H    0.1  105  103 M
 
 H    A 3   A 3 
      1013
Now Ka3     1010
 HA2   HA2  10  3
   
P X  10
57.
E
KE
Metal (Work function  E0 )
E  E0   KE max
hc
 4.41  1019  KE

6.63  1034  3  108
 4.41  1019  KE
9
300  10
So,  KE max  6.63  1019  4.41  1019
 2.22  1019 J  222  1021 J
58.
59. Conceptual
60. H 2 N  NH 2 , H 2 N  OH , CH 3CHO

MATHEMATICS
61. A  48, B  25, C  18
A  B  C  60, A  B  C  5
A B
A B C   A   A B  A B C
  A  B  48  25  18  5  60  36
No. of men who received exactly 2 medals
 A  B  3 A  B  C  36  15  21
62.
P C
1 6
4
5 7
6
63. Beautiful is relative term so, it is not well defined term. Therefore, it is not a set.
64. 2m  2n  56  m  6, n  3
65. Given set A with equation x  1  2 and set B with equation x  1  2 .
A : x   3,1
B : x  (, 1]  [3, )
Take, B  A  (, 3]  [3, )  R   3,3 also, A  B   3, 1
66. Since according to definition of R  xy  Ai

Ai , iff, 1  i  k it is clear R is an equivalence
relation. e.g. Let A  1, 2,3
R  1,1 , 1, 2  , 1,3 ,  2,1 ,  2, 2  ,  2,3 ,  3,1 ,  3, 2  ,  3,3
67. We are given that
R1   a, b   N  N : a  b  13
R2   a, b   N  N : a  b  13
Now for R1 :
(i) Reflexive relation
 a, a   N  N : a  a  13
(ii) Symmetric relation
 a, b   R1,  b, a   R1 a  b  13  b  a  13
(iii) Transitive relation
 a, b   R1,  b, c   R1   a, c   R1 :
Let 1,3  R1  3,16   R1 but 1,16   R1 : Not Transitive

For R2 :
(i) Reflexive relation
 a, a   N  N : a  a  13
(ii) Symmetric relation
 b, a   N  N : b  a  13
(iii) Transitive relation
 a, b   R2 ,  b, c   R2   a, c   R2
Let 1,3  R2 ,  3,14   R2 but 1,14   R2 : Not Transitive
68. We have, R  1,3 , 1,5  ,  2,3 ,  2,5  ,  3,5 ,  4,5 
R 1   3,1 ,  5,1 ,  3, 2  ,  5, 2  ,  5,3 ,  5, 4 
Hence, RoR 1   3,3 ,  3,5  ,  5,3 ,  5,5 
69. Transitivity property usually not hold in unions. For example, consider the set
A  1,2,3 and R  1,2  , S   2,3 .
Clearly R and S are transitive but R  S  1,2  ,  2,3 is not transitive,
So  S1  is not true.
For  S2  . Let  a, b  ,  b, c   R  S
  a, b  ,  b, c   R and  a, b  ,  b, c   S
  a , c   R,  a , c   S   a, c   R  S
 If R, S are trans., then R  S is also trans.
So,  S2  is true. For  S3  R, S are reflexive, so
 a, a   R,  a, a   Sa  A  a  A,  a, a   R  S and R  S .
And so R  S and R  S both reflexive.
Also for S4 , R and S a re symmetric.
  a, b   R  S   a, b   R or  a, b   S
  b, a   R or  b, a   S   b, a   R  S
 R  S is symmetric.
70. Since, R is an equivalence relation on set A, therefore  a, a   R for all a  A . Hence, R
has at least n ordered pairs.
71. x  y  18 mean  5 ………. (i)
1  4  16  25  x 2  y 2
10   25
6
x 2  y 2  164 …………. (ii)
 xi  x 8
By solving (i) and (ii), x  8 & y  10 M .D. x   
6 3
72. Number of students are,

 x  12   2 x  5   x 2  3x   x  20  2 x 2  2 x  4  20  x 2  x  12  0
  x  4  x  3  0  x  3

Marks 2 3 5 7
No. of 16 1 0 3
students
32  3  21 56
Average marks    2.8
20 20
73.
C .I . fi xi fi xi C .F .
0-6 a 3 3a a
6-12 b 9 9b ab
12-18 12 15 180 a  b  12
18-24 9 21 189 a  b  21
24-30 5 27 135 a  b  26
N   26  a  b   504  3a  9b 
504  3a  9b 309
Mean 
26  a  b 22
 243a  111b  3054
 81a  37b  1018 ……… (i)
Median class is 12 – 18
a  b  26
 a  b
Now, median  12  2  6  14
12
a  b  26  2a  2b
  4  a  b  18 …………. (ii)
2
On solving eqs. (i) and (ii), we get a  8, b  10
74. fi  62  3k 2  4k  2  62
 3k 2  16k  12k  64  0
16
 k 4 or  (rejected)
3
f x
Now,   i i  put k  4
f i
8  2 15   3 15   4 17   5 156
  …………. (i)
62 62
2
As,  2  fi xi2   fi xi 
2
 156 
  8  1  15  4  15  9  17  16  1  25    
 62 

2
2500  156 
    ……….. (ii)
62  62 
500  2   2   8
By (i) and (ii)  2   2  
62  
75. We have combined variance.
2
n1 2  n2 2 n1n2  x1  x2 
 
n1  n2  n  n 2
1 1
2
15.14  15. 2 15.15 12  14 
 13  
30 30  30
14   2 4
 13     2  10
2 4
76. E   x is a prime number  2,3,5,7
P  E   P  X  2   P  X  3  P  X  5   P  X  7 
 0.23  0.12  0.20  0.07  0.62
F   X  4  1,2,3
P  F   P  X  1  P  X  2   P  X  3  0.15  0.23  0.12  0.5
E  F   X is prime as well as less than 4  2,3
P  E  F   P  X  2   P  X  3  0.23  0.12  0.35
Therefore, the required probability is
P E  F   P E  P F   P E  F 
 0.62  0.5  0.35  0.77
77.   xi P  x  xi   3.24
78.    xr .P  x  xr 
1
79. P  X  xi   1  81a  1  a 
81
Mean    xi P  X  xi   5.48
N 3Ck
80. P  X  k   1   1
k 1 k
e 3c   2  c  log3  log 2 
81. B and C will contain three digit numbers of the form 9 k  2 and 9k   respectively. We
need to find sum of al elements in the set B  C effectively. Now,
S  B  C   S  B   S  C   S  B  C  where S  k  denotes sum of elements of set k .
Also B  101,109,.....992
100
 S  B  101  992   54650
2
Case – I : If   2 then B  C  B
 S  B  C   S  B
Which is not possible as given sum is
274  400  109600 .
Case – II : If   2
Then B  C  
 S  B  C   S  B   S  C   400  274
110 110 110
 54650   9k    109600  9  k     54950
k 11 k 11 k 11
 100
 9 11  110     100   54950
 2 
 54450  100  54950  5
2
82. k  2n  n  26  64
83. Given S  4,6,9 and T  9,10,11......1000
Where, A  a1  a2  ......  ak : K  N  &ai  S
Here by the definition of set ' A ' A  a : a  4 x  6 y  9 z
Except the element 11, every element of set T is of the form 4 x  6 y  9 z for some
x, y, z W  T  A  11
84. Given a set 1,2,......50
Possible choices of P are
2,3,5,7,11,13,17, 23,29,31,37,41,43 and 47 .
So, we can calculate no. of elements in R1 as
 2,20  ,  2,21 ...... 2,25 

3,30  ,........3,33 
5,50  ,.......5,52 
 7,70  ,..... 7,72 
11,110  ,......11,111 
Every number of P n should lie in the given set 1,2,.....50 .
n  R1   6  4  3   2  10   36
Similarly for relation R2 ,
 2,20  ,  2,21 ...... 47,470 ,  47,471 

n  R2   2  14  28
Required difference
n  R1   n  R2   36  28  8
85. R   2,4  ,  4,3 ,  6,2  ,  8,1 : Range  1, 2,3, 4
86. 70  75  30  x  72  100  x  65
87. Mode =3 Median 2 Mean : Mode  3  6   2  5   8
88. Given   9
4x
Let a student obtains x marks out of 75. Then his marks out of 100 are . Each
3
4
observation is multiply by .
3
4
 New SD,    9  12   2  144
3
1 d
89. 0 1   3  d 1 .......... i 
4
1  2d 1 3
0 1   d  ...... ii 
4 2 2
1  4d 3 1
0 1   d  ..... iii 
4 4 4
1  3d 1
0 1    d 1 ....... iv 
4 3
1 1
From  i  ,  ii  ,  iii  and  iv     d 
3 4
1
Minimum value of d  
3
6  3d 5
Mean x  
4 4
1
90. P  X  xi   1  k
14
 
P  2nd moment about origin  E x 2  xi2 P  X  xi 
 P  76k  14 P  76

only.
.
SRI CHAITANYA IIT ACADEMY, INDIA 24‐12‐23_ Sr.S60_Elite, Target & LIIT‐BTs _Jee‐Main_GTM‐03_Q.P
6. Use Blue / Black Point Pen only for writing particulars / marking responses on the Answer Sheet. Use of
pencil is strictly prohibited.
7. No candidate is allowed to carry any textual material, printed or written, bits of papers, mobile phone any
electron device etc, except the Identity Card inside the examination hall.
9. On completion of the test, the candidate must hand over the Answer Sheet to the invigilator on duty in the
Hall. However, the candidate are allowed to take away this Test Booklet with them.
Admission Number:
24‐12‐23_Sr.Super60_Elite, Target & LIIT‐BTs _ Jee‐Main_GTM‐03_Test Syllabus

PHYSICS : EXTRA SYLLABUS
CHEMISTRY : EXTRA SYLLABUS
MATHEMATICS : EXTRA SYLLABUS

correct.
1. Two tubes of radii r1 and r2 , and lengths l1 and l2 , respectively, are connected in series
and a liquid flows through each of them in streamline conditions. P1 and P2 are pressure
l
differences across the two tubes. If P2 is 4P1 and l2 is 1 , then the radius r2 will be equal
4
to:
r
1) r1 2) 2r1 3) 4r1 4) 1
2
2. Statement I : When a Si sample is doped with Boron, it becomes P type and when doped by
Arsenic it becomes N-type semi conductor such that P – type has excess holes and N-type
has excess electrons.
Statement II : When such P-type and N-type semiconductors are fused to make a junction, a
current will automatically flow which can be detected with an externally connected
ammeter. In the light of above statements, choose the most appropriate answer from the
options given below.
1) Statement I is true and Statement II is false
2) Statement I is false and Statement II is true
3) Statement I and statement II are true
4) Statement I and statement II are false
3. A simple telescope, consisting of an objective of focal length 60 cm and a single eye lens of
focal length 5 cm is focused on a distant object in such a way that parallel rays emerge from
0
the eye lens. If the object subtends an angle of 2 at the objective, the angular width of the
image is
0
4) 1/ 6 
0
1) 100 2) 24 3) 500
4. A solid body of constant heat capacity 1 J / 0 C is being heated by keeping it in contact with
reservoirs in two ways:
i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat.

ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same
amount of heat.
In both the cases body is brought from initial temperature 100 0 C to final temperature
2000 C . Entropy change of the body in the two cases respectively is:
1) ln 2, 2 ln 2 2) 2 ln 2, 8 ln 2 3) ln 2, 4 ln 2 4) ln 2 , ln 2
5. This question has Statement – I and Statement – II. Of the four choices given after the
Statements, choose the one that best describes the two Statements.
Statement I: Very large size telescopes are reflecting telescopes instead of refracting
telescopes.
Statement II: It is easier to provide mechanical support to large size mirrors than large size
lenses.
3) Statement I and statement II are true and statement II is correct explanation for statement I
4) Statement I and statement II are true and statement II is not the correct explanation for
statement I
Statement I : If the Brewster’s angle for the light propagating from air to glass is B, then

Brewster’s angle for the light propagating from glass to air is  B
2
Statement II : The Brewster’s angle for the light propagating from glass to air is
 
tan 1  g where  g is the refractive index of glass.
In the light of the above statements, choose the correct answer from the options given below:
3) Both Statement I and II are false
7. A single slit of width b is illuminated by a coherent monochromatic light of wavelength  .
If the second and fourth minima in the diffraction pattern at a distance 1 m from the slit are
at 3 cm and 6 cm respectively from the central maximum, what is the width of the central
maximum? (i.e. distance between first minimum on either side of the central maximum)
1) 1.5 cm 2) 3.0 cm 3) 4.5 cm 4) 6.0 cm
Statement I : For diamagnetic substance  1    0, where  is the magnetic
susceptibility.
Statement II : Diamagnetic substances when placed in an external magnetic field, tend to
move from stronger to weaker part of the field.
In the light of the above statements, choose the correct answer from the options given below.
3) Statement I and statement II are false
9. Match List I with List II:
List – I List – II
A Gauss’s Law in electrostatics I   d B
 .dl  
E
dt
 
B Faraday’s Law II  B.dA  0
C Gauss’s Law in Magnetism III   d E
 .dl  0ilC  0 0
B
dt
D Ampere-Maxwell Law IV   q
 E.ds 
0
1) A  IV , B  I , C  II , D  III 2) A  I , B  II , C  III , D  IV
3) A  III , B  IV , C  I , D  II 4) A  II , B  IV , C  I , D  III
10. Match List I with List II:
A Intrinsic semiconductor I Fermi-level near the valence band
B n-type semiconductor II Fermi-level in the middle of valence and
conduction band
C p-type semiconductor III Fermi-level near the conduction band
D Metals IV Fermi-level inside the conduction band
1) A  III , B  I , C  IV , D  II 2) A  I , B  II , C  III , D  IV
3) A  II , B  III , C  I , D  IV 4) A  II , B  I , C  III , D  IV
11. Which of the following is incorrect about a simple microscope?
1) It forms erect and real image
2) It forms erect and virtual image
3) The object is kept at a distance of less than one focal length from the lens
4) It brings the object closer to the eye than near point
12. If the polarizing angle of a piece of glass for green light is 54.740 , then the angle of
minimum deviation for an equilateral prism made of same glass is
(Given: tan 54.740  1.414 )
1) 450 2) 54.740 3) 600 4) 300
13. In Fraunhoffer diffraction at single slit of width d with incident light of wavelength
5500 A0 , the first minimum is observed at angle of 300 . The first secondary maximum is
observed at an angle  , equal to
1 1  1  1  1  3   3
1) sin   2) sin   3) sin  4) sin 1  
 2  4  4  2 
14. The graph between susceptibility  and 1/T for paramagnetic material will be represented
by, (T = Absolute temperature)
Y Y Y Y
   
X X X X
1) O 1/ T
2) O 1/ T
3) O 1/ T
4) O 1/ T
15. The length, breadth and mass of two bar magnets are same but their magnetic moments are 3
M and 2M respectively. These are joined in parallel with similar poles on same side and are
suspended by a string. When oscillated in a magnetic field of strength B , the time period
obtained is 5s . If the poles of either of the magnets are reversed then the time period of the
combination in the same magnetic field will be
1) 2 2s 2) 5 5s 3) 1s 4) 3 3s
16. An iron rod is placed coaxially inside a solenoid on which the number of turns per unit
length is 600. The relative permeability of the rod is 1000. If a current of 0.5 A is passed
through the solenoid, then the magnetization of the rod will be
1) 2.997 102 A / m 2) 2.997 103 A / m
3) 2.997 104 A / m 4) 2.997 105 A / m
17. Pick out the correct statements
i) Susceptibility of a diamagnetic substance is high and negative
ii) In paramagnetic substance, the intrinsic magnetic moment is not zero
iii) When a paramagnetic substance is heated, it becomes ferromagnetic
iv) Ferro magnetic material becomes paramagnetic above a certain temperature
1) i and iii 2) iii and iv 3) ii and iii 4) ii and iv
18. A sinusoidal voltage of r.m.s value of 200 volt is connected to the diode and capacitor C in
the circuit, so that half wave rectifications occurs. The final potential difference in volt
across C is:
Ev  200 ~ C
Volt
1) 500 2) 200 3) 283 4) 141

19. Select the output Y of the combination of gates, for three sets of inputs
A  1, B  0; A  1, B  1 and A  0, B  0 respectively:
1)  0,1,1 2)  0,0,1 3) 1,0,0  4) 1,1,1

20. A cube of each side r has edges parallel to x, y and z axis of rectangular coordinate
system. During the propagation of electromagnetic wave the electric field E is parallel to
y  axis and the magnetic field is parallel to x  axis. The rate at which energy flows through
each face of the cube is (E and B are instantaneous values)
r 2 EB
1) in parallel to x  y plane face and 0 in others
2 0
r 2 EB
2) in parallel to x  y plane face and 0 in others
0
r 2 EB
3) in all faces
0

r 2 EB
4) in parallel to y  z plane face and 0 in others
0
21. An engine operates by taking a monatomic ideal gas through the cycle shown in the figure.
400
The percentage efficiency of the engine is . Find the value of K
K
3P0
B C
P0
A D
V0 2V0
22. A person cannot see the objects beyond 100 cm. The focal length of the lens in cm required
x2
to correct this vision is x . Find the value of .
25
23. A microscope has an objective of focal length 1.5 cm and an eye piece of 2.5 cm. If the
distance between objective and eye piece is 25 cm, is the value of magnification produced
for relaxed eye is 50 n. Find the value of n approximated to nearest integer.
24. A single slit of width a is illuminated by violet light of wavelength 400 nm and the width
of the primary diffraction maximum is measured as y . When half of the slit width is
covered and illuminated by red light of wavelength 700 nm, the width of the primary
ny
diffraction maximum . Find the value of n
4

25. Unpolarized light of intensity 64 W m 2 passes through three polarizers such that the
transmission axis of the last polarizer is crossed with that of the first. The intensity of final
emerging light is 3 W m 2 . The intensity of light transmitted by first polarizer will be

26. Two magnets are suspended by a given wire one by one in a horizontal uniform magnetic
field. In order to deflect the first magnet through 450 , the wire has to be twisted through
5400 whereas with the second magnet, the wire requires a twist of 3600 for the same
deflection. Then the magnetic moments of the first and second magnets are in the ratio n :14 .
Find the value of n
27. An N  type silicon sample of length 6 102 m has width 4 103 m and thickness
25 105 m . When a voltage is applied across the length of the sample, the current in
sample is 4.8 mA. If the free electron density in sample is 1022 m3 , then the time taken for
the electrons to travel the full length of the sample in seconds is t. Find 400 t.
28. For the zener diode circuit shown, the current through the zener in mA is I. Find 2I
29. A point source of electromagnetic radiation has an average power output of 800 W. The
V
maximum value of electric field at a distance 4.0 m from the source in is 1000K . Find
m
the value of K approximated to nearest integer.
30. A parallel plate capacitor made of circular plates each of radius R  6 cm has capacitance
C 100 pF . The capacitor is connected to a 230V a.c. supply with an angular frequency of
300 s 1 . The rms value of displacement current in  A is I . Find the value of 10I .

correct.
31. Rutherford’s experiment on scattering of alpha particles showed for the first time that atom
has:
1) electrons 2) protons 3) nucleus 4) neutrons
32. For lead storage battery pick the correct statements
A) During charging of battery, PbSO4 on anode is converted into PbO2
B) During charging of battery, PbSO4 on cathode is converted into PbO2
C) Lead storage battery, consists of grid of lead packed with PbO2 as anode
D) Lead storage battery has ~ 38% solution of sulphuric acid as an electrolyte
1) B, D only 2) B, C, D only 3) A, B, D only 4) B, C only
33. If 0 and  be the threshold wavelength and the wavelength of incident light, the speed of
photo-electrons ejected from the metal surface is:
2h 2hc 2hc  0    2h  1 1 
1)  0    2)  0    3)   4)   
m m m  0  m  0  
34. An octahedral complex with molecular composition M.5 NH3.Cl.SO4 has two isomers, A
and B. The solution of A gives a white precipitate with AgNO3 solution and the solution of
B gives white precipitate with BaCl2 solution. The type of isomerism exhibited by the
complex is:
1) Linkage isomerism 2) Ionisation isomerism
3) Coordinate isomerism 4) Geometrical isomerism
35. Statement I : For the physical equilibrium H 2O  s   H 2O  l  on increasing temperature
and increasing pressure more water will form.
Statement II: Since forward reaction is endothermic in nature and volume of water is
greater than that of the volume of ice.
1) If both the statements are TRUE and statement II is the correct explanation of statement I
2) If both the statements are TRUE but statement II is not the correct explanation of
statement I
3) If statement I is TRUE and statement II is FALSE
4) If statement I is FALSE and statement II is TRUE
36. 100 mL of 0.1 M HCl is taken in a beaker and to it 100 mL of 0 .1 M N a O H is added in
steps of 2 mL and the pH is continuously measured. Which of the following graphs
correctly depicts in change in pH ?
1) 2)
3) 4)
37. The compound formed in the borax bead test of Cu 2  ion is oxidising flame is
1) Cu 2) CuBO2 3) Cu  BO 2 2 4) None of these
38. Which of the following curves represents the Henry’s law?
log m
log m
log P log P
1) 2)
log m
log m
log P log P
3) 4)

39. According to Einstein’s photoelectric equation, the graph between kinetic energy of
photoelectrons ejected and the frequency of the incident radiation is:
K .E . K .E .
v v
1) 2)
K .E . K .E .
v v
3) 4)
EnzymeA
C12 H 22O11  H 2O 
 C6 H12O6  C6 H12O6
Sucrose Glu cos e Fructose
40.
EnzymeB
C6 H12O6   
 2C 2 H 5OH  2CO2
Glu cos e
In the above reactions, the enzyme A and enzyme B is respectively are
1) Amylase and Invertase 2) Invertase and Zymase
3) Zymase and Invertase 4) Invertase and Amylase
41. Aqueous solution of a salt Y  is alkaline to lithmus. On strong heating, it swells-up to give
a glassy material. When conc. H2SO4 is added to a hot concentrated solution of Y  , while
crystals of a weak acid separate out. Hence, the compound (Y) is
1) Na2SO4.10H2O 2) Ca2P6O11.10H2O
3) Na2B6O11 4) Na2B4O7.10H2O
42. Two oxides of metal contains 50% and 40% of metal M respectively. If the formula of the
first oxide is MO2. Then formula of the second oxide will be:
1) M2O5 2) MO3 3) M2O 4) MO2

43. Metallic bond is a bond between
1) Valence electron and positively charged metal ions
2) The ions of two different metals
3) a metal and non metal
4) Two metal atoms
44. Which of the following used to reagent is used to test for phosphate ion is?
1) Nessler’s reagent 2) Ammonium molybdate
3) Baeyer’s reagent 4) Fentos reagent
45. Statement – I : D.D.T is a non-biodegradable pollutant
Statement – II : D.D.T insecticide used in agriculture
1) Both statements are correct 2) Both statements are wrong
4) Statement I is true statement II is false
46. Match the weight of reactants given in Column I with weight of products marked (?) given
Column II
A 2H2  O2  2H2O p 0.56 gm

1 .0 g m 1 .0 g m ?
B N2  3H2  2NH3 q 1.333 gm
1 .0 g m 1 .0 g m ?
C  r 1.125gm
CaCO3   CaO  CO 2
1 .0 g m ?
D C  2H2 CH4 s 1.214 gm
1g 1g ?
1) A  q, B  p,C  r, D  s 2) A  s, B  p,C  q, D  r
3) A  r , B  s,C  p, D  q 4) A  r , B  q,C  s, D  p

47. Match List – I with List – II
(Mixture) (Purification Process)
A Chloroform and Aniline I Steam distillation
B Benzoic acid and Naphthalene II Sublimation
C Water and Aniline III Distillation
D Naphthalene and Sodium chloride IV Crystallization
1) A  IV , B  III , C  I , D  II 2) A  III , B  I , C  IV , D  II
3) A  III , B  IV , C  II , D  I 4) A  III , B  IV , C  I , D  II
48. Assertion: Thin layer chromatography is an adsorption chromatography.
Reason: A thin layer of silica gel is spread over a glass plate of suitable size in thin layer
chromatography which acts as an adsorbent.
2) Both A and R are true but R is NOT the correct explanation of A
49. The Lassaigne’s extract is boiled with dil HNO3 before testing for halogens because
1) A g C N is soluble in HNO3 2) Silver halide are soluble in HNO3
3) NaCN and Na2S are decomposed by HNO3 4) Ag2S is soluble in HNO3
50. Which of the vitamins the one whose deficiency causes rickets (bone deficiency) is
1) Vitamin E 2) Vitamin B 3) Vitamin D 4) Vitamin C
is above 10 and less than 10.5 round off is 10 andIf answer is from 10.5 and less than 11 round off is 11).
51. Two elements A and B which form 0.15 moles of A2B and AB3 type compounds. if both
A2B and AB3 weigh equally, then the atomic weight of A is__times of atomic weight of B.
52. The number of 4 f electrons in the ground state electronic configuration of Gd 2  is ______
[Atomic number of Gd  64 ]
53. A 5.0 ml solution of H2O2 liberated 1.27 gm of iodine from an acidified KI solution. The
percentage strength of H2O2 is x . The value of x is (Round of the nearest integer)

54. Henry’s law constant for CO2 is water is 1.6 108 Pa at 298K . The quantity of CO2 is 500
ml of soda water when packed under 2.5 atm CO2 pressure at 298 K is x gm. Then x is
(Round off the nearest Integer).
55. During estimation of Nitrogen present in an organic compound by KJeldahl’s method, the
ammonia evolved from 0.5 gm of the compound in KJeldahl’s estimation of nitrogen,
neutralized 10 ml of 1 M H2SO4 .The percentage of nitrogen in the compound is x .
What is x ?
56. H3A is a weak tri basic acid with K a1  10  5 , K a 2  10  9 and Ka3  10  13 . The value of
PX of 0.1 M H3A solution.
Where P X   log10 X and
 A3 
X  
is
 HA2 
 
57. The work function of sodium metal is 4.41 1019 J . If photons of wavelength 300 nm are
incident on the metal, the kinetic energy of the ejected electrons will be
 h  6.63  10 34 Js; c  3  108 m / s  ________ 1021J .
58. Re d hot Cu
CH 3  C  CH     0
  X , the no of  bonds in X is
2000 C
59. In carius method of estimation of halogen’s 250 mg of an organic compound gave 141 mg of
A g B r . The percentage is x . Then x is
60. Amongest the following the total number of compounds which does not give Lassaigne’s
test for nitrogen
NH 2
i) ii) H2N  CH2  COOH iii) H2N  NH2

iv) H2N  OH v) Urea vi) PhNO2 vii) CH3CHO
N
viii) CH3CN ix)

correct.
61. An organization awarded 48 medals in event ‘A’, 25 in event ‘B’ and 18 in event ‘C’. If
these medals went to total 60 men and only five men got medals in all the three events, then,
the number of men received medals in exactly two of three events is k then 10k is
1) 100 2) 90 3) 210 4) 150
62. In a class, 18 students offered physics, 23 offered chemistry and 24 offered Mathematics. Of
these, 13 are in both chemistry and Mathematics, 12 in Physics and Chemistry, 11 in Physics
and Mathematics, and 6 in all the three subjects.
Statement I : The number of students offered Mathematics but not Chemistry  11
Statement II : The number of students who offered exactly one of the three subjects = 11
3) Statement I and statement II both are true
4) Statement I and statement II both are false
63. The group of beautiful girls is_____
1) A null set 2) A finite set 3) An infinite set 4) Not a set
64. Two finite sets have m and n elements. The total number of subsets of the first set is 56
more than the total number of subsets of the second set. The value of m  n is
1) 6 2) 9 3) 3 4) 18
65. Let A   x  R : x  1  2 and B   x  R : x  1  2 . Then which one of the following
statements is true?
1) A  B   3,1 2) B  A  R  (  3, 3 ]
3) A  B   3, 1 4) A  B  R  [1, 3]

66. Let a set A  A1  A2 ......  Ak , where Ai  A j   for i  j ,1  i , j  k . Define the
relation R from A to A by R   x , y  : y  Ai if and only if x  Ai ,1  i  k  . Then, R
is:
1) reflexive, symmetric but not transitive
2) reflexive, transitive but not symmetric
3) reflexive but not symmetric and transitive
4) reflexive, symmetric and transitive
67. Statement I : R1   a , b   N  N : a  b  13 is an equivalence relation
Statement II : R2   a , b   N  N : a  b  13 . is an equivalence relation
68. If R be a relation '  ' from A  1, 2,3, 4 to B  1,3,5 , i.e.,  a, b   R  a  b , then
RoR 1 is
1) 1,3  , 1,5  ,  2,3  ,  2,5  ,  3,5  ,  4,5 
2)  3,1 ,  5,1 ,  3, 2  ,  5, 2  ,  5,3  ,  5, 4 
3)  3,3  ,  3, 5  ,  5, 3  ,  5,5 
4)  3,3  ,  3, 4  ,  4,5 
69. Let R and S be two relations on a set A. Consider the following statements:
S1: R and S are transitive, then R  S is also transitive
S2 : R and S are transitive, then R  S is also transitive
S3 : R and S are reflexive, then R  S is also reflexive
S4 : R and S are symmetric then R  S is also symmetric
Then correct statements are:
1) S1, S2 , S4 2) S2 , S3, S4 3) S1, S2, S3 4) S1, S2, S3, S4

70. Let R be an equivalence relation on a finite set A having n elements. Then the number of
ordered pairs in R is:
1) less than n 2) greater than or equal to n
3) less than or equal to n 4) only one
71. Let the mean of 6 observation 1, 2 , 4 , 5 , x and y be 5 and their variance be 10. Then three
times of the mean deviation about the mean of the given data is equal to
8 7
1) 2) 3) 3 4) 8
3 3
72. If for some x  R , the frequency distribution of the marks obtained by 20 students in a test is
Marks 2 3 5 7
Frequency 2x  5
 x 12 x 2  3x x
Then 20 times of the mean of the marks is:
1) 32 2) 30 3) 25 4) 56
73. Consider the following frequency distribution:
Class: 0 – 6 6 – 12 12 – 18 18 – 24 24 – 30
Frequency : a b 12 9 5
309
If mean and median  14 , then the value 2 a  3b is equal to________.
22
1) 46 2) 36 3) 8 4) 10
74. Let  be the mean and  be the standard deviation of the distribution.
xi 0 1 2 3 4 5
fi k2 2k k2 1 k2 1 k2 1 k 3
Where fi  62 .
Statement I : Value of k  3
Statement II : 5   2   2   40 where  x  denote the greatest integer  x ,
 
75. The mean and variance of a set of 15 numbers are 12 and 14 respectively. The mean and
variance of another set of 15 numbers are 14 and  2 respectively. If the variance of all the
30 numbers in the two sets is 13.
Statement I : Value of  2  10
n112  n2 22 1 2
Statement II : Combined variance   
2
  x1  x2 
n1  n2  n1  n2 2
76. A random variable X has the following probability distribution:
X : 1 2 3 4 5 6 7 8
p X : 0.15 0.23 0.12 0.10 0.20 0.08 0.07 0.05
For the events E   X is a prime number and F   X  4 , then the value of

100 P  E  F  is
1) 87 2) 77 3) 35 4) 50
77. For a biased dice, the probabilities for the different faces to turn up are given below:
X 1 2 3 4 5 6
P(X) 0.10 0.32 0.21 0.15 0.05 0.17
Then the mean of the above distribution is:
1) 3. 24 2) 3.34 3) 4.24 4) 4.34
78. A random variable x takes the values 0 ,1, 2 , 3 and its mean is 1.3. If P  x  3  2 P  x  1
and P  x  2  0.3 , then P  x  0   P  x  1 is
1) 0.4 2) 0.1 3) 0.5 4) 0.2

79. A random variable X has the following probability distribution:
X x 0 1 2 3 4 5 6 7 8
P X  x a 3a 5a 7a 9a 11a 13a 15a 17a
1
Statement I: The value of 'a' is
81
Statement II: Mean of random variable X is 6
80. The range of a random variable x  1, 2,3...... and the probabilities are given by
3CK
P x  k    k  1,2,3..... and C is a constant. Then 2C 
K!
1 log e  log 2 
1) 2log3  log2 2) log  log 2  3) 4) log2  log3
2 log 3e
is above 10 and less than 10.5 round off is 10 andIf answer is from 10.5 and less than 11 round off is 11).
81. Let A  {n  N : n is a 3 – digit number]
B  9k  2 : k  N 
And C  9k   : k  N for some   0    9
If the sum of all the elements of the set A   B  C  is 274  400, then 3 is equal
to_______.
k
82. If A  1, 2,3 , the number of reflexive relation in A is k then is
4

83. Let S  4,6,9 and T  9,10,11,.....,1000 . If
A  a1  a2  ......  ak : k  N , a1, a2 , a3 ,....., ak  S , then twice the sum of all the elements
in the set T  A is equal to___________.
84.  
Let R1 and R2 be relations on the set 1,2,..........,50 such that R1  { p , p n : p is a prime
and n  0 is an integer} and R 2   p , p  : p is a prim e and n  0 or 1 .

n
Then 3 times the number of elements in R1  R2 is__________.

85. Let A be set of first ten natural numbers and R be a relation on A , defined by
 x, y   R  x  2 y  10 . Then number of elements in range of R is
86. In a class of 100 students there are 70 boys whose average marks in a subject is 75. If the
x
average marks of the complete class is 72, and the average marks of the girls is x then is
5
87. In a moderately skewed distribution, the values of mean and median are 5 and 6
respectively. The value of mode for such distribution is k then k2 is_______
88. The marks of some students were listed out of 75. The SD of marks was found to be 9.
Subsequently the marks were raised to a maximum of 100 and variance of new marks was
calculate. Then twice the new variance is:
89. The probability distribution of X is:
X 0 1 2 3
P X  1 d 1  2d 1  4d 1  3d
4 4 4 4
For the minimum possible value of d , 120 times the mean of X is equal to___________.
90. The probability distribution of the random variable X is as follows:
X 3 2 1 0 1 2 3
P X  5k 3k k k k 2k k
If 2nd moment about origin of the random variable X is P , then14P is______



Time: : 03.00Pm to 06.00Pm GTM-02 Max. Marks: 300
KEY SHEET
PHYSICS
1) 2 2) 2 3) 3 4) 2 5) 3
6) 1 7) 1 8) 4 9) 1 10) 4
11) 1 12) 4 13) 1 14) 3 15) 4
16) 2 17) 4 18) 1 19) 2 20) 2
21) 6 22) 15 23) 4 24) 6 25) 2
26) 20 27) 363 28) 3 29) 5 30) 6
CHEMISTRY
31) 2 32) 1 33) 2 34) 3 35) 1
36) 4 37) 3 38) 4 39) 4 40) 3
41) 3 42) 4 43) 3 44) 1 45) 2
46) 2 47) 1 48) 4 49) 1 50) 2
51) 4 52) 3 53) 12 54) 8 55) 4
56) 42 57) 13 58) 2 59) 40 60) 3
MATHEMATICS
61) 4 62) 2 63) 2 64) 2 65) 1
66) 2 67) 1 68) 2 69) 3 70) 4
71) 1 72) 1 73) 2 74) 4 75) 4
76) 3 77) 2 78) 4 79) 3 80) 2
81) 1 82) 9 83) 5 84) 0 85) 4
86) 825 87) 1 88) 14 89) 3 90) 0
Sec: Sr.Super60_ Elite, Target & LIIT‐BTs Page 1

SOLUTIONS
PHYSICS
128  1000 g N 2  50 g
1. 
100  100  100cm 3
50  50  50cm3
2n 2 19.6
2. Time taken by mango   = 2 second
g 9.8
Distance = vt
5
6  2  3.33 m
18
3. Conceptual
4. Slope = a = - 5 = - μg
Acceleration of the body is the slope of the v-t graph
10
From the graph, slope of line   5
2
So deceleration is 5 m / s 2    5 / 10  0.50
k2 2
5. For a solid sphere , 
R2 5
9.8sin 30
a a  3.5 m / sec 2
2
1
5
Time of ascent is given by
v  u  at
0  1  3.5t
1
t s
3.5
Time of decent
1
t sec.
3.5
Due to symmetry of motion.
2
Total time, T   0.57 s
3.5
pr 3  M 1/2 L3/2   L3/2 
6. T  k 3/2 Dimensions of RHS  3/4
 M 1/8 L0T 3/2
s  MT 
2
Dimensions of L.H.S  Dimensions of R.H.S

 option A
GM
7. The correct option is B
R
Gravitational potential at any inside point is given as
V 
GM
2R 3  
3R 2  r 2 . ….(i)
R 11GM
for r  ,V  
2 8R
M R
Subtracting potential due to cavity of mass M c  and Rc 
8 2

Gravitational potential at center is obtained by substituting r  0 in equation (i)
3GM c

2 Rc
M
3G
11GM  3GM c  11GM 8  V   GM
V    
8R  2 Rc  8R 2
R R
2
8. Conceptual
9. Conceptual
10. y  A sin t   
A A
Here, y  , A sin t    
2 2
 5
So,   t    or
6 6
A
So, the phase difference of the two particles when they are crossing each other at y 
2
5  2
in opposite directions are   1   2   
6 6 3
 mv  4 3
11. Pf  0 ˆj Pi  mv0iˆ  mv0 ˆj
2 5 5
Impulse acting on ball
     4 1 
J  Pf  Pi  J  mv0   i  j 
 5 10 
 p  p2  3 9
12. w   1   v2  v1    10  0.3   10
6 5
 2  4 4
10  0.4
6
0.5  10  0.1
6
T2  T1 
nR nR
 5R     35
105 3.5
U  n    105
 2  nR 4
22 R
  11 105 
nc
nR
105 3.5 c
7
13. u1  u 2 v1.v2  0
u i  gt j  .u  i   gt j   0
1 2
14.
Taking torque about point C
T
 60  20  50  80  100
2
 3 T  100  800
 300 N

15. xCM   xdm

 dm
 x  
1
x  x 2 dx
xCM  0
   x  x  dx
1
2
9
x CM  m
20
v2  u 2
16. Area under a versus x gives
2
 
2
v2  10
  45  v  10 m / s
2
17. 6  2  3  1   3  6 v
x is the maximum extension in the spring
1 1 1 1
 9 1   200  x 2   312   6  2 
2 2
2 2 2 2
 1  e  m1   m2  em1  m1  M
18. V2    u1    u2 Given & M >>m
 m1  m2   m1  m2  m2  m
 m2 
 e
m m 1  e  u   m1  u
 m  M  or  2  0 V2 
m 1  m2  2
M m1 1 2 1 
 m1 
m1 
 1 1 8V  V
V2  1  e  u1  eu2  1    2V    V    3V
 3 3 3
19. In the general expression h  2T cos 
rdg
r 2T 1
 radius of curvature of meniscus formed so h = or R  .
cos  Rdg h
u2
20. Maximum range up the inclined plane  Maximum range down the inclined
g (1  sin  )
u2
plane  The maximum possible distance between the two bullets after they hit
g (1  sin  )
u2 u2 2u 2
the inclined plane  d max   , d max 
g (1  sin  ) g (1  sin  ) g cos 2 
21. Conceptual
22. r  2i  3 j F   pk

  r  F  2 j  3 j   pk   
a 3 3
    x3  5 x  5  3  15
b 2 2
23. 200  f  m  1.5
300  f  m  2.5
In solving m = 100kg
1 1 1
24. cos 2   cos 2   cos 2   1 cos 2   1 
 
4 2 4
  60 0 with z - axis  30 0 with x - y plane
at 3t
25. from figure, a cos 300  at and a sin 300  at  tan 300  
at at
v2 dv v t 3 2
at  3t or
1
 3t  v 2  3t , But at 
dt
 3t or 0
dv  3  tdt or v =
0 2
t
3
 t 4  3t or t 3  4 Given t  22/3 s  n  2
4
Fl 100  l  120l l  l 100 10
26. e l1  l  l2  l   1  
Ay Ay Ay l2  l 120 12
11  11 
Given l2  12  l1  l    l1  l 10
l1
10  10 
l
12l1  12 l  11l1  10 l l1  2 l  l  1  x  2 10 x  20
2
1 1
27. Applying Bernoulli's theorem: P1   gh   v 2  P2  0   (2v) 2
2 2
3 363
Putting the values, 4100  800{ v 2  10}  v  m / s  x  363
2 6
28. Let x mole of the gas be dissociated, x moles of atomic becomes 2x moles of a
monatomic gas after dissociation.
3 5
Internal energy of n moles of an ideal gas = nRT for monatomic gas = nRT for
2 2
diatomic
gas so, (internal energy of 2x moles of monoatomic gas + internal energy of (4 –x) moles
3
of a diatomic gas) – internal energy of 4 moles of a diatomic gas = nRT  given 
2
 2 x  
3RT   5 RT   5 RT  3
  4  x    4    RT on solving x = 3 moles
 2   2   2  2
29. Taking disc element, the required volume is
5 R 3
R R
  r 2 dy    R  y 2  dy 
2
R/2 R/2
24
30. Applying conservation of energy between initial and final states
2 2
1  v  1  mg   mg  6mg 2
2m    k    2mg  , Solving we get v   6m / sec
2 2 2  k   k  k

CHEMISTRY
31. Charcoal
1
32. Screening effect 
Ionization Potential
33. Last electron enters into p - Orbital
34. Both A and R are true but R is not the explanation for A.
35. Only two optically inactive isomers
Br CONH2
1 COOH
4 2
5 3
6
36. CHO COCl

2
37. CO has a single structure i.e., resonance hybrid of the above three structures.
3
38. 2 nodes are possible for 3s orbit.

39. More the number of keto group more is the stability order.
Incase of II structure as the positive charge is present on adjacent carbons it is unstable.
Z2
40. E 2
n
r1 for H  n 2
rn  =
rn n 2
Z
nh 2h
mvr  
2 2
h
V
 ma0
1
K .E   mv 2
2
h2

32 2 ma02
41. 8 mole atoms – 1 mole
0.25 mole atoms ?
0.25 1
 3.125 102
8
42. Equivalent weight is not fixed
43. Conceptual
44. Conceptual

45.
46. Hoffmann elimination followed by reductive ozonylysis.
1
47. Heat of combustion  Stability
48. In order of generate electrophile HNO3 acts as base.
49.
50. Pd / BaSO4 is for syn addition

Br
2
1
3
5
4
51. Cl 7
52.  E2  Elimination 
53. Cr24  1s 2 2s 2 2 p 6 3s 2 3 p 6 3d 5 4 s1
For s - orbital electrons l  m  0

 2  2  2  2  2  1  1 = 12
56.14 0.347
54.  1
161.4 0.347
43.86 2.4366
 7
18 0.347
 x  1, y  7
 x  y  1 7  8
55. C4 H10  48  10  58
1000 13
  32  3.586 kg  4 kg
58 2
12.6
Optical purity =  100  42%
56. 30
PV PV
57. 1 1
 2 2
T1 T2
760  V1  749  13 22.4 

273 290
760  V1 736   22.4 

273 290
736 273
V1   22.4 
760 290
V1  0.96842  0.9413  22.4
 0.911 22.4
28  20.42
%N 
22, 400  wt.of organic sample
28  20.42

224  0.2
28  20.42 14  20.42
 
224  0.2 224
 12.76
=13
58. Conceptual
59. H 2   H P  H R  (Enthalpy of product-enthalpy of reaction)
= 10  50 
=40
60. A 
O3 / H 2O
Aldehyde 
on. vap
 A  448 mL at STP
 Cholorocompound  1.53g 
As A undergoes ozonolysis so A is unsaturated Let A is Cn H 2n1Cl
A  vapour
1.53 g 448 mL at STP
1.53
 22400  22400 mL = 76.5g n = 3
448

MATHEMATICS

61. tan 1    sin    cot 1   cos      sin     cos 
2
2  sin   cos     2, 2 
n
  1  1   1 
 n  1   n  2   n  22   n  2k 1  
62. lim     ....  
 n   n   n
n  n 
 
      
 1 
1 1 1 1 1  k   1 
e 
1  2 ..... k 1 2  21 k 
e 2 2 2
e  2 
1
1
2
1
 1  x 2 2
63. f 1  x   e x    e . x3
2 x 
2 1 2 1
x  1   1 2 
3
 1   2
 x  e  3   e  2  3 x  
 x x2 4
f 11
   e   x e
x2
2 x   2  2   
x
1
 1 1  x2  4 6 
 f 11  x   e  x     e  x6  x4 
 4x 4x x   
 x 1
e  1  x 2  2   2  1
 f 11  x    1   e  4 3 2    ,   2
4x  x  x  x  4
64.
x y
Equation of L is  1
2 4
2x  y  4  0 1
1
Equation of L1 is y – 1 =  x  2
2
2y  2  x  2
x  2y  4  0  2
Solving 1 & 2  , 
4 12
5 5 
65. Projection of OA on (4, 3, 0) ; (0, 4, 0)
12 12
The SD is equal to 
4 3 2 2 5
66. Any point  2  1,3  2,6  3
Given 4 2  9 2  36 2  9

3
49 2  9 
7
g  x   x 1
67. fog  x   x  3  x
 f g  x      
2
x 1  3 x 1  5
 f  x   x 2  3x  5
 f  0  5
1 1 1 0 0 1
1 1
68. A  4 6 8  2 2 8  2  A 
2
100 100 99 0 1 99
Adj  2 A   2 A  2 A  23 A
2
1  1 
  23 A   23 A
2
 26.  23    16  4  12
4  2 
69. Use   0 &  3  0
1 2 3
4 3 4  0  3  16  2  4  32   3 16  24   0
8 4 
3  16  8  64  24  0
72
5  72  0   
5
1 2 3
4 3 4  0  27  3  16  2  36  4  32   3 16  24   0
8 4 9 
21
5  21  0  
5
70. 1  t  .2  0
  1  2
Taking dot with 
 .  .1   .2
2
4  6  20   .t  0 10  t 
1
10  t 16  9  25  t 
5
1 1
      2  2    
5 5

 i  2 j  4k   15  4i  3 j  5k   9i  135j  15k
5  .  i  j  k    9i  13 j  15k  .  i  j  k   9  13  15  7
2
71. {x+b} = {x} where b is an integer

1000
 x   
 x  b  1000  x
 x 
b 1 1000 1000
  x    x  x
72.
Statement 2 is a property of an A.P so it is true.
a1  a 4  a7     a16  147
  a1  a16    a4  a13    a7  a10   147
 3  a1  a16   147  a1  a16  49
 a1  a6  a11  a16   a1  a16    a6  a11   2  a1  a16   98
73.
 x 1  x  0
1  x 0  x 1

74. f  x  
x 1 x  2
1  x 2 x3
Clearly f is discatinuous at x  0,1,2 no.of points = 3
75. Conceptual
76. A (1, 0, 7) B (1, 6, 3)
Midpoint of AB= (1, 3, 5) lies in the line
DR’s of AB (0, 6, -4) the line passing through A and B is perpendicular to the given line
hence B in the mirror image.
Statement II is also true but not a correct explanation of I as there are infinitely many
lines passing through the midpoint of the line segment and one of the lines is
perpendicular bisector
77.
 x  f  x
78. f   Put x  y  1
 y  f  y
f  x  h  f  x
f 1  x   lt
h0 h
f  x  h  xh
1 f  1
f  x  x 
f  x  . lt  f  x  . lt
h0 h h 0 h
 h
f  1    f 1
f  x f  x 1 2 f  x
. lt 
x
  . f 0   f 1 1  2
x h0 h x x
x
79. lt f  2  h   f  2 
h 0
lt a  14  2  h  48  3  22
h 0
a  14  45
a  2011
80. AT  A and B T   B
 A  B  A  B    A  B  A  B 
A2  AB  BA  B 2  A2  AB  BA  B 2
AB  BA
 AB    1 AB
T K
BT AT   1 AB
K
 BA   1 AB
K
 AB   1 AB
K
K is odd
4 cos 3   3cos  cos 3
81.  4 cos 2   3  
cos  cos 
0 0 0 0
cos 27 cos 81 cos 243 cos 729 cos 7290
ꞏ ꞏ ꞏ  1
cos 90 cos 27 0 cos 810 cos 2430 cos 90
82. We have x > 1
1
 log8  x 2  x   2.log 1  x  1  0
3 2
1 1
 log 2  x  x   2 log 2  x  1  0
2
3 3
log 2 2  log 2  x 2  x   6.log 2  x  1  0
2  x  1 2  x  1
6 5
log 2 1 1
x  x  1 x
Put x – 1 = y as y > 0
2 y5 2 y5  y  1 2 y5  2 y  y 1
1  0 0 0
y 1 y 1 y 1
2 y  y 4  1   y  1
0
y 1
2 y  y 2  1   y 2  1   y  1  2 y  y  1  y 2  1  1 
  0  y  1  0
y 1   y  1 
 y  1  2 y  y 2  1  y  1  1 y 1
 0   0  y 1  0
y 1 y 1
y  1 x  2
83. f   x   3x 2  3  0  f  x  is increasing
a
 f  3  27 f   0   3; a 1  r   3
1 r
2
a 2  81 a9 r
3
84. Let p  x   a0 x4  a1 x3  a2 x2  a3 x  a4
p1 1  0 p1  2  0
p  x 1
Also lim 2
 1  a4  0, a3  0 and a 2  1, a1  1, a0 
x0 x 4
1 4
 p  x   x  x3  x 2 p  2  0
4
85.
 3 1  5  2   7  3  9  4   11 5   13  6   15  7 
86.
17  8   19  9   2110 
10
   2r  1 r  825
r 1
87. f 1  1
f  2  f  f 1   f  2  f 1   f 1  f 1  2
f  3  f  f  2    f  3  f  2    f  2  f 1
2+1 = 3
1 20 1
Thus f  n   n  f  r   1  2  3  ......  20 

30 r 1 30
1 20  21 
 . 7  1
30 2 7
x  2 y 1 z  6
88. Lines are  
3 2 2
   
x  6 y 1 z  8 c  a b d 
  S.D     
3 2 0 bd
  
b  3iˆ  2 ˆj  2kˆ, d  3iˆ  2 ˆj a  2iˆ  ˆj  6kˆ, c  6iˆ  ˆj  8kˆ
4 2 14
3 2 2
3 2 0 16  12  168 196
    14
î ˆj kˆ 4iˆ  6 ˆj  12kˆ 14
3 2 2
3 2 0
Tr 1 1 1 1 1
 T1  a; T2  a T3  2 a..........T7  6 a
89. Tr 3 3 3 3
a 1
6
 a3
3 243
a a a2 32 1
Tr  Tr 1  r 1  r  2 r 1  2 r 1  2 r 3
3 3 3 3 3

1 1 1

r 1
Tr .Tr 1  3   3  5  ........
3 3 3
3 27
   3.375
1 8
1 2
3
3
 2x  1 
 
Lt f  x   Lt  x 
 sin  x log 2   log 1  x log 4 
x 0 x 0 2
90.  log 2    .log 4

 x log 2  x 2 log 4
 log 2 
3
1
  log 2
 log 2  2 log 2  2
1 
  log 2   0
2 

1. Immediately fill in the Admission number on this page of the Test Booklet with Blue/Black Ball Point Pen only.
2. The candidates should not write their Admission Number anywhere (except in the specified space) on the Test
Booklet/ Answer Sheet.
5. There are three parts in the question paper 1,2,3 consisting of Physics, Chemistry and Mathematics having 30
questions in each subject and subject having two sections.
value (Example i,e. If answer is above 10 and less than 10.5 round off is 10 and If answer is from 10.5 and less
than 11 round off is 11).

SRI CHAITANYA IIT ACADEMY, INDIA 22‐12‐23_ Sr.Super60_Elite, Target & LIIT‐BTs_Jee‐Main_GTM‐02_Q.P
6. Use Blue / Black Point Pen only for writing particulars / marking responses on the Answer Sheet. Use of pencil is strictly
prohibited.
Admission Number:
22‐12‐23_Sr.Super60_Elite, Target & LIIT‐BTs _ Jee‐Main_GTM‐02_Test Syllabus

PHYSICS :1st YEAR SYLLABUS
CHEMISTRY : 1st YEAR SYLLABUS
MATHEMATICS : 1st YEAR SYLLABUS

correct.
-3
1. The density of a material in SI units is 128 kg. m in certain units in which the unit of length
is 50 cm and the unit of mass is 50g, the numerical value of density of the material is
1) 640 2) 320 3) 40 4) 410
2. A NCC parade is going at a uniform speed of 6km/h under a mango tree on which a monkey
is sitting at a height of 19.6 m. At any particular instant, the monkey drops a mango. A cadet
will receive the mango whose distance from the tree at time of drop is
(Given g = 9.8 m/s2)
1) 5 m 2) 3.33 m 3) 19.8 m 4) 24.5 m
3. Choose the correct relationship between Poisson ratio (σ), bulk modulus (K) and modulus of
rigidity (η) of a given solid object
3K  2 6 K  2 3K  2 6 K  2
1)   2)   3)   4)  
6 K  2 3K  2 6 K  2 3K  2
4. A block of mass 2 kg is given a push for a moment horizontally and then the block starts
sliding over a horizontal plane. The graph shows the velocity – time graph of the motion.
The co – efficient of kinetic friction between the plane and the block is g  10m / s 2  
1) 0.02 2) 0.50 3) 0.04 4) 0.40

-1
5. A sphere of mass 2 kg and radius 0.5 m is rolling with an initial speed of 1 ms goes up an
Inclined plane which make an angle of 300 with the horizontal plane, without slipping. How
long will the sphere take to return to the starting point A?
1) 0.60 s 2) 0.52s 3) 0.57 s 4) 0.80 s

6. Given below are two statements: one is labelled as Assertion (A) and other is labelled as
Reason (R)
Assertion (A): Time period of oscillation of a liquid drop depends of surface tension (S), if
3
density of the liquid is ρ and radius of the drop is r, then T  K  r 3/2 is dimensionally
S
incorrect, where K is dimensionless.

Reason (R) : Using dimensional analysis we get R.H.S having different dimension than
thatof time period. In the light of above statements, choose the correct answer from the
1) Both (A) and (R) are true and (R) is the correct explanation of (A)
2) Both (A) and (R) are true but (R) is not the correct explanation of (A)
3) (A) is true but (R) is false
4) (A) is false but (R) is true
From a solid sphere of mass M and radius R, a spherical portion of radius   is removed as
R
7.
2
shown in figure. Taking gravitational potential V = 0 at r   , the potential at the center of
the cavity thus formed is____(G = universal gravitational constant)

GM GM 2GM 2GM

1) 2) 3) 4)
R 2R 3R R
8. The incorrect statement is
1) work done by all forces acting on a system is equal to change in kinetic energy
2) work done by a conservative force is negative change of potential energy
3) work done depends on frame of reference
4) work done by friction is always negative
Statement – I: Pressure in a reservoir of water is same at all points at the same level of
water. (assume water is incompressible)
Statement – II: The pressure applied to enclosed water is transmitted in all directions
equally.
1) Both statement -I and statement - II are true
2) Statement –I is false but statement –II is true
3) Both statement -I and Statement -II are false
4) statement- I is true but statement -II is false
10. Two particles execute SHM of the same amplitude and frequency along the same straight
line. If they pass one another when going in opposite directions, each time their
displacement is half their amplitude, the phase difference between them is:
   2
1) 2) 3) 4)
3 4 6 3
11. A ball collides with a wall and after collision moves parallel to the wall as shown in the
figure. The impulse acting on the ball during the collision is
 4 1   4 1 
1) mv 0   i  j  2) mv 0   i  j 
 5 10   5 5 
 3 1   3 1 
3) mv0   i  j  4) mv0   i  j 
 5 10   5 5 
12. A process 1  2 using diatomic gas is shown on the P-V diagram below.
P2  2 P1  106 N / m 2 ,V2  4V1  0.4 m 3 .The molar heat capacity of the gas in this process
will be
35R 25R 35R 22 R

1) 2) 3) 4)
12 13 11 7

13. At high altitude, a body at rest explodes into two equal fragments with one fragment
receiving horizontal velocity of 10 m/s. Find the time taken (in second) by the velocity
vectors of the fragments to make 900 with each other (g = 10 m/s2).
1) 1 2) 2 3) 3 4) 4
14. An object of mass 8kg is hanging from one end of a uniform rod CD of mass 2 kg and length
1 m pivoted at its end C on a vertical wall as shown in figure. It is supported by a cable AB
such that the system is in equilibrium. The tension in the cable is (Take g = 10 m/s2)
1) 240 N 2) 30 N 3) 300 N 4) 90 N
15. Find the x-coordinate of centre of mass of a uniform plate bounded by the parabola
y  x 2 and x  y 2 , where x and y are in meter
1 3 9 9
1) m 2) m 3) m 4) m
2 20 10 20
16. A car is moving on a horizontal road whose acceleration versus position graph is drawn
below. At the initial moment of time when particle was at x  0 , its speed was 10 m / s .
Find the speed of the particle when it reaches at x  8 m .
1) 84 m / s 2) 10 m / s 3) 110 m / s 4) 11 m / s
17. Two blocks of mass 3kg and 6kg respectively are placed on a smooth horizontal surface
they are connected by a light spring of force constant k = 200 Nm-1. Initially the spring is
unstretched and velocities of 1 ms-1 and 2 ms-1 are imparted in opposite directions to the
respective blocks as shown in figure. The maximum extension of the spring will be
1) 15 cm 2) 20 cm 3) 25 cm 4) 30 cm
18. A heavy sheet of wood and a light ball is moving towards each other as sown in the figure.
What will be the speed of the ball after collision ?
1) 3v 2) 2v 3) v 4) v/2
19. A long capillary tube of radius r is initially just vertically completely immersed inside a
liquid of angle of contact 00. If the tube is slowly raised then relation between radius of
curvature of meniscus inside the capillary tube and displacement (h) of tube can be
represented by
1) 2)
3) 4)
20. A man standing on an inclined plane of inclination θ with the horizontal, fires one bullet up
the plane and another bullet down the plane with speed ‘u’. Both the bullets get stuck to the
plane after hitting the inclined plane. The value of maximum possible distance between the
two bullets after they hit the inclined plane is:
2u 2 2u 2 u2 u2
1) 2) 3) 4
g g cos 2  g g cos 2 
21. The centre of mass of a solid hemisphere of radius 8 cm is x cm from the centre of the flat
surface. Then value of 2x is _____.
22. A force of - Pk acts on the origin of the coordinate system. The torque about the point
  a
b
x
(2, -3) is P ai  b j , the ratio of is . Find the value of 5x.
2

23. When a horizontal force of 200N is applied on a body placed on a rough horizontal surface
The acceleration produced in it is 1.5 m/sec2. When the force is 300N the acceleration
produced in the same body is 2.5 m/sec2 then the mass of the body is n × 25kg then n = ___
24. A vector makes angles 600 and 450 with x and y axes respectively. If the angle between the
vector and x – y plane in degrees is 5n, find n.
25. A particle is moving on a circular path of radius 1m. It starts from rest, its tangential
acceleration is 3t . The time after which the acceleration of the particle makes an angle of
300 with centripetal acceleration is (n)2/3s. Find the value of n.
26. The length of wire becomes l1 and l2 when 100 N and 120 N tensions are applied
1
respectively. If 10l2  11l1, the natural length of wire will be l1. Find the value of 10x is___
x
27. An ideal fluid density 800 kg m-3, flows smoothly through a bent pipe (as shown in figure)
that tapers in cross – sectional area from a to a/2. The pressure difference between the wide
x 1
and narrow sections of pipe are 4100 Pa. At wider section, the velocity of fluid is ms
6
for x = _____ (Given g = 10 m s-2)
3
28. Heat Q  RT is supplied to 4 moles of an ideal diatomic gas at temperature T. how many
2
moles of the gas are disassociated into atoms if temperature of gas is constant?

R
29. A solid sphere of radius R is cut into two parts at a distance of from the centre as shown
2
n R3
in the figure. The volume of the upper part (spherical cap) is given as . Find the value
24
of n
30. Block ‘A’ is hanging from a vertical spring and is at rest. Block ‘B’ strikes the block ‘A’
with ‘v’ and sticks to it. Then the value of ‘v’ (in m/s) for which the spring just attains
natural length is

correct.
31. Which form of carbon is most reactive form
1) Diamond 2) Charcoal 3) Graphite 4) Fullerene
32. The consequence of high screening effect of inner electrons of the nucleus for an element
“X” is
1) A decrease in the ionization potential

2) An increase in the ionization potential.
3) No effect in the ionization potential
4) An increase in the attraction of the nucleus to the electrons
33. Hypothetically, Assume that possible l -values for a given “n” values are l =0,1,2,……..n.
Then find out Zn belongs to which block in the periodic table if all other rules for filling
electrons in an atom are followed (n=principal quantum number, l= azimuthal quantum
number)
1) s 2) p 3)d 4)f
34. For the following Assertion and Reason, the correct option is
Assertion (A): When Cu (II) and sulphide ions are mixed, they react together extremely
quickly to give a solid.
2
Reason (R): The equilibrium constant of Cu (aq )
 S(aq
2
)
 CuS(s) is high because the
solubility product is low.
1) A is false and R is true.
2) Both A and R are false.
3) Both A and R are true but R is not the correct explanation for A.
4) Both A and R are true and R is the correct explanation for A.
35. Total number of stereoisomer formed by the given compound
1)2 2) 3 3) 4 4) 8
36. IUPAC name for the compound should be
Br CONH 2
COOH
CHO COCl
1) 4-Formyl-2-chloroformyl-3-carbamoyl-5-bromohexanoic acid.
2)5-Bromo-4-carbamoyl-2-chloroformyl-3-formylhexanoic acid
3)2- Chloroformyl-3-carbamoyl-4-formyl-5-bromohexanoic acid
4) 5- Bromo-3-carbomoyl-2-chloroformyl-4-formylhexanoic acid.
37. Resonance in carbonate ion  CO32 
Which of the following is true?
1) All these structures are in dynamic equilibrium with each other

2) Each structure exists for equal amount of time.
3) CO32 has a single structure i.e., resonance hybrid of the above three structures.
4) It is possible to identify each structure individually by some physical or chemical method
38. Which of the following is the correct plot for the probability density  2  r  as a function of
distance ‘r’ of the electron from the nucleus for 3s orbital?
1) 2)

3) 4)
39. The correct stability order of the following three quinone is
O O
O O
O II III
I
1) III  II  I 2) I  III  II 3) I  II  III 4) III  I  II

40. The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is
[a0 is Bohr radius]
h2 h2 h2 h2
1) 2) 3) 4)
4 2ma02 16 2ma02 32 2ma02 64 2ma02
41. How many moles magnesium phosphate, Mg3  PO4 2 , will contain 0.25 mole of oxygen
atoms?
1) 2.5  102 2)0.02 3) 3.125  102 4)1.25  102
42. Assertion: Equivalent weight of HCl is always 36.5.
Reason: Equivalent weight of HCl can be greater than 36.5
1) Both A and R are true and R is the correct explanation of A.
2) Both A and R are true, but R is not the correct explanation of A.
3) A is true, but R is false.
4) A is false, but R is true.

43. Read the following statements
Statement-1: During the reaction: N 2  g   2 H 2  g   N 2 H 4  l  volume contraction takes
place.
Statement-2: Volume contraction always takes place when product is in liquid state.
1) Statement 1 and 2 are true. Statement 2 is the correct explanation for statement 1.
2) Statement 1 and 2 are true. Statement 2 is not the correct explanation for statement 1.
3) Statement 1 is true. Statement 2 is false.
4) Statement 1 is false, statement 2 is true.
44. S-I: Boron never forms B 3 ion

S-II: In IIIA elements Boron is having very high I.P values hence B 3 ion does not exists
1) Statement I and II are true. Statement II is the correct explanation for statement I.
2) Statement I and II are true. Statement II is not the correct explanation for statement I.
3) Statement I is true. Statement II is false.
4) Statement I is false, statement II is true.
45. Which compound would give 5-keto-2-methyl hexanal upon ozonylysis?
CH 3
CH 3
CH 3
CH 3
CH 3 H 3C
CH CH
1) 2) 3
3) 3
4)
46. The major product(s) obtained in the following reaction is/are:

 i  KOBu
 ii 2eq O3 / Me2 S
Br
CHO and OHC-CHO

OHC OHC
1) CHO 2)
O'Bu
OHC OHC CHO
3) CHO 4)
47. The correct order of heat of combustion for following alkaldienes is:
a) b) c)
1) c  b  a 2) a  c  b 3) b  c  a 4) a  b  c
48. Benzene on nitration gives nitrobenzene in presence of HNO3 and H 2 SO4 mixture, where:
1)both H 2 SO4and HNO3 act as an acids
2)both H 2 SO4and HNO3 act as a bases
3) HNO3 acts as an acid and H 2SO4 acts as a base
4) HNO3 acts as a base and H 2SO4 acts as an acid.
49. The structure of A and B formed in the following reaction are:  Ph  C6 H 5 
O
AlCl3  2 eq  Zn / Hg
+ O
  A  B
H 2O HCl
O
O
Ph
OH
A= OH , B= Ph
O
1) O
Ph
A= , B= Ph
O
OH
2)
O O
Ph
, OH
A= OH B=Ph
3) O
Ph
Ph
,
B=
A=
4) O

50. Match List-I with List-II
List-I List-II
(Chemicals) (Use/preparation/Constituent)
a)Alcoholic potassium hydroxide (i) electrodes in batteries
b) Pd / BaSO4 (ii) obtained by addition reaction
c)BHC/(Benzene hexachloride) (iii) used for   elimination reaction
d)Cold alk KMnO4 (iv) Bayer’s reagent.
Choose the most appropriate match:
1)  a    ii  ,  b    i  ,  c    iv  ,  d    iii  2)  a    iii  ,  b    i  ,  c    ii  ,  d    iv 
3)  a    iii  ,  b    i  ,  c    iv  ,  d    ii  4)  a    ii  ,  b    iv  ,  c    i  ,  d    iii 
51. How many total number of substituent’s are present in the following compound?
Br
Cl
52. Find the value of x.
Br
Br
Br
x Alc.KOH
Aromatic Compound
Br
Br
Br
53. Find out the possible Number of electrons having l  m value equal to zero in Cr.
54. For the Empirical formula xZnSO4 yH 2O of the following minerals that have the following
composition. ZnSO4  56.14, H 2O  43.86 .Find out x  y  ___

55. In a rocket motor filled with butane, how many kg of liquid oxygen should be provided for
each kg of butane for its complete combustion?
C4 H10  13 / 2O2  4CO2  5H 2O
56. Optical activity of an enantiomeric mixture is 12.60 and the specific rotation of (+) isomer
is 300 .The optical purity is_______%.
57. A sample of 0.2 g of an organic compound when analysed by Duma’s method yields 22.4
mL of N2 gas which is collected over KOH solution at 170 C and at a pressure of 749 mm Hg.
The percentage of nitrogen in the given organic compound is_____(Nearest integer)
a) Aqueous tension of water at 290 k is 13 mm Hg
b) R=0.0821 L atm/ k.mole.
58. Find number of statement/s which is/ are incorrect
a) Bonding molecular orbital is denoted as  MO   A   B
b) Anti bonding molecular orbital is denoted as  AMO   A  B
c) The bond order of acetylide ion is same as CN  
_
d) If 2s and 2p orbital’s of carbon are not mixing then C2 molecule is diamagnetic
e) If dipole moment of chlorobenzene is 1.8 Debye then metadichlorobenzene will have
dipole moment of √2 D (D = Debye)
f) As temperature increase pH of water decreases
59. According to the following figure the magnitude of enthalpy change of the reaction
A  B  C  D in kJ/mole is equal to _____(nearest integer)
a
A B
b CD
a = 50, b = 40, c = 10
60. A chloro compound “A”,
i) Forms aldehydes on ozonolysis followed by the hydrolysis.
ii) When vaporized completely 1.53 g of A, gives 448 ml of vapour at STP.
The number of carbon atoms in a molecule of compounds A is______

correct.
 3 5 7   2 4  
61. If tan 1         ....   cot 1    1    ...   then the maximum
 3! 5! 7!   2! 4!  2
   
value of  equals to
1 1 1
1) 2) 1 3) 4)
2 2 2
n
   1  1   1  
 n  1  n    n  2  ..... n  k 1   
62. lim n nk
n   2  2   2 
 1 
21 k 
 1 
1) 2 2) e  2  3) 2 1   4) e2
 2k 
1 1
e x
2
 x x2  1  ex  2 
63. f  x  e e . If f  x    .
"
 1      3  then  ,   
 x x x4  x2 
1  1   1   1 
1)  ,2  2)  , 2  3)   , 2  4)   , 2 
4  4   4   4 
64. Let L be the line passing through the point P (1,2) such that its intercepted segment between
the Co – ordinate axes is bisected at P. If L1 is the line perpendicular to L and passing
through the point (-2, 1) then the point of intersection of L and L1 is
 3 23   4 12   11 29   3 17 
1)  ,  2)  ,  3)  ,  4)  , 
 5 10  5 5   20 10   10 5 
65. One vertex of a rectangular parallelopiped is at the origin O and the lengths of its edges
along x,y and z axes are 3, 4 and 5 units respectively. Let P be the vertex (3,4,5). Then the
shortest distance between the diagonal OP and an edge parallel to z axis, not passing through
O or P is
12 12 12
1) 2) 12 5 3) 4)
5 5 5 5

x 1 y  2 z  3
66. The coordinates of one of the point on the line   . Which is at a distance
2 3 6
of 3 units from the point 1, 2,3 is
 13 23 3   13 5 39   1 5 3   1 5 39 
1)  , ,  2)  , ,  3)  , ,  4)  , , 
 7 7 7  7 7 7   7 7 7   7 7 7 
67. For x  R , two real valued functions f(x) and g(x) are such that, g  x   x  1
and fog  x   x  3  x . Then f(0) is equal to

1) 5 2) 1 3) -3 4) 0
 1 1 1
 8  then the value of
68. Let A be a non – singular matrix such that A1   4 6
100 100 99 
 
det  Adj  2 A    det  2 A  is
1) 8 2) 12 3) 16 4) 28
69. If the system of equations x  2 y  3z  3,4 x  3 y  4 z  4,8 x  4 y   z  9   has infinitely
many solutions, then the ordered pair   ,   is equal to
 72 21   72 21   72 21   72 21 
1)  ,  2)  ,  3)  ,  4)  , 
 5 5   5 5   5 5   5 5 
70. Let   4i  3 j  5k and   i  2 j  4k . Let 1 is a parallel to  and  2 be perpendicular to

 . If   1   2 Then the value of 5 2 . i  j  k is
1) 11 2) 6 3) 9 4) 7
71. Assertion: If {x} and [x] represent fractional part and integral part of x then
 
1000 x  b
 x   x
b 1
1000
Reason: {x}=x-[x] and [x+I] = [x] + I where I is an integer
1) Assertion is true, reason is true, and reason is correct explanation for assertion
2) Assertion is true, reason is true, and reason is not correct explanation for assertion
3) Assertion is true, reason is false
4) Assertion is false, reason is true
72. Statement 1: a1, a2 , a3 ,     a16 is an A.P such that a1  a4  a7     a16  147 ,then
a1  a6  a11  a16  98
Statement 2: In an A.P the sum of the terms equidistant from the beginning and the end is
always same and it is equal to the sum of first and last term
1) Statement 1 is true statement 2 is true, statement 2 is a correct explanation of statement 1
2) Statement 1 is true statement 2 is true; statement 2 is not a correct explanation for
statement 1
3) Statement 1 is true statement 2 is false
4) Statement 1 is false statement 2 is true
73. Assertion(A): If a,b,c are in A.P., the system of equations
3x  4 y  5 z  a ,
4x  5 y  6z  b ,
5x  6 y  7 z  c ,
is consistent.
Reason(R): If A  0 , the system of equations AX=B is consistent.
1) Assertion is true, reason is true, and reason is correct explanation for assertion
2) Assertion is true, reason is true, and reason is not correct explanation for assertion
3) Assertion is true, reason is false
4) Assertion is false, reason is true
74. Let f  x    x   1  x for -1  x  3 where [x] is the integral part of x. Then the number of
values of x in [-1, 3] at which f is not continuous is.
1) 0 2) 1 3) 2 4) 4
75. Let A and B two 2×2 matrices. Consider the statements
I: AB  O  A  O or B = O
II: AB  I 2  A  B 1
2
III:  A  B   A2  2 AB  B 2 Then
1) I and II are false, III is true 2) II and III are false, I is true
3) I is false II and III are true 4) I and III are false, II is true

76. Statement I: The point A (1,0,7) is the mirror image of the point B (1,6,3) in the line
x y 1 z  2
 
1 2 3
x y 1 z  2
Statement II: The line   bisects the line segment joining A (1,0,7) and B
1 2 3
(1,6,3)
1) Statement I is false. Statement II is true
2) Statement I is true, statement II is true
3) Statement I is true, statement II is true statement II is not correct explanation for
statement I
4) statement I is true, statement II is false
77. If f 1  1, f ' 1  3 then the derivative of f f f  x   f  x      2 at x  1
1) 12 2)33 3) 6
4) 15
 x  f  x
78. Let f be differentiable function satisfying the condition f    for all x, y   0   R
 
y f  y 
and f  y   0 .If f 1 1  2 then f 1  x  is equal to
f  x 2 f  x
1) 2 f  x  2) 3) 2x f  x  4)
x x
3  x 2 x2
79. f  x   and if f  x  has a local maximum at x = 2, then greatest
 a  14   x  48  x  2
value of a is
1) 2013 2) 2012 3) 2011 4) 2010
80. Let A and B be two 3x3 matrices of real number where A is symmetric and B is skew
symmetric. If  A  B  A  B    A  B  A  B  and  AB    1 AB then the possible
T K
values of K are
1) 2 2) 3 3) 8 4) 4
81.  4cos2 90  3 4cos2 270  3 4cos2 810  3 4cos2 2430  3 
1 1
82. Number of integers ≤10 satisfy the inequalities 2log 1  x  1   is____
3 log 2 8
2 x x
83. The sum of infinite terms of a decreasing GP is equal to the greatest value of the function
f  x   x3  3x  9 in the interval [-2,3] and the difference between the first two terms in
f   0  . If the common ratio of the GP is p/q then find the value of p + q? (where p,q are in
their lowest form)
84. Let p(x) be a polynomial of degree 4 having extremum at x = 1,2 and lim  1  p  x    2
2 x 0  x 
then the value of p(2) is _____
85 Let f be a twice-differentiable function such that f " x    f  x  , f '  x   g  x  ,
2 2
h  x    f  x     g  x   . If h  5  4 , then h 10  is ……
86. If   denote the greatest integer   then
 1    2    3   ......   120  is equal to
       
87. Let f be defined on the natural numbers as f(1) = 1 and for n > 1,
1 20 
f  n   f  f  n  1   f  n  f  n  1  then the value of 
30 r 1
f  r  is  , then
7
x  2 y 1 z  6 x  6 1 y z  8
88. The shortest distance between the lines   and   is equal to
3 2 2 3 2 0
__________
1
89. If Tr be the r th term of a sequence, for r = 1,2,3…..If 3Tr 1  Tr and T7  , then the
243

value of  Tr .Tr 1 is ____
r 1
 2 x  1
3
90. If f  x  is continuous every where, then the integral part of

sin  x log 2  log 1  x log 4 
2
f(0) is ______



KEY SHEET
PHYSICS
1) 1 2) 3 3) 1 4) 3 5) 4
6) 1 7) 2 8) 4 9) 4 10) 3
11) 2 12) 3 13) 2 14) 3 15) 4
16) 2 17) 4 18) 3 19) 1 20) 3
21) 4 22) 4 23) 5 24) 5 25) 3
26) 3 27) 7 28) 2 29) 54 30) 2
CHEMISTRY
31) 2 32) 4 33) 3 34) 1 35) 4
36) 3 37) 3 38) 2 39) 1 40) 1
41) 3 42) 2 43) 2 44) 1 45) 3
46) 1 47) 4 48) 4 49) 1 50) 2
51) 600 52) 12 53) 2 54) 2 55) 5
56) 84 57) 5 58) 5 59) 25 60) 3
MATHEMATICS
61) 1 62) 2 63) 1 64) 1 65) 2
66) 2 67) 3 68) 3 69) 3 70) 3
71) 4 72) 3 73) 2 74) 4 75) 1
76) 1 77) 3 78) 2 79) 3 80) 2
81) 7 82) 9 83) 2 84) 55 85) 5
86) 1 87) 4 88) 2 89) 300 90) 1002

SOLUTIONS
PHYSICS
1. G    M 1 L3T 2 
 c    LT 1 
 h   ML2T 1 
x y z
 M o LT
1 o
   M 1 L3T 2   LT 1   ML2T 1 
x  z  0 3x  y  2 z  1
2 x  y  z  0
1 3 1
x  ,y   ,z 
2 2 2
1
 x  y x
   16
 z 
2. t is independent of 
3.
u2 u2
Rmax  Rmin 
g/ 2 2g
Rmax
2 2
Rmin
l010a.e  kx  kx 
4. L  10 log dB  10[log10a  log e  kx ]  10  a  dB
l0  2.3 
20  2t
7. a
2
a  10  t
v 2
 dv   10  t dt
0 0
2
 t2 
v  10t  
 2 0
 20  2  18m / s
9. By equation of continuity
1 A  A2VB
AV
VB  4VA  16 m / s
Bernoulli’s theorem at point A and B
1 1
PA  VA2  PB  VB2
2 2
1 1
2.8  105   900   4   PB   900  16 
2 2
2 2
PB  172  10 N / m
3 2
 172kNm2   43  4 kNm2  K0  4
q q 3q
10.  Curved surface =  (1  cos  ) 
0 2 0 4 0

11. At t  , i 
2R
 3
At t  0, i  
R 4R
R R
2
R
R
2
 4R 2
Ratio = 
2 R 3 3
12. Conceptual
dl
dl  10000;V  30
14. Sol.  5000;V  15 dt
dt
L  3 mH for 4  6ms
V
L  3 mH , For 0  2ms
dl
dt
 
15. M  iA(  K )  4 (0.5) 2 K
 
M   K
     
T  M  B     K   10 i  10 J
 
2
mR 1 1
l  2  0.5  
2
2 2 4
T 10
   40 rad / s 2
I 1
4
 bx 
L L
0  dm  x 0  a  L  dx.x
16. xcm  L  L
 bx 
0 dm a  a  L  dx
17. i  iz  iL
iz  i  iL
19.

E  1 107  200 109

20. n   11011
hc 6.6 1034  3  108
1011
Number of electrons ejected = 3  108
10
q
v 
4 0 r

10 8
1.6 1019   9 109
 3V
4.8 102
3
21. For minima the minimum path difference is , then
2
3
 2   y2  y 
2
2
7
y
12
22. n  2 r
n2
 2 a0
z
4
2  2 a0
1
23. If the time of penetration is  t
Then resistance force
mv0
F
t
For reaction at the end
3a a
F  Mg 
4 2
mv0 3a Mga
 
t 4 2
3 mv0
t 
2 Mg
24. Impulse =change in momentum=J
Initially mass m is at rest .
 it will gain the momentum J.
At maximum compression the block and car will move with same velocity and
J2
momentum J.  KE of system at this instant 
2  5m 
J2
Initial KE=  From conservation of energy
2m
J2 1 2 J2 1 4J 2
 kx   kx 2 
m 2 2  5m  2 2  5m 
4J 2 2J
or x  
5mk 5mk
3RT 3R  373 3R  T
25. Vms  V   3V 
M M M
T T
 3 3  T  3  373  1119 K
373 373
T  o C   846o C
F sin 
26. area AB  tan 
F cos 
area AB
27. By MEC,
KEi  PEi  KE f  PE f
2
GMm 11 GMm 1  R 
0  0  K 
2R 8 R 2 2
7GMm
K
R3
13.6 nh
28. E  2 &L
n 2
29. K1  K 2  5.5MeV
And P1  P2
30. I A  I cm  md 2

CHEMISTRY
31.
 CH 3 3 C  CH3 3 C OH
CO2 Et

CH 3 MgBr
etherH 2O

H 2 SO4 , 00 C
COCH 3
 CH3 3 C  CH3 3 C

AlCl3
CH 3COCl

32. Glucose, Fructose and Mannose produces same osazone.

33. At C1 and C3 carbon dotted line species must be on the same side and opposite to C2
carbon species.
34.
Cl Br
Cl
HBr (Electrophilic addition)
A) Cl Br
HBr
peroxide (free radical addition)
Anhy HI
SN 2 OH I
B)
cone
HI I
OH
1
SN
SOCl2 CH 3  CO  Cl
O
C) CH 3  C  OH
Re dp Cl  CH 2  CO  OH
Cl2

CH 3 CH 3 CH 3
D)
H OH 
Na
 H ONa 
CH3Cl
SN 2
H OMe
14CH 3 14CH 3 14CH 3
CH 3 CH 3
Red p
I H 
MeONa
SN 2
 H OMe
I2
SN 2 14CH 3 14CH 3
35. cyclohexanol is more soluble in water. 1-hexanol can form inter molecular H-bond
with water
36.
OH


H 2O / H
 
OH O O
CHO
H HO
H OH
37. Reaction is Benzoin condensation.

38. In case of NI 3 , the lone pair moment adds on the resultant of the N  I moments but in
case of NF3 , the lone pair moment on N partly cancels the resultant N  F moments.
39. NaCl reacts with conc. H 2 SO4 to give colourless fumes of HCl which on treatment with
MnO2 get oxidized to yellowish green coloured Cl2 gas.
40. Only 10 aliphatic amines are prepared. 3, 4 cannot be prepared because of steric
hinderence.
41.
42. N 2 H 4 is not a chelating and ambident ligand.
43. Total 3 isomers 2 are cis and 1 are trans
2MnO2  2 K 2CO3  O2  2 K 2 MnO4  CO2
44.
X Y 
2 KMnO4  Cl2  2 K 2 MnO4  2 KCl
Z 
45. 2CuSO4  4 KI  Cu2 I 2  I 2  2 K 2 SO4
KI  I 2  KI 3
46. Both NO2 and NO3 gives brown gas with H 2 SO4 .

47.
Y Y
Z Z
P Z
P Y
Z Z
Y Y
Non zero zero
48. The   bonding molecular orbital possess two nodal palnes.
34.5
49. V.P  with  in temperature. T f  i.K f .m  1 2   1000  3
46  500
50.
N 2  3H 2  2 NH 3
Initial: 1 3 0
Af eq 1  x 3  3x 2x
Out of 4 moles 2 moles are reacted
1  x  3  3x  2  x  0.5
Total moles at eq  1  x  3  3x   2 x  3
1 p
PNH 3   p 
3 3
Q1 T1 Q T Q  Q2 T1  T2
51.    1 2  1 2  1 
Q2 T2 Q1 T1 Q1 T1
w T1  T2 w 1000  800
     w  600 J
Q1 T1 3000 1000
2.303 a
52. When t  t114 , a  a 0 / 4 t1/ 4  log 0
K a0 / 4
When t  t1/10 , a  a0 / 10 then
2.303 a0 t1/ 4 2.303 K
t1/10  log  20  log 4   20
K a0 / 10 t1/10 K 2.303  log10
log 4 2 log 2 2  0.3  20
  20   20   12
log10 log10 1
cell constant 1.15
53. conductivity (K)    5  10 3 s cm 1
Resistance 230
K 1000 5  103 103
Equivalent conductivity   eq     2 ohm 1cm 2 eq 1
normality 2.5
103 
2 2
0.059  Fe 2  0.059
54. EE  0
log  1.67  log
103    0.1
4 4
n  H    PO2  4
0.059
E  1.67  log107  1.67  0.103  1.567  2
4
55.  NH 2 S  Fe3  FeS  S  NH 4
HI  Fe 3  FeI 2  I 2 Sn 2  Fe 3  Fe 2  Sn 4

3
CN   Fe 3  Fe  CN 3  
excess
CN 
  Fe  CN 6 
yellow ppt
3
NaNO2  Fe  NO reaction SO2  Fe 3  Fe 2  SO42
S 2O32  Fe 3  Fe 2  S 4O62 SCN   Fe 3  Fe  SCN 3
Blood red
3 
NaIO3  Fe  H  No reaction
56. Equivalents of Na2C2O4  Equivalents of KMnO4
= total equivalents of KMnO4  excess equivalents of KMnO4 reacted with H 2C2O4
1000  W  2
 45  0.02  5  10  0.1 2  W=0.1675gm
134
0.1675
% purity =  100  83.75%  84%
0.2
57. Conceptual
58. a, b, c, e, f are true.
1.86  1 1000
59. T f  K f .molality  M  acid observed   120
0.155  100
Eq. acid = Eq. base
0.75 1 Normal 150
 1000  25   M  acid  Normal  150  i=   1.25
M acid 5 observed 120
i  1      0.25  %  25
  .. ..
60. O H ,  N HCOR,  CH  CH  COOH ,  N H 2 ,  N O


MATHEMATICS
61. f  x has minimum value at x  1
62. x  v  2 xdx  dv
2
1 ev v 2
f t    dv
2  v 2  2v  2 2

1 v 1 2v  2 1 ev
 e  dv 
2  v  2v  2  v  2v  2 2
2 2 2 v 2  2v  2

t
1  1 1
2 2
ex et
 f t    4     
2  x  2 x  2  0 2  t  2t  2 2 
2 4 2
2 5
e 1 et t e 7e 1
f 1   f t  
1
 f 1 1   f 1  f 1 1  
 t 4  2t 2  2 
2
10 4 25 50 4
x2 y 2
63.   1, x 2  y 2  8 x  0
9 4
x2 x2  8x 6
 1  13 x 2  72 x  36  0  x  6,
9 4 13
But x  6 is acceptable

A 6, 2 3  B  6, 2 3 
Equation of circle is x 2  y 2  12 x  24  0
dy  x  2    y  2 
64. 
dx  x  2    y  2 
Put x  2  h, y-2=k
dk h  k dv 1  v
 met k=vh  v+h 
dh h  k dh 1  v
1V 1
 dv   dh
1V 2
h
 tan 1  v   log 1  v 2   log h  c
1
2
1  y  2  1   y  2 2 
tan    log 1    log  x  2   c ______ 1

 x2 2   x  2 
2
 3, 2   0  0  c  c  0
Also (1) passes through  P  2, 3
1 1  1  1
tan 1    log 1  2   log p 2 tan 1    log 1  p 2 
 p 2  p   p
The equation of the circle is  x  h    y  k   k 2
2 2
65.
If passes through  1,1 then  1  h   1  k   k 2
2 2

1
h 2  2h  2k  2  0    0  4  4  2 k  2   0 k 
2
2 2
66. e
2
67. A  x  y  z  12, x  1, y  1, z  1 11C2  55
No. of ways of getting one correct  7C1 6!1  

1 1 1 1 1 1
68.       7C1  265 
 1! 2! 3! 4! 5! 6! 
No. of ways of getting two correct  7C2 5! 1        7C2  44 
1 1 1 1 1
 1! 2! 3! 4! 5! 
No. of ways of getting three correct  7C2 4!1       7C3  9 
1 1 1 1
 1! 2! 3! 4! 
Required no. of ways  7C1  265   7C2  44   7C3  9 
e x  cos x  e x  1   1  cos x 
2 2
1 3
69. Lt  Lt  2      1 
x 0 
x2  x   x 
x 0 2
2 2
71. n  S   55 n  A   5C2  1  1  3  or 1  2  2    1500
1500 12 1
p   2
55 25  p
n x n x 200  25  300  10
72. x 1 1 2 2   16
n1  n2 500
n1  12  d12   n2  22  d 22 
d1  x1  x  9, d 2  x2  x  6  
2
 67.2
n1  n2
74. Lt f  x   f  0 
x 0
5  a  10  a  5
75. Total non empty subsets – Subsets with product is odd.
1
76. Area  PR  QS
2
x 2  4 x  30
77. Let f  x   3  2  3  x  7  37
x  8 x  18
f  x1   f  x2   3 , but x1  x2
 f is not one-one
m 3
78. tan 600   m  0 or 3
1  3m
y  2  3  x  3  y  3 x  2  3 3  0
79.
r
C1C2  2r
2  r  2r  r 2  4r  4  0

80. f  x   tan  2 x  3   tan  2 x  2   tan 1  2 x  4   tan 1  2 x  3
1 1
 tan 1  2 x  5   tan 1  2 x  4   tan 1  2 x  6   tan 1  2 x  5 

 tan 1  2 x  6   tan 1  2 x  2 
 r 3
81. Focus of parabola is  3, 5  let  be angle between focal orders then tan  
2 s11 5
15
tan  
8
82.
lt
e x0
 2  2 cos x cos 2 x   x  3  e    e
3 3 9
x2
4 1  2x 1   x 1
83. e 2 x  4e x  58   2x  0  e  2 x   4  e  x   58  0
 e   e 
x
e e
 x 1 
 e  x  p   p  2  4 p  58  0 p  4 p  60  0
2 2
 p  10  p  6   0
 e 
85. PA  PB is minimum when R lies on AB  PA  PB  AB  5
1
86. a .b  a .c  0 , b.c=
2
1 0
2
a b  a c 1 a  b  a  c  11 2 1  1  1 1  1
0
2
87.
D
 /4
B A 2
 2
a 2b 3c
88.   116  0 , both roots are common    a  3, b  4, c  5
3 8 15
ABC is right angle triangle
89. a 1  r  r 2   70
4 a , 5ar , 4 ar 2  A.P
1
5r  2  2r 2  r  2,
2
If r  2, a  10
 2x  x 
2

1  x  1   3   ...........
1000
90.
 1  x  1 x  
2
 x 
 1  x  1 
1000

 1 x 
2
 1 
 1  x   1  x 
1000 1002
 
1 x 

only.
.
Admission Number:

correct.
1. The dimensions of length are expressed as G xC y h z . G is the universal gravitational

1
 x  y x
constant, c is the speed of light and h is the Planck’s constant. The value of   is
 z 
1) 16 2) 4 3) 32 4) 8
2. An adjustable smooth thread AB is attached between two circumferential points A and B of
a loop of radius R. Time taken by a bead between these points is t. The variation of ‘t’ with
angle  which AB made with the vertical is best represented by
A

g
t
t

1)  2)
t
t
 
3) 4)

3. A particle is projected from ground aiming the maximum horizontal range. Ratio of
maximum to minimum radius of curvature of the projectile path is
1) 2 2 2) 2 3) 2 4) 3 2
4. Intensity of sound in a medium is varying as I 0 .10 a .e  KX . X  0 is the position of the
source. Loudness (L) represented with X is ( I0 is the reference intensity)

L
10a L
10a
2.3a x x
1) 2) 2.3k
L
L 10a
10a
x x
2.3a 2.3k
3) k 4) a
5. Statement –I : A coin is lying at rest on ground when you approach it while riding on
bicycle according to you there is kinetic friction acting on the coin.
Statement –II : Kinetic friction acts when there is a relative motion between the contact
surfaces.
1) Statement – 1 is true , Statement – 2 is true : Statement – 2 is correct explanation
for statement-1
2) Statement – 1 is true , Statement – 2 is true : Statement – 2 is not a correct
explanation for statement-1
3) Statement – 1 is true , Statement – 2 is false
4)Statement – 1 is false, Statement – 2 is true
6. Assertion: current through a pure inductor is wattles.
Reason : Phase difference between e.m.f across a pure inductor and current through it is 900

or .
2
1) If both assertion and reason are true, and the reason is the correct explanation of
assertion.
2) If both assertion and reason are true, and the reason is not a correct explanation of
assertion.
3) If the assertion is true but the reason is false
4)If the assertion is false but the reason is true
7. Consider the arrangement shown. Velocity of the block at t=2 s is
F=2tN
t in seconds
2kg
1) 20 m/s 2) 18 m/s 3) 16 m/s 4) 17 m/s

8. Statement – 1: In resonance column experiment using same tuning fork, the difference
between successive resonant lengths decreases as the diameter of the tube is decreased.
Statement – 2: End correction decreases as the diameter of the tube decreases.
1) Statement – 1 is true , Statement – 2 is true : Statement – 2 is correct explanation
for statement-1
2) Statement – 1 is true , Statement – 2 is true : Statement – 2 is not a correct
explanation for statement-1
3) Statement – 1 is true , Statement – 2 is false
4)Statement – 1 is false, Statement – 2 is true
9. Oil of specific gravity 0.9 enters a horizontal pipe with a bend as shown if figure with speed
4 m/s and pressure 2.8  105 N / m 2 . The pressure of oil at the point B is 43 K 0 ( kNm  2 ) .
The numerical value of K0 is
1) 1 2) 2 3) 3 4) 4
10. A point change q is placed inside a cone as shown in figure. Electric flux passing through the
curved surface of the cone is
q q
1) 2) 3) 3q 4) 2q
2 0 4 0 4 0 3 0
11. Ratio of reading of ammeter A, at t  to that at t = 0 is (switch S is closed at t = 0 )
1) 4/3 2) 2/3 3) 1 4) 1/3

12. The output of the following logic gate combination is 1 with the given input that shown,
with one unknown input x ,then
1) X is 0 only 2) X is 1 only
3) Output is independent of X 4) Output 1 is not possible for any value of X
13. Match the E.M. waves with their applications
Column I Column II
A) Radio waves (i) Purification
B)Infrared (ii) Security scanner
C) Ultraviolet (iii) Broadcasting
D) X- rays (iv) Television remote control
1) A(i); B(ii); C(iii); D (iv) 2) A(iii); B(iv); C(i); D (ii)
3)A(iii); B(iv); C(ii); D (i) 4) A(iii); B(ii); C(i); D (iv)
14. A single element has the current voltage function graphed in figure. Determine the element.
1) 1 mH inductor 2) 1  F capacitor 3) 3 mH inductor 4) 3  F capacitor

15. A conducting ring of mass 2 kg and radius 0.5 m is placed on the smooth horizontal plane.
The ring carries a current i = 4A. A horizontal magnetic field B = 10t is switched on at t = 0
as shown in figure. The ring is left free to move at t = 1 second. The angular acceleration of
the ring is at t = 1 second is
1) 5 rad / s 2 2) 15 rad / s 2 3) 20 rad / s 2 4) 40 rad / s 2

bx
16. A uniform thin rod AB of length L has linear mass density   x   a  , where x is
L
measured from A. If the CM of the rod lies at a distance of  7  L from A, then a and b are
 12 
related as :
1) a  2b 2) 2a  b 3) a  b 4) 3a  2b
17. In the circuit, the current through Zener diode is (the break down voltage is 10V)
1) 10mA 2) 6.67mA 3) 5mA 4) 3.33mA

18. In P-N junction with open ends.
Statement –I : There is no systematic motion of charge carries
Statement –II : Holes and conductor electrons systematically go from P-side and from the
n-side to the p-side respectively.
Statement –III : There is no net charge transfer between the two sides.
Statement –IV : There is a constant electric field near the junction.
1) Statement I, II are correct 2) Statement II, III are correct
3) Statement III, IV are correct 4) Statement IV, I are correct
19. An object is placed in front of a convex mirror at a distance of 50cm. A plane mirror is
introduced covering the lower half of the convex mirror. If the distance between the object
and plane mirror is 30 cm it is found that there is no parallax between the images formed by
the two mirrors. What is the radius of curvature of the convex mirror?
1) 25 cm 2) 30 cm 3) 15 cm 4) 20 cm
20. A silver ball of radius 4.8 cm is suspended by the tread in the vacuum chamber. UV light of
waves length 200 nm is incident on the ball for some times during which a total energy of
1107 J falls on the surface. Assuming on an average one out of 103 photons incident is
able to eject electron. The potential on sphere will be
1) 1V 2) 2 V 3) 3 V 4) Zero
21. Two coherent point sources s1 and s2 vibrating in phase emit light of wavelength  The
separation between the sources is 2  . The smallest distance from s2 on a line passing
7
through s2 and perpendicular to s1 s2, where a minimum intensity occurs is then p
3p
value is________
22. de- Broglie wavelength of an electron revolving in second orbit of hydrogen atom ( a0 is
Bohr radius ) is K a0 then K value is ________
23. A cube of wood of side a and mass M is resting on a rough horizontal surface. A bullet of
mass (m <<M) and moving with velocity v0 strikes the block (cube). And gets embedded in
it. Assume that wood offers constant resistance force and cube cannot slide due to high
friction. The time of penetration of bullet into the cube so that the normal reaction passes
xv0 m
through D is then find x  y
yMg
24. A cart of mass 4m holds a block of mass m which is attached to the former by means for
spring of spring constant K, as shown in the figure. All surfaces are frictionless. Now a sharp
impulse J is given to block m as shown. The maximum compression of spring will be
2J
x mk
the x value is
25. The r.m.s. speed of the molecules of a gas at 100o C is v. The temperature at which the r.m.s.
speed will be 3v is k  282 C then k value is

26. For the given diagram, the ratio of shearing stress to normal stress at cross -section AB will
1
be (   30o ) then x value ___________
x
27. A small ball of mass m is released from a height R above the surface of a planet of mass M
and radius R as shown in fig. There is a narrow grove inside the planet in which a spring of
spring constant K and natural length R is fixed at it is one end O. If the mass ‘m’ moves R/2
xGMm
distance inside planet before coming to rest value of K will be then find x value____
3
R
28. The angular momentum of an electron in Bohr’s hydrogen atom whose energy is 3.4eV is
nh
then n value__________
2
29. A nucleus with mass number 220 initially at rest emits an  particle. If the Q value of the
reaction is 5.5MeV, the kinetic energy of the  particle is x  105 eV ______
30. A uniform rod of mass m is bent into the form of a semicircle of radius R. The moment of
inertia of the rod about an axis passing through ‘A’ and perpendicular to the plane of the
semi circle kmR 2 , then k value is :
A

correct.
31. The reaction of compound P with C H M gB r (excess) in  C2 H 5 2 O followed by addition of

3
H O gives Q. The compound Q on treatment with H S O at 0 C gives R. Compound R

0
2 2 4
with C H C O C l in presence of anhydrous A lC l in C H C l followed by treatment with H O

3 3 2 2 2
produces compound S. The no. of carbons and degree of unsaturation in S is __ and ___
respectively. The compound P is
 CH3 3 C
CO2 Et
1) 17 and 7 2) 17 and 6 3) 16 and 5 4) 16 and 6

32. Which of the following pair of compound gives same osazone product ?
a) Glucose and fructose b) fructose and mannose c) Glucose and mannose
1) Only a 2) Only c 3) Both a and c 4) a, b and c
33. Which of the following is correct regarding chlorination of methane in the presence of
sunlight
1) Involves the formation of carbon cation
2) Involves the formation of carbanion
3) Involves the formation of carbon free radical
4) Gives only one type of product
Column-I Column-II
Cl
HBr
I
HBr
Peroxide II
A) P) I and II are identical
Anhdrous
I
HI
Conc
II
B) HI Q) I and II are different

O SOCl2
I
CH 3  C  OH
Red P / Cl2
II
C) Catalytic R) Mechanism of formation of I
and II are same
CH 3 Na Me  Cl
a I
H OH
Re d P MeONa
b II
CH 3
I2
D) S) Mechanism of formation of I
and II are different
1) A  P, S ; B  P, S ; C  Q, S ; D  P, R
2) A  P, R; B  P, S ; C  Q, R; D  P, S
3) A  P, S ; B  P, R; C  Q, R; D  P, R
4) A  P, Q; B  Q, S ; C  P, R; D  Q, S
35. Statement –I : Cyclohexanol is less soluble in water than 1-hexanol.
Statement –II : 1-hexanol can form inter and intra molecular H-bond with water.
1) Both I and II are true and II is explaining I
2) Both I and II are true and II is not explaining I.
3) Statement I is true and II is false
4) Both the statements are false.
 P


H 2O / H
36. In the given reaction . (P) Will be

OH OH
+ +
1) OH 2) CHO
OH
+
HO   CH 2 4  CHO  CHO
3) 4)
2
CHO 
KCN / C2 H5OH / H 2O

 s
37. In the given reaction product (s) would be
OH
COO  and CH
CH 2OH CN
1) 2)
O OH O O
C  CH C__C
3) 4)
38. Assertion (A) : Among nitrogen halides N X , the dipole moment is highest for
3 N I3 and
lowest for N F3 .
Reason (R): Nitrogen halides NX 3 , have trigonal pyramidal structure.

1) A and R are true and R is explaining A
2) A and R are true and R is not explaining A
3) A is false R is true
4) A is true and R is false
39. Assertion (A) : NaCl reacts with concentrated H2SO4 to give cololurless fumes with pungent
smell. But on adding MnO2 the fumes become greenish yellow.
Reason (R): MnO2 oxidizes HCl to Cl2 which is greenish yellow gas.
2) A and R are true and R is not explaining A
3) A is false R is true
40. Which of the following amines can be prepared by Gabriel-phthalimide reaction.
i) Triethyl amine ii) n-butyl amine iii) t-butyl amine iv) neo pentyl amine v) Aniline
1) Only ii 2) Only ii and v 3) Only ii,iii,iv 4) i, ii, iii, iv and v

41. Compound ‘A’ is dibasic acid COOH  CH 2 5 COOH
A HI
excess
 CH3   CH 2 5  CH3
B
 i  C2 H 5ONa  i  C2 H 5 3 N NaOH
O C D E
O
 ii  H  C  O  C2 H5  ii  CH 2  CH  C  CH3
Compound E is
H 3C OH CH 3
O O
1) CHO 2) CHO
O
O
O
3) 4)
42. Assertion (A) : In N2H4 any one nitrogen atom can coordinate with central metal or both
can co-ordinate with central metal i.e., it can act as a chelating ligand.
Reason (R) : N2H4 is an ambident neutral ligand.
1) Both A and R are true and R is explaining A.
2) Both A and R are false.
3) A is true and R is false.
4) A is false and R is true
43. The ratio of cis and trans isomers of the complex Pt  NH3   H2O Cl Br  is
1) 5:4 2) 2:1 3) 7:2 4) 3:1

44. Silver ions are added to a solution with  Br    Cl    CO32    AsO43   0.1M . Which
       
compound will precipitate with lowest  Ag   ?

 
13
1) AgBr (ksp  5 10 ) 2) Ag 2CO3 ( K SP  8.1  10  12 )
3) AgCl ( K SP  1.8  10  10 ) 4) Ag 3 AsO 4 ( K SP  1  10  22 )
45. On adding KI solution in excess to solution of CuSO4 we get a precipitate ‘P’ and another
liquor ‘M’. Select the correct pair
1) P is CuI, M is I2 2) P is CuI2, M is I2
3) P is CuI, M is KI3 4) P is CuI2, M is KI3

 
46. Assertion (A) : NO3 ion cannot be distinguish from NO2 ion by brown ring test
 
Reason (R) : Both NO2 and NO3 ions evolve brown NO2 gas with conc. H 2 SO4 acid.
2) A and R are true but R is not explaining A
4) A is false and R is true
47. Two compounds PX 2Y3 and PX 3Y2 . (where P= phosphorous atom and X , Y  monovalent
atoms). If all X atoms replaced by Z atoms and electro negativity order is X  Y  Z . Then
incorrect statement(s) is / are
i) The dipole moment of product obtained from PX 2Y3 is non zero.
ii) The dipole moment of product obtained from PX 2Y3 is zero
iii) The dipole moment of product obtained from PX 3Y2 is zero
iv) The dipole moment of product obtained from PX 3Y2 is non zero
1) i,ii 2)ii,iii 3)i,ii,iv 4) ii,iv

48. The incorrect statement regarding molecular orbital(s) is
1) If a nodal plane lies in the inter nuclear axis then corresponding orbitals is Pi   bonding
M.O.
2) If there is a nodal plane perpendicular to the inter nuclear axis and lying between the
nuclei of bonded atoms then corresponding orbital is anti bonding M.O.
3) The  bonding molecular orbital does not contain nodal planes containing the inter
nuclear axis.
4) The   bonding molecular orbital possess three nodal planes containing the inter nuclear
axis.
49. Pure water freezes at 273K and 1 bar. The addition of 34.5gm of ethanol to 500g of water
changes the freezing point of the solution. Use the freezing point depression constant of
water as 2 K Kgmol  1 . The figure show below represent plot of vapour pressure (V.P)
versus temperature (T). Among the following the option representing change in freezing
point is : (Mol. Wt of ethanol =46 gm mol 1).
V .P  bar  Water V .P  bar  Water

1 1
ice ice
1)
270 273 T/K 2)
271 273 T/K
Water Water
1 1
V .P  bar 
V .P  bar  ice ice
270 273 T/K 271 273 T/K 

3) 4)

50. A mixture of N2 and H2 in the molar ratio. 1:3 attains equilibrium when 50% of mixture has
reacted. If P is the total pressure of the mixture the partial pressure of NH3 formed is P/Y.
Then the value of Y is ____
1) 2 2) 3 3) 1 4) 4
51. An engine absorbs heat at a temperature of 1000K and rejects heat at 800K. If the engine
operates at maximum possible efficiency, the amount of work performed by the engine for
3000J heat input is ___J.
52. An organic compound undergoes first order decomposition. The time taken for the
decomposition to 1/4 and 1/10 of its initial concentration are t1/4 and t1/10 respectively. What
is the value of
t1/4   20 ? (take log 2  0.3 )
t1/10  10
53. The resistance of 2.5N solution of acetic acid is 230 ohm, when measured in a cell having a
cell constant of 1.15cm 1 . The equivalent conductance is ___ ohm  1cm 2 equiv  1
54. Consider the following cell reaction 2 Fe  s   O 2  g   4 H   aq   2 Fe  2  aq   2 H 2 O  l  ;
E 0  1.67V ;  Fe 2   10 3 M , P  O2   0.1 atm and P

H
 3 , the cell potential at 250 C is __V
 
(Round off to nearest integer).
55. How many of the following species / reagent can reduce Fe  3  aq  in to Fe  2  aq  at
2  
normal conditions  NH4 2 S , HI , Sn  aq  , CN  aq  , NaNO2 , SO2, Na2S2O3, SCN  aq  ,
acidified NaIO3

56. An impure sample of sodium oxalate  Na2C2O4  . Weighing 0.20 g is dissolved in aq.
Solution of H 2 SO4 and solution is titrated at 70 0 C required 45 ml of 0.02M KMnO4

solution. The end points is over run, and back titration is carried out with 10 ml of 0.1M
oxalic acid solution. Find % purity of Na2C2O4 in the sample
( round off to nearest integer )
57. calculate the P H of 105 M in HCl solution is ______

58. How many of the following statements are correct
a) White phosphorous is more reactive then red phosphorus
b) Sulphuric acid is more viscous than water
c) Among chalcogens tendency of catenation is maximum for sulphur
d) Among hydrides of nitrogen family NH3has highest boiling point
e) Cl2 gas bleaches the articles in presence of moisture.

f) HClO is stronger acid than HBrO
59. 1.0 g of a mono basic acid HA in 100gm water lowers the freezing point by 0.155K. If 0.75
gm of same acid requires 25ml of N/5 NaOH solution for complete neutralization then %
degree of ionization of acid is  K f ( H 2O)  1.86kg mol 1 
60. How many of the following species are ortho para directing
COOH , CHO, OH ,  NO2 ,  NH 2 ,  NHCOR ,
COR, CF3 ,  SO3 H ,  NR3 , CONHCH 3 .

correct.
2
61. The Minimum value of the function f  x    e x t dt is ________
0
1) 2 e 1 2) 2 e  1 3) 2 4) e  e  1
 
t  2 5 
x
62. If f  t    e x   dx then f 1   f 1   _______
1
 
2
0  x4  2 x2  2 
 
3e 1 7e 1 7e 1 2e 1
1)  2)  3)  4) 
10 4 50 4 50 2 5 2
x2 y 2
63. The circle x  y  8 x  0 and hyperbola   1 intersect at the points A and B. The
2 2
9 4
equation of the circle with AB as its diameter whose
I: centre is  6,0
II: radius is 2 3
1) I is true, II is true 2) I is false, II is true
3) I is true, II is false 4) I is false, II is false
dy x  y  4
64. If the solution curve of the differential equation  passes through the point
dx x y
 3,2 and  P  2,3 , P  0 then

1
1) 2 tan 1    log p 2  1
 p
  1
2) tan 1    log p 2  1
 p
 
 p2  1 
3) 2 tan 1  1  2

 p  1   log p  2 p  2
 
 4) 2tan 1  1 
 p   log  p 
   

65. Consider a family of circles which are passing through the point  1,1 and are tangent to the
x-axis. If  h, k  are the coordinates of the centre of the circles then set of values of K is
given by the interval
1 1 1 1 1
1) 0  K  2) K  3)   K  4) K 
2 2 2 2 2
66. In a group of 100 persons 75 speak telugu and 40 speak hindi. Each person speaks at least
one of the two language. If the number of persons who speak only telugu is  and the
x2 y2
number of persons who speak only hindi is  then the eccentricity of the ellipse  1
 2
2
is ______
595 119 129 595
1) 2) 3) 4)
12 12 12 12
x y -z 
 3 where x, y, z  N . If det  adj  adjA   2 3 then the no. of such
8 4
67. Let A  1 2

1 1 2
matrices A is _______
1) 220 2) 45 3) 55 4) 110
68. 7 people have their bags outside temple and after returning picked one bag each at random.
In how many ways at least one and atmost three of them get their correct bags.
1) 7 C6 9  7 C5 44  7 C1 265 2) 7 C6 265  7 C5 9  7 C4 44
7 7 7 7 7 7
3) C39  C2 44  C1265 4) C39  C2 44  C1265
2
ex  cos x
69. Lt  ______
x0 sin2 x
3 5
1) 2 2) 3 3) 4)
2 4
70. Let R be the set of real numbers
Statement-I: A   x , y   R  R / y  x is an integer is an equivalence relation on R
Statement-II: B   x, y   R  R / x   y for some rational number   is an equivalence
relation on R.
1) Statement-I is true, Statement-II is true, Statement-II is the correct explanation of
Statement-I
2) Statement-I is true, Statement-II is true, statement-II is not the correct explanation of
Statement-I
3) Statement -I is true, Statement-II is false.
4) Statement-I is false, Statement-II is true.
71. 5 different marbles are placed in 5 different boxes randomly given that each box can hold
any number of marbles. If p is the probability that exactly two boxes remain empty, then find
1
the value of   where  .  represents GIF
 p
1) 4 2) 3 3) 1 4) 2
72. The mean of two samples of sizes 200 and 300 were found to be 25 and 10 respectively their
standard deviations were 3 and 4 respectively. The variance of combined sample of size 500
is ________
1) 64 2) 65.2 3) 67.2 4) 64.2
73. Let A and B be two symmetric matrices of order 3
Statement-I: A  BA and  AB  A are symmetric matrices.
Statement-II: AB is symmetric matrix if matrix multiplication of A with B is commutative
Statement-I
Statement-I
3) Statement-I is true, Statement-II is false.
 log 1  5 x   log 1  ax 
 , x0
74. Let f  x    x
 x0
 10
Be continuous at x  0 . Then a=_______
1) 10 2) -10 3) 5 4) -5
75. Let S  1,2,3,....50 the number of non empty subsets A of S. such that the product of
elements in A is even is __________

1) 2 25 2 25  1  50
2) 2  1
25
3) 2  1
50
4) 2  1

76. The area of the quadrilateral PQRS with vertices P  2,1,1 Q 1,2,5 R  2, 3,5 S1, 6, 7
is equal to ________
1) 8 38 2) 9 38 3) 54 4) 7 38
where a  0 and n is a positive integer then f  f  x   x

1
77. Statement-I: If f  x    a  x n
 n
x2  4x  30
Statement-II: The function f  x   is not one-to-one
x2  8x 18
1) I is true, II is false 2) I is false, II is false
3) I is true, II is true 4) I is false, II is true.
78. A straight line L through  3, 2  is inclined at an angle 600 to the line 3 x  y  1 . If L also
intersects the x-axis then the equation of L is ______
1) y  3x  2  3 3  0 2) y  3x  2  3 3  0
3) 3y  x  3  2 3  0 4) 3y  x  3  2 3  0
79. Let a circle be inscribed in the quadrant of a circle of diameter 4 then
Statement-I: The radius of inscribed circle is the positive root of the equation r 2  4r  4  0
Statement-II: Distance between their centers = 2 (radius of circle inscribed)
Statement-I
Statement-I
3) Statement-I is true, Statement-II is false.
80.       
If f  x   cot 1 4 x 2  10 x  7  cot 1 4 x 2  14 x  13  cot 1 4 x 2  18 x  21  cot 1 4 x 2  22 x  31 
 185 
then f 1  0   ________
 16 
1) 3 2) 4 3) 1 4) 5
If angle between two focal chords of a parabola  y  5  8  x  1 which are tangents to the
2
81.
circle x 2  y 2  9 is tan 1   where a and b are relatively prime then a  b  _ _ _ _ _ _

a
b
x3
82. If the value of lt
x 0
 3  2 cos x cos 2 x x 2  e a then a  _________
83. The number of real solutions of the equations e 4 x  4 e 3 x  58 e 2 x  4 e x  1  0 is ________

n
in the expansion of  2   are equal then the value of n is __

x
84. If the coefficient of x7 and x8
 3
85. As z  x  iy varies the minimum value of z  4i  z  3 is __________
86. Let a ,b ,c be unit vectors such that a is perpendicular to the plane of b and c . If
 b , c   60 then
0
a  b  a  c  ________
 z2 
87. If z is such that z  2i  2 2 then Arg    , then K = __________
 z2 k
88. If the equations ax 2  2bx  3c  0 and 3 x 2  8 x  15  0 have a common root where a,b,c are the
lengths of the sides of ABC then sin 2 A  sin 2 B  sin 2 C  _ _ _ _ _ _ _ _ _ _ _
89. Three numbers are in G.P whose sum is 70 if the extrems be each multiplied by 4 and the
mean by 5. They will be in A.P then the sum of numbers is _______
in 1  x   2 x 1  x   3 x 2 1  x   ........... is KC50 then k = ______
1000 999 998
90. If the coefficient of x50



KEY SHEET
PHYSICS
1) 4 2) 2 3) 2 4) 4 5) 4
6) 4 7) 1 8) 3 9) 3 10) 2
11) 3 12) 3 13) 1 14) 3 15) 3
16) 1 17) 2 18) 4 19) 2 20) 2
21) 10 22) 1 23) 300 24) 2 25) 9
26) 1 27) 9 28) 4 29) 2 30) 0
CHEMISTRY
31) 1 32) 1 33) 1 34) 2 35) 3
36) 2 37) 3 38) 2 39) 2 40) 1
41) 1 42) 4 43) 4 44) 2 45) 4
46) 1 47) 4 48) 2 49) 2 50) 2
51) 33 52) 2 53) 28 54) 1 55) 4
56) 9 57) 3 58) 66 59) 136 60) 10
MATHEMATICS
61) 1 62) 1 63) 2 64) 2 65) 1
66) 2 67) 2 68) 1 69) 3 70) 2
71) 3 72) 1 73) 1 74) 4 75) 3
76) 2 77) 4 78) 1 79) 2 80) 2
81) 2 82) 2 83) 0 84) 5 85) 6
86) 21 87) 63 88) 11 89) 28 90) 8

SOLUTIONS
PHYSICS
1. let the scooterist velocity be v. Then
2000
1000  10  100   v  100  100v  2000  v  20 m / s
100
2. Conceptual
3. In the given equation     b x ;
 b  
   . But  is mass per unit length and x is distance, therefore
 x
ML1
b    ML2T 0 .
L
4. Point A is at rest w.r.t. motion, hence, v at A = 0. At point B there are two horizontal
velocities, hence vB  2v
5. mg  2TL   r 2 Ldg  2TL   r 2dg  2T .
This relation is independent of L.
6. Conceptual
7. Both are diatomic gases are C p  Cv  R for all gases.
K11l2  K 22l1 K  0  2  3K  100  1
8.    60 C
K1l2  K 2l1 K  2  3K  1
1 1 2 3C  2C
9.   
C C  C C C  C  C 
 1  3 
2C2  2CC  C 2  0  C  C  
 2 
10. R increasing with increasing temperature V  IR
I 1 1
Slope of graph   ; Slope of T1 is more i.e. is more, hence R1 is less. This
V R R1
concludes that T1 will be less than T2 as R1 is less than R2 .
RR 4 2 8 4
11. Req  1 2  Parallel     
R1  R2 42 6 3
V0 100 V 100 V 100

12. iR    5, iL  0   10 and iC  0  5
R 20 X L 10 X C 20
Current, i  iR2   iC  iL   52  52  5 2 amp.

2

13. f max   mg , amax   g
g
If A is the amplitude amax  A 2  4 2 A 2   g . Therefore, A 
4 2 2
14. Total time taken to travel distance d is:
d d n n  d 3
 d 1 2  ; n2  3n1  neff  n2
2n1 2n2  2n1n2  neff 2
15.  A1  A2    A3  A4   A12  A34
16.
nh h
mvr  ,  ;
2 mv
1 h
2 r  ________  i 
mv
h
 ________  ii 
mv
2 r h  mv 1
Divide (ii) by (i),    1:1
 mv  h 1
2 1 
17. VB  V A     2dx   3dy     2  2  1  3 1  2      2  3  1V
1 2 
18. In the graph given, slope of curve 2 is greater than the slope of curve 1.
P P
      2  1
 V 2  V 1
 He   O2
19. Conceptual
20. Conceptual
21. Here, m A  0.5 Kg ; mB  1kg
Force on block A, T   mA g       1

Force acting on block B, F  T   mA g    mA  mB  g        2 
F   m A g   m A g   m A g   mB g
From (1) and (2) F  3 m A g   mB g   g  3m A  mB 
 0.4  10   3  0.5  1  10 N
v GM 1 2GM
22. Given, v0  e 
2 Rh 2 R
hR
GMm  GMm  1 2
    mv
When the satellite stops, K  0 2R  R  2
v  gR

T T 200  400
23. Here the number of molecules is same. Hence, T final  1 2   300 K
2 2
24. Magnetic induction at O due to coil Y is given by,
2 I  2r 
2

BY  0         1
4  2
32
 2r   d 
2
 
Similarly, the magnetic induction at O due to coil X is given by
 2 Ir 2
BX  0          2
4  2 2 3 2
 r   d 2  
B 1
From eq. (1) & (2), Y 
BX 2
kq
25. Field inside the shell is zero, so potential is uniform. V   9 kilovolts
2R
26. Due to balanced wheatstone bridge ABCEA BE  8 can be taken out
RAC 
1  2  2  4   18  2
1  2    2  4  9
Now, ACFGHA is a balanced Wheatstone bridge, hence HC=10Ω can be taken out
6  18 108
RAG    4.5 
6  18 24
V 6.5
Hence, current i   1A
Rtotal 6.5
27. Conceptual
28. Conceptual
29. Given that: x  40cos  50 t  0.02 y 
  40  50    sin  50 t  0.02 y  

dx
Particle velocity vP 
dt
1
Putting y=25 cm and t  s
200
vP  10 2
30. All the four corners will be at the same potential W  qV  0

CHEMISTRY
31. Key 1
 C6 H 5 NH 3  OH 

C6 H 5 N  H 2O 

1 0 0
1-  
( Dissociation occurs in presence of NaOH and thus dissociation of C6 H 5 NH 2 will
suppress)
[C6 H 5 NH 3 ] 108  [OH  ]
Kb  
[C6 H 5 NH 2 ] [0.24]
Kw 1014
 Kb for C6 H 5 NH 2  
K a forC6 H 5 NH 3 2.4  105
 1014  0.24
[OH ]   0.01
2.4  105  108
 NaOH  102 M
32. Key 1
33. Key 3
34. Key 2
35. Key 3
This compound has only one ionizable Cl  out of given three
[Co( NH 3 ) 4 Cl2 ]  Cl 
[Co( NH 3 ) 4 Cl2 
36. Key 2
(four unpaired electrons)
(in the presence of weak ligand, it also has four unpaired electrons).
37. Key 3
38. Key 2
Diazonium ion acts as an electrophile in coupling reaction. The greater the electron-
withdrawing power, the higher is the electrophilicity.
39. Key 2
40. Key 1
If keto group is at β-position w. r. t. COOH heating causes decarboxylation

41. Key 1
On cis-alkene there is syn addition of two-OH groups forming meso compound.
42. Key 4
43. Key 4
No complex formation
As
NH3 combines with H  of acid the changes to NH 4 which have no doner site.
44. Key 2
45. Key 4
order of electron gain enthalpy Cl > F > O
Second electron gain enthalpy for an element is always positive.
46. Key 1
Conceptual
47. Key 4
According to Drago's rule lone pair on phosphorus resides in almost pure s-orbital, hence
due to non-directional nature, its overlapping tendency is greatly reduced in comparison
to a lone pair present in hybrid orbital, which is directional as present in N H 3 .
48. Key 2
49. Key 2
Stability order is I > II > III > IV, because I is neutral, when in II all atoms with complete
octet system while is III is more stable than IV because in case of IV oxygen is positive
with incomplete octet system.
50. Key 2
51. Key 33
Mole at equilibrium (a-x) 2x

Given original volume = (75/100) or existing volume at equilibrium
Or Initial volume = (75/100) x mole at equilibrium

52. Key 2
The ratio constant for the first-order reaction is given by
2.303 a
k log
t ax
Now, when three-fourth of the reaction is complete,
and when the reaction is half completed, x = a/2
Dividing (1) by (2), we get
Thus, the time taken for three-fourth of the reaction to be completed is double than that
taken for half of the reaction
53. Key 28
17
The energy needed to see object = 10 J
hc
Photon energy used to see object = 
54. Key 1
where n = order of reaction

55. Key 4
56. Key 9
Total energy released during combustion of
3.5 g gas = (mS) x ∆T
= 2.5 x 0.45 = 1.125 kJ
 Heat released by 1 mole of gas combustion.
28
  1.125 (molar mass of gas = 28)
3.5
 9kJ mol 1
57. Key 3
58. Key 66
The redox changes are
59. Key 136

 m=136 g mol1
60. Key 10

MATHEMATICS
2 2
61. 1. a  7a  1  0 and a  8a  1  0 which satisfied by 6 negative integers
62. 1
Sol. Relation R is reftexive and symmetry as xRy is true then yRx is also true
63. 2
Sol. Any number having exactly 4 factors is of the form m  p 3 (p prime) or m = p.q (where
p & q are
distinct primes) So we have 5 C2  2  12 such factors
64. 2
65. 1
Sol. The given equation is dx – x  ydx  xdy   x 5 y 4  ydx  xdy 
66. B
Sol. The distance of the point (1,3) from the line 3x+4y=5 is 2. Now the value of
sec2   2cosec 2 is
greater than 2 hence two such lines will be possible.
67. B
Sol. Incircle and circumcircle have same centre but radius is half of that of circumcircle
Also c = - (2g + 2f + 2) as (1, 1) lies on circumcircle
68. A
Sol. FA = 4, FB = 5
1 1 1 20 80
We know that   a  4a 
a AF FB 9 9
69. C
2
b
Sol. For the given ellipse a 2  16 . Therefore e  1 
a2
So, the foci ellipse are (  ae,0)i.e.   16  b 2 ,0 

 
For the hyperbola e is given by
70. B
Sol
71. C
Sol. Let assume that ‘A’ wins after n deuces, n   0,   
2 2 1 1 5
Probability of a deuce  .  . 
3 3 3 3 9
(A wins his serve then B wins his serve or A loses his serve then B also loses his serve)
Now
probability of 'A' winning the game

72. A
Sol.
73. A
Sol. 3 median = 2 Mean +Mode
74. D
Sol. There will be ( n1  1) gaps created by n1 men. Now women have to be seated only in
these gaps.
Thus number of such sitting arrangements
75. C
Sol. Since x2 y3 z4 is occurring in the expansion of (x +y +z)n , so n should be 9 only.
Now
Coefficient of
76. B
Sol.
No. of positive solutions = No. of division of n 2 n2  7  9  13  819
77. D
Sol.
78. B
Sol.
79. B
Sol.
80. B
Sol.
81. 2
Sol. Image of A(1, 2) in y = x will be So B is (2, 1)
Image of A 1, 2  in x  2 y  1  0 will lie on BC.
9 2
So another point on BC  , 
5 5
Equation of BC is 3x  y  5  0
82. 2
Sol. a, b, c are in
Again a, b, c are in H.P.
From (A) and (B)

83. 83
Sol. The combined equation of the tangents drawn from (0,0) to
This equation represents a pair of perpendicular straight lines

If coefficient of
84. Shortest distance will take place through along the common normal
Equation of normal for y 2  4 x at ( at 2 , 2t ) is y  tx  2t  3 t
If it passes through (6, 0), then t 3  4t  0  t  0, t  2
 Min distance  20  25  2 5  5  5
85. Shaded region is the required one.

 Required Area
86. 21
Sol. Hint: trace of
then
87. 63
Sol. When everyone gets a different number  6 C4 . 4
When only two persons siting opposite to each other get same number  6 C3.3 C1.2. 2
When two pairs of persons sitting opposite to each other get same number  6 C2 . 2
So = N = 630
88. 11
Sol.

89. 28
Sol.
90. 8
Sol.

Sec: SR.IIT_*CO-SC(MODEL-A,B&C) Date:21-12-23
Time: 3HRS Max. Marks: 300
Name of the Student: ___________________ H.T. NO:
21-12-23_SR.STAR CO-SUPER CHAINA(MODEL-A,B&C)_JEE-MAIN_GTM-1_SYLLABUS

PHYSICS: TOTAL SYLLABUS
CHEMISTRY: TOTAL SYLLABUS
MATHEMATICS: TOTAL SYLLABUS
MISTAKES
SUBJECT JEE JEE TOTAL
SYLLABUS Q'S EXTRA SYLLABUS Q'S Q'S
MATHS
PHYISCS
CHEMISTRY
Narayana IIT Academy 21-12-23_SR.IIT_*CO-SC(MODEL-A,B&C)_JEE-MAIN_GTM-1_Q’P
PHYSICS MAX.MARKS: 100
SECTION – I
This section contains 20 multiple choice questions. Each question has 4 options (1), (2), (3) and (4) for its answer,
out of which ONLY ONE option can be correct.
Marking scheme: +4 for correct answer, 0 if not attempted and -1 if not correct.
1. A particle is moving parallel to x-axis as shown in the figure such that at all instants the
y - component of its position vector is constant and is equal to ‘b’. The angular velocity
of the particle ‘P’ about the origin at the given instant is
v v v 2
(A) cos  (B) sin  (C) sin  (D) vb
b b b
2. An object is placed on the surface of a smooth inclined plane of inclination . It takes

time t to reach the bottom. If the same object is allowed to slide down a rough
inclined plane of same inclination  ,so as to move the same distance, it takes time
‘nt’ to reach the bottom where n is a number greater than 1. The coefficient of friction
 is given by:
(A)   tan   1 
1 
(B)   cot   1 
1 
 
 n2   n2 
1 1
(C)   tan  1  2  (D)   cot  1  2 

1 2 1 2
 n   n 
3. A system of two bodies of masses ‘m’ and ‘M’ being interconnected by a spring of
stiffness k,in its natural length, moves towards a rigid wall on a smooth horizontal
surface as shown in figure with a K.E. of system ‘E’. If the body M sticks to the wall
after the collision, the maximum compression of the spring will be:
k
m
M
mE 2mE 2m  M  E 2ME

(A) (B) (C ) (D)
Mk  M  m k mk  M  m k
SR.IIT_*CO-SC Page. No. 2

4. A sphere of mass m1 = 2kg collides with a sphere of mass m2 = 3kg which is at rest.
Mass m1, after the collision will move at right angle to the line joining centers of two
spheres at the time of collision, assuming colliding surfaces are smooth, if the
coefficient of restitution is
4 1 2 2
(A) (B) (C) (D)
9 2 3 3
5. From a circular disc of radius and mass 9 , a small disc of radius /3 is removed
from the disc. The moment of inertia of the remaining disc about an axis perpendicular
to the plane of the disc and passing through is
A) B) C) D)
6. A planet of radius R has an acceleration due to gravity of g s on its surface. A deep
smooth tunnel is dug on this planet, radially inward, to reach a point P located at a
R
distance of from the centre of the planet. Assume that the planet has uniform density.
2
The kinetic energy required to be given to a small body of mass m, projected radially
outward from P, so that it gains a maximum altitude equal to the thrice the radius of the
planet from its surface, is equal to
O P
63 3 9 21
A) mg s R B) mg s R C) mg s R D) mgs R
16 8 8 8

7. Water rises to a height of 10cm in a capillary tube and mercury falls to a depth of 3.42
cm in the same capillary tube. If the density of mercury is 13.6g/cc and the angle of
contact of mercury and water are 135 and 0 respectively, the ratio of surface tension of
water to mercury is
(A)1 : 0.15 (B)1 : 3 (C)1 : 6.5 (D)1.5 : 1
8. Two moles of ideal helium gas are in a rubber balloon at 30℃. The balloon is fully
expandable and can be assumed to require no energy in its expansion. The temperature
of the gas in the balloon is slowly changed to 35℃. The amount of heat required in
raising the temperature is nearly (take 𝑅=8.31 𝐽/𝑚𝑜𝑙.𝐾)
A) B) C) D)
9. Two rigid boxes containing different ideal gases are placed on a table. Box A contains
one mole of nitrogen at temperature while box B contains one mole of helium at
temperature The boxes are then put into thermal contact with each other and heat
flows between them until the gases reach a common final temperature (Ignore the heat
capacity of boxes and heat exchange will happen only between boxes). Then, the final
temperature of the gases, in terms of is
A) B) C) D)
10. A particle free to move along the x-axis has potential energy given by
U ( x )  k [1  exp(  x 2 )] for   x   , where k is a positive constant of appropriate
dimensions. Then
A) At point away from the origin, the particle is in unstable equilibrium
B) For any finite non-zero value of x, there is a force directed away from the origin
C) If its total mechanical energy is k/2, thenits kinetic energy at the origin is k.
D) For small displacements from x = 0, the motion is simple harmonic.
11. The electric potential in a medium of dielectric constant ‘unity’ is   x, y, z   ax2 where
‘a’ is a constant of suitable dimensions. The total charge contained in a cube of
dimensions L  x, y, z  L is
A) zero B) 2a 0 L3 C) 16a 0 L3 D) 4a 0 L3

12. Three large plates are arranged as shown. How much charge will flow through the key
when it is closed?
A) B) C) D) Q
13. Assertion : A current I flows along the length of an infinitely long straight and thin
walled pipe. Then, the magnetic field at any point inside the pipe is zero.
 
Reason :  B .d l  o I
Read the assertion and reason carefully to mark the correct option out of the options
given below:
A) Both Assertion and Reason are true and the reason is the correct explanation of the
assertion.
B) Both Assertion and Reason are true but reason is not the correct explanation of the
assertion.
C) Assertion is true but Reason is false.
D) Assertion and Reason both are false.
14. The graph for an alloy of paramagnetic nature is shown in fig. the curie constant
is nearly.

0.4
0.3
0.2
0.1
O
0 2 4 6 7
1/T(in 10–3K–1)
A) B) C) D)

15. The natural frequency of the circuit shown in the figure is (Assume that equal currents i,
i exist in the two branches, and the charges on the capacitors are equal)
q q
i+ – + – i
C C
L L L
1 1 1 1 1 1 1 4
A) B) C) D)
2 LC 2 3LC  LC  3LC
16. In the circuit shown in the figure, the ac source gives a voltage V  20 cos(2000 t ).
Neglecting source resistance, the voltmeter and ammeter reading will be ( 2  1.4 )
A) 0V, 0.47A B)1.68V, 0.47A C) 0V, 1.4 A D) 5.6V, 1.4 A

17. A plane Electromagnetic Waves travelling along the positive -direction has a
wavelength of 3 mm. The variation in the electric field occurs in the -direction with an
amplitude66Vm-1. The equations for the electric and magnetic fields as a function of
and are respectively
A)
B)
C)
D)
18. In young’s double slit experiment, the distance between the slits varies with time as
d  t    2d 0  d 0 sin wt  , where d0 and ‘w’ are positive constants. The difference between
the largest and the smallest fringe width obtained over time is___
(D= distance between slits and screen  d &  = wavelength of light used)
D D 2 D D
A) B) C) D)
2d 0 3d 0 3d 0 6d 0

19. Match the following.
A) Moderator e) Absorbs heat
B) Control rods f) Prevents radiation from getting exposed
to the outside
C) Radiation shielding g) Absorb neutrons
D) Coolant h) Slow down neutrons
A) A – h, B – g, C – f, D – e B) A – g, B – h, C – f, D – e
C) A – h, B – g, C – e, D – f D) A – h, B – f, C – e, D –g
20. A student performs an experiment to determine the Young’s modulus of a wire, exactly
2 m long, by Searle’s method. In a particular reading, the student measures the extension
in the length of the wire to be 0.8 mm with an uncertainty of at a load of
exactly 1.0 kg. The student also measures the diameter of the wire to be 0.4mm with an
uncertainty of
Take (exact). The Young’s modulus obtained from the reading is
A) B)
C) D)
SECTION-II
(NUMERICAL VALUE ANSWER TYPE)
This section contains 10 questions. The answer to each question is a Numerical value. If the Answer in the
decimals , Mark nearest Integer only. Have to Answer any 5 only out of 10 questions and question will be
evaluated according to the following marking scheme:
Marking scheme: +4 for correct answer, -1 in all other cases.
21. In the system shown all the surfaces are frictionless while pulley and string are massless.
Mass of block 𝐴 is 3𝑚 and that of block 𝐵 is 𝑚. If the acceleration of block 𝐵with
respect to ground after system is released from rest is ‘a’, then the value of ‘10 a’ is (take
g=10 m/s2and 5 2  7.0 )

22. An arrow sign is made by cutting and re-joining a quarter part of a square plate of side
'L=1m' as shown. The distance OC, where 'C' is the centre of mass of the arrow, is
…..(in cm)
23. A uniform solid cylinder of mass and radius is placed on a rough horizontal board
of same mass, which in turn is placed on a smooth surface. The coefficient of friction
between the board and the cylinder is =0.3. If the board starts accelerating with
constant acceleration as shown in the figure, then the maximum value of , so that the
cylinder performs pure rolling is…...(in m/s2) given g=10m/s2
24. The variation of lengths of two metal rods and with change in temperature is shown
in Figure. If the coefficients of linear expansion for the metal = n 106 / 0C ,then
the value of ‘n’ will be, in nearest integer, given C)
25. Two strings and of a sitar produce a beat frequency of When the tension of the
string is slightly increased the beat frequency is found to be If the frequency of
is then the original frequency of was (in Hz)
26. The potential difference across 8 ohm resistance is 48 volt as shown in the figure. The
value of potential difference across X and Y points will be…..(in Volts)
X
3
20 30 60
24 8 48V
1
Y

27. Two thin symmetrical lens of different nature have equal radii of curvature of all faces
R = 20 cm. The lenses are put close together and immersed in water. The focal length of
the system is 24 cm. The difference between refractive indices of the two lenses is ……
1 4
× . Refractive index of water is .
9 3
4
=
1 3
2
28. When a monochromatic point source of light is at a distance 0.2 m from a photoelectric
cell, the saturation current and cut-off voltage are 12.0 mA and 0.5 V. If the same source
is placed at 0.4 m away from the same photoelectric cell, then the saturation current,
now, will be …..(in mA)
29. In the circuit given if the current through the Zener diode is I Z . Find the value of 6 I Z (in
mA)
30. In a circuit for finding the resistance of a galvanometer by half deflection method, a 6 V battery
and a high resistance of 11 kΩ are used. The figure of merit of the galvanometer is
60μA/division. In the absence of shunt resistance, the galvanometer produces a deflection
of θ=9 divisions when current flows in the circuit. The value of the shunt resistance (in  ) that
can cause the deflection of θ/2, is closest to____

CHEMISTRY MAX.MARKS: 100
SECTION – I
31. If  0 is the threshold wavelength for photoelectric emission from a metal surface,  is
the wavelength of light falling on the surface of metal and m is the mass of electron, then
the maximum speed of ejected electrons is given by
1 1
 2h 2  2hc
  0   
2
(A)    0     (B) 
m   m 
1 1
 2hc   0     2  2h  1 1   2
(C)    (D)     
 m   0   m  0  
32. Given below are two statements: One is labelled as Assertion A and the other is labelled
as Reason R.
Assertion A: In TlI3 which is isomorphousto CsI3 , the metal is present in +1 oxidation
state.
Reason R: Tl metal has fourteen f electrons in its electronic configuration.
In the light of the above statements, choose the most appropriate answer from the
(A) Both A and R are correct and R is the correct explanation of A
(B) A is not correct but R is correct
(C) Both A and R are correct R is NOT the correct explanation of A
(D) A is correct but R is not correct

33. Assertion (A): Bond dissociation energy of F2 is smaller than that of Cl2
Reason (R): Bond dissociation energy is in order F2  Cl2  Br2  I 2
A) Assertion is True, Reason is True; Reason is correct explanation for Assertion
B) Assertion is True, Reason is True; Reason is NOT a correct explanation for Assertion
C) Assertion is True, Reason is False
D) Assertion is False, Reason is True

34. 5 g of urea is dissolved in one kg of water. If the solution is cooled to 0.200o C , then
how many grams of ice would separate? (Kf of water = 1.86)
(A) 200 (B) 225 (C) 325 (D) 175
35. A brown precipitate insoluble in excess of aqueous ammonia solution is formed by
(A) FeCl3  NH 4 Cl  (B) FeCl3  NH 4 OH 
(C) CrCl3  NH 4OH  (D) Hg 2 Cl2  NH 4 OH 
36. For the first-order reaction A g   2Bg   Cg  , the total pressure after time t from the start
of reaction with A is P and after infinite time, it is P . Then the rate constant of the
reaction is
1 P 1 2P
(A) ln (B) .ln
t P t 3  P  P 
1 2P 1 2P
(C) .ln (D) ln
t 3P  P t P  3P
37. The order of E oM 3

/ M 2
value is
(A) Co >Mn> Fe (B)Mn> Co > Fe (C) Co > Fe >Mn (D) Mn> Fe > Co
38. A chromatography column packed with silica gel (used as stationary phase) is used for
the separation of
The column was then eluted with a mixture of Dichloro methane-nitromethane in 80:20
ratio, the sequence of elution of A, B and C is
A) A, B, C B) C, B, A C) B, C, A D) A, C, B

39. Consider the following reactions, What is A
A) Acetylene B) propyne C) ethylene D) 2-butyne

40. In azomethane, H3CNNCH3, what are the molecular geometries around the carbon and
nitrogen atoms, respectively?
A) Tetrahedral at carbon, bent at nitrogen
B) Tetrahedral at carbon, linear at nitrogen
C) Square planar at carbon, bent at nitrogen
D) Square planar at carbon, linear at nitrogen
41. An organic compound contains only carbon, hydrogen, nitrogen, and oxygen. It is
61.71% C, 4.03% H, and 16.00% N by mass. What is its empirical formula?
A) C5H4NO B) C9H7N2O2 C) C10H8N2O C) C11H8NO2
42. Consider the following sequence of reactions:
The functional groups undergoing change in the conversion of A to E respectively are
A) COOH,  NC,  CONH 2 ,  NH 2 ,  CH 2 OH

B) COOH,  CN,  CONH 2 ,  NH 2 ,  CH 2 OH
C) COOH,  CONH 2 ,  CN,  CH 2 NH 2 ,  CH 2 OH
D) CONH 2 ,  COOR,  NC,  NHR,  CHOH

43.
The compounds which can be reduced with formaldehyde and conc. AqKOH, are
A) only II and V B) only I and V
C) only II and III D) only I, II and IV
44. What is the X for given reaction?
A) B) C) D) All
45. The correct sequence of reactions to get ‘Q’ as the only product from ‘P’ is
A) (i) H2&Pt catalyst (ii) C2H5Cl & AlCl3
B) (i) Mg in ether (ii) aqueous alcohol (iii) C2H5Cl & AlCl3
C) (i) Mg in ether (ii) C2H5Cl & AlCl3
D) (i) C2H5Cl & AlCl3 (ii) Mg in ether (iii) aqueous alcohol

46. When acid ‘X’ is heated to 230°C, along with CO 2 and , a compound ‘Y’ is formed. If
‘X” is HOOC  CH 2 2 CH  COOH 2 , the structure of ‘Y’ is
(A) HOOC  CH 2 3 COOH (B)
(C) CH3CH 2 CH  COOH 2 (D)
47. The best reagents and conditions to accomplish the following conversion is?
A) (i) LiAlH4 in ether, (ii) 3 moles of CH 3 I followed by heating with AgOH
B) (i) LiAlH4 in ether; (ii) P2O5 and heat
C) (i) 20 % H2SO4& heat, (ii) P2O5 and heat
D) H2 and Lindlar catalyst
48. Two students did a set of experiments on ketones ‘X’ and ‘Y’ independently and
obtained the following results.
The ketones ‘X’ and ‘Y’ are respectively

(A) 2-ethylcyclobutanone and 3-ethylcyclobutanone
(B) 2-methylcyclopentanone and 3-methylcyclopentanone
(C) 3-methylcyclopentanone and 2-methylcyclopentanone
(D) 3-methyl-4-penten-2-one and 4-methyl-1-penten-3-one
49. RNA and DNA are chiral molecules, their chiralityis due to:
A) D-sugar compound B) L-sugar component
C) chiral bases D) Chiral phosphate ester groups
50. The electronic configurations of bivalent europium and trivalent cerium are
(atomic number : Xe = 54, Ce = 58, Eu = 63)
(A)  Xe 4f 2 and  Xe  4f 7 (B)  Xe 4f 7 and  Xe  4f 1
(C)  Xe 4f 7 6s 2 and  Xe  4f 2 6s 2 (D)  Xe 4f 4 and  Xe  4f 9
SECTION-II
51. Geraniol is an acyclic unsaturated alcohol C10H18O, a terpene found in rose oil, adds two
moles of bromine to form a tetrabromide, C10H18OBr4 . It can be oxidized to a ten-carbon
aldehyde or to a ten-carbon carboxylic acid. Upon vigorous oxidation, geraniol yields:
Number of double bonds in Geraniol is?

52. The solubility product of AgC 2 O 4 at 25o C is 2.3  1011 M 2 . A solution of K 2 C 2 O 4
containing 0.15 moles in 500 ml water is shaken at 25o C with Ag 2 CO3 till the following
equilibrium is reached.

Ag 2 CO3  K 2C 2O 4  Ag 2 C2O 4  K 2 CO3
At equilibrium, the solution contains 0.035 mole of K 2 CO3 . Assuming the degree of
dissociation of K 2 C 2 O 4 and K 2 CO3 to be equal, The solubility product of Ag 2 CO3 is
x 1012 , then x is ______
53. The volume, in mL, of 0.02 M K 2 Cr2 O 7 solution required to react with 0.288 g of ferrous
oxalate in acidic medium is ______. (Molar mass of Fe = 56 g mol1 )
54. Equivalence conductance at infinite dilution of NH 4 Cl, NaOH and NaCl are
129.8, 217.4 and 108.9  1 cm 2 mol 1 , respectively. If the equivalent conductance of 0.01 N
solution of NH 4 OH is 9.532  1cm 2 mol1 , then the percentage (%) dissociation of NH 4 OH

at this temperature is
55. Glycine (C2H5O2N) is the simplest of amino acids. Molecular formula of the linear
oligomer synthesized by linking ten glycine molecules together via a condensation
reaction is CwHxOyNz
What is the value of w+x+y+z?
56. For silver, C p  JK 1 mol1  = 23 + 0.01T. If the temperature (T) of 3 moles of silver is
raised from 300 K to 1000 K at 1 atm pressure, the value of H will be close to
57.
If X is the moles of KOH consumed per mole of reactant in reaction-1 and Y is the
moles of KOH consumed per mole of reactant in reaction 2, what is (X+Y)

58. Lovastatin, a drug used to reduce the risk of cardio vascular diseases has the following
structure
The number of stereogeniccenters present in lovastatin is
59. How many of the following are an example of aromatic electrophilic substitution
reactions
60. Total number of paramagnetic species in which unpaired electron(s) is/are present in
 * M.O.
O 2 ;O 2 ;O 2 ; N 2 ; N 2 ; N 22  ; B2

MATHEMATICS MAX.MARKS: 100
SECTION – I
61. The number of points in the complex plane that satisfy the conditions z  2  2,
z 1  i   z 1  i   4 (where i  1 ) is
A) 0 B) 1 C) 2 D) more than two
62. Let A and B are square matrices of same order satisfying AB  A and BA  B , then
A 
2019 2020
 B 2019 is equal to:
A) A  B B) 2020  A  B  C) 22019  A  B  D) 22020  A  B 
x 2  3x x 1 x  3
4 3 2
63. If px + qx + rx + sx + t = x 1
2
2 x x 3 then p is equal to
x2  3 x  4 3x
A) –5 B) –4 C) –3 D) –2.
64. 2
If  is the root of the equation x  x  2  0 then the value of

6 3  22    is
  3  3  2
5 4 3
equal to:
A) 3 B) 6 C) 9 D) 12
65. Number of 4 digit numbers of the form N = abcd which satisfy following three
conditions:
i) 4000  N  6000 ii) N is a multiple of 5
iii) 3  b  c  6 is equal to:
A) 12 B) 18 C) 24 D) 48
66. If a and b are chosen randomly by throwing a pair of fair cubical dice, then the
2
 a x  bx x
probability that lim    6 equals:
x 0
 2 
4 2 3 1
A) B) C) D)
9 9 9 9

67. A random variable X takes values x = 0, 1, 2, 3, ... with probability proportional to
x
1
  . Then ____ (E(X) denotes the mean of the random variable X)
5
A) P(X = 0) = 16/25 B) P(X =1) = 4/25
C) P(X  1) = 7/25 D) E(X) = 25/32
 x 2  2 x  a , x  1
68. Let f  x    , then number of positive integral value(s) of ‘a’ for
 6  x, x 1
which f  x  has local minima at x = 1, is ______
A) 6 B) 7 C) 8 D) 9
69.
1
  12  
Let g  x   f 2x 2  1  f 1  x 2 x  R, where f " x   0x  R , g(x) is necessarily
4
increasing in the interval
 2 2
A)   ,  B)   2   2 
,0    , 
 3 3  3   3 
 2
C)  1,1 D)  ;  
 3 
  
70. Let a, b, c are three vectors having magnitudes 1,2,3 respectively satisfy the
    
relation a  b  .c  6 . If d is a unit vector coplanar with b and c such that b.d  1 then
   2    2
 
the value of a  c .d  a  c  d is  
9 9
A) 9 B) 3 C)  D)
2 2
71. Three straight lines mutually perpendicular to each other meet in a point P and one of
them intersects the x-axis and another intersects the y-axis, while the third line passes
through a fixed point (0, 0, c) on the z-axis. Then the locus of P is
A) x 2  y 2  z 2  2cx  0 B) x 2  y 2  z 2  2cy  0
C) x 2  y 2  z 2  2cz  0 D) x 2  y 2  z 2  2c  x  y  z   0

72. Area of the trapezium whose vertices lies on the parabola y 2  4 x and its diagonals pass
25
through 1, 0  and having length unit each, is
4
75 625 25 25
A) sq.units B) sq.units C) sq.units D) sq.units
4 16 4 8
73. Let f be a differentiable function defined in 0,1 such that f  f  x    x and f  0   1. If

1
p
  x  f  x 
2018
the value dx  where p and q are relatively prime then the value of
0
q
 p  q  , is ___________
A) 2023 B) 2024 C) 2020 D) 2022
74. Let f(x) = Maximum {x2, (1  x)2, 2x(1  x)}, where 0  x  1. Determine the area of the
A
region bounded by the curves y = f(x), x-axis, x = 0 and x = 1 is then find the value
54
of A.
A) 30 B) 36 C) 32 D) 34
75. Let y  y  x  satisfy the differential equation
 y3 
 2xy  x 2
y 
3 
2 2
 
 dx  x  y dy  0 . If y 1  1 and the value of  y  0    ke k  N 
3

Then k is _________
A) 3 B) 4 C) 1 D) 2
76. If a1,a2 ,......, a4001 are in arithmetic progression and

1 1 1
  .....   10 and a2  a4000  50 . Find the value of a1  a 4001 .
a1a 2 a 2a 3 a 4000a 4001
A) 20 B) 32 C) 60 D) 30
n n
1 r
77. If an   n C n  n C , then the number of ordered pairs  p, q  such that
, b 
r 0 r r 0 r
ap
c p  cq  1, where c p  , is:
bp
A) 0 B) 1 C) 2 D) 3

If  sin 250  3 cos 850  sin850   a  b cos 500 , a, b, are rational numbers in their
2
78.
simplest form, then a  b  
A) 0 B) -2 C) -1 D) 3
 3  x; x 1
 x  1; x  0  2
79. . Let f  x    and g  x   x  2x  2; 1  x  2 then:_______
2  x ; x  0  x  5; x 2

Note: [k] denotes greatest integer function less than or equal to k.
A) lim g  f  x    2 B) lim g  f  x    3
x 0 x 0
C) lim f  f  x     0 D) lim g  g  x     1

x 0 x 0
x
dt
80. Let f be real-valued function such that e 2 x
f x   x  3   for all x   1,1 and
0 t6 1
let y  g  x  be a function whose graph is reflection of the graph of y  f  x  w. r. t. line
y = x, then g '  3  is equal to:
1 1 1
A) 1 B) C) D)
2 4 8
SECTION-II
x 3  1 x 5 3x  2
 
81.  
Let A   y  1 6x 2  2 z  1  and A  3 . If f  x   tr . B 1 and B = adj(A), then
 2 3 9x  6
 
global maximum value of f  x  in x  0,6 is:
82. Let f : A  A where A  1,2,3, 4,5,6,7 , then number of function f such that
 
f f  f  x    x x  A , is:

83. The point on the line r   2i  6j  34k   t  2i  3j  10k  , t  R which is nearest to the line
   
r  6i  7 j  7k   4i  3j  2k ,   R is a , b, c  , then the value of a  b  c is equal to
x 2 dx
1 1
ex I
84. Let I1   dx and I 2   3 . Then 1 is
0
1 x 0 e
x

2 x 3

e.I 2
85. A, B, C are the vertices of triangle with right angled at A and P  4,0  ;Q  0,6  are two
given points. If the ratio of the distances from each vertex of triangle to P, to that of Q is
 r 
2:3, the centroid of triangle ABC lies on a circle with radius ‘ r ‘ then   is equal to
 13 
_____ [ . ] represent GIF
86. Let A,B,C,D are four points, in I, II, III, IV quadrants respectively, lying on the circle
 x  12   y  22  25 so that AC and BD are perpendicular diameters. Circles, touching

the given circle, are drawn at A, B, C, D with radii 1, 2, 3, 4 respectively. If A is  4,6  ,
then the sum of the coordinates of centers of the circles corresponding to B and D is
87. Mr. A either walks to school or take bus to school everyday. The probability that he
takes a bus to school is 1/4. If he takes a bus, the probability that he will be late is 2/3. If
he walks to school, the probability that he will be late is 1/3. The probability that Mr. A
p
will be on time for at least one out of two consecutive days is , where p and q are co-
q
prime, find the value of  q  p  .
cos ec 2x  2019  f x  

88. If   cos x 2019 dx   C , then find the value of f    g  0  . (where C
 g  x 
2019
4
is constant of integration)
89. Let y1, y2 , y3.....yn be n observations. Let wi  lyi  k , i  1,2,3....n where l , k are
constants. If the mean of yi ' s is 48 and their standard deviation is 12, the mean of wi ' s
is 55 and standard deviation of wi ' s is 15, then values of l + k + 0.75 should be ______
90. Let f be an invertible function from R  R satisfying the equation
     
f 3  x   x 3  2 f 2  x   2x 3  1 f  x   x 3  0 . Then the value of f '  8  f 1 '  8  , is:

Sec: SR.IIT_*CO-SC(MODEL-A,B&C) GTM-2 Date: 24-02-23
Time: 3 HRS JEE-MAIN Max. Marks: 300
KEY SHEET
PHYSICS
1 B 2 B 3 D 4 A 5 A
6 D 7 D 8 B 9 C 10 A
11 A 12 C 13 B 14 A 15 C
16 B 17 B 18 B 19 A 20 C
21 15 22 3 23 2 24 3 25 3
26 2 27 4 28 3 29 8 30 0
CHEMISTRY
31 B 32 D 33 A 34 A 35 D
36 D 37 D 38 D 39 A 40 C
41 D 42 A 43 D 44 C 45 C
46 D 47 A 48 B 49 D 50 A
51 3 52 4 53 5 54 1 55 5
56 9 57 8 58 9 59 6 60 6
MATHEMATICS
61 B 62 A 63 A 64 C 65 B
66 B 67 B 68 D 69 C 70 C
71 B 72 A 73 C 74 A 75 B
76 C 77 C 78 B 79 B 80 B
81 5 82 11 83 2 84 0 85 90
86 2 87 540 88 3 89 18 90 1
Narayana IIT Academy 24-12-23_SR.IIT_*CO-SC(MODEL-A,B&C)_JEE-MAIN_GTM-2_KEY&SOL
SOLUTIONS
PHYSICS
1. Given velocity in x-direction
vx  2 m / s
Acceleration
E   e  8m   e  
a     . j  8 jm / s 2
m e m
d 1
 time, t   s
vx 2
1
Now y  .a y t 2
2
vy  u y  ayt
1
v y  0  8   4m / s
2
v y 4m / s
Now tan    2
vx 2 m / s
  tan 1  2
2. In the diagram balance all the forces
3.
0i
B
2 d

 4 107   100 
2  3.14  4
B  5  106 T (Southwards)
4.   MB  cos2  cos 1 
MB
  MB  cos 600  cos 00  
2
Torque  MB sin 60  2.sin 600
0
 3
SR.IIT_*CO-SC Page NO: 2

5.
3L
   B xdx
2L
5
  .Bl 2
2
6.  0  283v,   320 s 1
z  R2   X L  X C 
2
 7
 X L  XC 
  tan 1    45
0
 R 
7. Conceptual
 
2
I max  I1  I 2
8.
 
2
I min I1  I 2
Simplifying the required answer will be 2
9. Using
z2
En  13.6 2 ev
n
 3
2
E1  13.6 2 ev  122.4 ev
1
 3
2
E3  13.6 2 ev  13.6 ev
 3
E  108.8 ev
10. Conceptual
11. Vernier constant = Least count
1 MSD
= ,  10

1cm
1 MSD 
20
u2
12. H max   10m....... 1
2g
u2
Rmax   20m,   450  angle of projection 
g
13. I  I com  md 2
2
mL2  L 
  m 
6  2
2
I  ml 2
3

14. Energy   TE 3R   TE 2 R
GmM  GmM 
  
3R  2R 
GmM
Energy 
6R
DV
15. P   B.
V
 6  108 N / m2
16. Alongbc w  0
Q  V  60 J
Along ab u  180  60  120 J
also Q  250 J
So
w  250  120  130 J
Hence wab  wbc  130  0  130 J
fR
17. CV 
2
CP  CV  R
fR  f 
CP  R  CV  R   R 1  
2  2
CP

CV
18.
2
T ,   3 

19.
1 mm 1
  v  0   kx 2
2
.
2 2m 2
20. Energy lost = Energy utilised for rise in temperature.
21.
F
ac 
z
F
FBD of 1Kg f  1
3

f
 110    f  15 N
z
3R
22. x
8
23.
24. PT 2  C
 RT 2
.T  C
V
T 3V 1  C
Differentiating
1 dV 3
. 
V dT 1
y  y 
25. find &    2 fy0
t  t max
wave velocity  f 
26. Find equivalent resistance then distribute current.
 Am 
sin  
27. Use m   2 
sin  A / 2 
m  4
h
28. 
2mav
P m q
  82 2 3
 mP q P
29.
1 2 F
S at , a 
2 m
T  4t
K dv K
30. U  2 ,F    3
2r dr r
2
mv K K
 3  mv 2  2
r r r
Total energy = KE + PE = 0
CHEMISTRY
31. Conceptual
Z
32. Vn 
n
n2
rn 
Z
Z2
En 
n2
33. PCl5  PCl3  Cl2
1 0 0
(1   )  
Law of conservation of mass
WPCl5  Weq m
M PCl5  (1   ) M eq
M PCl5 D
(1   )  
M eq d
pKa1  pKa2
34. ( pH )i  9
2
( pH ) f  3  2 log 3
 pH  6  2 log 3
CP , m
35. CP 
M
R
 CP , m 
(  1)
R
CP 
M (1)
36. According to slow step
r (t )  K 2  N 2O2  H 2 
Where K1 
 N 2O2 
 NO 
2
 N 2O2   K1  NO 2
So, r (t )  K 2 K1  NO   H 2 
2
So, r (t )  K 2 K1  NO   H 2  K
2
1 d H2  1 d [ NO] d[ N 2 ]
    K [ NO]2 [ H 2 ]
2 dt 2 dt dt
Where K  K1 K 2
37. K GC*
C*  70
2 70
m    5  104
280 1000  0.5

38. Leaving group ability  its stability
Weak base are good leaving group
39. Conceptual
40. d xy , d yz , d zx are also known as non-axial d-orbitals
41. (A) 3 p  3 p  3 p  3d  3d  3d (  bond strength)
(B) Ethanol  C2 H 5OH  < ethylene glycol
 CH 2  CH 2 
 
 | |   Glycerol
 OH OH 

 CH 2  CH  CH 2 
 
 | | |  (Viscosity)
 OH OH OH 

(C) XeO3 F2  XeF4  XeF2 (All have zero dipole moment)
(D) more electro positive element in 13th group is Al
42. Conceptual
43. Conceptual
44. dien is tridentate
w
45. 0.05  0.92   0.0821  300
28
w  0.0522
w%  17.46
46. ror  stability of C 
47. Conceptual
48. Conceptual
49. SN 2
50. Cannizaro followed by esterification
moles of NaOH
51. 
2
moles of NaOH
WH 2 SO4   98
2
(0.75  98)
 3
25
1 1000 4.9
52. 0.19   
75 86 x
x4
53. Conceptual
54.
55. Both 2nd and 4th group cations are precipitated in the presence of H 2 S  NH 4OH

56. Mono halo product =1
Di halo products =2
Tri halo products=2
Tetra halo products=2
Penta and hexa are each one  total =9
57.
58. Conceptual
59.
OH
HO O
O
OH OH
Ascorbic acid
60. Conceptual
MATHS
61. f ( x)  x  3ax  3  a  1 x  1
3 2 2
f  ( x)  3x 2  6ax  3  a 2  1
 3( x - (a  1))( x - (a -1))
So f  ( x)  0  x  a  1 or a  1
 a  1 (2,4) if a  (3,3)
And a  1 (2,4) if a  (1,5)
 a  (1,3)
62. Apply by parts
 tan 4 x

 4  tan 3 x 
;0  x 
 3  2

 
63. f ( x )  b  2; x 
 2
 a
|tan x| 
(1 | cos x |)
b
;  x 
 2

tan 4 x tan 4 h

4 tan 3 x 4 (  cos 3 h )
lim f ( x)  lim f    lim  
x

x
 3  
h 0 3
2 2
So, =1=b+2
B=-1
a
|tan x|
lim f ( x)  lim (1 | cos x |) b
 
x x
2 2
a
 |tan x|
lim (1  cos x) b

x
2
a
 cot h
lim(1  sinh) b
h 0
a a
cosh
 lim(1  sinh) b
e b
h0 sinh
a
1 e  b  2
b
a=0, b  1
a+b  1
64. f ( x)   x3   x  (tan x)sgn x
f (- x)  f ( x)
   x3   x  tan x sgn x
  x3   x  (tan x)(sgn x)
 2  x 2    x  0x  R
   0 and   0
 [a]2  5[a ]  4  0
and 6{a}2  5{a}  1  0
 (3{a}  1)(2{a}  1)  0
 [a ]  1,4 and {a}  1 / 3,1 / 2
1 1 1 1
 a  1  ,1  ,4  ,4 
3 2 3 2
35
Sum of value a 
3
dy
65.  1  xe yx
dx
dy
 e y  e y  xe x
dx
dy dt
Let e y  t  e y 
dx dx
dt
  t  xe x
dx
dt
 t  xe x
dx
 Integrating function  e 
1dx
 ex
 Solution is t  e x    xe x  e x dx
x2
t  ex   C
2
x2
exy    C
2
 y 0  0  C  1
x2  x2 
e xy
1  y  x   x  ln 1  
2  2 
 
66. ‘P’ is orthogonal matrix. i.e., P.P  P .P  I
 
And Q 2023  PAPT PAPT     PAPT   
 PT Q 2023 P  PT  PA2023 P P
T
1 1  1 11 1  1 2 
Similarly A     A 2
  0 1 0 1   0 1 
 0 1     
1 3 
A3   
0 1
 1 2023 
A2023  
0 1 
 1 2023 
 PT Q 2023 P  
0 1 
 1 2023 
The inverse of PT Q 2023 P is 
0 1 
67. For multiple of 3, the sum of all digits should be divisible by 3.
So, number can be formed 0,1,2,3 (sum is 6 which is divisible by 3) or 0,2,3,4 (sum is 9which is
divisible by 3). 0 and 1 cannot be on highest digit in the number. Therefore, number of 4 digit
numbers
 2  3! 3  3!  30
 1  x2 
68. We have f ( x )   sin
1
 
log 4 x  sin  1
 4x 

Clearly domain of f ( x) is x  1 only, so f (1) 
2 
0  sin 1   
4 6
 
Hence range of f ( x)   
6
69. Draw the graph of f  x  .

x x
70. f ( x)    f (t ) tan tdt   tan(t  x) dt
0 0
x x
f ( x)    f (t ) tan tdt   tan(t )dt
0 0
(Use f (a  x)  f ( x))
x x
f ( x)    f (t ) tan tdt   tan(t )dt
0 0
Differentiating w.r.t x, we have
f  ( x)   f ( x ) tan x  tan x
dy
 (tan x) y  tan x
dx
dy
 (tan x) y   tan x
dx
IF  e 
tan xdx
 elog sec x  sec x

Therefore, solution is y. sec x   secx tan xdx  c
Or y sec x   sec x  c
Or y  c cos x  1
Curve passes through (0,0)
 0  c  1 or c  1
 y  cos x  1
Therefore, the maximum value of y is 0.
71. We have
In S1 units place can have 1 or 3 or 5
4!
In S 2 it is 5C3  3   5C2  9  2  6   600
2!

72. Line passes through the points a1  (3, 2,1) a
 
a2  (3,6,5), b1  2iˆ  3 ˆj  kˆ
  
b1  2iˆ  ˆj  3kˆ, a2  a1  6iˆ  4 ˆj  4kˆ
   

 a2  a1  b1  b2 
SHORTEST DISTANCE   

b1  b2 
iˆ ˆj kˆ
 
b1  b2 2 3 1  10iˆ  8 ˆj  4kˆ
2 1 3
   
 
 a2  a1   b1  b2  60  32  16  108
 
b1  b1  100  64  16  180
108 108 18
S .D   
180 6 5 5

73. Let the probability of the faces 1,3,5 or 6 p for each face.
Hence the probability of the faces 2 or 4 is 3p. therefore
1
4 p  6 p 1 p 
10
1
P(1)  P(3)  P(5)  P(6) 
10
3
P(2)  P(4) 
10
P( total of 7 with a draw of dice )  P(16,61, 25,52,43,34)
1 1  3 1  3 1
 2    2    2  
 10 10   10 10   10 10 
2  6  6 14 7
  
100 100 50
74. Apply condition for externally touching circle.
2
 x  2  ( y  2) 2
75. The hyperbola    1
 4  2
3
a  2, b  2, e 
2
If D be the centre, then
DC  ae  6 and DA  a  2
b2
AC  6  2 and BC   1
a
1 3
Now area of ABC  ( AC )( BC )  1
2 2
76. Use r  a  0
r b  0
Where v  xiˆ  yjˆ  zkˆ
And x  iˆ  21
 A  P  A  B 1
77. For option (1) P     PA 
B P  B 3
 A  2
Similarly P    P  A   
 B  3
1 5
A P  
A 1  P   3  6 1
B
P    
 B  1  P  B  5 3
6
 A  P  A   A  B 
P 
 A B P  A  B

1
PA 3
 
P  A  B 1  1  1
3 6 18
6 3
 
6  3 1 4
1 1
 
2
78. x 2  5 2  (20) 4 x  x 2  5  20 x 2
 x 4  5  2 5 x 2  2 5 x 2  x 4  5  x8  25
So,  8   8  50
79. If we write the elements of A + A, we can certainly find 39 distinct elements as
1  1,1  a 1 ,1  a 2 ,.......1  a 18 ,1  77,a 1  77, a 2  77,....a 18  77,77  77 .
It means all other sums are already present in these 39 values, which is only possible in case when all
numbers are in A.P.
Let the common difference be ‘d’
77  1  19d  d  4
18
18
So,  a1   2a1  17d   9 10  68  702
i 1 2
80. PQ is focal chord.
Quadrilateral PTQR is square
PQ  TR
Area  8
2
Given 27 pqr  ( p  q  r )
3
81.
But using AM  GM
pqr
  ( pqr )1/3
3
 p  q  r (using (i) and (ii))
Also, 3 p  4q  5r  12
 sin  3x 2  4x  1 
   3x 2  4x  1  x 2  1
 3x  4x  1 
2
82. lim    2
x 1 2x  7x 2  ax  b
3
3x 2  4x  1  x 2  1
 lim  2
x 1 2x 3  7x 2  ax  1

2  x  1
2
 lim  2
x 1 2x 3  7x 2  ax  b
So f  x   2x 3  7x 2  ax  b  0 has x  1 as repeated root, therefore f 1  0 and

f  1  0 gives a  b  5 and a  8
So, a  b  11
Let f  x   x  ax  bx  c
3 2
83.
f   x   3x 2  2ax  b  f  1  3  2a  b
f   x   6x  2a  f   2   12  2a
f   x   6  f   3  6
 f  x   x 3  f  1 x 2  f   2  x  f   3
 f  1  a  3  2a  b  a  a  b  3 .......1
 f   2   b  12  2a  b  2a  b  12 ....... 2 
From 1 and  2 
3a  15  a  5  b  2
 f  x   x 3  5x 2  2x  6
 f  2   8  20  4  6  2
84. According to given data
7
 x  62 
2
i
i 1
 20
7
7
   X i  62   140
2
i 1
So for any X i ,  Xi  62   140

2
 X i  50 i  1, 2,3,.....7
So no student is going to score less than 50.
85.
0 0 1
1
Area  x1 99  x1 1
2
x2 99  x2 1
1
  x1  99  x2   x2  99  x1  
2

1
Area   x1  x2  99 
2
Area is an integer. Then both x1 and x2 are simultaneously either even or odd.
Hence,
10
C2 10 C2  2 10 C2  90
1
log x
86. Area bounded   ex
1/ e
1
1   log x  
2 1
 x2 x2 
    e  log x  
e  2  1  2 4 1
e e
 1 
1 
1 1   2 e2 
     e 2  
e  2   2e 4 

 
1 1 e 1
   
2e 2e 4 4e
e 5
 
4 4e
87. K  3! 3! 1  36
P  6! 3! 6!  6! 7  5040
88. Conceptual
4 4
Vice   10  r    10 
3 3
89.
3 3
dV 4 2 dr
  310  r 
dt 3 dt
2 dr
 4 10  r 
dt
dr
At r  5,50  4  225
dt
dr 50

dt 4  225 
1
 cm/min
18

90. Let z  iy
 y 4  a1 y 3i  a3 y 2  ia3 y  a4  0
 y 4  a2 y 2  a4  0  1 and  a1 y 3  a3 y  0
 y  0 or y 2  a3   2 
a1
From (1) and (2)
a32 a
2
 a2 3  a4  0
a1 a1
a3 aa
  1 4 1
a1a2 a 2 a3


MISTAKES
MATHS
PHYISCS
CHEMISTRY
SECTION – I
1. A uniform electric field E = (8 m/e) V/m is created between two parallel plates of length
1m as shown in the figure (where m = mass of electron and e = charge of electron). An
electron enters the field symmetrically between the plates with a speed of 2 m/s. The
angle of deviation   of the path of the electron as it comes out of the field will
be_______
A) tan 1  4  B) tan 1  2  C) tan 1 1/ 3 D) tan 1  3
2. Four point charges equal to –Q are placed at four corners of a square and a point charge
q is at its centre. If the system is in equilibrium the value of q is
A) 
Q
4

1 2 2  B)
Q
4

1 2 2  C) 
Q
2
1 2 2  D)
Q
2

1 2 2 
3. A horizontal overhead power line is at a heightof 4 m from the ground and carries a
current of 100 A from east to west. The magnetic field directly below it on the ground is
 0  4  10 7 TmA1 
A) 2.5  10 7 T northwards B) 2.5  10 7 T southwards
C) 5  106 T northwards D) 5  106 T southwards
4. A magnetic needle lying parallel to a magnetic field requiresW units of work to turn
through 600 . The torque needed to maintain the needle in this position will be
3
A) 3W B) W C) W D) 2W
2

5. A metallic rod of length l is tied to a string of length 2 l and made to rotate with angular
speed  on a horizontal table with one end of the string fixed. If there is a vertical
magnetic field ‘B’ in the region, the emf induced across the ends of the rod is
5 2 Bl 2 3 4
A) .Bl 2 B) C) .Bl 2 D) .Bl 2
2 2 2 2
6. A sinusoidal voltage of peak value 283V and angular frequency of 320rad/s is applied to
a series LCR circuit. Given that resistance R  5 , inductance L = 25 mH and
capacitance C  1000  F . The total impedance and phase difference between the voltage
across the source and the current will nearly be (respectively)
A) 10 and tan 1   B) 10 and tan 1  

5 8
 3 3
C) 7 and tan 1  
5
D) 7 and 450
 3
7. The magnetic field of a plane electromagnetic wave is given by

B  3  10 8 sin  200  y  ct    iˆ T . Where C= 3  108 m/s is the speed of light. The
corresponding electric field is


A) E  9 sin  200  y  ct   k v / m

B) E  106 sin sin  200  y  ct   k v / m

C) E  3 108 sin  200  y  ct   k v / m

D) E  9 sin  200  y  ct   k v / m
8. Two coherent sources of light interfere. The intensity ratio of two sources is 1:4. For this
I max  I min
interference pattern if the value of 
I max  I min
 I max  Maximum intensity value 

 
 I min  Minimum intensity value 
A) 1.5 B) 1.25 C) 0.5 D) 1

9. The energy required for the electron excitation in Li   from the first to the third Bohr
orbit is
A) 12.1 ev B) 36.3 ev C) 108.8 ev D) 122.4 ev
10. STATEMENT – 1: Energy is released in nuclear fission.
STATEMENT – 2: Total binding energy of the fission fragments is larger than the total
binding energy of parent nucleus.
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 NOT a correct

explanation for Statement – 1.
C) Statement – 1 is True, Statement – 2 is False.
D) Statement – 1 is False, Statement – 2 is True.
11. In a Vernier callipers, each cm on the main scale is divided into 20 equal parts.
10Vernier scale divisions are equivalent to 9 main scale divisions.The value of Vernier
constant will be ______  102 mm
A) 5 B) 50 C) 500 D) 200
12. A boy can throw a stone up to a maximum height of 10m. The maximum horizontal
distance that the boy can throw the same stone up to will be ________ (in m)
A) 10 B) 10 2 C) 20 D) 20 2
13. Moment of inertia of a square plate of side l about the axis passing through one of the
corners and perpendicular to the plane of square plate is given by
Ml 2 2 2 Ml 2
A) B) Ml C) Ml 2 D)
6 3 12
14. Minimum energy required to move a satellite of mass m from an orbit of radius 2R to 3R
is, M = Mass of the planet
GMm GMm GMm GMm

A) B) C) D)
12 R 3R 8R 6R

15. The bulk modulus of a liquid is 3  1010 N / m 2 . The pressure required to reduce the volume
of liquid by 2% is
A) 3 108 N / m 2 B) 9 108 N / m 2 C) 6 108 N / m 2 D) 12 108 N / m 2
16. A sample of an ideal gas is taken through cyclic process abca as shown in the figure. The
change in internal energy of the gas along the path ca is -180J. The gas absorbs 250J of
heat along the path ab and 60J along the path bc. The work done by the gas along the
path abc is
A) 120J B) 130J C) 140J D) 100J

C 
17. The ratio of specific heats  P    in terms of degree of freedom ( f ) is
 CV 
( CP = Specific heat capacity at constant Pressure

CV = Specific heat capacity at constant Volume)
f 2 f 1
A) 1  B) 1  C) 1  D) 1 
3 f 2 f
18. A simple harmonic motion is represented by y  5 sin  3 t   3 cos  3 t  cm
The amplitude and time period of the motion are

3 2 2 3
A) 5cm, S B) 10cm, S C) 5cm, S D) 10cm, S
2 3 3 2
19. Two identical blocks A and B each of mass m resting on the smooth horizontal floor are
connected by a light spring of natural length L and spring constant k. A third block c of
mass m moving with a speed v along the line joining A and B collides elastically with A.
The maximum compression in the spring is
m mv m mv
A) v B) C) D)
2k 2k 2k k

20. A large number of water drops each of radius r combine to form a drop of radius R. If
the surface tension is T and mechanical equivalent of heat is J, the rise in heat energy per
unit volume will be
3T 2T 3T  1 1  2T  1 1 
A) B) C)    D)   
J rJ J r R J r R
SECTION-II
21. The coefficient of static friction between two blocks is 0.5 and table is smooth. The
maximum horizontal force that can be applied to move the block together is ________N.
22. The centre of mass of a solid hemisphere of radius 8cm is x cm from the centre of the
flat surface. Then value of x is ________.
23. The centre of a wheel pure rolling on a plane surface moves with a speed of v0 . A particle
on the rim of the wheel at the same level as the centre will be moving with xv0 .The
value of x is ________.
24. An ideal gas is expending such that PT 2  C . If gas expands from 0K, the co-efficient of
x
volume expansion of the gas is . The value of x is _____________.
T
A transverse wave is described by the equation y  y0 .sin 2  ft   . The maximum

x
25.
 
 y0
particle velocity is equal to four time the wave velocity if   . The integer part of
2
 is __________.(to the nearest integer )

26. In the given circuit diagram, a wire is joining point B and D. The current in this wire is
__________ A.
27. A prism of refractive index  and angle of prism A is placed in position of minimum
angle of deviation. If minimum angle of deviation is also A, then in terms of refractive

index the value of A is  cos 1   . The value of    is__________.
 
28. An  - particle and a proton are accelerated from rest by a potential difference of 100 v.
P
After this their de Broglie’s wavelength are  and P . The ratio to nearest integer is _.

29. Consider the situation shown in figure. The plates of the capacitor have plate area A and
are clamped in the laboratory. A dielectric slab is released from rest with length a inside
the capacitor. Neglecting any effect of friction or gravity the time period of periodic
motion of slab will be   x 

  l  a  lmd  , Then x =___________.(K is dielectric

  0 AE 2  K  1 
constant of the slab.)
30. A particle is moving in a circular path of radius a under the action of an attractive
K
potential U   . Its total energy is ____________.
2r 2

SECTION – I
31. Which of the following statement is incorrect ?
A) The order of the ionic character of the following chlorides is
BeCl2  MgCl2  CaCl2  SrCl2  BaCl2
B) The reducing nature of alkaline earth metals in aqueous medium follows the
increasing order as Be  Mg  Ba  Ca  Sr
C) Berylium is not readily attacked by acid because of the presence of an oxide film on
the surface of Be.
D) Sodium hydroxide is produced on a large scale by the electrolysis of aqueous
solution of NaCl using Hg as cathode
32. Which of the following product in a hydrogen atom, dependent of the principal quantum
number ‘n’ ? [Given : En  Total energy of electron; rn  Radius of Bohr orbit ; vn 
velocity of electron]
 1  1
A) vn2 . rn B) En . rn C)   . vn2 D) vn .  
 En   rn 
33. At a given temperature the dissociation equilibrium of PCl5 ( g )  PCl3 ( g )  Cl2 ( g ) , the
variation of (1   ) against (D/d) is represented on :
(Given :   dod of PCl5 , D = initial vapour density, d = vapour density at equilibrium)
A) B)
C) D)

34. Find the  pH (initial pH – final pH) when 100 mL of 0.01M HCl is added in a solution
containing 0.1 m moles of NaHCO3 solution of negligible volume
 ka 1  107 , ka2  1011 for H 2CO3 
A) 6  2 log 3 B) 6  log 3 C) 6  2 log 2 D) 6  2 log 3
35. For an ideal gas having molar mass M, specific heat at constant pressure can be given as:
 RM R RM R
A) B) C) D)
(  1) M (  1) (  1) M (  1)
36. For the reaction 2 H 2  2 NO  N 2  2 H 2O the following mechanism has been suggested :
[equilibrium constant  K1 (fast)]
N 2O2  H 2 
K2
 N2O  H 2O (slow)
N 2O  H 2 
K3
 N 2  H 2O (fast)
According to following mechanism is correctly matched.
1 d  H2 
A)   K [ NO]  H 2  where K  K 2  K3
2 dt
1 d [ NO]
B)   K [ NO][ H 2 ] where K  K 2  K1
2 dt
d[ N2 ]
C)  K [ NO ] [ H 2 ] where K  K 2  K1
dt
1 d [ NO]
D)   K [ NO]2 [ H 2 ] where K  K 2  K1
2 dt
37. Resistance of 0.2 M solution of an electrolyte is 50  the specific conductance of the

solution is 1.4 Sm1 the resistance of 0.5 M solution of the same electrolyte is 280  the
molar conductivity of 0.5 M solution of the electrolyte in Sm2 mol 1 is
A) 5  103 B) 5  103 C) 5  102 D) 5  104

38. Which of the following order of leaving group is incorrect ?
A) F   Cl   Br   I 
B)
C) CH 3  OH  CH 3  O 
D) Cl   H 
39. Statement-1: The 5th period of periodic table contains 18 elements not 32.
Statement-2: The order in which the energy of available orbitals 4d, 5s and 5p increases
is 5s < 4d < 5p and the total number of orbitals available are 9 and thus 18 electrons can
be accommodated.
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 not a correct explanation for

statement-1.
C) Statement-1 is true, statement-2 is false
D) Statement-1 is false, statement-2 is true
40. Which of the following compound used maximum number of non-axial d-orbitals in
hybridization of central atom ?
A) XeF4 B) XeO3 F2 C) XeF5 D) XeO3
41. Which of the following option is CORRECT ?
A) 3 p  3 p  3 p  3d  3d  3d (  bond strength)
B) Ethanol < Glycerol < ethylene glycol (Viscosity)
C) XeO3 F2  XeF4  XeF2 (Dipole moment)
D) In 13th group Al has least S.R.P

42. Assertion (A) : 1 mole of FeC2O4 is oxidised by 0.6 mole of MnO4 in acidic medium
Reason (R) : MnO4 oxidises both Fe 2 as well as C2O42
A) Both A and R are true and R is the correct explanation of A
B) Both A and R are true but R is not the correct explanation of A
C) A is true but R is false
D) A is false but R is true
43. Which one amongst the following are good oxidizing agents ?
a) Sm2 b) Ce 2 c) Ce 4 d) Tb 4
Choose the most appropriate answer from the options given below:
A) c only B) d only C) a and b only D) c and d only
44. The IUPAC name dichloride bis ( ethane 1,2-di amine) cobalt (III) chloride referred for
A) Co  dien 2 C 2  C B) Co  dien  C  2  C
C) Co  en 2 C 2  C D) Co  en 2 C   C  2
45. In Dumas method for estimation of nitrogen, 0.3 g of an organic compound gave 50 mL
of nitrogen collected at 300k temperature and 715 mm pressure. Calculate the
percentage composition of nitrogen in the compound. (Aqueous tension at 300k =
15mm)
A) 40% B) 70% C) 17.5% D) 6%
46. The correct reactivity order of following C  C / C  C bonds towards H  is
A) 4 > 3 > 2 > 1 B) 3 > 2 > 1 > 4 C) 1 > 3 > 4 > 2 D) 1 > 3 > 2 > 4

47. Test Reagent
1) Lucas test a) C6 H 5 SO2Cl
2) Hinsberg test b) Alc KOH  CHCl3
3) Iso cynide test c) An hyd ZnCl2  Conc.HCl
4) Molisch test d) Alc  -naphthol+Conc. H 2SO 4
A. 1-c, 2-a ,3-b, 4-d B) 1-a, 2-c, 3-b, 4-d
C. 1-c, 2-a, 3-d, 4-b D) 1-c, 2-b, 3-a, 4-d
48. P-Cresol reacts with chloroform in alkaline medium to give the compound A which adds
HCN to form, the compound B. The latter on acidic hydrolysis gives chiral carboxylic
acid. The structure of carboxylic acid is
A) B)
C) D)
49.
Me H KSH
A , then A is ?
Et D (major)

A) B)
C) D)
50.
A) B)
C) D)
SECTION-II
51. 500 mL of 2M impure H 2 SO4 sample can completely react with 1.5 L of 1 M NaOH
solution. Then find ‘  ’ ?
Weight of H 2 SO4
Where   .
25

52. 1 gram of arsenic dissolved in 86 gm of benzene brings down the freezing point to
-1
5.310 C from 5.500 C . If k f of benzene is 4.9 k. molality , then atomicity of molecule is ?
[Given atomic weight of As = 75 amu]
53. How many of the compound will evolve methane gas on treatment with CH 3 MgBr
(A) H 2O (B) CH 3  OH (C) CH 3COCH 3 (D) CH3COOH
(E) CH 3CONH 2 (F) CH 3COCl (G) CH 3COOCH 3 (H) CH 3Cl
(I) Ph  OH
54. C6 H12 

Br2
CCl4
 C6 H12 Br2
X (Optically inactive single product with 2 chiral center)
How many structures of “X” can satisfy above reaction condition ?
55. How many of the following ions gives precipitate with H 2S+NH 4 OH
Pb 2 , Cu 2 , Al .3 , Ca 2 Mn 2 , Zn 2 , Cd 2 , Cr 3 , Mg 2
56. How many isomers obtained by the halogenation of ethane?
57. The number of monobromo products (including stereoisomers) obtained during heating
of methylcyclobutane with Br2 .
58. The condensation of two amino acids, glycine and () alanine yields total products x.
Find value of x. Assume that only dipeptides are formed.
59. How many of the following compounds contains carboxylic acid group ? Ascorbic acid,
Phthalic acid, Cinnamic acid, acrylic acid, aspirin, formic acid, picric acid, adipic acid.
60. Total number of  -keto monocarboxylic acids (including stereoisomers) which on
heating give .

SECTION – I
61. The set of values of a for which all extremum of
 
function f ( x)  x3  3ax 2  3 a 2  1 x  1 lie in the interval  2,4 is
A)  3,4  B)  1,3 C)  3, 1 D)  2, 1
 sin 101x   sin x  dx equals

99
62.
sin 100x  sin x  cos 100x  sin x 

100 100
A) C B) C
100 100
cos 100x  cos x  sin 100x  sin x 

100 101
C) C D) C
100 101
 tan 4 x
 
4 tan 3 x 
; 0 x
 3  2

 
63. The function f ( x )  b  2 ; x then the sum  a  b  if f is
 2
 a
|tan x| 
(1 | cos x |)
b
;  x 
 2

continuous at x  is
2
A) –1 B) 1 C) 2 D) –2
64.    
Let f ( x)  [a]2  5[a]  4 x3  6{a}2  5{a}  1 x  (tan x)sgn x, be an even function for

all x  R   2n  1 , n  Z then sum of all possible values of a is (where [.] and {.}
2
denote the greatest integer function and fractional part functions respectively)
70 35 35 70
A) B) C) D)
3 6 3 6

65. Let y  y  x  be the solution of the differential equation
dy
dx

 1  xe yx ,  2  x  2, y  0   0 then the minimum value of y  x  , x   2, 2 is 
equal to
 
A) 2  3  log e 2  
B) 1  3  log e  
3 1
 
C) 1  3  log e  3 1   
D) 2  3  log e 2
 3 1 
 
66. If P   2 2  , A   1 1  and Q  PAPT then the inverse of PT Q2023 P is
 0 1
 1 3  
 
 2 2 
 1 2023   2023 0  1 0   1 2023 

A)  B)  C)  D) 
0 1   1 1  
 1 2023  0 1 
67. The number of numbers between 2000 and 5000 that can be formed with the digits
0,1,2,3,4 (repetition of digits is not allowed) and are multiple of 3 is :
A) 24 B) 30 C) 36 D) 48
  1  x2 
68. The range of the function f ( x )  cos 1
 
log 4 x   sin 
2  4
1
x
 is equal to

           
A)  0,   B)  ,   C)  ,  D)  
 2 2 2 2 2 6 2 6
 e  x  k, x0

69. If f  x    e x  1, 0  x  1 is one-one and monotonically increasing for all
 ex   ,
2
x 1

x  R, then difference of maximum value of k and minimum value of  is
A) 0 B) 1 C) 2 D) 3

70. f ( x ) is differentiable function which satisfies the equation
  
f ( x)    f (t ) tan tdt   tan(t  x) dt where x    ,  , passes through  0,0 , then
x x
0 0
 2 2
the maximum value of f ( x ) is
A) 1 B) –1 C) 0 D) 2
71. Four digit numbers are formed using the digits from the set 0,1, 2,3, 4,5 repetition of
digits is allowed then
Statement- S1 The number of such numbers formed that are odd is 480
Statement- S 2 The number of such numbers formed such that it contains exactly three
different digits is 360
A) S1 , S 2 are true B) S1 , S 2 are false
C) S1 is true and S 2 is false D) S1 is false and S 2 is true
x  3 y  2 z 1 x3 y 6 z 5
72. The shortest distance between the lines   and   is
2 3 1 2 1 3
18 22 46
A) B) C) D) 6 3
5 3 5 3 5
73. Two loaded dice each have the property that 2 or 4 is three times as likely to appear as
1 or 3 or 5 or 6 on each roll. When two such dice are rolled, the probability of obtaining
a total of 7 is
1 1 7
A) B) C) D) none of these
8 7 50
74. Circles of radii 36 and 9 touch externally. The radius of the circle which touches the two
circles externally and also their common tangent can be
A) 4 B) 5 C) 17 D) 18

75. Consider a branch of the hyperbola x 2  2 y 2  2 2 x  4 2 y  6  0 with vertex at the
point A. Let B be one of the end points of its latus rectum and C be the focus of the
hyperbola nearest to the point A, the area of the triangle ABC is :
3 3 3 2
A) B) 1 C) 1 D)
2 2 2 3
76. The vector r , which is normal to both a  4iˆ  5 ˆj  kˆ and b  iˆ  4 ˆj  5kˆ and r  c  21
Where c  3iˆ  ˆj  kˆ is_____
A) 7(iˆ  ˆj  kˆ) B) 7(iˆ  ˆj  kˆ) C) 7(iˆ  ˆj  kˆ) D) 7(2iˆ  ˆj  2kˆ)
1 1
77. Let A and B be two independent events such that P  A   and P  B  . Then, which
3 6
of the following is TRUE?
A 2  A  1 A 1  A  1
A) P    B) P    C) P    D) P  
B 3  B  3  B  3 A B 4
Let  ,  be two roots of the equation x 2  (20)1/4 x  (5)1/2  0 . Then    is equal to

8 8
78.
A) 10 B) 50 C) 5 D) 30
79. Let A  1,a1 ,a 2 ,.......a18 ,77 be a set of integers with 1  a 1  a 2  ......  a 18  77 . Let the
set A  A  x  y : y  A contain exactly 39 elements. Then, the value of
a 1  a 2  .....  a 18 is equal to_____.
A) 600 B) 702 C) 800 D) 200
80. Normals of parabola y2  4x at P and Q meets at R  x 2 ,0 and tangents at P and Q meets
at T  x1 ,0  . If x 2  3, then the area of quadrilateral PTQR is____
A) 4 B) 8 C) 12 D) 6

SECTION-II
81. Let p,q,r  R  and 27pqr  (p  q  r)3 and 3p  4q  5r  12 then the value of
8p  4q  7r is equal to_______.
sin  3x 2  4x  1  x 2  1
82. If lim  2, then the value of  a  b  is equal to______
x 1 2x 3  7x 2  ax  b
83. Let f : R  R be a differentiable function such that

f  x   x 3  x 2 f  1  xf   2   f   3 , x  R . Then f  2  is equal to
84. Suppose a class has 7 students. The average marks of these students in the mathematics
examination is 62, and their variance is 20. A student fails in the examination if he/she
gets less than 50 marks, then in worst case, the number of students that can fail is_____
85. Consider the set of all triangles OPQ where O is the origin and P and Q are distinct
points in the plane with non-negative integral coordinates  x, y  such that 5 x  y  99 .
Number of such distinct triangles whose area is a positive integer is______
log x K
86. If the area of the region bounded by the curves y  ex log x and y  is , find the
ex 4e
value of [K] (Where  represents greatest integer function) _____
87. Using all of 0,0,0,1,1,1,-1,-1,-1, a set ‘S’ of all 3  3 matrices formed.If number of
symmetric matrices in S is K and number of matrices having trace zero in S is P, then the
P
value of  K is
10
x  2 y 1 z  3
88. The foot of perpendicular of the point (0,2,7) on the line   is
1 3 2
4
( ,  ,  ) then the value of is


89. A spherical iron ball of radius 10cm is coated with a layer of ice of uniform thickness
that melts at a rate of 50 cm3/min. When the thickness of the ice is 5cm, the rate at which
1
the thickness (in cm/min) of the ice decreases is , find the value of k.
k
90. If the equation z 4  a1 z 3  a 2 z 2  a3 z  a 4  0 where a1 , a2 , a3 , a4 are real coefficient

different from zero has purely imaginary roots then find the value of the expression
a3 aa
 1 4.
a1a2 a2 a3

Sec: SR.IIT_*CO-SC(MODEL-A,B&C) GTM-11(N) Date: 09-01-24
KEY SHEET
PHYSICS
1 B 2 D 3 C 4 A 5 B
6 C 7 A 8 A 9 D 10 B
11 A 12 C 13 A 14 B 15 A
16 A 17 D 18 C 19 C 20 B
21 45 22 50 23 12 24 5 25 3
26 30 27 245 28 6 29 3 30 4
CHEMISTRY
31 D 32 C 33 B 34 D 35 D
36 A 37 A 38 C 39 D 40 D
41 A 42 A 43 B 44 A 45 A
46 D 47 C 48 B 49 C 50 A
51 4 52 9 53 3 54 75 55 4
56 1 57 446 58 15 59 8 60 4
MATHEMATICS
61 C 62 C 63 A 64 D 65 C
66 B 67 D 68 A 69 C 70 B
71 C 72 C 73 D 74 A 75 C
76 D 77 A 78 D 79 A 80 C
81 7 82 4 83 0 84 0 85 26
86 22 87 3 88 2 89 32 90 12
Narayana IIT Academy 09-01-24_SR.IIT_*CO-SC(MODEL-A,B&C)_JEE-MAIN_GTM-11(N)_KEY&SOL
SOLUTIONS
PHYSICS
1. The given formula of young’s modulus of elasticity,
5. Current through capacitor

dq d  dV 
ic    CV   C  
dt dt  dt 
di c  d2V 
C 2 
dt  d t 
Current through inductor
iL  1  ic
diL di
 c
dt dt
di di
 VL  L L  L c
dt dt
d2V
VL  LC 2
 3  2  10 6 2
dt
VL  12  V
7. According to given condition    1 t  n for minimum t , n  1
So,    1 tmin  
 
tmin    2
  1 1.5  1
8. For a vibrating string
p T
f 
2l 

p= no.of loops or no.of of harmonics
For the given problem.
p T = constant or p1 T1  p2 T2
T2 p Mg 5 M 25
Or  1 or  or 
T1 p2 9g 3 9 9
Or M = 25 kg
9. Given, forward resistance, R1  50
11. Given, frequency, f  500 MHz  5 108 Hz

h h
14. mg   Fb   balance torque about the edge 
2 3
2
18. No of reactions   6.023 1026
235
2  6.023  1026  200 106  1.6 1019
Power output 
235  30  24  60  60
 63.2 MW
19. For closed organ pipe resonating frequency is odd multiple of fundamental frequency
 2n  1 1.5  20
 number overtones heard = 7
21. Let the total height of building be x .
2d 2 x 2d
3x 2 
23. V  E0   dx   2 dx   12dE0  U  qV  12qdE0
0 d 0
d 
P 2  P 10  Q   1  2  10
24. s  ;
Q 3 10Q 1 3 10  Q
10
 20  2Q  30.  Q  5; P  
3
25. Velocity component along x direction after time t is
vx  4v02  v02
qE
vx  t
m
26.

  c
 90     c
4
 cos   ...... 1
5
sin    sin 
sin  sin 2 
 sin    1  cos 2   cos 2   1 
 2
sin 2  4 sin 2  .4 4
 cos   1    1 
2 5 5 5
1 sin 2  .4 1 1
    sin 2   sin  
5 5 4 2
  30 0
28. Let ‘x’ be the elongation in the springs and  be angular displacement.
 l   l 
 = l  K    x   
 2   2 
ml 2 Kl 2
   
12 2
 6K  m
     T  2
 m  6K
  1 m3 2
29. F   w
2
2 3
 n  hc
30. 1.388 103   
 At  
1.388  103  550

1240 1.6  1019
 4  10 21
CHEMISTRY
2 3
33. Cr is reducing and Mn is oxidizing when both have d 4 configuration. Oxidation state of iron in
ferrates is +6  FeO4  Actinoid contraction is greater from element to element than lanthanide
2
.
contraction,.
34. No. of unpaired electrons in the above species are
4 3
 Fe  CN 6  0  Fe  CN 6  1
3 2
Cr  NH 3 6   3  Ni  H 2O 6   2
Strength of ligands CN  ( strong) > NH3(strong)>H2O(weak)
35. When electron density is pushed from metal atom into -bond, the CO bond is weaken as electrons
enter into anti bonding orbital of CO. With 2 unit negative charge on metal atom -back bonding
from metal to CO increases maximum electron density.
36. Conceptual
40. G o  H o  TSo  E o  (PV)  TSo
Assuming ideal gas behavior,
G o  E o  (n)RT  TSo

Using the value of R=1.987 cal mol.K, and the fact that 2 mol of gas (D) is produced from 3 mol
(2A+B),
(n)RT  ( 1 mol) (1.987 cal / mol.K)(298 K)  592 cal
G o  2500 cal  (592cal)  (298 K)(10.5cal / K)  3.09kcal  0.04 kcal
Since the value of G o is positive , the indicated reaction cannot be spontaneous.
41.
m  2 x  6 y  8 z
eq  2 x  6 y  8 z
8
42. The correct sequece of carbocation is II  I  III  IV . Electron releasing stabilizes the carbocation
by dispersal of positive charge. More the number of alkyl groups, the greater the dispersal of positive
charge and the more stable is the carbocation.
43. Propane will not formed as in wurtz reaction two molecules of each reacting species combine with
the removla of NaX.
44. None of the ozonolysis products is chiral.
45. Presence of electron releasing roups,
46. Only aliphatic – OH is substituted by Cl  . This is because in phenol the C  O bond is stabilized by
resonance.
47. In phenyl magnesium bromide, phenyl is attached with that

48. It is a   keto acid which undergoes decarboxylation in very mild
49.
50. Sugars that have free aldehyde, a ketone, a hemiacetal or a
Hemiacetal can be easily reduced by oxidising agent such as Tollen’s reagent

51. SOCl2 , PCl3 , PCl5 , RPBr3
52. HNO2, SO2, NO, N2O3, NO2, SeO2, TeO2 can act both as oxidizing and reducing agents
53. 3 & 7 are non-polar
Note : iii, iv,& vi are non-polar molecules
54. 80 gm SO2 18 gm water H2O
?  4.5
6.581015 z 2
55. Number of revolutions per second =
n3
56. 2 H 2 S  2 H 2 g   S 2  g 
2
 0.1x   0.1x 
   
Let x be the degree of dissociation K c  
v   2v 
2
 107
 0.1  0.1x 
 
 v 
3
x
 106 x = 0.01
2v
Degree of dissociation  
0.1x
 x
0.1
1% dissociation of H 2 S
MV 4  500
57. Number of moles of Na  reduced at cathode   2
1000 1000
Mass of amalgam = Mass of 2 mol Na + Mass of 2 mol Hg  2x23  2x200  446g
 A0 
58. Amount of A left in n1 halves =
2n1
Amount of B left in n2 halves =  A0   4  B0 
22  2n1  n2
n1  n2  2
T  n1  t1/ 2 A similarly T  n2  t1/ 2 B
So, n1  3n2
59.
60. a,b,c,g

MATHS
62. x ydx   x  y  dy
2 3 3
y 2  x 2 ydx  x3dy  y 3 dy
y 2
y6 y6
x 2 y 3 dx  x 3 y 2 dy 1
 dy
y6 y
 x3  1
 d  3    dy  c
 3y  y
x3
 my  c
3 y3
9 y 2  x 2 ydx  x 3 dy   3 y 5 dy
3x 2 ydx  3x3 dy  y 3 dy
63.
f 1  h   f 1  f  1
64. lim 
h 0 h  h  3
2
3
65
66. No of elements A  B  9
No of subsets of A  B with at least two & atmost seven elements is
9 C2  9 C3  9 C4  9 C5  9 C6  9 C7  492
67. V1.V2  0
abc  0
a  2, b  0, c  2 6 ways
a  1 , b  0 , c  1 6 ways
a  2, b  1, c  1 3 ways
a  2, b  1, c  1 3 ways
18 ways
68.  f  x   x  3  a  1 x  1
3 2
 f   x   3 x 2  3  a 2  1  0x  R for f  x  to be invertible

69. f  x  y   f  x. f  y 
f  x  h  f  x
f '  x   lim
h 0 h
Get f  x   2 x
64 f  6 
Area enclosed 
3 3
70.  c is a root of x 2  ax  b  0  c 2  ac  b  0
71.  Let common difference of A.P. be d.
72. Let eccentricity of conjugate hyperbola be e’
2 3 4 4 1 1 10
73. Required probablity        
6 7 6 8 7 3 21
 21  10
73.
74. There is no change in the standard deviation if each observation is increased by a constant number
whilemean is increased by that number
New mean 30  2  32 and standard deviation =2

75.
76. n (A) = 3
77.

 1   11 7   
78. Centroid is M  , , 
 3 3 3 
A  2,3, 5
   5 2   
DR’S AM  , , 
 3 3 3 
For AM to be equally inclined
  7,   10
f f  x   axcbx ; f  2   1 f   2   0; 2ac 4th  1 …(1)
2
79.
80.
h
10
r
dv
 3m3 / min
dt
r 5 1
 
h 10 2
h
r
2
1 h2  h3
v    h 
3 4 12
dv 3 h dh 2
 .
dt 12 dt
dh 4

dt 3
  
1 1 ex
81. I   x 1 3 x dx   dx  1 e  e2 x  e2  dx
1 e e  1 e  e  e .e 
x 2 x

x
  tan t 
1 2
dt
0
82. lim
x 
1  x2
1
 tan x 
2
 tan x 
1
1 2
x 1
 lim  lim x2
x  2x x  x
2 1 x 2
 2

4
cos ec
1
83. B  t 1  t  dt
1
2
84.
85. P 1  P  2      P  6   1
0.1  0.32  0.21  0.15  0.05  k  1
P 1 or 2   0.1  0.32  0.42
10 5
P  req   
42 21
86. 2n
C3r  2 nCr  2
2n  4r  2  n  2r  1
n   11 
10  r 5   10    22
 nr  2  5

87. Point of intersection is  0, 2 
0
A 2   x   dx

0
x  
3/ 2
2   12
3
2 
  9    3
2
 4 
88. tan 1  2 
 4r  3 
 
1
 1 
 tan 
3
 r2  
 4
 
 1 
 tan 1 
1
 1 r2  
 4
 1  1
 r  2r  2  
 tan 1     
 1   r 1  r  1  
   
  2  2  
 1  1
 tan 1  r    tan 1  r  
 2  2

 4   1  1 

r 1
tan 1  2    tan    tan 2
 45  3  2 2
1
2 b2 1
89. 1 1 
3 16 2
1  b2  1
1   
3  16  4
b2 3
1 
16 4
b2 1
  b2  4
16 4
Length of minor axis = 4

2 16 1
1 1 2 
3 b 2
 16  3
1  2  
 b  4
16 1
  b 2  64
b2 4
b8
Length of minor axis = 8
f  x  y f  x  y
90.   4 xy
x y x y
f  x  y f  x  y
  x  y2    x  y 
2

x y x y
f  x  y f  x  y
  x  y    x  y  K
2 2
x y x y
f t  2
x y t t  K f  t   Kt  t 3
t
1
1 1 1
A  x 2  x 3  dx    f  x   x3  Kx f 1  1
0
3 4 12
f  x   x3 K 0

KEY SHEET
PHYSICS
1 D 2 B 3 B 4 C 5 B
6 B 7 B 8 A 9 B 10 B
11 D 12 A 13 C 14 B 15 D
16 D 17 A 18 A 19 A 20 A
21 4 22 6 23 25 24 141 25 32
26 3 27 2 28 352 29 400 30 125
CHEMISTRY
31 C 32 D 33 A 34 A 35 B
36 B 37 B 38 C 39 B 40 A
41 C 42 B 43 A 44 D 45 C
46 D 47 C 48 D 49 B 50 D
51 5 52 5 53 5 54 44 55 3
56 10 57 6 58 5 59 25 60 3
MATHEMATICS
61 D 62 A 63 C 64 A 65 D
66 B 67 D 68 C 69 A 70 A
71 A 72 A 73 B 74 A 75 D
76 C 77 C 78 D 79 B 80 B
81 3 82 3 83 4 84 5 85 6
86 2 87 3 88 3 89 26 90 68
SOLUTIONS
PHYSICS
V  VG 10  1
1. IG   100  104   R  900 
R R
R
2. m    . 4 r 2 dr   KR 4
0
GM G   KR 4  KR3G
V0   
r 2R 2

3. a m / s2
2
Velocity after t  2
v  m/ s
V2 2
Then aN  
R 1
aN   m / s
2 2
2
 
So net according aT      2 
2
2
2
aT   4
4

aT  1  4 2
2
So N = 4
1
T Vrms T' M
4. Vrms  ,   2
M Vrms T M'
1  2  1 1  1.25  1.5   1 1 
5.    1      1   
f  1  R1 R2  100  1   20 40 
1.5 1 1.5 7 1.5  6 9
 1     1   .
1 6 1 6 7 7
6. An   r 2
An    r0 n 2 
2
An   r02 n 4
A
So n n  4 n n
A1
7. No.of divisions = 50
Pitch = 0.5 mm
0.5 mm
Least count   10  m
50
1
8. 
LC
 LC 1
 
0 8L 2C 4

 0 .
4
V V V
9.  6  6
2 4 4
V V V
   3
4 8 8
  
10. For stable equilibrium, m is parallel to B (i.e.,   00 ) and for unstable equilibrium, m is

antiparallel to B (i.e,   1800 ).
 
PE in the unstable equilibrium, U 2  m.B  m B cos 00  m B
 (0.40 J / T )  0.16T   0.064 J
Amount of work done to displace the solenoid from its stable to unstable orientation, i.e.,
U 2  U f  0.064 J   0.064 J   0.128 J
K A2  x 2 1  1/ n 2
11.  2
 2
 n2  1
U x 1/ n
12. Vrel  VF   VB
4
 9  3  VB
3
 VB  4.5 m / s
KQ
13. A) V  3R 2  r 2 
3 
2R
R KQ  2 R 2  11KQ 11V
At r  , V   3R   
2 2R 3  4  8R 8
KQ V
B) V  
2R 2
KQ R KQ V
C) E  3  
R 2 2R 2 2
KQ V
D) E  
4R 2 4
14. Here the diode is in forward bias. So we replace it by a connecting wire,
30  VA 0  VA 0  VA
  0
10 10 10
3V
3 A
10
VA  10 V
15. In photoelectric experiment, speed of fastest emitted electron is given by
1 2 hc
mvmax  w
2 
1 hc
Case-I : mv 2  w ……… (i)
2 
1 hc
Case-II : mv '2  w
2 3 / 4
1 4hc
mv '2  w ……. (ii)
2 3
From eqn. (i) & (ii)
4 w
v '2  v 2 
3 3
4
Hence v '  v
3
16. If the insect is not sliding, mg sin   f
mg sin    mg cos 
3
 tan    
4
The height h from the bottom

h  R  R cos 
4 R
h  R  R     0.20 m .
5 5
3
17.  U  nCv  T  1   8.314  100  1247.1 J
2
18. Conceptual
19. In x direction : Applying conservation of momentum
mu  2mv cos 30
u u
v 0

2 cos 30 3
v u 2
Also e  0
 
u cos 30 3u 3
2
 e
3
d d
20. 
s u  10
u  10  s  u  1 s m / s
4
I0 I 1 I
cos 2 60  0     0
4
21. I
2 2  4  512
1 CC 
22. Loss   1 2  V 2
2  C1  C2 
1
  30  1012  (20) 2  6  109 J
2

23. Conservation of linear momentum, pi  pf
2  4  2  1  m2  v2
m2 v2  6 ….. (1)
v v
The coefficient of restitution, e  2 1
u1  u2
v v
1  2 1  v2  5 ms 1
u1  u2
By (1), m2  5  6
 m2  1.2 kg
m v  m2 v2 2  1  1.2  5 8 25
vcm  1 1   
m1  m2 2  1.2 3.2 10
x  25
24. VL  VC  0
200
VR 
2
3
 4T  3  4T   r 
25.  P1   r   P2   
 r   r/22
 4T  8T
  P1   8  P2 
 r  r
T
 P2  8P1  24
r
v v
26. up  rel and down  rel
2R 2R
U ˆ U ˆ
27. F  i j  6iˆ  8 ˆj
dx dy
 a  3iˆ  4 ˆj
 ax  3
s x  6
2 6
t  2s
3
2
1 1  4 
28. u  0 Erms2
  8.8  1012     352  1013
2 2  2
29. Conceptual
hc
30. By photoelectric equation,    K max
2
1240
K max   1.25  1.25
500
2mK
r
eB
2mK
B  125  107 T .
er

CHEMISTRY
31. Highest amount of heat will be released when equivalents of acid and base are equal i.e., complete
neutralization takes place.
1
32. For second order reactions t1/ 2 
K[A]0
 Half-life decrease with increase in concentration of reactants.
ln 2
For 1st order, t1/ 2  and is independent of concentration
k
dx
For zero order,  K (both are same)
dt
10 10 10
 P  100   P  200  P  10000
33.  0   10 180 ;  0   10 180 ;  0   10 180
 P A   P B   P C 
100 18 200 18 10000 18
 P   P   P 
 0   0   0 
 P  A  P  B  P C
P
Relative lowering of V.P. = 0  Xsolute .
P
35. (KE)1  hv1  hv0
(KE)2  hv 2  hv0
As (KE)1  2  (KE)2
 hv1  hv0  2(hv2  hv0 )
or hv0  2hv2  hv1
or v0  2v1  v1 .
= 2  (2  1016 )  (3.2  1016 ) = 0.8 1016 Hz  8 1015 Hz .
36. Number of milli equivalents of
MnO 4  0.02  3  250  15
Number of milli equivalent of
I2  0.11 250  25
Thus, here MnO 4 is limiting reagent
 Number of milli equivalents of I 2 formed = Number of milli equivalent of MnO 4
15
= 15 or number of equivalent of I 2 formed =  0.015
1000
0.015
 Number of moles of I 2 formed =  0.0075 .
2
38. Statement-I is correct: Pure aniline is colourless liquid
Statement-II is incorrect: Aniline becomes dark brown due to action of air and light (oxidation)
39. Cross Cannizzaro is simply redox process.
41.
5-amino-4-hydroxymethyl-2-nitrobenzaldehyde

 2
43. 2Cu  Cu  Cu
The stability of Cu 2 (aq) rather than Cu  (aq) , is due to the much more negative  hyd H of
Cu 2 (aq) then Cu  (aq) , which more than compensates for the second ionization enthalpy of Cu.
48. During electrolysis of CuSO4 solution using Cu electrodes, the cell reaction is
 Cu   2e
Anode: Cu(S) 
Cathode: Cu  (aq)  2e   Cu(s)
The loss in weight of anode is equal to gain in weight of cathode.
50. Pentaamine chloride cobalt (III) chloride
[Co(NH 3 )5 Cl]Cl 2 [Co(NH3 )5 5Cl  ]2  2Cl 
Gives 3 ions in aqueous solution.
51. Given compound undergoes free radical bromination under given conditions, replacing H by Br.
C is chiral carbon.
(III) has two chiral centres and can have two structures.
(IV) has also two chiral centres and can have two structures.
It has plane of symmetry thus, achiral,

Thus, chiral compounds are five, I, IIIA, IIIB, IVB and V

54. Weight of C in 0.792 gm CO2
12
=  0.792  0.216 .
44
0.216
% of C in compound = 100  43.90%  44 .
0.492
56. G 0  H0  TS0  54.07  103 J  298 10 J = 57.05 103 J
Also, G 0  2.303RT log K .
G 0 57.05 103
 log K  =  10 .
2.303 RT 5705
57. A decapeptide has nine peptide (amide) linkage as
Therefore, on hydrolysis, it will absorb nine water molecules.

Hence, total mass of hydrolysis product = 796  18  9  958
958  47
 Mass of glycine in hydrolysis product =  450 .
100
450
 Number of glycine molecule in one molecule of decapeptide=  6.
75
58.
59.
AB2  AB(g)  B(g)
500  
500  x x x
Require = 600 torr = 500 – x + x + x, x = 100 torr
P  P 100 100
K p  AB B   25 torr .
PAB2 400
60.

MATHS
1
x 2 sin    2 x
 x x
61. lim .
1  x   e
1/ x
x 0 x
 1  x  2 4
 lim  x sin    2    0  2     
x 0
  x   e 1  x  11 x 2  ....  e  e e
 2 24 
62.
Median CD = 3r
1
2  3   2  4    2r 
2 2 2
3r 
2
36 r 2  18  32  4 r 2  40 r 2  50
5 5
r2  r
4 2
Diameter AB = 2r  5

2 x 1  sin x 
63. I  dx

1  cos 2 x

2 x 1  sin x   2   x 1  sin x  
x sin x
Using King and add 2I  

1  cos x 2
dx  4 

1  cos 2 x
dx
 
x sin x
I  2  f  x dx  I  4 dx
 0
1  cos 2
x

sin x
Using Kind and add 2 I  4  dx
0
1  cos 2 x
Put cos x  t
1
dt 
2 I  4   I  4  2.
1
1 t2 4
64.  f 1  3  g  3  1
 Point   3,1
1 1 1
g f  x    g  3   
f x  f  1  4
1
 Tangent  y  1   x  3  x  4 y  1  0 .
4
65.  f   x   ax  x  1  f   2  6  a3
2
3x
f   x   3  x2  x   f  x   x3  C
2
  3
f x   x2  x  
 2
f  2  2  C  0 .

66.
Point A be the intersection of AC & AB i.e.  4,5  & B be the intersection of AB & BC i.e.  3, 2 
 1 7
Mid–point of AC will be   , 
 2 2
7
7  1
Equation of diagonal BD is y   2  x    7 x  y  0
2 1  2
2
23 2  ax  by1  c1 
Distance of A from diagonal BD  d   50d 2   23   d  1 
50  a 2  b 2 
50 d 2  529 .
67. Let F, H and B be the sets of television watchers who watch football, Hockey and Basketball
respectively.
Then, according to the problem, we have
n U   500, n  F   285, n  H   195,
n  B   115, n  F  B   45,
n  F  H   70, n  H  B   50,
And n  F   H   B   50,
Where U is the set of all the television watchers.
Since, n  F   H   B   n U   n  F  H  B 
 50  500  n  F  H  B   n  F  H  B   450
We know that,
n  F  H  B  n  F   n  H   n  B  n  F  H   n  H  B  n  B  F   n  F  H  B
 450  285  195  115  70  50  45  n  F  H  B 
 n  F  H  B   20
Which is the number of those who watch all the three games. Also, number of persons who watch
football only  n  f  H   B 
 n  F   n  F  H   n  F  B  n  F  H  B
 285  70  45  20  190
The number of persons who watch hockey only
 n  H  F   B
 n  H   n  H  F   n  H  B   n  H  F  B   195  70  50  20  95
And the number of persons who watch basketball only
 n  B  H   F 
 n  B   n  B  H   n  B  F   n  H  F  B   115  50  45  20  40
Hence, required number of those who watch exactly one of the three games
 190  95  40  325 .

68.  The image of ‘z’ in the line y  x is i z .
i z i  
 The image of the given curve is arg  
 i z 1  4
 z 1 
 arg  
 z i  4
z 1  
  arg    arg  z    arg  z  
 z i  4
 z i     z1   z2  
 arg    arg     arg    .
 z 1  4   z2   z1  
69. x  sin 1  sin10   3  10
y  cos 1  cos10   4  10
 yx .
70. I ) Sum of perimeters = 3  24  12  6      
 
 1 1   1 
72 1   ........   72    144
 2 4  1
 1 
 2
II) log 2  a  b  c  d   4
  a  b  c  d   24
abcd 4
   a  b  c  d   a  b  c  d  8 .
4
1 1  x  1  y 
A  x  A  y   1  x  1  y  
1
71.  
  x 1   y 1 
1 1  xy  ( x  y 
 1  xy  ( x  y )   
( x  y) 1  xy 
 x y 
1 
1  xy 
1
 x y 
 1      A z 
 1  xy    x  y 1 

 1  xy
 
72. The given equation can be written as
 x 2     1 x  5  0
  ,  are the roots of this equation.
 1 5
    and  
 
  4
But, given  
  5
2  2 4
 
 5
   1
2
10

   
2
 2 4  2
 4
  
 5 5 5

   1
2
 10 4
    2  12  1  4
5 5
  2  16   1  0
It is a quadratic in  , let roots be 1 and 2 , then 1  2  16 and 12  1
1 2 12  22  1  2   212 16   2 1
2 2
      254 .
2 1 12 12 1
73. A  1, 2,3, 4,5,6, 7
Case–I: When exactly 4 values follows f  i   i
1 1
C4  3!    70
7
 2! 3! 
Case–II: When exactly 5 values follows f  i   i
7
C 5  1  21
Case–III: When all 7 values follows f  i   i
Number of function = 1
Total functions = 70 + 21 + 1 = 92.
74.     
Let l  0iˆ  0 ˆj  0kˆ   aiˆ  bjˆ  ckˆ   aiˆ  bjˆ  ckˆ 
iˆ ˆj kˆ
 aiˆ  bjˆ  ckˆ  1 2 3  4iˆ  5 ˆj  2kˆ
2 2 1

 l   4iˆ  5 ˆj  2kˆ and 
l1 1    iˆ   11  2  ˆj   7  3  kˆ
P is intersection of l and l1
 4  1   , 5   1  2  , 2   7  3
By solving there equation   1, P  4, 5, 2 
Let Q  1  2 , 2 ,1   
Then, PQ   5  2 , 5  2 ,1   
 
PQ. 2iˆ  2 ˆj  kˆ  0  2  4  4  1    0
1  7 2 10 
 Q , , 
9  9 9 9 
7 2 10
                y  .
5
 9 9 9  9
75. Let DC  CB  BA  AD  k
 Coordinates of B are  k , k  ,
Which lie on y x
 k  k
 k  2
 BC  k   2
Also, let CG  GF  FE  EC  k1
 Coordinates of F are   2  k1 , k1  ,

Which lie on y   x
Then k1    2
 k1 
 k12   4   2 k1
Or k12   2 k1   4  0
2   4
 4 4  k1 1 5
 k1  Or 
2  2
2
FG 5 1
Or 
BC 2
2 4  2 4 8
sin 2 sin 2 sin 2 sin 2 sin 2 sin 2
76. S 7  7  7  S 7  7  7
 2 4  2 4
sin 2 sin 2 sin 2 sin 2 sin 2 sin 2
7 7 7 7 7 7
  2 4    2 4 
S  4  cos 2  cos 2  cos 2   4  1  2 cos  cos  cos 
 7 7 7   7 7 7 
 1
 4 1  2    5 .
 8
4 1
77. Let X denote the number of aces. Probability of selecting an ace, P  
52 13
1 12
Probability of not selecting an ace, q  1  
13 13
 1   12  24
P  X  1  2       
 13   13  169
1 1 1
P  X  2   
13 13 169
24 2 2
Mean  Pi X i    .
169 169 13
 
20
78. Let 8  3 7  I  f, where f = fractional part and I = integral part
 
20
Also let 8  3 7  g then 0  g  I
   8  3 7  
 2 820  20 C2.818. 3 7    
20 20 2 20
Here I  f  g  8  3 7 
 ......  20 C20 3 7

 I  f  g = even integer
But 0  f  g  2
So, I  1  even Integer  f  g  1
 I  odd Integer
79. Possible favourable outcomes will be getting by, either all outcomes are positive or any two are
negative.
3 1
Now, p  P  All positive   
6 2
2 1
q  p  both negative   
6 3
So, required probability
5 2 3 4 1
1 1 1 1 1 521
 C5    5C2      5C4     
5
.
2  3  2  3   2  2592
80. Arranging the data in ascending order of magnitude, we obtain
Height (in 150 152 154 155 156 160 161
cm)
Number of 8 4 3 7 3 12 4
students
Cumulative 8 12 15 22 25 37 41
frequency
Here, total number of items is 41, which is an odd number.
th
41  1
Hence, the median is  21st item.
2
From cumulative frequency table, we find that median i.e., 21st item is 155.
81. Putting x = y = 1, we get f 1  2
Putting y = 1, f  x   x  1
 f 1  x   x  1
 f  x  f 1  x    x 2  1  f  2  f 1  2   22  1  3
xdy  ydx
82.  dy
y2
x
   d     dy
 y
x
   yc
y
At x  1, y  1  c  2 .
3
x  3   y  2
y
y2  2 y  3  0
 y2  3y  y  3  0
 y  y  3   1  y  3  0
  y  1 y  3  0
 y3  y  0 .
 2 4  x3 4 x 
83. f  x   a  x    f  x   a     b passes through  0, 0  and 1, 2 

 3 3 3 
-2 O
2
 b  0, a  2
2x 2
f  x 
3
 x  4
  2
2

Required area   2  4 k 4.
2 2

1  7 cos x
2
sec 2 x 1
84. We have  7 2
dx   7
dx  7  dx
sin x cos x sin x sin 7 x
I    sin x   sec2 x  dx  7   cos ecx  dx integrating by parts
7 7
    
I II
tan x cos x dx
  7 tan xdx  7 
 sin x   sin x   sin x 
7 8 7
tan x cos x dx tan x

  7  7  C
 sin x   sin x   sin x   sin x 
7 7 7 7
Hence, g  x   tan x
So, g   x   sec2 x
And g   x   2sec2 x tan x

 g   0   1 and g     4
4
 
Hence, g   0   g     1  4  5 .
4
85. Given
R
P C  P  
C  C 
Now, P   
 R  P A P R   P B P R   P C P R 
           
 A B C 
1 

3    4
 0.4   6
1 4 1 5 1 
   
3 10 3 10 3    4 
86. Let P divides AL in the ratio  :1
P divides DB in the  :1
  
Let AB  a , BC  b

 b 
a   
 3    a   b  
 1  1
   3;   3
BD 4
  2
PQ 2

87.
H h H
  r
H  h R
r R H
 r 2h  R2
hH  h
2
 V  
3 3H 2
dV  R 2   R2
   
  H  h  H  3h 
2
 H  h  2 h H  h 
dh 3 H 2  3H 2
H H
 Vmax when h    3 .
3 h
88. Required circle
x2 y2

 1   x2  y 2  0
16 9

Using x2 coefficient  y 2 coefficient.
1 1 7
     
16 9 288
288
Required circle x 2  y 2  .
25
89. In the given figure there are 8 squares and we have to place 6X’s this can be done in
8.7
8
C6  8C2   28 ways
1.2
But these include the possibility that either headed row or lowest row may not have any X. These
two possibilities are to be excluded.
 Required number of ways = 28 – 2 = 26.
90. Now, A  adjA  A I

x 3 2
Now, A1 y 4
2 2 z
 x  yz  8  3  z  8   2  2  2 y 
 xyz   8 x  4 y  3z   28
 60  20  28  68
From equation (i), A  adjA   68I .

KEY SHEET
PHYSICS
1 4 2 3 3 4 4 3 5 4
6 1 7 2 8 4 9 2 10 4
11 4 12 1 13 2 14 1 15 1
16 1 17 1 18 1 19 4 20 3
21 223 22 10 23 6 24 10 25 48
26 5 27 15 28 4 29 5 30 26
CHEMISTRY
31 2 32 2 33 4 34 3 35 3
36 3 37 1 38 1 39 2 40 4
41 2 42 4 43 1 44 3 45 2
46 2 47 2 48 2 49 4 50 4
51 16 52 3 53 45 54 6 55 12
56 6 57 4 58 5 59 25 60 6
MATHEMATICS
61 4 62 2 63 3 64 4 65 1
66 2 67 1 68 2 69 1 70 4
71 4 72 1 73 3 74 4 75 2
76 1 77 1 78 2 79 2 80 4
81 18 82 4 83 6 84 3 85 3
86 16 87 20 88 8 89 5 90 1
SOLUTIONS
PHYSICS
E
1. F   E 1V 1T 1 
VT 
  1,   1,   1
2.   MB sin 
 sin 
 1 sin 1  sin 900
  
 2 sin  2  / 2 sin  2
 2  300
Angle of rotation  90  30  600
 mg 1
3. 0.1  2 xdx   mv02
2
2
v0  10 m / s
v02
4. aA 
R
VA  2v0
v A2
RC   4R
aA
l l
5. T  2 ; TM  
g g
Mgl
l 
YA
Mgl
l'  l 
YA
1 A
 l '  l 
Y Mgl
 TM2  A
  2  1
T  Mg
V 
6. WAC  PV0 0 
 V0 
 P0 
2 PV
0 0 ln  0 0 ln  3 
  PV
P
P
P 0
3
7. PV  RT
 T  V 2 
P  R 0 
 V 
dP
0
dV
T
  02    0
V

8.
2 
t
T 12
T
t
24
 /2
 Rd cos 
9.  dE 0
 4 0 R 2


E sin  2
4 0 R
0
2 
E 
4 0 R 2 2 0 R
10. Since equivalent capacitance increases, charge on capacitor 4 increases. By KVL, charge on
capacitor 2 decreases.
11. Req  2
For max power, R  2
Eeq
 3  Eeq  6V
2
6
I   1.5 A
4
P0  1.5   2  4.5W
2
12. Inductance can be set to be analogous to mass, as it poses inertia to current change in the electrical
circuit.
2V0  3t

13. VC   1  e 2 RC

3  
2V0  1 
VC  1
3  3 
4V
VC  0
9
V 2V
I C  0
2R 9R
14. V  5  3sin t 
32 59
Vrms  52  
2 2
   ^
15. E  B is along positive Z-direction. B is along positive j direction.

 
16.   2 c   c 
2 4
1
 sin  c     2

D
17.  max 
d0  a0
D
 min 
d 0  a0
1 1 1 1 1 1  225
18.  R 2  2   R    
  n1 n2    9 25  16 R
19. Req  6
v0
I  4A
Req
20. Reading = 17+8  0.1-2  0.1=17.6 mm
21. L.C. =0.01 mm, zero error =-0.08 mm
Reading = 2 mm+15  0.01+0.08=2.23 mm
22. f T
2T  0.50  4  10  T  10 N
dv
23. mv P
dt
m v 2 2v0
P
t 2 v0
2
R
I   dI 
R 0
24.  0 r 4 dr
2 0 R 5 2 0 R 4
T 
5R 5
v0
25. v1 
2
2  x0  2 x0  4 x0
Radius of curvature at farthest position  
3 3
2
 v0 
 2  3v02
amin  
4 x0 16 x0
3
2 s cos 
26. Height of capillary rise =
 gR
2 S A cos 
When in A 5cm=
 AgR
2 S cos 
When in B h= B
 BgR
S B  2 S A and  B  2  A
2  2 S A  cos 
h  5cm
2  AgR

27. f 0  50 Hz
v
f0 
4l
300
50   l  1.5m
4l
A T  1.5
 A
28. B TA
A 1

B 2
TA  2eV , TB  0.5eV
B  EB  TB  4eV
29. Va  5  10  10  1  Vb
 Va  Vb  5V
30. Energy released = Change in B.E. (7.6 × 4) – [4 × 1.1] = 26 MeV
CHEMISTRY
31. 1.6 gm oxide looses 0.16 gm
80 gm oxide looses = 8 gm ‘O’ = ½ mole of ‘O’
TiO 1   TiO 3  Ti 2O3
 2  2
 2
32. Zn + Ni Zn+2 + Ni
+2
o o
Eº = ENi2 / Ni – EZn2 / Zn
= –0.23 – (–0.76) = + 0.53 V
Positive value shows that the process is spontaneous.
Rest of all (I) (II) (III) combination have negative Eº value.
(I) Eº = –0.44 – (–0.23) = –0.21 V
(II) Eº = –0.76 – (–0.23) = –0.53 V
(III) Eº = –0.76 – (–0.44) = –0.32 V
33. HINT: Assume rate law
r = K[H3AsO4]x [H3O+]y [I–]z
Solving by the help of various experiments
x = 1, y = 2 and z=1
total order = 4
34. Tf  i Kf m
3.72  i  1.86  0.4
i  5 n  4
35. Hint: Due to extra stability of half-filled f-subshell.
36. Co  NH 3 4  ONO 2  Cl =linkage isomers
Co  NH 3 4  NO2  Cl   NO2 = ionization isomers

37. C3H8(g) + 5O2 3CO2(g) + 4H2O()
6  B.E. (C  O) 
8  B.E. (C — H)   8  B.E.(O — H) 
CH =  2  B.E. (C — C) –  
 3 | R.E. | of CO2 
 5  B.E. (O  O)   
 4   vap H(H2 O) 
38. X(g)  X+(g) + e–
If I.E. is ionisation enthalpy, then
N0 2E1
 (I.E.) = E1  I.E. =
2 N0
X(g) + e–  X–(g)
If  egH is electron gain enthalpy, then
E2
 2N0(E.A.) = – E2   egH = – .
2N0
39. (A) High spin, d 6 :CFSE=  0.6  2  0.4  4   0  0.4 0

(A) Low spin d 5 ; CFSE =  0.6  2  0.4  5  0  2.00
(B) Low spin d 4 ; CFSE=  0.6  2  0.4  4   0  1.6 0
(C) High spin d 7 ; CFSE=  0.6  2  0.4  5  0  0.8 0
Magnitude of CFSE is maximum in (B)
40.
41. Mass of organic compound = 0.2 g

N
Unused acid required = 40 mL NaOH
10
N N N
40 mL NaOH  40mL H 2 SO4  20mL H 2 SO4
10 10 5
N N
Acid used for absorption of ammonia= (60-20)mL H 2 SO4  40mL H 2 SO4
5 5
1
1.4   40
1.4  N1  V 5 1.4  40
Percentage of nitrogen =    56%
W 0.2 0.2  5
no of bond's in between two atom
42. Bond order of CO2 (by resonance method)=
no of resonating structures
4
i.e., bond order in CO32 (by resonance method)=  1.33
3
1
Bond length 
Bond order

43.
44. The compound A, despite a tertiary alcohol, cannot be readily converted into chloride because OH
is present at bridge head. The compound C, allyl alcohol can be readily converted into allyl chloride,
whose formation is responsible for white cloudiness.
45. t  BuO   astearic ally hindered base will give Hofmann elimination as major product. Where
as EtO  will give Saytzeffs product.
46.
solution : B
eletrophilic

aromaticsubstitution 
N  NCl 

47.
 
(A) is not possible because CH 3  CH 2  C H 2 is less stable than CH 3  C H  CH 3
(C) is not possible because acetophenone and CH 3OH cannot be formed.
OH
(D) is not possible because |
CH 3  CH  CH 3
48.

49.
50.
2Ze2
51. Velocity of an electron in He+ ion in an orbit = .....(i)
nh
n 2 h2
Radius of He+ ion in an orbit = ....(ii)
42me2 Z
By equations (i) and (ii),
u 83 Z2me4
Angular velocity () = = ....(iii)
r n3 h3
8  (22 / 7)  (2)  (9.108  10 )  (4.803  10 10 )4
3 2 28
= = 2.067 × 1016 sec–1.
(2)3  (6.626  1026 )3
CH3 CH3 Br
| | |
Br2 /hv
52. CH3  C  CH 2  CH3   CH3  C  CH  CH3 (One chiral carbon)
| | 
CH3 CH3
2
53. Number of possible dipeptides is (3) = 9
5x9=45
54. Basicity of H3PO4, H3PO3 and H3PO2 are 3, 2, 1 respectively.
 sum of basicity = 6
55. On dilution (addition of water) pH of the buffer solution will not change therefore x=0 and
x+12=12
56.  M   aq.   e   M  s   2 : E 0  0.52V
M  s   M 2   aq.  2e : E 0  0.34V
____________________________________________________
2M   M  s   M 2 E 0  0.52  0.34  0.18

At equilibrium Ecell  0
0.06
Hence E 0  log k
2
0.18  2
log k  6
0.06

57. The molar masses of glucose and urea are 60 g/mol and 180g/mol respectively.
50mL  1.2 g
The number of moles of urea   0.01 mol
100mL  60 g / mol
50mL  2.4 g
The number of moles glucose 
100mL  180 g / mol
 0.0067 mol
0.01 mol  0.0067 mol 1000mL
The molar concentration    0.167 M
50 mL  50 mL 1L
The osmotic pressure is   CRT
 0.167 M  0.08 L.atm / mol / K  300 K
 4 atm
58.

59. For a zero order reaction ……(1)

 A 0
For zero order reaction K  ……..(2)
2t 1
2
Since  A 0   2M, t 1  1h, K  1

2
 from equation (1)
0.25
t  0.25 h
1
60.
Total six Cl Ax  P  Cleq bonds at 900 to each other

MATHS
61. Consider statement-2
y
z  2i 1
P
1
C
O x
Q is the point on the circle which is nearest to the origin

OQ  OC  CQ = 5  1
Statement-2 is true
Consider statement-1
Point on the circle with maximum argument is 1
1 1
If COP    sin    tan  
5 2
4
and arg  z0   2  tan 1  
3
Statement-1 is false.
63. Let 1 ,  2 and  3 be the roots of f  x   0
Such that 1   2   3 and g  x  takes all values from  6,  
g  x    x  1  6  6
2
 3  7,  2  8, 1  9
 a  b  c  719
Minimum value of a  b  c is 719
1   2   3  a  a  24  a  24
1 2   2 3   31  b  b  191
1 2 3  c
 c  504
a  b  c  719
64. Make 1 group of 2 persons, 1 group of 4 persons and 3 groups of 3 persons among 15 persons
15!
(except 2 particular persons) hence the number of ways by grouping method is
2!4! 3! 3!
3
Now we add that 2 persons in the group of 2 persons and thus the number of arrangements of these
15! 15!
groups into cars and autos is  2! 3! 
2!4! 3! 3! 4! 3!
3 3
65. E1 be the event that the answers is guessed

E2 be the event that the answer is copied
E3 be the event that the examinee knows the answer and
E be the event that the examinee answer correctly
1 1
P  E1   , P  E2  
3 6
Assume that event E1 , E2 and E3 are exhaustive P  E1   P  E2   P  E3   1
1
 P  E3  
2
E 1  E  1 E
P    , P    , P    1 (Probability of answering correctly by knowing)
 E1  4  E2  8  E3 
E
P  E3  P  
E 
P 3    E3  
24
 E E  E   E  29
P  E1  P    P  E2  P    P  E3  P  
 E1   E2   E3 
66. P  E   P  2   P  3  P  5   P  7   0.62
P  F   P 1  P  2   P  3  0.50
P  E  F   P  2   P  3  0.35
PE  F   PE  PF   PE  F 
 0.62  0.50  0.35  0.77
68. Let BP=x from the similar triangle properly
A
m2
B P O
m1
AO m2 mm
  AO  1 2
m1 x x
d  AO  m1m2 dx

dt x 2 dt
m
When x  1
2
d  AO  2m
 2 m/s
dt 5
1
dx
72. Let I   1
0  5  2 x  2 x 1  e 
2 2 4 x
By kings rule
e 2  4 x dx
1
I  2
0  5  2 x  2 x 1  e 
2 2 4 x
Adding 1 and 2
1
11 1
1 1 1  x
dx dx dx 1
2I      ln 2 2
0 5  2 x  x
5  2 x  2x 2 2
0 11  1
2
2 11 11  1
0
 x    x  
4  2 2  2 0
1  11  1  1  11  1 
2 ln    ln  
2 11  11  1  11  11  1 
 
2
1 11  1
I ln
2 11 10
73.
y
y  x  x3/2
O 1 x
y  x x 3/2
 y  x
2
 x 3  y  x   x 3/ 2
 y  x  x3/2
y  x  x3/2  1, y  x  x3/2  2
1 is an increasing function
2 meets x=axis at x=0,1
1
  x  x    x  x dx
3/ 2 3/ 2
Required area
0
1 1
2  4
 2  x dx  2  x 5/ 2   sq.units
3/ 2
0 5 0 5
74.  x  x   x  which is periodic with period 1.
Statement 2 is true.
Consider Statement 1.
f  x   sin  3 x  3 x   sin 3 x 
1
Using Statement 2, period of f  x  is .
3
Statement 1 is false.

st
75. Let 1 term be ‘a1’ and common ratio be ‘r’ then a1  a1r  a1r 2  .....  a1r 100  125
a1 1  r101 
  125  let 0  r  1
1 r
1  1  
101
    1
101
1 1 1 1 1 a1  r    here 1  1
 a   a   2  ...  100    
  a1r a1r a1r 1  r 
r 1 i 1 1
r

1  r  101

1

125

125

125

125

1
a1r 100
1  r  a1r 100
a1  a1r 50   a51   25 5
2 2 2
n log83
 1   1 
76. Last term expansion is cn     
n
 2   33 9 
8
 1 n / 2   1 log3 5
 3log 3 2
  1 
n

  5/3   3 3
 2  3 
5
5 1
 35log3 2  3log3 2  2 5     n  10
2
4
   1  10 2 1 10
6
th
 5 term from the beginning 10
c4 3
2    c4 2 . 22  c4
 2
77. Let the two numbers be a, b
2  4  10  12  14  a  b
x  8
7
 a  b  14  1
x  xi  2 2
 2
 
 n 
 16
i
N  
 460  a  b  16  64   7  a 2  b2  100  2
2 2
From 1 and 2, a  b  2  3
a  8, b  6
78. a sin      sin      b sin      sin    
 2a sin  cos   2a sin  cos 
 a tan   b tan 
 
2a tan 2b tan
2  2 1
2  2
1  tan 1  tan
2 2

b tan c
   2
a tan  b tan  c  tan  2
2 2 2 a
  
From 1 and 2 we get tan
2
a 2
 b 2  c 2   bc  1  tan 2 
 2

2 tan
2  2bc
 sin  
   a  b2  c 2
2
1  tan 2  
2
79. Let  tan x  sin x    tan x  sin x    tan x  sin x   .....

y  tan x  sin x   y
 y 2  y   tan x  sin x   0
1  1  4  tan x  sin x 
y
2
Again z  x3  x3  x3  ....
 x3  z  z 2  z  x3  0
1  1  4 x3
z
2
lim
1  1  4  tan x  sin x 
 lim

4  tan x  sin x   1  1  4 x3 
x 0
1  1  4 x 3 x 0

4 x 3 1  1  4  tan x  sin x  
 sin x sin x 
   1  1 4x 3
sin x 1  cos x  1  1  4 x3
 lim 
cos x 1 
 lim 
x 0

x3 1  1  4  tan x  sin x  x 0
 x3 cos x 1  1  4  tan x  sin x 
sin x 2sin 2 x / 2 1 1  1  4 x3 1 11 1

 lim . . .  1. .1. 
x 0 x 4 x / 4 cos x 1  1  4  tan x  sin x 
2
2 11 2
80. Put x  2cos  , y  3sin 
 1 1 
sin 1 1  cos 1  cos   sin   2 
 2 2 
   
sin 1 1  cos 1  cos      2 
  4 
 
1  cos      2  1
 4
   
1  cos      1  cos      1
 4  4
   3
  cos 1 1  2    cos 1  1    
2 2 2 2
81. Let z  x  iy
z  z1   x  10    y  6  i
z  z2   x  4    y  6  i
 z  z1  
arg  
 z  z 2  4
 6  y  6  
 tan 1  2

  x  10  x  4    y  6   4
 x2  y 2  14 x 18 y  112  0
z  7  9i   x  7    y  9   x 2  14 x  y 2  18 y  130  112  130  18
2 2 2

83. The polynomial is every where differentiable function
 The points of extremum can only be the no.of derivative

The derivative of polynomial is P1  x   a  x  1 x  3   a x 2  4 x  3 
P 1  6
x
 x3 4
P  x    P1  x  dx  6  a   2 x 2  3 x    6
1  3 3
Also P(3)=6  a  3
P  x   x3  6 x2  9 x  2
 P  2   4, P1  0   9
 P  2   P1  0   7  6
x  0 y 1 z  0 x 1 y  0 z  0
84. L1      L2    
1 1 1 , 2 1 1
Any point on L1 and L2 be   ,   1,   and  2   1,  ,   respectively.
2  1       1   
 
2 1 2
On solving   1,   3
A   3, 2,3 , B  1,1,1
AB  4  1  4  3
a  c b d 
 
85. Shortest distance 
bd
Here a  3i  8 j  3k , b  3i  j  k , c  3i  7 j  6k , d  3i  2 j  4k
270
 Shortest distance   270  3 30   30    3
270
n  2 n 2 n  n  1
1   x   1  x 
n
86.  ....
1! 2!
= 1  8 x  24 x 2  ....
 n  8 and  2 n  n  1  48
   2, n  4
Now P  , n    2, 4 
Any line through P cuts the circle x 2  y 2  4 at A and B
 PA.PB  S11  4  16  4  16
87. Combined equation of pair of lines AC and BC can be obtained by homogenizing the hyperbola
x2 y 2
  1 with the help of AB
16 25
A
B
C  0, 0  x

2
x 2 y 2  x cos   y sin  
  
16 25  p 
 1 cos 2  2  1 sin 2   2 xy sin  cos 
x2    y    0
 16 p2   25 p2  p2
ACB  900
1 cos 2  1 sin 2 
    0
16 p2 25 p2
20
 p
3
Also Cl=P always perpendicular to given variable line hence variable line always touches a circle of
20
radius p  r   3r  20units
3
f 1  x  g  x   g 1  x  f  x  dx
88. I 

 f  x   g  x   g  x  f  x   g  x 2 
f 1  x  g  x   g1  x  f  x 

 g  x 
2
dx
 f  x  f  x
  1 1
 g  x  g  x
f  x 2tdt
Put 1  t 2  I   2
g  x t  2 t
2  t 
I tan 1  c
2  2
 f  x 1 
 2 tan 1   c
 2g  x  2 
 
 f  x   g  x  
 2 tan 1  c
 2 g  x  
m  2, n  2
m2  n2  8
89. f  x    x    x  2
  x    x  2   x    x   2  2
So f  x  is constant function
It is continuous every where  p  0
 3x 4  2 
q  lt  3
 x  3x  4 
x  8
pq25
90. f  x   10 x   x  1 x  2  x  3 x   
f 12   f  8 
 f 12   f  8   19840  1
19840

11-01-24_SR.STAR CO-SUPER CHAINA(MODEL-A,B&C)_JEE-MAIN_GTM-13(N)_SYLLABUS

MISTAKES
MATHS
PHYISCS
CHEMISTRY
Narayana IIT Academy 11-01-24_SR.IIT_*CO-SC(MODEL-A,B&C)_JEE-MAIN_GTM-13(N)_Q’P
SECTION – I
1. In a new system of units, if energy (E), velocity (V) and time (T) are chosen as
fundamental quantities, then dimensional formula of force is  E V  T   . The value of
     is
1)2 2)-2 3)0 4)1

2. A bar magnet is held perpendicular to a uniform magnetic field. If the couple acting on
the magnet is to be halved by rotating it, then the angle by which it is to be rotated is
1) 300 2) 450 3) 600 4) 900
3. A block is moving on horizontal rough surface. It has velocity v0 i at x  0.1 m. The
1
coefficient of friction varies with distance (x) from origin as   . The minimum
2x 2
value of v0 so that it never stops is
1) 5 m/s 2) 7 m/s 3) 8.5 m/s 4) 10 m/s
4. A disc (m, R) is doing pure rolling motion on a horizontal surface. Radius of curvature
of trajectory followed by point A at topmost point is
A
1) R 2) 2R 3) 4R 4) 2 2R
5. A pendulum made of a uniform wire of cross-sectional area A has time period T. When
an additional mass M is added to its bob, the time period changes to TM . If the Young’s
1
modulus of the material of the wire is Y, then is equal to (g = gravitational
Y
acceleration)
 TM  2  Mg   TM  2  A   T 2  A  T  2  A
1)    1 2) 1     3) 1     4)  M   1
 T   A   T   Mg   TM   Mg  T   Mg

6. An ideal monatomic gas undergoes isothermal expansion from state A to B. Work done
by the gas from A  B is double of work done by the gas from A  C . Pressure at point
C is
P0 P0 2 P0 2 P0
1) 2) 3) 4)
3 3 3 3
7. One mole of an ideal gas undergoes a process given by, T  T0  V 2 where T is the
temperature and V is the volume of gas. Volume of gas when its pressure is least is
[Where T0 and  are positive constants].
2T0 T0
1) T0 / 2 2) T0 /  3) 4) 2
 
8. Two particle A and B are performing SHM with amplitude A0, and time period T about
A0
the same mean position. At t = 0, A is at mean position and B is at distance from
2
mean position and is going towards mean position. At what time they will be at
maximum separation? (At t = 0, direction of velocities of A and B are same)
T T T T
1) 2) 3) 4)
12 8 30 24
9. Electric field at point ‘P’ due to long rod having uniform charge density,  as shown is
   
1) 2) 3) 4)
4 0 R 2 2 0 R 2 0 R 2 0 R

10. For the circuit shown in figure, dielectric slab of dielectric constant k = 2 is inserted in
space between the plates of capacitor 1. As the dielectric is inserted
1) Potential across capacitor 1 increases
2) Charge on capacitor 3 decreases
11. For the circuit shown in figure, value of resistance R is adjusted so that power delivered
to resistor, R is maximum and is equal to P0. Value of P0 is
1) 6.0W 2) 3.0W 3) 9.0W 4) 4.5W

12. STATEMENT-1 : Inductance plays same role in the electrical circuits as mass plays in
the mechanical circuits.
STATEMENT-2 : Greater the value of inductance, harder it is to change the current in
the circuit.
(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
(3) Statement-1 is True, Statement-2 is False
(4) Statement- 1 is False, Statement- 2 is True

13. For the RC circuit shown in figure, switch 'S' is closed at time t = 0. Current I at
2 RC
t ln 3 , is
3
V0 2V0 2V0 V0
1) 2) 3) 4)
9R 9R R 4R
14. A sinusoidally varying source voltage is given as a function of time as shown. RMS
value of voltage is
59 7
1) V 2) V 3) 4V 4) 6V
2 2
15. Magnetic field associated with electromagnetic wave whose electric field is given by

E  2.1sin  3  108 t  1.8Z  i N / C , is
^
 
1) B  1.26 108 sin  3 108 t  1.8Z  j T 2) B  1.8 108 sin  3 108 t  1.8Z  j T
^ ^
 
3) B  0.7 108 sin  3 108 t  1.8Z  j T 4) B  1.26 108 sin  3 108 t  1.8Z  j T
^ ^
16. A ray of light travelling from glass to air is incident at angle i. Maximum angle of

deviation suffered for any angle of incidence is , then refractive index of glass is
2
1) 2 2) 3 3) 2 4) 4/3

17. In a standard YDSE, the distance between the slits varies with time
 t  as  dt   d0  a0 sin t  . The difference between the largest fringe width and the smallest
fringe width over time is given as (  is wavelength of monochromatic light used, D is
the screen distance from slits. D >>d)
2 Da0 2 Dd0 D  Da0

1) 2) 3) 4)
d02  a02 d02  a02 d0  a0 d02  a02
18. In a hydrogen atom, electron jumps from 4th excited state to 2nd excited state.
Wavelength of photon emitted is [R : Rydberg constant]
225 225 100 100

1) 2) 3) 4)
16R 4R 21R 4R
19. In the circuit with ideal diodes as shown, current (in A) through battery is
1) 3 2) 5 3) 6 4) 4
20. In a Vernier calipers, one main scale division is 1mm and 9main scale divisions are
equal to 10 vernier scale divisions. When nothing is put between jaws of the calipers,
zero of the Vernier scale lies to the right side of zero of the main scale and the 2nd
division of the Vernier scale coincides with a main scale division. While measuring
inner diameter of a hollow cylinder the zero of Vernier scale lies between 1.7cm and
1.8cm of the main scale. Also, 8th division of Vernier scale coincides with a main scale
division, inner diameter of the cylinder is
1) 17.2mm 2) 17.4mm 3) 17.6mm 4) 17.8mm

SECTION-II
21. Pitch of a screw gauge is 1mm and its cap is divided into 100 divisions. When nothing is
placed between studs of the screw gauge, zero of circular scale is 8 divisions above the
reference line and zero of the main scale is not visible. Now, when a cylindrical wire is
placed between its studs the main scale reading is 2 divisions and 15th division of
circular scale coincides with reference line. Diameter of the wire is x mm. Find the value
of 100 x.
22. If coefficient of friction between all the surfaces is 0.50, then force, F (in N) required to
move the block of mass 4 kg is [g = 10 m/s2]
23. A block of mass 'm' placed on smooth horizontal surface is acted upon by a horizontal
force as shown, delivering constant power 'P'. If velocity of block changes from v0 to
qmv02
2v0, then time taken is ,find the value of q x n is
nP
r
24. Mass density of a disc is given by    0 , where  0 is constant, r is distance from
R
centre and R is radius of disc. Moment of inertia of disc about an axis passing through
n 0 R 4
centre and perpendicular to plane of disc is . Find the value of n  m is
m
25. The minimum and maximum distances of a planet from sun, revolving around the sun,
are x0 and 2x0 . If the maximum speed is v0 , then minimum acceleration of planet during
nv02
motion is . Find n  m
mx0

26. The height of liquid column raised in a capillary tube of certain radius when dipped in
liquid A vertically is 5cm. If the tube is dipped in a similar manner in another liquid B of
surface tension and density double the values of liquid A, the height of liquid column
raised in liquid B is xcm, then find the value of x _____
27. In an organ pipe successive resonance are obtained at 250 Hz, 350 Hz and 450 Hz. If the
speed of sound is 300 m/s, then length of organ pipe (in m) is x, then the value of 10x is
_____. (ignore end correction)
28. When photon of energy 4.0eV strikes the surface of a metal A, the ejected
photoelectrons have maximum kinetic energy TAeV and de-Broglie wave length A . The
maximum kinetic energy of photoelectrons liberated from another metal B by photon of
energy 4.50eV is TB  TA  1.5 eV . If the de-Broglie wave length of these photoelectrons
B  2A , then the work function of metal B is __ eV.
29. If at an instant the value of i is 1 A and it is increasing at the rate of 1 A / s then Va  Vb at

that instant will be______volt
30. Two lighter nuclei combine to from a comparatively heavier nucleus by the relation
given blow:
2
1 X  21 X  42Y
The binding energies per nucleon for 21 X and 42Y are 1.1 MeV and 7.6 MeV respectively. The
energy released in the process is ______ MeV

SECTION – I
31. Titanium oxide  TiO2  is heated in stream of hydrogen to give water and new oxide
Ti x O y . If 1.6gm TiO 2 produces 1.44 gm Ti x O y , (atomic mass Ti = 48, O = 16), The sum
of ' x '& ' y ' is
1) 3 2) 5 3) 7 4) 8
32. The standard reduction potential for Zn+2/Zn; Ni+2/Ni and Fe+2/Fe are –0.76V, –0.23V, –
0.44V respectively. The reaction X + Y+2  X+2 + Y will be non-spontaneous when :
X Y
(I) Ni Fe
(II) Ni Zn
(III) Fe Zn
(IV) Zn Ni
1) I, II, IV 2) I, II, III 3) II, III, IV 4) I,III,IV
33. Consider the following chemical reaction and the corresponding kinetic data showing
the initial reaction rate as a function of the initial concentrations of the reactants:
H3AsO4(aq) + 2H3O+(aq) + 3I-(aq)  HAsO2(aq) + I3-(aq) + 4H2O(liq)
Initial Rate  10–5

[H3AsO4] [H3O+] [I–]
(M/sec)
3.7 0.001 0.01 0.10
7.4 0.001 0.01 0.20
7.4 0.002 0.01 0.10
3.7 0.002 0.005 0.20
Using the data, establish the correct reaction composite order.
1) 1 2) 2 3) 3 4) 4

34.. A 0.4m aqueous solution of Na x A has freezing point 3.720 C . If K f  H 2 O  is 1.86K kg
mol 1 . The value of ‘X” is (salt is 100% ionized).
1) 2 2) 3 3) 4 4) 6
35. Which of the following below electronic configuration of lanthanides is related to the
formation of stable +2 oxidation state.
1)  Xe 4 f 7 ,5d 1 ,6s 2 2)  Xe 4 f 14 ,5d 1 ,6s 2 3)  Xe 4 f 7 , 6s2 4)  Xe 4 f 1 ,5d 1,6s 2
36. Co  NH 3 4  NO2 2  Cl exhibits:
1) linkage isomerism, geometrical isomerism and optical isomerism
2) linkage isomerism, ionization isomerism and optical isomerism
3) linkage isomerism, ionization isomerism and geometrical isomerism
4) ionization isomerism, geometrical isomerism and optical isomerism
37. The enthalpy of combustion of propane (C3H8) gas in terms of given data is :
Bond energy (kJ/mol)
 C—H  O=O  C=O  O—H  C—C
+x1 +x2 +x3 +x4 +x5
Resonance energy of CO2 is –z kJ/mol and  Hvaporization [H2O(l)] is y kJ/mol.
1) 8x1 + 2x5 + 5x2 – 6x3 – 8x4 – 4y – 3z
2) 6x1 + x5 + 5x2 – 3x3 – 4x4 – 4y – 3z
3) 8x1 + 2x5 + 5x2 – 6x3 – 8x4 – y – z
4) 8x1 + x5 + 5x2 – 6x3 – 8x4 – 4y + 3z
N0
38. atoms of X (g) are converted into X+ (g) by absorbing E1 energy. 2N0 atoms of X (g)
2
are converted into X–(g) by releasing E2 energy. Calculate ionisation enthalpy and
electron gain enthalpy of X(g) per atom.
2E1 E2 E2 2E1
1) I.E. = ,  egH = – 2) I.E. = – ,  egH =
N0 2N0 2N0 N0
E1 E2 N0
3) I.E. = ,  egH = – 4) I.E. = ,  egH = – 2N0
2N0 2N0 2E1 E2

39. For an octahedral complex, which of the following d-electron configuration will give
maximum CFSE?
1) high spin, d 6 2) low spin, d 5 3) low spin, d 4 4) high spin, d 7
40. In ICl2 , ICl2 and ICl4 sum of the bond pairs and lone pairs on eachiodine atom in the
given ionic species are
1) 2,2 and 4 2) 2,3 and 2 3) 4,5 and 4 4) 4,5 and 6
41. 0.2 g of an organic compound was analysed by kjeldahl’s method. Ammonia evolved
was absorbed in 60mL N / 5 H 2 SO4 . Unused acid required 40 mL of N/10 NaOH for
complete neutralisation. Find the percentage of nitrogen in the compound.
1) 70 % 2) 56 % 3) 46 % 4) 66 %
42. The correct order of increasing C  O bond length of CO, CO32 , CO2 is
1) CO32  CO2  CO 2) CO2  CO32  CO
3) CO  CO32  CO2 4) CO  CO2  CO32
43. The major product in the following reaction.
1) 2)
3) 4)

44. Which of the following can give immediate turbidity on treatment with Lucas Reagent?
1) 2) CH3  CH 2  CH 2  OH
3) CH 2  CH  CH  OH 4)
|
CH3
45.
Product (P) and (Q) respectively
1) 2)
3) 4)
46.
1) 2)
3) 4)

47.
A, C, D are
CH2CH2CH3 OH CH3 CH CH3 OH
A) , ,CH3COCH3 B) , ,CH3COCH3
CH3 CH CH3 COCH3 CH3 CH CH3 OH

OH
C) ,CH3OH D) , , CH3
, CH CH3
48. Which is the product formed when cyclohexanone undergoes aldol condensation
followed by heating?
1) 2) 3) 4) O O
49. Which of the following reactions will not give propane?
50. The IUPAC name of the following compound is:
1) 3- ethyl – 4 – methylhex – 4 – ene 2) 4,4 – diethyl – 3 – methylbut – 2 – ene

3) 4-methyl – 3 – ethylhex -4-ene 4) 4-ethyl-3-methylhex-2-ene

SECTION-II
51. The angular velocity of an electron occupying the second Bohr orbit of He+ ion is
2.067 × 10x rad/sec, what is x___________
CH3
| Br2 /hv
52. CH3  C  CH 2  CH3   isomeric monobromo compound ‘X’ (major)
|
CH3
The position of the bromine atom in the major product ‘X’ is___ (as per IUPAC
Nomenclature)
53. Number of dipeptides possible using alanine, glycine and tyrosine is ‘X’ then what is the
value of 5x?
54. Sum of basicity of H3PO4, H3PO3 and H3PO2 is equal to
55. A buffer solution is formed by mixing 100 mL 0.01 M CH3COOH with 200 mL 0.02
MCH3COONa. If this buffer solution is made to 1.0 L by adding 700 mL of water, pH
will change by a factor of ‘x’ then what is the value of x+12?
56. Given the half-cell reactions:
M   aq.  e   M  s  ; E 0  0.52V
M 2   aq.  2e   M  s  ; E 0  0.34V
If equilibrium constant of the reaction
2 M   aq.  M  s   M 2   aq.
2.303RT
Is ‘K’ find value of log10 K  ?(Given that  0.06 )
F

57. The osmotic pressure of a solution in atm obtained on mixing each 50 mL of 1.2 % w/v
urea solution and 2.4 % w/v glucose solution at 300 K is:(R = 0.08 litre atm K 1mol 1 )
58. Consider the following compounds and count number of compounds which can produce
tribromo derivatives on reaction with Br2 / H 2O .
(i) (ii) (iii) (iv)
(v) (vi) (vii) (viii)
OH
COOH
59. The time for half life period of a certain reaction A  products is 1 hour when the initial
concentration of the reactant ‘A’ is 2.0 mol L1 . What time does it take for its
concentration to come from 0.50 to 0.25 mol L1 is X  10 2 h x is, if it is a zero order
reaction?
60. Total number of bonds in PCl5 which are at 900 to each other, is:

SECTION – I
61. Statement-1: The point Z lies on the circle Z  2  i  1 . The point Z0 on the circle with
4
maximum argument is given by Z 0  2  cos   i sin   where tan   and
5
Statement-2: Point Z on the circle Z  2  i  1 nearest to the origin has modulus  5 1
1) Statement 1 is true, statement 2 is true, statement 2 is a correct explanation of
statement 1
2) Statement 1 is true, statement 2 is true, statement 2 is not correct explanation of

statement 1
3) Statement 1 is true, statement 2 is false
4) Statement 1 is false, statement 2 is true.
 bc b 2  bc c 2  bc
62. If a  cot 80 0 , b  cot 600 and c  cot 40 0 , then the value of a 2  ac ac c 2  ac
a 2  ab b 2  ab ab
is equal to
1)2 2) 1 3)4 4)3
63. f  x  is a cubic polynomial x3  ax 2  bx  c such that f  x   0 has three distinct integral
roots and f  g  x    0 does not have real roots, where g  x   x 2  2 x  5 then the minimum
value of a  b  c is
1)504 2) 532 3)719 4)764
64. In how many ways can 17 persons depart from railway station in 2 cars and 3 autos
given that 2 particular persons depart by the same car (4 persons can sit in a car and 3
persons can sit in an auto) is
15! 16! 17! 15!

1) 2) 3) 4)
2!4! 3!  2! 4! 3! 2!4! 3! 4! 3!
3 2 3 3 3

65. In a test an examinee either guess or copies or knows the answer to a multiple choice
question with four choices of which only one option is correct. The probability that he
makes a guess is 1/3 and the probability that he copies the answer is 1/6, the probability
that his answer is correct given that he copied is 1/8, then the probability that he know
the answer to the question given that he correctly answered it is
1)24/29 2)23/29 3)22/29 4)21/29
66. A random variable X has the probability distribution

X: 1 2 3 4 5 6 7 8
P  X  :0.15 0.23 0.12 0.10 0.20 0.08 0.07 0.05
For the events E   X is a prime number and F   X  4 then probability of P  E  F 
1)0.87 2)0.77 3)0.35 4)0.50
67. If g  x   max y 2  xy , 0  y  1 then minimum value of g  x  for real x is a  2 b (a, b are
natural numbers) then a + b = ______
1) 5 2) 11 3) 24 4) 2
68. A lamp of negligible height is placed on the ground m1 mt away from a wall. A man m2
m1
mt tall is walking at a speed of mt / sec from the lamp to the nearest point on the wall.
10
When he is midway between the lamp and the wall, the rate of change in the length of
his shadow on the wall is
5m2 2m2 m2 m2

1) m/s 2) m/s 3) m/s 4) m/s
2 5 2 5
  2x  
  1  2cos  k 
69. If f ( x)     3   , then number of points where
k 1  3 
 
 
 xf ( x)  xf ( x )  ( x  1) x 2  3 x  2 is non-differentiable in x  (0,3 ) is equal to
(where [.] denotes greatest integer function)
1) 5 2) 6 3) 4 4) 8
70. A flight of stairs has 10 steps. A person can go up the steps one at a time, two at a time
or any combination of 1s and 2s. The total number of ways in which the person can go
up the stairs is
1) 75 2) 79 5) 85 4) 89
   
71. Let k , l , m, n are four distinct unit vectors in a three-dimensional space such that
          1   A
k . l  l . m  m.k  n.l  n.m  . If the value of k . n can be expressed as , where A, B
11 B
are coprime positive integers, then the unit digit of the value of (A+B) is ________
1) 0 2) 4 5) 6 4) 8
1
dx
72. The value of   5  2 x  2 x 1  e
0
2 24 x

is
       
2 2 2 2
1 11  1 1 11  1 1 11  1 1 11  1
1) ln 2) ln 3) ln 4) ln
2 11 10 11 10 2 11 10 11 10
The area bounded by the two branches of curve  y  x   x3 and the straight line x  1 is
2
73.
1) 1/5 sq.unit 2) 3/5 sq.unit 3) 4/5 sq.unit 4) 8/5 sq.unit
74. Statement-1: Period of f  x   sin 3x cos 3x   cos 3x sin 3x where [ ] denotes the greatest
2
integer function, is
3
Statement-2: Period of  x where {} denotes the fractional part of x, is 1
1) Statement-1is True, Statement-2 is True; Statement-2 is a correct explanation

forStatement-1
2) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation

for Statement-1
3) Statement-1 is True, Statement-2 is False
4) Statement-1 is False, Statement-2 is True

101 101
1
75. Let a1 , a2 , a3 ,....., a101 are in GP with a51  25 and  ai  125 , then the value of   a  equals
i 1 i 1  i 
to
1)5 2)1/5 3)1/25 4)1/125
n log3 8
 1   1 
76. The last term in the binomial expansion of  3 2   is  3  then the 5th term from
 2 3 9 
the beginning is
1 1
1) 10 C6 2) 2.10 C4 3) .10 C4 4) .10 C4
2 3
77. The mean and variance of 7 observations are 8 and 16 respectively. If five observations
out of them are 2, 4, 10, 12, 14 then the product of the two observations is
1)48 2)49 3)36 4)24
 
78. If  a  b  sin       a  b  sin     and a tan  b tan  c then the value of sin  is equal to
2 2
2ab 2bc 2bc 2ab
1) 2) 3) 4)
a  b2  c 2
2
a  b2  c 2
2
a  b2  c 2
2
a  b2  c 2
2
1   tan x  sin x    tan x  sin x    tan x  sin x   ...

79. The value of lim 
x 0
1  x3  x 3  x3  ...
1)1 2)1/2 3)1/4 4)-1

 x2 y2   x y 
80. sin 1     cos 1    2  is equal to
 4 9  2 2 3 2 
  3
1) 2)  3) 4)
2 2 2
SECTION-II
81. Let z1  10  6i and z2  4  6i . If z is any complex number such that the argument of

 z  z1  /  z  z2  is and if z  7  9i is k then k2=
4
82. The value of tan 780 tan 420  tan120 tan 480 

83. Consider a polynomial p  x  of the least degree that has a maximum equal to 6 at x=1
and a minimum equal to 2 at x=3 then the value of p  2   p '  0   7 is
84. A line with direction ratios (2,1,2) intersects the lines r   j    i  j  k  and
 
r  i   2i  j  k at A and B respectively then length of AB is
x 3 y 8 z 3 x3 y7 z 6
85. If the shortest distance between the lines   and   is
3 1 1 3 2 4
 30 units then the value of  is
If 1   x   1  8 x  24 x 2  ..... and a line through P  , n  cuts the circle x2  y 2  4 in A and B

n
86.
then PA.PB 
87. If a variable line xcos  y sin   p  is parameter  which is a chord of the hyperbola
x2 y 2
  1 subtends a right angle at the centre of the hyperbola and always touches a
16 25
circle of radius ‘r’ then ‘3r’ is
f ' x g  x  g ' x f  x  f  x  g  x 
88. Let   f  x  g  x  f  x g  x  g 2  x
dx = m tan 1 
 n g  x  
  c where m, n  N and c

is constant of integration then m2  n2 
89. If P is the number of discontinuity points of f  x    x    x  2 where  is the G.I.F
 3x 4  2 
and q is the limiting value of lt   then p  q  2 is
x 
 x 8
 3 x  4 
90. If f  x  is a polynomial of degree 4 with leading coefficient “1” satisfying
 f 12   f  8  
f 1  10, f  2   20 and f  3  30 then   is
 19840 

KEY SHEET
PHYSICS
1 C 2 A 3 B 4 C 5 A
6 C 7 C 8 D 9 B 10 D
11 D 12 A 13 A 14 A 15 B
16 D 17 D 18 C 19 A 20 B
21 28 22 25 23 9 24 27 25 296
26 160 27 5 28 3 29 20 30 110
CHEMISTRY
31 C 32 C 33 C 34 B 35 B
36 B 37 A 38 D 39 B 40 A
41 B 42 C 43 A 44 A 45 D
46 D 47 A 48 B 49 A 50 B
51 2 52 7 53 50 54 4 55 73
56 62 57 7 58 8 59 2 60 5
MATHEMATICS
61 C 62 C 63 B 64 A 65 C
66 D 67 B 68 C 69 B 70 A
71 C 72 A 73 C 74 D 75 B
76 D 77 D 78 D 79 B 80 D
81 21 82 351 83 11 84 3 85 0
86 37 87 25 88 2 89 –3 90 16
SOLUTIONS
PHYSICS

v p0 

1. Let OP = r. Angular speed about the origin =  =  , where vp0 = The component of
r 
velocity of P w.r.t. O perpendicular to OP.

v sin
  where r = b cosec 
r
v sin2

b
2. Time taken in moving distance S along smooth inclined surface with inclination  is 2S
ts 
g sin 
And time taken in moving distance S along smooth inclined surface with inclination
 is t  2S
 n ts
g  sin    cos  
r
Using the expression obtained above

1 1  1
 1   cot  cot  = 1 - 2  =  1   tanθ .
 n  n2 
3. The velocity of both the bodies m & M are equal. If the block M sticks to the wall, the block m
will continue to move which compresses the spring through x. The K.E. of the block m will be
converted into the potential energy of the spring as it compresses the spring.
Conservation of energy yields
1 2 1 2 m
mv  kx  x  v
2 2 k
1
where  M  m  v 2  E
2
2E 2mE
 v  x 
 M  m  M  m k
4. by conservation of linear momentum along the line of collision (along the line joining the centers of
two spheres), m1u sin   m2 v
v2  v1
Since, Coefficient of restitution for oblique collision, e  
u2  u1 along the line of collision
v m 2
e  1 
u sin  m2 3
5.
or …(i)
Here,
Substituting in Eq. (i), we have

6. The gravitational potential at a point Q (OQ  x) is given by :
  3 1 x2  
 s  
 g R 2 
, when x <R 
V(x)  
 2 2 R  

 R 
 g s R   , when x >R 
 x 
The energy required to project the body, to a maximum altitude of 3R from its surface, is :
  9
m  VB R  VP   mg s R .
 x  8
 2 x  4R 
2T cos  hr  g h
7. h T  or T 
r g 2 cos  cos 
TW h cos  2 1
 1 
THg h2 cos 1  2
Putting the values, we obtain 1 : 6.5
8.
.
9. When two gases are mixed together then
Heat lost by the Helium gas = Heat gained by the Nitrogen gas
Box A Box B
1 mole N 2 1 mole He
Temperature = T 0 Temperature = 7 T 0
3
By solving we get
2
10. Potential energy of the particle U  k (1  e  x )
Force on particle F  dU

2
 k [e  x  (2 x )]
dx
2  x4 
F   2kxe  x  2kx 1  x 2   ...... 
 2 ! 
For small displacement F  2kx
 F   x i.e. motion is simple harmonic motion
11.   x, y, z   ax 2

E  2ax i
 

  2aLi L i  2aL
E.ds 2

3

s
x  L  
    
E.ds   2a(L  i  . L2 i  2aL3
s
x  L 
 E.ds = 0

each of the other faces
  q
  E.ds  4aL3 
0
 q  4a 0 L3
12.
Solving, we get
Charge flown through
.
13. Here, both Assertion and Reason are correct, and reason is the correct explanation of assertion.
14. .
15.
i i
C C
L 2L 2L L
16. Z  (R ) 2  ( X L  X C ) 2 ;
R  10 , X L  L  2000  5  10 3  10 
1 1
XC    10  i.e . Z  10 
C
2000  50  10 6
V
Maximum current i0  0  20  2 A
Z 10
2
Hence irms   1 .4 A
2
and Vrms  4  1 .41  5.64 V
17. The equation of electric field occurring in -direction
Therefore, for the magnetic field in -direction

D
18.    for  max , ' d ' will be min  2d 0  d 0  d 0
d
And for  min , ' d ' will be max  2d 0  d 0  3d 0
D D
  max  and  min 
d0 3d 0
D  1   2 D 
  max   min  1     
d0  3   3d 0 
19. Conceptual
20. Young’s modulus
Further,
Or
21. and R=ma  T=4ma
.
Hence acceleration of B w.r.t. ground is 2 2 m/s2.
22.
 M  L   M 4L 
( M  0)       
 4 3   4 6  L
x 
M 4
23. Let the frictional force be in the forward direction, then
and
and
For pure rolling,

For pure rolling, 𝑓𝐿=𝜇𝑀g
𝜇𝑀g⇒𝑎 3𝜇g amax  9m / s 2
24. Slope of line
i.e., (i)
Similarly, for line
(ii)
Dividing Eq.(i) by Eq.(ii),
(iii)
From Eq.(iii),
25. As the tension in string Y is increased hence it’s frequency will increase. But as given, beat
frequency is decreased so , in the beginning nX  nY  4  300  nY  4  nY  296 Hz
26. The given circuit can be redrawn as follows
3
X
i
10
20 30 60
A
i
6
24 8 48 V
B
Y
1 i
24  8
Resistance between A and B  6
32
48
Current between A and B = Current between X and Y  i   8A
6
Resistance between X and Y  (3  10  6  1)  20 
 Potential difference between X and Y = 8  20 = 160 V.
1 1 1 1 (1  )  2  (1  4)  2  31  4
27. = +  = R  = R  =
f eq f1 f2 f1    4   2R
   2   
=   2   =  3 2 
1
f2    R   2R 
1 3(1   2 ) 2R R 20 5
= 1 – 2 = = = =
24 2R 24  3 12  3 36 9
28. As the current depends on the number of photons incident. Now by inverse square law,
or .
29. Current in R1=5/500=10mA and current in R2=10/1500=20/3 mA , hence the current in Zener
Diode=10/3 mA
30. Since there is no shunt resistance , I  9  60 A
6 1000
  540  106  G 
11000  G 9
RS 1000 11000 S
Since for half deflection , G     S  110
RS 9 11000  S

CHEMISTRY
1
1 hc hc  2hc   0     2
31. mv 2max    v max   
2  0  m   0 
32. TI3 is T  I3
CSI3 is Cs  I3
Thallium shows T  state due to inert pair effect.
33. Conceptual
5 / 60
34. f  K f .m  0.2  1.86 
m Solvent
 M Solvent  0.775Kg
 Mass of water separate as ice = 1  0.775  0.225Kg
35. FeCl3  NH 4 OH  Fe  OH 3  excess  insoluble
NH OH 4
 Brown ppt.
36. A   2B  C
t=0 P0 0 0
t=t P0  x 2x x
t  0 2P0 P0
From question : P  2P0  P0
P
 P0  
3
and P   P0  x   2x  x
P  P0 3P  P
x 
2 6
1 P 1 o
P0
Now, K  .In A  .In
t PA t P0  x
1 P / 3
 .In
t P 3P  P

3 6
1 2P
 .In
t 3  P  P 
37. Fe has less positive value of E oM3 / M2 since Fe3  3d 5  is more stable than Fe 2  3d 6  . Also, the order
is Co (1.97) >Mn (1.57) > Fe (0.77).
50. Eu 2 :  Xe 4f 7 ;Ce3 :  Xe  4f 1
51. Conceptual
52. 
Ag 2CO3  s   C2 O 24  Ag 2C 2O 4  s   CO3
2
0.15
Initial  1000 0
500
 0.3M
0.035
Final 0.3  x x  1000
500
 0.23M  0.07 M

K sp  Ag 2 CO 3 
2
 CO3   Ag 2 
Now, K eq    
K sp  Ag 2 C 2 O 4 
2
C 2 O 4   Ag  
2
0.07 K sp  Ag 2 CO3 
Or, 
0.23 2.3  10 11
 K sp  Ag 2CO3   7  1012
53. M.eq. of K 2 Cr2 O 7  M. eq. of FeC2 O 4
FeC 2 O 4  Cr2 O 72  
 Fe3  CO 2  Cr 3
0.288
V  0.02  6   3  1000
144
V = 50 mL.
54.  o eq NH OH    o eq NH Cl   o eq NaOH   o eq NaCl
4 4
129.8  217.4  108.9  238.3ohm 1cm 2 mol 1

 9.532
Now,   eqo   0.04  4%
 eq 238.3
56. Given n = 3
T1  300;T2  1000
C p  23  0.01T
The relation between H and C p is
T2
H   nC p dT ….. (i)
T1
After putting all variable values in eq. (i)

1000
H 
300
  23  0.01T  dT
1000
 0.01T 2 
 3  23T 
 2  300
= 3 [16100+ 4550] = 3  20650 = 61950 J
= 61.95 kJ
= 62 kJ
60. O 2 ;O 2 ; O 2 ; N 2 ; N 22 
MATHS
61. Centre of the circle z  2  2 i.e., 2 lie on z 1  i   z 1  i   4 ,
Hence given line z 1  i   z 1  i   4 pass through the centre of circle i.e., intersect at two points.
 Number of solutions = 2.
62. A  A2  A 3  ....... and B  B 2  B 3  ..........
A 
2020
 A  B 
2019 2020
 B 2019
 A  B 3  2  A  B  A  B   22  A  B 
 A  B 2020  22019  A  B 

2 3 4
1 1 1 1
63. p  q    r    s   t  
x x x x
3 1 3
1 1 1
x x x
1 2 3
 1 2 1 1 
x x x
3 4
1 2 1 3
x x
as x   , we get
1 1 1
p  1 1 1  4 .
1 1 3
64. 2    2  0 ………1
So,

6  3  2 2    

6 2  2  1 
5
  3  3   4 3 2
 2
 3 2
 3  3  1 
6    1
2
6 6
   3
 2
   1 3
   2
2
65.
2  a x 1 b x 1 
lim  
x 0 x  2 
66. e 
 e ln ab  ab  6
a,b   1,6 ,  6,1 ,  2,3 3,2
4 1
 P E   
36 9
x
1
67. Let P(X = x) =    , x  0
5
We have
4

5
68. a 1  7
a 8
"
69. f x  0
 f ' isinc.fn
To find : where g is nec. Inc
g is inc  g '  0

1
 
2 1 2
 
 .f ' 2x  1  4x   f 1  x  2x   0
4 2
 x f '  2x  1  f ' 1  x   0
2 2
Case 1 : x  0  1 f '  2x  1  f ' 1  x 

2 2
2 2
 2x  1  1  x
 2  2 
 x   ,     ,     2
 3   3 
 2 
1   2  x   ,   ……….(3)
 3 

Case II : x  0   3 f ' 2x  1  f ' 1  x
2
  2

2 2
 2x  1  1  x
 2 2
 x   ,    4
 3 3
 2 
 3   4     ,0    6 
 3 
 g is inc in x   5    6 
 2   2 
 x    , 0    ,  
 3   3 
    
70. Let the angle between a and b is  /and a  b and c is 
a  b  .c  6  sin  cos   1
  
So  a , b , c are mutually perpendicular
 2 2
Req  a  c d  9
71. Let L1 , L2 , L3 be the mutually perpendicular lines and P  x0 , y0 , z0  be their point of concurrence. If
L1 cuts the x-axis at A(a, 0, 0), L2 meets the y-axis at B(0, b, 0) and C(0, 0, c)  L3 , then
x0  x0  a   y0  y0  b   z02  0

x02   y0  b  y0  z0  z0  c   0 
x0  x0  a   y02  z0  z0  c   0
Eliminating a and b from the equations, we get
x02  y02  z02  2cz0  0
72. t1t2  1
25
 t1  t2 
2

4
1
t1  , t2  2
2

73.
74. We have,
f(x) = max{x2, (1 – x2), 2x(1 – x)}
 1  x 2 , for 0  x 1/ 3

f(x) = 2x 1  x  , for 1/ 3  x  2 / 3
 2
 x , for 2 / 3  x 1
Hence the area bounded by the curve y = f(x); x–axis and the lines x = 0 and x = 1 is given by
1 2
3 3 1
 1  x  dx   2x 1  x  dx    x dx
2
= 2
0 1 2
3 3
17
= sq. unit.
27
A 17
   A = 34 Sq. unit.
54 27

75.
76.

77.
  3
 
2
78. LHS = 3 sin 550  3 sin 50  3 cos 2 250  1  cos 500
2
3
a  b 
2
79. f  0    2; g  2    3
80.
adjA
81. B=adj(A)  B  A  B  A A 1
A
1
B  3A 1, B 1  A
3
f x  
3

1 3
x  6 x 2  9x  9 
f '  x   x 2  4x  3
1
Global maximum at x = 6   72  72  54  9   21
3
82. A  1, 2,3, 4,5,5, 7
Case I: All elements of set A satisfy f  x   x
In this case number of functions =1
Case II: 4 elements of set A satisfy f  x 
Total number of functions 7 C4 .2  7

e.g ,: f  4   4, f  5  5, f  6   6, f  7   7
Now for elements 1, 2,3 we have two options mapping
Case III: 1 elements of set A satisfy f  x   x
For remaining six elements make groups of  3,3
6!
Hence, total functions 7 C1   2  2  280
3!.3!.2!
Hence, total functions are 351.
83.    
L1 : r  2i  6 j  34k  t 2i  3 j  10k , t  R
L : r   6i  7 j  7k     4i  3 j  2k  ,   R
2
P is a point on L1 and Q is a point on L2

PQ.L1  0
PQ.L2  0
3x 2dx
1
1
84. I2   , put t  x 3
 
x 3 3
3 0 e 2x
1
1 dt
 
3 0 e 2  t 
t
1
1 dt I
 
3 0 e 2  t 
t  1  3e
I2
PA 2 PB 2 PC 2
85.  ;  ; 
QA 3 QB 3 QC 3
 12 12  12
 A, B, C lies on circle with diameter ends  ,  ;  12, 12  , this circle radius is 13
 5 5 5
4
Required circle radius is 13
5
86.
87.

88.
89. Let Y : y1 , y2 , y3 ,..., yn , W : w1 , w 2 , w 3 ,..., w n

Given, W  Y  k
Use  2  W    2 2  Y  ,   W     Y   k
90. f '  x   3x 2
1 1
f '  8   192 and g '  8   
f '  2 12
 f ' 8  f   '  8   192  12

1 1
 16

KEY SHEET
PHYSICS
1 B 2 B 3 A 4 A 5 B
6 B 7 B 8 B 9 B 10 A
11 A 12 A 13 A 14 A 15 B
16 B 17 C 18 A 19 C 20 B
21 5 22 3 23 15 24 25 25 4
26 2 27 2 28 60 29 4 30 40
CHEMISTRY
31 B 32 A 33 C 34 C 35 A
36 D 37 C 38 D 39 C 40 D
41 B 42 B 43 D 44 A 45 B
46 D 47 A 48 A 49 A 50 C
51 4 52 5 53 5 54 3 55 6
56 5 57 2130 58 4 59 68 60 10
MATHEMATICS
61 B 62 C 63 D 64 C 65 C
66 D 67 C 68 A 69 B 70 B
71 C 72 A 73 B 74 D 75 C
76 C 77 B 78 B 79 D 80 C
81 3 82 97 83 6 84 4 85 1
86 0 87 64 88 84 89 96 90 9
SOLUTIONS
PHYSICS
1. I max  I1  4 I1  2 I1 4 I1  9I1
I min  I1  4 I1  2 I1 4 I1  I1
I1  I1 10 5 2  1
  
9 I1  I1 8 4   1
  2  1
 2
  2
 1
1
2. Total mechanical energy = - (potential energy)
2
[for circular orbits under central forces]
SQ T .M .E.A :T .M .EB
GMm1 GMm2
= : 
2r1 2 r2
= m1r1 : M 2 r2 = (4m) (4r) : (3m) (3r) = 16 : 9
 
3. F AB  F CD
1
4. I   0cE 2
2
 2 L  2M
5. New magnetic moment M '  m  2 R   m  
   
  
2u 50
6. Time of flight =  = 5 sec
g 10

r  f
foot ball
 
 2iˆ  5 ˆj  5 = 10iˆ  25 ˆj

r  f
player
   
 5iˆ  8 ˆj  2iˆ  4 ˆj  6 ˆj  2iˆ  3 ˆj = 9iˆ  21 ˆj
distance = 12  42 = 17
C
7. V2  , Q2  2C V2
C  2C
8. Conceptual
9.
N – mg = 0, f= N
f = ma  a =  g = 4 m / s2
22  02
u=0, =2   u = 2as  s 
2 2
= 0.5 m
2 4

V0 4V0
10. (i) V  t V t
T /4 T
1/ 2
 T /4 2 
4V0  0
 t dt 
 V0
 Vrms = V 2  =  T /4  =
T  3.
dt 
 0 
11. In equilibrium, mg = Fe
FB  V 0 g and mass = volume x density
4 4
  R 3  r 3  0 g   R3  w g
3 3
 27
Given , relative density, 0 
w 8
  r 3  27
 1     W  W
  R   8
r3 8
 1 
R 3 27
r3 8 r3 8 19
1 3   3  1 
R 27 R 27 27
8
 r = 0.89 R = R.
9
12.   BA   0 ni  A = 0 n  kt e  t  A
d d
e   0 nAk  te t 
dt dt
= - 0 nAk t   a  e  e  t  = - 0 nAk  e  t 1   t  
 t
e  0 nAk  t
i=   e 1  t  
R R 
At t = 0 ,i  - ve
at t = 1 sec ,i  0
13. Surface energy per nucleon,

r 2 A2/3 1 a1
bs  
A A
 1/3  1/3
A A
 r  r0 A1/3 
A is incorrect
Contribution to binding energy by columbic forces is given by
a Z  Z  1
bc  2 1/3
A
B is incorrect
Volume energy, bv  A
C is correct
For (D), if we consider only surface energy contribution then option is correct.
Contribution then option is correct.
For € only 3 interactions contribute to surface energy.
14. Zero error = - (10 – 7) x 0.1 = - 0.3 mm diameter = 77.0 + 8 x 0.1 – (- 0.3) = 78.1 mm

15.
Effectively , 25 cm column of water from top of right vessel entered the left a = mgh
(h is height reduced of the COM)
= (16) (25) 10 3 g (25) x 10 2 = 1J
16. In the case of Vrms  of mixture,
N1  N 2
M mix  ( N1 , N 2 are number of molecules of gaseous 1 and 2) which is Harmonic mean
N1 N 2

M1 M 2
m  m2 n1M 1  n2 M 2
But M mix  1  which is the Arithmetic mean is used in velocity
n1  n2 n1  n2
of sound expression.
17. If the particle covers a further phase of 60C , it will be at the extreme.
  = 360 – 60 = 300 .
18. PdV = nCv dT
nRT
 dV = nCv dT
V
3 dT
=
2 T
 V 2 = CT 3 , where C is a constant.
1 q1 q2
19. F
4 0 r d2
9mR 2  mR 2 4mR 2  9mR 2  mR 2  1  
20. = -   = -    4 
2  18 9  2  9  2 
9mR 2  mR 2  9   9mR 2 mR 2 8mR 2
= -   2   =  = = 4m R 2
2  9  2 2 2
21. In the given circuit we can find the voltage across Zener diode which is reverse
biased
10 1K
Voltage across Zener diode (V) V  = 5 Volt
1K  1K
Thus the voltage across the Zener diode is less than the breakdown voltage
(given as 6V) i.e. V  VZ So, VO  V = 5 Volt
h h
22.  
p 2mE

23. A  4 3iˆ  3 3 ˆj  5kˆ
As incident vector A makes I angle with normal z – axis and refracted vector R makes r
. angel with normal z – axis with help of direction cosine
 
1  Az  1
 5 
i  cos    cos  
 A
   
2 2
 4 3  3 3  52 
 
 5 
cos 1    i  60
 10 
By shell’s law, we have
2 sin 60 = 3 x sin r  r = 45
Difference between i and r = 60 - 45 = 15
dV
24. V 2  100 x ; 2V = 100
dx
dV
V = a = 50 , F = ma
dx
25. Mass of element,
1  R 2 d
dm   R  Rd  dm 
2 2
 /2
R 2
 2R 
 x dm  2
d 
 3
cos  

xcm  = 0  /2
 dm R 2
0 2 d 

 /2
2R  cos  d 2R 2
 0
 /2
=  
3 3  
 d
0
4R
=  x = 4.
3
26. At t = 0 , x  0 , y = 0 , and velocity of particle is positive
 =  rad
27. When disc slides then acceleration , a1 = g sin  and, distance travelled
1 1
S  ut1  a1t12 = g sin  .t12 … (i)
2 2
Again , when disc do pure rolling
g sin  2  1 
a2  = g sin   I disc  mr 2 
I 3  2 
1 2
mr
1 1 2 g sin  2
 S = ut2  a2t22 = . g sin  .t22 = t2 … (ii)
2 2 3 3
From equation. (i) & (ii),
S g sin  t12 g sin  t22
=1= /
S 2 3
t 3
 2 =
t1 2
28. We have li  l f
Mgl YA T
 l  T   M=
YA g
2 1011  3106  2 105  50
 M=
10
 M = 60 kg
2402
29. P
36
2
P1  240
18
2
P2  240
18
2402
Ptotal  P1  P2 
9
P 1

Ptotal 4
30. Using the principal of calorimetry
M ice L f  mice  40  0  Cw  mstream Lv  mstream 100  40  Cw
 M  540   m 1 100  40  = 200  80  200 1 40
 600M = 24000
 M = 40g

CHEMISTRY
31. IP1 of Nitrogen is more than Oxygen
32. Reason: Pb  I bond initially formed during the reaction does not release enough energy to unpair
6S 2 electrons
33. Melting point/K : HI  HF  HBr  HCl
34. Magnetic moment more for d 5 configuration.  d1 ,d 3 ,d 4 ,d 5 
3 2 3 2
I) Fe  3d  3d 4  3d 6  3d 5
5
35. II) Cr III) Co IV) Mo
36. CH 4  0,  NF3   NH 3   H 2O
37.
OH
COOH COOH COOH
NO2 CH3
CN I
IV
II III
Electron with drawing nature: -NO 2 > -CN

38.
39. Formation of (P) allylic substitution; Formation of (Q) electrophillic aromatic substitution
40. Elemination with NaNH 2 , followed by reaction with Hg  OAC 2 / H 2O; NaBH 4
41.
O

OH Br MgBr
  
V
eth
 er

  
OMgBr
P 
Br3
 ether
Mg

W  X
Na2CrO
2 7
O
O H3O
CH3 OCl
O
OH
 Z O C CH3
V Y

42.
43. Formation of A – Nucleophilic substitution

B – EAS, Formation of C = Clemenson’s reduction.
44. 20 amines reacts with Hinsberg’s reagent forms alkali insoluble product
45. Protein is formed by  -amino acid, B is not amino acid
46. x5  y6  x : y  6 :5
47.
E1  1.312  106 J / mol
E1 1.312 106
E2  2
 J mol 1
2 4
E   E2  E1 
1 3  1 3 
48. S  S xy3   S x2  S y2   50    60   40   40 Jk 1mol 1
2 2  2 2 
G  H  T S
H
G  0  T   750 k
S
49. Higher the B.P of solvent, normally higher is its K b value
50.  Ag   C     k sp ,  Ag   0.1  1.7  1010  Ag    1.7  109
2 2
k sp   Ag    cro42  ,1.9  1012   Ag   103  Ag    1.9  109  1.9  103
In AgCl  Ag   is low  so AgCl undergoes ppt.
0.06 PH 2
51. E  E0  log 2 
 Assu min g 2 H   2e   H 2 
n  H 

0.06 1
0.18  0  log 2
2  H  
 H    103 M

C6 H 5 N H 3  H 2O  C6 H 5 NH 2  H 3O 
C  x  M xM
x 103
h   0.04
C 1
40
 h  4%

52.
O
Cl I
NH2 SH H2N NH2
53. 4.9  5
54. RMg X reacts with acidic hydrogens forms alkane
E.g: OH ,  SH , COOH ,  NH 2 , C  CH
55. For each CH 2OH one mole of HCHO is consumed

56. Sulphur containing amino acid =2
Basic amino acid =2
Essential amino acid  6
226
M 5
2
1 492
57. Mole of O2 consumed   20 Mole of NaClO3 required = 20
0.082  300
Mass of NaClO3  20 106.5  2130 gm
58. X   NH 4 3 PO4 .12 MoO3
15th group and SP 3 hybridized atoms in X  4,  3, N  atoms  1 P  atom 
59. TB  TF  105
TB  100    0  TF   5
Tb  TF  5 Kb m  K F m  5
5 5 w  1000
m  2 ; w  68.4 gm
K b  K F 2.5 342  100
60.
SO2 g   NO2 g   SO3  NO g 
t0 1 1 1 1
t  eq 1  x 1 x 1  x 1 x
Kc 
 SO3  NO 
 16 
1 x
2
 SO2  NO2  1  x 
2
3
 x  moles  0.6 mole
5
At equi nSO2  0.4nNO2  0.4 nSO3  1.6 nNO  1.6
0.4
% of nNO2   100  10%
4

MATHS
61. If z  i  z  i = 8,
PF1 + PF2 = 8  z max  4

62. Has non – trivial solutions
  0
p p 1 p 1
  p 1 p p2 0
p 1 p  2 p
R2  R2  R1 ; R3  R3  R1
p p 1 p 1
1
 1 1 3 0  p
2
1 1 1
 Exactly one real value of p.
63. n(S )  100
n( E )  36
36
 PE   36%
100
64. A ' s  4 ; L ' s  2; E ' s  1; K ' s  1; H ' s  1
5!
a) 6 C4
2!
5!
b)
4!
c) AAAAELLKH
5! 5!

2! 4!
4! 5!
d) 
2! 4!
 2n  1 .2n.  2n  1 n  4n  1
2
2 n 1
65. Total ways  C3  
1.2.3 3
Let the three numbers a,b,c are drawn, where a  b  c and given a,b and c are in AP.
 2b  a  c....  i 
It is clear from Eq. (i) that a and c both are odd or both are even
 Favorable ways  n 1 C2  n C2

 n  1 n  n  n  1  n 2
1.2 1.2
n2 3n
 Required probability  
n  4n  1  4n 2  1
2
3
 Statement -2 is false

In statement 1, 2n  1  21
 n  10
3  10 30 10
 Required probability   
4 10   1
2
399 133
 Statement -1 is true.
66. Let f ( x )   ( x  1)( x  1)   ( x 2  1)
Integrating on both sides, we get
 x3 
f ( x)     x   c
 3 
 2
Now, f (1)  1  1  c …… (1)
3
2
Similarly, f (1)  3  3  c ….. (2)
3
 From (1) and (2) , we get
2c = 2  c = 1 and  = 3
 x3 
 f ( x)  3   x   1  x 3  3x  1
 3 
dS dr 1
67. S  4 r 2 ; 8 
dt dt  r
4 dv dr dv
Now, v   r 3   4 2  4r   r
3 dt dt dt
68. We have
 
V1.V2  a  b  c  0
But a, b, c  {-2, -1, 0, 1, 2}
Now (i) if a = 1, b = -1, c = 0, number 3!=6
(ii) if a = 2, b = -2, c = 0, number = 3!=6
3!
(iii) if a = 1, b = 1, c = -2, number = 3
2!
3!
(iv) if a = -1, b = -1, c = -2, number = 3
2!
Total = 18
70. g(5) = 1 f(1) = 5
f  1
g   5   
 
 f  1 
3
x  x16  x8  2 x16  3x8  6  dx

24 1/8
71.
Put 2 x 24  3 x16  6 x8  t
   54 ,   9 ,   8

72. Differentiate with respect to x.
y  x   y  x   2 x
dy
 y  2x
dx
I .F  e x
ye x  2 xe x dx
ye x  2  xe x  e x   C
y  2  x  1  ce  x
At x = 1, y(1) = 1,  c = e
Y = 2(x-1) +e1-x
73. 72 x3  108 x 2  46 x  5  0
Let root be a  d , a, a  d
3 1
3a  a
2 2
5 1
a  a2  d 2   d 
72 3
2
Difference .
3
74. Mean of a, 2a, 3a, ….. 50a
25a  26a 51
Median =  a
2 2
 xM  50  a 
51 51
 2a  a
50 2 2
51 51
+ ..... 25a  a  26a  a  .....
2 2
51  49a 47 a 45a a
 50a  a 2    ....    2500
2  2 2 2 2
(1  3  ....  49)a  25
25
1  49 a  25
2
25x25a=25
A=4
  k      
sin       k  1   
 6 4   6 4 
13
75. 2
    k 
k 1
sin    k  1  .sin   
4 6 4 6 
   13   
 2 cot  cot   
 4  6 4 
2  3 1 .

76. If elements are not repeated, then number of elements in A1  A2  A3 .....  A30 is 30x5
30  5
But each elements is used 10 times, so S   15
10
If elements in B1, B2, ….., Bn are not repeated, then total number of elements is 3n but each elements
3n 3n
is repeated 9 times, so S   15 
9 9
n = 45.
78. Pn  Pn 1   n   n    n 1   n 1 
  n  2  2      n  2   2   
 4  n  2   n  2 
 Pn  Pn 1  4 Pn  2
 P15  P14  P16  P15   16
P13 P14
1
80.  4 tan 2
x   a  b  c  a  b  c 
2
81.
2 2 2 2
x  a  ab  ac  ab  b  bc  ca  cb  c ….. (1)
y   a  b  c 2  a  b 2  c
2
 
2 2 2 2
y  a  ab 2  ac  ba  b  bc 2  ca 2  cb  c ----- (2)
z   a  b  c 2  a  b  c 2
2
  ---- (3)
2 2
 x  y  z 3 a  b  c
2
 2 2 2

(As (1+  +  2 =0)
2 2 2
x  y z
 2 2 2
3
a b c
82. We have f  x   ax3  9 x 2  9 x  3
 f   x   3ax 2  18 x  9  3  ax 2  6 x  3
As f  x  is strictly increasing on R, So a  0 and Disc < 0  36 -12 a  0  a  3
So, a  [3,  )
Hence, number of integral values of a, a  [-5, 100] are 98.
x y z x y 1 z
83. The lines are   ,  
1 1 1 1 1 0
S .D 

C  a . bd  
bd
h   2 k  2 2    2  1
2
84.  
1 1 2
h  1  2 ----- (1)
k  2 1 ---- (2)
Put value of  from (1) and (2)

2
 h 1 
 ; 4  y  1   x  1 .
2
k 1  
 2 
Hence, a = -1, b = 4 and c = 1.
85.
Let ABC be the given equilateral triangle. Then C must lie on the y – axis.
Let C   0, a  , Also, AC = AB. Therefore,
1  a2  2 or 1+a2 = 4 or a  3
 1 
Then, the centroid of ABC is  0, .
 3
But in an equilateral triangle, the circumcenter coincides with the centroid. Therefore, the
 1 
circumcenter is  0, .
 3
Also, Radius of circumcircle = C1B
2
 1  1 2
1  0    0    1  
2

 3 3 3
Therefore, the equation of circumcircle is
2 2
 1   2 
 x  0   y     
2
 3  3
2y 1 4
or x 2  y 2   
3 3 3
2y
or x 2  y 2  1  0
3
1 1 1 1
86.  1  x7  4 dx   1  x 4 7
 dx
0 0
y  f ( x)  (1  x 7 )1/4  y 4  1  x7   x  1  y 4 
1/7
Hence functions are inverse of each other

1 1 1 0
I   f  x  dx   g  y  dy   f  x  dx   xf   x  dx
0 0 0 1

1
Hence I    f  x   xf   x   dx  xf  x ]10  f 1  0

0
Aliter:
b b
As,  f ( x)dx   g  y  dy   bd  ac  where g  f 1

a c
(property of definite intergral)
  = 0.
1 0 1
 1 x   1 x  0   1  x7  7 1  x 4
4 7 7 4 4
So,
0 1 0
87. Clearly required area

0 4
 16  x  dx    x  4
2
 2
dx
4 0
4
 x  4
0 3
 x3 
 16 x   
 3 4 3
0
64 64
 64    64 square units.
3 3
3 r
88. Tr 1 9 Cr a 9 r b r x 2
 r  3 (for term independent of x)

 Term independent at x is equal to
9
Cr a 6 b 3
To get the maximum value of a6b3 use AM  GM
 a2   b 
3   3  1
3  
3  a b 
6 3 3
i.e,.    6 
6  3 
  a 6b 3  1  Answer = 9 C3
max
89. lim x
a x 2
 x4  1  2x2 
x 
x  1/ 2
2
 x4  1  2x
 lim x
a  x4  1  x2 
 2
  
x  1/ 2
 x  x 1  2x 
4
 

 1 
 1 4  4
a
 x 
 lim x 1 1/ 2

x 
 1 
1  1  4   2
 x 
1  1 1
 1 
 lim x1  . 4  ........  1
2 2 x 
 2 x 
 For limit to exist and has value non – zero, we must have   1  4    3
1 1 1
and L   
2 2 2 4 2
1
L2 
32

  3  32  96
L2
90. Derangement
So, D (4) = 9


MISTAKES
MATHS
PHYISCS
CHEMISTRY
SECTION – I
1. Two coherent sources of light interfere. The intensity ratio of two sources is 1: 4 . For this
I max  I min 2  1 
interference pattern if the value of is equal to , then will be
I max  I min  3 
A) 1.5 B) 2 C) 0.5 D) 1
2. Two satellites A and B, having masses in the ratio 4 : 3, are revolving in circular orbits
of radii 3r and 4r respectively around the earth. The ratio of total mechanical energy of
A to B is.
A) 9 : 16 B) 16 : 9 C) 1 : 1 D) 4 : 3
3. An infinite current carrying uniform wire passes through point O and in perpendicular to
the plane containing a current carrying loop ABCD as shown in the figure. Choose the
correct option(s)
A) Net force on the loop is zero
B) Net torque on the loop is zero
C) As seen form ‘ O ' ’, the center of mass of the loop moves towards ‘O’.
D) As seen from ‘ O ' ’, the center of mass of the loop moves away from ‘O’.
4. The amplitude of electric field at a distance r from a point source of light of power P is
(taking 100% efficiency)
P P P P
A) B) C) D)
2 r 2c 0 4 r 2c 0 8 r 2c 0 2 r 2 c 0

5. An iron rod of length L and magnetic moment M is bent in the form of a semicircle.
Now its magnetic moment will be
2M M
A) M B) C) D) M 
 
6. In a football game, a player wants to hit a football from the ground to one of his
teammates, who is running on the field. Take hitter position as origin and receiver’s
initial positions as 2iˆ  3 ˆj , where iˆ and ĵ are in the plane of field. Football’s initial
velocity vector is 2iˆ  5 ˆj  25kˆ and in the subsequent run receiver displacement is 5iˆ and
8 ˆj , then 2iˆ  4 ˆj and then 6 ˆj . How far is the receiver from the football when football

lands on ground? (assume g  10kˆ )
A) 10 B) 17 C) 26 D) 13
7. In the given circuit diagram, switch was connected to position 1 for long time. At t = 0,
switch is shifted from position 1 to position 2. Find the final charge on capacitor 2C.
CV CV 2CV 4CV
A) B) C) D)
6 3 3 3
8. STATEMENT – 1: Two teams having a tug of war always pull equally hard on one
another. (Ignore mass of rope)
STATEMENT – 2: The team that pushes harder against the ground, in a tug of war,
wins.
B) Statement – 1 is True, Statement – 2 is True; Statement – 2 is NOT a correct
C) Statement – 1 is True, Statement – 2 is False.
D)Statement – 1 is False, Statement – 2 is True

9. A bag is gently dropped on a conveyor belt moving at a speed of 2 m/s. The coefficient
of friction between the conveyor belt and bag is 0.4. Initially, the bag slips on the belt
before it stops due to friction. The distance travelled by the bag on the belt during
slipping motion, is :
[Take g = 10 m / s 2 ]
A) 2 m B) 0.5 m C) 3.2 m D) 0.8 ms
10. The voltage time (V – t) graph for triangular wave having peak value V0 is as shown in
figure.
T
The rms value of V in time interval from t = 0 to is :
4
V0 V0 V0 2V0
A) B) C) D)
3 2 2 
11. A hollow spherical shell at outer radius R floats just submerged under the water surface.
27
The inner radius of the shell is r. If the specific gravity of the shell material is w.r.t
8
8
water, the value of r is : (Given 3 19  )
3
8 4 2 1
A) R B) R C) R D) R
9 9 3 3
12. A very long solenoid of radius R is carrying current I(t) = Kteat  k  0 , as a function of
time  t  0 . Counter clockwise current is taken to be positive. A circular conducting coil

of radius 2R is placed in the equatorial plane of the solenoid and concentric with the
solenoid. The current induced in the outer coil is correctly depicted,

A) B)
C) D)
13. For a nucleus ZA X having mass number A and atomic number Z
A. The surface energy per nucleon  bs  = a1 A2/3

Z  Z  1
B. The coulomb contribution to the binding energy bc  a2
A4/3
C. The volume energy bv  a3 A
D. Decrease in the binding energy is proportional surface area.
E. While estimating the surface energy, it is assumed that each nucleon interacts with 12
nucleons,
( a1 , a2 and a3 are constants)
Choose the most appropriate answer form the options given below;
A) C , D only B) B , C, E only C) A , B , C , D only D) B , C only
14. In a Vernier calipers having 10 VSD, the Vernier constant is 0.1 mm. when the jaws are
closed, zero of Vernier lies to the left of zero of main and 7th VSD coincides with a main
scale division. When a cylinder is placed between the jaws the main scale reading was
7.7 cm and Vernier scale read 8 divisions. What is the diameter of the cylinder?
A) 78.1 mm B) 77.5 mm C) 77.8 mm D) 78.5 mm
15. Two cylindrical vessels of equal cross – sectional area 16 cm2 contain water upto heights
100 cm and 150 cm respectively. The vessels are interconnected so that the water levels
in them become equal. The work done by the force of gravity during the process, is
[Take , density of water = 103 kg/ m3 and g = 10 ms 2 ]
A) 0.25 J B) 1 J C) 8 J D) 12 J

3RT
16. In the expression for Vrms of a mixture of 2 gases at same temperature, Vrms  , the
M
molecular weight M is
A) Arithmetic mean of molecular weights
B) Harmonic mean of molecular weights
C) Geometric mean of molecular weights
D) Larger Molecular weight of the two
17. A particle executing SHM is given by equation x  A cos t    . When particle is at half
the amplitude with positive velocity initially, the initial phase  is
  5 5
A) B) C) D)
6 3 3 6
18. A monatomic ideal gas sample is given heat Q. One half of this heat is used as work
done by the gas and rest is used for increasing its internal energy. The equation of
process in terms of volume and temperature is
V2 V2
A) = constant B) = constant C) VT 3 = constant D) V 2 T = constant
T3 T
19. Five point charges each + q , are placed on five vertices of a regular hexagon of side L.
The magnitude of the force on a point charge of value – q placed at the centre of the
hexagon (in newton) is
3 q2 q2 q2
A) Zero B) C) D)
4  0 L2 4  0 L2 4 3 0 L2
20. A disc has mass 9 m and radius R. A hole of radius R/3 is cut from it as shown in the
figure. The moment of inertia of remaining part about an axis passing through the centre
‘O’ of the disc and perpendicular to the plane of the disc is :
40 37
A) 8m R 2 B) 4m R 2 C) m R2 D) m R2
9 9

SECTION-II
21. In the circuit shown below, the Zener diode is ideal and the zener voltage is 6V. The
output voltage V0 (in volts) is _______________
22. The energy (in eV) that should be added to an electron, to reduce its de – Broglie
wavelength from 110 9 m to 0.5 109 m will be x times the initial energy.
23. The X – Y plane be taken as the boundary between two transparent media M 1 and M 2 .
M 1 in Z  0 has a refractive index of 2 and M 2 with Z < 0 has a refractive index of 3.
A ray of light travelling in M 1 along the direction given by the vector

A  4 3 iˆ  3 3 j  5kˆ , is incident on the plane of separation. The value of difference
between the angle of incident in M 1 and the angle of refraction in M 2 will be
__________ degree.
24. A body of mass 500g moves along x axis such that its velocity varies with displacement
x according to relation v  10 x m/s. Force acting on the body is _____________ N
25. The disc of mass M with uniform surface mass density is shown in the figure. The centre
xa xa
of mass of the quarter disc (the shaded area) is at the position , where x is .
3 3
(Round off to the Nearest integer) [a is the radius of disc as shown in the figure]

26. A progressive wave travelling along the positive x - direction is represented by y  x , t  =
A sin  kx  t    . Its snapshot at t = 0 is given in the figure.
x
For this wave, if the phase  is then x is ___________.     0, 2
2
27. A circular disc reaches from top to bottom of an inclined plane of length ‘L’. When it
slips down the plane (without friction) , it takes time ‘ t1 ’. When it rolls down the plane
t2 3
with pure rolling , it takes time t2 . The value of is . The value of x will be ______.
t1 x
28. A thin rod having a length of 1 m and area of cross section 3 106 m 2 is suspended
vertically from one end. The rod is cooled form 210C to 160C . After cooling , a mass
M is attached at the lower end of the rod such that the length of rod again becomes 1 m.
Young’s modulus and coefficient of linear expansion of the rod are 2 1011 Nm2 and
2 105 K 1 , respectively. The value of M is _________ kg. (Take g = 10 m s 2 )
29. A uniform heating wire of resistance 36  is connected across a potential difference of

240 V. The wire is then cut into half and potential difference of 240 V is applied across
each half separately. The ratio of power dissipation in first case to the total power
dissipation in the second case would be 1 : x , where x is ___________.
30. M grams of steam at 100C is mixed with 200 g of ice at its melting point in a thermally
insulated container. If it produces liquid water at 40C [heat of vaporization of water is
540 cal/g and heat of fusion of ice is 80 cal/g], the value of M is ____________ grams.

SECTION – I
31. In which of the following arrangements the order is NOT according to the property
indicated against it?
A) Al 3  Mg 2  Na   F   increasing ionic size
B) B  C  N  O  increasing first ionization enthalpy
C) I  Br  F  Cl  increasing electron gain enthalpy (with negative sign)
D) Li  Na  K  Rb  increasing metallic radius
32. Assertion : PbI 4 doesnot exist
Reason : Pb  I bond initially formed during the reaction does not release enough energy
to unpair 6S 2 electrons
A) Both Assertion and Reason are correct and the Reason is a correct explanation of the
Assertion
B) Both Assertion and Reason are correct but Reason is not a correct explanation of the
Assertion.
C) The Assertion is correct but Reason is incorrect
D) The Assertion is incorrect and Reason is correct
33. The incorrect order among the following is
A) Bond dissociation enthalpy (kj/mole): Cl2  Br2  F2  I 2
B)Melting point /K : I 2  Br2  Cl2  F2
C) Melting point/K : HF  HI  HBr  HCl
D) Electron gain enthalpy (kj/mole): Ne  Ar  Kr  Xe  He

34. Which of the following is not arranged in correct sequence?
A) MO, M 2O3 , MO2 , M 2O5  decreasing order of basic nature (M = d-block metal)
B) Sc, Ti,V , Cr , Mn  increasing order of highest possible oxidation state
C) d 5 , d 3 , d 1 , d 4  increasing magnetic moment
D) Mn 2 , Fe2 , Cr 2 , Co 2  decreasing stability
35. Which of the following is incorrectly matched? (Accoring to CFT)
Complex Number of unpaired electrons
I)  FeF6 
3
5
2
II) Cr  en 3  2
3
III) Co  NH 3 6  4
2
IV)  Mn  H 2O 6  5
A) III B) I C) IV D) II
36. The correct order of dipole moment is:
A) NH 3  NF3  CH 4  H 2O B) CH 4  NH 3  NF3  H 2O
C) H 2O  NF3  NH 3  CH 4 D) CH 4  NF3  NH 3  H 2O
37. The correct order of acidic strength character of the following compounds is
OH COOH COOH COOH
NO2 CH3
CN
I
II III IV
A) IV > III > II > I B) III > II > I > IV
C) II > III > IV> I D) I > II > III > IV

38.
 i C H CO O ,HBr

 ii
6 5

2 2
CoF2

P
Br (major product)
Major product ‘P’ of above reaction, is:
A) B) C) D)
39.
(P)
Cl2 / hv
Cl2 / AlCl3
(Q)
Identify major product of both reactions P and Q respectively
CH2 Cl
Cl
B)
A)
Cl
Cl
Cl Cl
D)
C)
Cl Cl
Cl

40. Which of the reagents shown below would accomplish the following transformations?
OH
Br
1. A
2. B
a) A is H 3O  ; B is BH 3  THF ; H 2O2 / NaOH
b)A is NaOH ; B is BH 3  THF ; H 2O2 / NaOH
c) A is HBr in ether ; B is Hg  OAC 2 / H 2O; NaBH 4
d) Ais NaNH 2 ; B is Hg  OAC 2 / H 2O; NaBH 4
41.
OH
O
||
1V H3CCl

W 
PBr3
X  
Mg
2 H3O
ether
Y C  Z
Na2CrO
2 7
Product Z of above reaction is :

O
|| O
OCCH3 ||
OCCH3
A) B)
O
O ||
|| OCCH3
OCCH3
C) D)

42. The products of the following reaction are,
CH2OD CH2OH
A) B)
OD
COOD
C C
C) D) HO
A)A,B B) A,C C) A,D D) C,B
43. Consider the following sequence of reactions,
COOH
ZnHg
SO
 
Cl2
A2.
 
1.AlCl3
H O
BC 
onc.HCl, heat
C
3
CH2C6H5
The end product ( C ) is :
O
A) B) O
C6H5
C6H5
C) D)

44. The compound X  C7 H 9 N  reacts with benzene sulphonyl chloride to give
y  C13 H 13 NO2 S  which is insoluble in alkali. The compound X is :
NH2
NHCH3
CH3
A) B)
NH2
NH2
C) D)
H3C
CH3
45. The amino acid that cannot be obtained by hydrolysis of proteins is-
CH2COO
H NH3
A) HOOCCH 2CH  N H 3  COO 


B) SH
 
H2CCH COO
NH3
NH

C) N D) N H 3  CH 2  4 CH  NH 2  COO
46. A certain amount of a reducing agent reduces x mole of KMnO4 and y mole of K 2Cr2O7 in
different experiments in acidic medium. If the change in oxidation state in reducing
agent is same in both experiments, x : y is
A) 5:3 B) 3:5 C) 5:6 D) 6:5
47. The ionisation enthalpy of H atoms is 1.312  106 J / mol the energy required to excite the
electrons in the one of mole of hydrogen atoms from n  1 to n  2 is
A) 9.84  105 J / mol B) 8.51 105 J / mol
C) 6.56  105 J / mol D) 7.56  105 J / mol

1 3
48. x2  y2  xy3 ; H r  30kJ
2 2
Entropies of x2 , y2 and xy3 are 60,40 and 50J K 1mol 1 respectively for the reaction. The
temperature in kelvin at which the above reaction attains equilibrium.
A)750 B) 650 C) 550 D) 880
49. Match the following if the molecular masses of X,Y,Z and W are same
[Solutions]
Column-I Column-II
(Solvent and its boiling point) (Molal elevation constant, Kb )
A) X 100 C 
0
P) 0.68
B) Y  27 C 
0
Q) 0.53
C) Z  253 C  0
R) 0.98
D) W 182 C 
0
S) 0.79
A) A  P; B  Q; C  R; D  S B) A  Q; B  P; C  R; D  S
C) A  P; B  Q; C  S ; D  R D) A  S ; B  Q; C  R; D  P
50. A solution is 0.1M in Cl  and 0.001M in CrO42 . Solid AgNO3 is gradually added to it.
Assuming that the addition does not change in volume and
Ksp  AgCl   1.7 1010 M 2 and Ksp  Ag2CrO4   1.9 1012 M 3. Select correct statement from the
following
A) Ag2CrO4 precipitates first because the amount of Ag  needed is low.
B) AgCl precipitates first because its K sp is high
C) AgCl will precipitate first as the amount of Ag  needed to precipitate is low.
D) Ag2CrO4 precipitates first as its K sp is low.

SECTION-II
51. Percentage of aniline hydrochloride hydrolysed in its M/40 solution at 250 C is
Given ;  EC H NH .HCl / H  0.18V ; 

2.303 RT
 0.06 
 6 5 2 2
F 
52. How many compounds will give positive Lassaigne’s test ?
OH NH2 NH2 NH2 SH
O O
Cl I NH2
NH2
53. ----------- is the magnetic moment of a di valent ion in aqueous solutionswith atomic
number 26. (nearest integer)
54.
CH2 OH
CHOHCH3MgBr xCH4
CH2 SH (Excess)
What is the value of ' x ' in the above reaction?
55.
HO OH
O
O
||
CH3 CCH3 xHCHOK
OH

 C
HO
HO HO OH
x moles of HCHO consumed .
Value of  x  will be

56. a) Glycine b) Valine c) Leucine d) Cysteine
e) Arginine f) Methionine g)Lysine h) Try Ptophan
Sulphur containing amino acid x
Basic amino acid  y
Essential amino acid  Z
Find the value of
57. NaClO3 is used, even in spacecrafts, to produce O2 . The daily consumption of pure O2 by
a person is 492 L at 1atm, 300K. How much amount of NaClO3 , in grams, is required to
produce O2 for the daily consumption of a person at 1 atm,
300K?  Na  23gm / mol, Cl  35.5gm / mol  According to the following reaction.
NaClO3  s   Fe  s   O2  g   NaCl  s   FeO  s 
R  0.082 L atm mol 1K 1
58. Consider the following reaction

Na3 PO4   NH 4  2 MoO4  HNO3  dil   ' X '  canary yellow PPT  then calculate total number of
atoms of 15th group elements which are SP 3 hybridized in compound ‘X’.
59. How many grams of sucrose (mol. Wt. = 342) should be dissolved up to the nearest
integer in 100 gm water in order to produce a solution with 1050C difference between the
freezing point & boiling point temperature at 1 atm?
Unit : k f  2 K .kg mol 1; kb  0.5K .kg mol 1 
60. At a certain temperature equilibrium constant  Kc  is 16 for the reaction.

SO2 g   NO2 g   SO3 g   NO g  .If we take one mole each of all the four gases in one litre
container, then the % moles of NO2 at equilibrium is

SECTION – I
61. If z is a complex number satisfying the equation z  i  z  i = 8, on the complex plane
then maximum value of z is
A) 2 B) 4 C) 6 D) 8
62. The system of equations
px+(p+1)y+(p-1)z = 0
(p+1)x+py+(p+2)z=0
(p-1)x+(p+2)y+pz=0 has a non – trivial solutions for
A) Exactly two real values of p. B) no real value of p.
C) Exactly one real value of p. D) infinitely many real value of P.
63. If two whole numbers are randomly selected and multiplied then the chance that the unit
place in their product is 0 or 5, is
A) 20% B) 16% C) 61% D) 36%
64. Consider the word ‘HALEAKALA’. The number of ways the letters of this word can be
arranged if
Column-I Column-II
a) All ‘A’ are separated p) 5
Keeping the position of each
b) q) 60
consonant fixed
c) All the vowels occur together r) 300
d) No two vowels are consecutive s) 900
Code:
A) a  s, b  p, c  p, d  r B) a  q, b  r , c  s, d  p
C) a  s, b  p , c  r , d  q D) a  q, b  r , c  s, d  s

65. Statement-I: Out of 21 tickets with numbers 1 to 21,3 tickets are drawn at random,
10
the chance that the numbers on them are in A.P is .
133
Statement-II: Out of  2n  1 tickets consecutively numbered, three are drawn at random,

the chance that the numbers on them are in A.P is  4n  10  /  4n 2  1 .
A) Statement- I is true, Statement-II is true, Statement-II is correct explanation of for

statement-I
B) Statement- I is true, Statement-II is true, Statement-II is not correct explanation of for
statement-I
C) Statement- I is true, Statement-II is false
D) Statement- I is false, Statement-II is true
66. A cubic polynomial y = f(x) is such that A(-1, 3) and B (1, -1) are the relative a
maximum and relative minimum points respectively. The value of f(2) is
A) 6 B) 4 C) 0 D) 3
67. If the surface area of a sphere of radius r is increasing uniformly at the rate 8cm2/sec,
then the rate of change of its volume is
A) Proportional to r B) Constant
C) Proportional to r D) Proportional to r2
 ^ ^ ^  ^ ^ ^
68. If V1  i  j  k; V2  a i  b j  c k where a, b, c  {-2, -1, 0, 1, 2}, then find the number of non-
 
zero vectors V2 which are perpendicular to V1
A) 18 B)24 C) 21 D) 12
69. Let L be the line passing through the point P (1,2) such that its intercepted segment
between the Co-ordinate axes is bisected at P. If L1 is the line perpendicular to L and
passing through the point  2,1 then the point of intersection of L and L1 is
 3 23   4 12   11 29   3 17 
A)  ,  B)  ,  C)  ,  D)  , 
 5 10  5 5   20 10   10 5 

70. Let f : R  R be defined by f ( x)  x3  3x  1 and g is the inverse of ‘f’ then the value of
g  (5) is equal to
A) 1 B) 1 C) 1 D) 1
6 36 6 36

1
x  2 x  3x  6  dx   2x 
1/8
71. If 24
x x
16 8 16 8 24
 3x  6 x
16 8 
 C (Where C is constant of

integration and  ,  are coprime numbers) then the value of       is…
A) 51 B) 61 C) 71 D) 81
x
72. Let y = y(x) satisfy the equation y( x)   y(t )dt  x2 . The value of y(e) is equal to
1
A) 2e-2+e1-e B) 2e-1+e1-e C) 2e+2+e1-e D) 2e+1+e1-e
73. If the roots of the equation 72 x3  108 x2  46 x  5  0 are in A.P, then the difference between
largest and smallest root lies in the interval
1 1   1 
A)  0, 
1
B)  ,  C)  ,1 D) 1, 2 
 2  2 2  2 
74. If the mean deviation about the median of the numbers a, 2a,…… 50a is 50, then
a equals:
A) 1 B) 2 C) 3 D) 4
13
1
75. The value of 
k 1    k  1     k 
is equal to
sin    .sin   
4 6  4 6 
A) 3  3 B) 2  3  3  C) 2  3  1 D) 2  2  3 
76. Suppose A1, A2, ….., A30 are thirty sets each having 5 elements and B1, B2, ….., Bn are n
30 n
sets each with 3 elements, let  Ai   B j  S and each elements of S belongs to exactly
i 1 i 1
10 of the Ais and exactly 9 of the Bj s . Then, n is equal to
A) 40 B) 54 C) 45 D) 27

 1 1 1
77. 
Let A be a non- singular matrix such that A   4 6 8  , then the value of
1
100 100 99 
 
det  Adj  2 A    det  2 A  is
A) 8 B) 12 C) 16 D) 28
78. Let  ,      be the roots of the quadratic equation x 2  x  4  0 . If Pn   n   n . n  N
P15 P16  P14 P16  P152  P14 P15

then is equal to
P13 P14
A) 15 B) 16 C) 17 D) 18
79. Let f  x    x   1  x for 1  x  3 where [x] is the integral prat of x. Then the number
of value of x in [-1,3] at which f is not continuous is .
A) 0 B) 1 C) 2 D) 4
80. cos 1  cos  5    sin 1  sin  6    tan 1  tan 12   is equal to :
(The inverse trigonometric functions take the principal values)
A) 3  11 B) 4  9 C) 4  11 D) 3  1
SECTION-II
2 i
81. Let   e 3
and a, b, c, x, y, z be non – zero complex number such that
a+b+c=x
a + b  + c  2 =y
a + b  2 +c  =z
2 2 2
x  y  z
Then the value of 2 2 2
, is equal to
a b c

82. If f  x   ax  9 x  9 x  3 is strictly increasing on R, then find the number of
3 2
integralvalues of ' a ' , a   5,100
x  0 y 1 z 1
83. The shortest distance between the lines x = y = z and   is then  =
1 1 0 
84. The locus of image of the point ( 2 , 2 ) in the line mirror x–y+1= 0 where  parameter is
 x  a  b  y  c  where a, b, c  I, then find the value of (a + b + c).

2
85. Two vertices of an equilateral triangle are (-1,0) and (1,0), and its third vertex lies above
2y
the x – axis. If the equation of its circumcircle is x2+y2-   =0, then find the value of
3
.
1 1 1 1
86.  1  x  dx   1  x  dx is equal to
7 4 4 7
0 0
Find the area of the region bounded by y   x  4  , y  16  x2 and the x  axis.

2
87.
88. Let a and b are two positive real numbers such that a2 + b = 2, then the maximum value
of term independent of x in the expansion of (ax1/6 + bx-1/3)9 is
89. If lim x  x 2  x 4  1  2 x  exist and has value non – zero finite number L, then find
x   
the value of 2 .
L
90. Let A  1, 2, 3, 4 and B  1, 2, 3, 4 . If f : A  B is an one – one function and f  x   x for
all x  A , then the number of such possible functions, is


MISTAKES
MATHS
PHYISCS
CHEMISTRY
SECTION – I
1. A galvanometer of resistance 100  contains 100 division. It gives a deflection of one
division on passing a current of 104 A . Find the resistance to be connected to it, so that it
becomes a voltmeter of range 10V.
500 100
A)  B) 500  C)  D) 900 
9 9
2. If mass density of earth varies with distance ‘r’ from centre of earth as   k r and ‘R’ is
radius of earth, then find the orbital velocity of an object revolving around earth at a
distance ‘2R’ from its centre.
 kR3G  kR3G  kR3G

A) B) C) D)  kR 3G
4 2 8
3. A particle moves clockwise in a circle of radius 1 m with centre at ( x, y )  (1 m, 0) . It

starts at rest at the origin at time t  0 . Its speed increases at the constant rate of
  
  m / s . If the net acceleration at t  2sec is
2
2
2
1  N 
2
then what is the value of N ?
A) 2 B) 4 C) 1 D) 6
4. The rms speed of hydrogen molecules at a certain temperature is 300 m/s. If the
temperature is doubled and hydrogen gas dissociates into atomic hydrogen, the rms
speed will become
A) 100 m/s B) 300 m/s C) 600 m/s D) 300 2 m / s
5. The power of a lens (biconvex) is 1.25 m 1 in a particular medium. Refractive index of the
lens is 1.5 and radii of curvature are 20 cm and 40 cm respectively. The refractive index
of surrounding medium is
9 3 4
A) 1.0 B) C) D)
7 2 3

6. Let An be the area enclosed by the nth orbit in a hydrogen atom. The graph of n  An / A1 
against n(n)
A) will not pass through the origin
B) will have certain points lying on a straight line with slope 4
C) will be a monotonically increasing non linear curve
D) will be a circle
7. A screw gauge has 50 divisions on its circular scale. The circular scale is 4 units ahead
of the pitch scale marking, prior to use. Upon one complete rotation of the circular
scale, a displacement of 0.5mm is noticed on the pitch scale. The nature of zero error
involved and the least count of the screw gauge are respectively
A) Positive, 0.1  m B) Positive, 10  m C) Negative, 10  m D) Negative, 0.1  m
8. In an LC oscillator, if values of inductance and capacitance become twice and eight

times, respectively, then the resonant frequency of oscillator becomes ' x ' times its initial
resonance frequency 0 . The value of ' x ' is
1 1
A) B) 4 C) 16 D)
4 16
9. A closed organ pipe and an open pipe of same length produce 6 beats per sec when they
are set into vibrations simultaneously with their fundamental frequency. If the length of
each pipe is doubled, then the number of beats produced is
A) 4 B) 3 C) 5 D) 7
10. A closely wound solenoid of 1000 turns and area of cross – section 2.0  104 m 2 carries a
current of 2.0 A. It is placed with its axis horizontal in a uniform horizontal magnetic
field of 0.16 T. If the solenoid is free to turn about the vertical direction. The amount of
work (in J) needed to displace the solenoid from its stable orientation to its unstable
8x
orientation is . Find the value of x ?
250
A) 5 B) 4 C) 3 D) 2

11. The ratio of the KE and PE possessed by a body executing SHM when it is at a distance
of 1/n of its amplitude from the mean position is
1
A) n 2 B) C) n 2  1 D) n 2  1
n2
12. A fish rising vertically up towards the surface of water with speed 3 ms 1 observes a bird
diving vertically down towards it with speed 9 ms 1 . The actual velocity of bird is
4
(Given water  )
3
A) 4.5 ms 1 B) 5.4 ms 1 C) 3.0 ms 1 D) 3.4 ms 1

13. Electric potential on the surface of a uniformly charged solid sphere is V. Radius of the
sphere is R(=1m). Match the following two columns. ( ' r ' is the distance from the centre
of the sphere)
Column I Column II
R V
(A) Electric potential at r  (P)
2 4
V
(B) Electric potential at r  2R (Q)
2
3R 3V
(C) Electric field at r  (R)
4 4
11V
(D) Electric field at r  2R (S)
8
A) A – P; B – R; C – Q; D – S B) A – S; B – P; C – Q; D - R
C) A – S; B – Q; C – R; D – P D) A – R; B – Q; C – P; D – S
14. In the figure, potential difference between A and B is (Diode is ideal)
A) 5 V B) 10 V C) zero D) 15 V
15. In a photoelectric experiment, with light of wavelength  , the fastest electron has speed
3
v. If the exciting wavelength is changed to , the speed of the fastest emitted electron
4
will become
3 4 3 4
A) v B) v C) less than v D) greater than v
4 3 4 3

16. An insect is at the bottom of a hemispherical ditch of radius 1 m. It crawls, up the ditch
but starts slipping after it is at height ' h ' from the bottom. If the coefficient of friction
between the ground and the insect is 0.75, then ' h ' is ( g  10 ms 2 )
A) 0.80 m B) 0.60 m C) 0.45 m D) 0.20 m
17. For one mole of mono atomic ideal gas temperature (T in Kelvin) and Volume (V in m3 )
relate as shown in the graph. What is the change in internal energy from B to D?
 25 1 1 
 R  3 J mol K 
A) 1250 B) 1290 C) 1300 D) 1350

18. Statement 1 : A large soap bubble expands while a small bubble shrinks, when they are
connected to each other by a capillary tube
Statement 2 : The excess pressure inside bubble is inversely proportional to its radius
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 not a correct

C) Statement -1 is True, Statement – 2 is False.
19. Two identical billiard balls are in contact on a smooth table. A third identical ball
strikes them symmetrically and comes to rest after impact. The coefficient of restitution
is
2 1 1 3
A) B) C) D)
3 3 6 2

20. Three elephants A, B and C are moving along a straight line with constant speed in same
direction as shown in figure. Speed of A is 5 m/s and speed of C is 10 m/s. Initially
separation between A & B is ‘d’ and between B & C is also d. When ‘B’ catches ‘C’
separation between A & C becomes 3d. Find the speed of B (in m/s).
A) 15 B) 20 C) 5 D) 25
SECTION-II
21. A beam of natural light falls on a system of 5 polaroids which are arranged in succession
such that the pass axis of each Polaroid is turned through 600 w.r.t. the preceding one.
1
The fraction of intensity of incident light that passes through the system is . Then
128 n
the value of n ?
22. A 60 pF capacitor is fully charged by a 20 V supply. It is then disconnected from the

supply and is connected to another uncharged 60 pF capacitor in parallel. The
electrostatic energy that is lost in this process by the time the charge is redistributed
between them is (in nJ) __________
23. A body of mass 2 kg moving with a speed of 4 ms 1 makes an elastic collision with
another body at rest and continues to move in the original direction but with one fourth
of its initial speed. The speed of the centre of mass of two body system after the collision
x
is ms 1 . Then the value of ' x ' is _________
10

24. In the LCR circuit shown, if RMS voltages across resistor, inductor and capacitor are
equal, the RMS value of voltage across resistor will be _________ volt.
25. In a cylinder piston arrangement, air is under a pressure P1 . A soap bubble of radius r
lies inside the cylinder. Soap bubble has surface tension T. The radius of bubble is to be
reduced to half. The pressure P2 to which air should be compressed isothermally is
BT
AP1  . Then the value of A  B is ________
r
26. A system of identical cylinders and plates is shown in figure. All the cylinders are
identical and there is no slipping at any contact. The velocity of lower and upper plates
are V and 2V, respectively, as shown in figure. Then the ratio of angular speeds of the
upper cylinders to lower cylinders is
27. The potential energy (in joule) of a body of mass 2kg moving in the x-y plane is given
by U  6 x  8 y . Where the position coordinates x and y are measured in metre. If the
body is at rest at point (6m, 4m) at time t = 0, it will cross the y-axis at time t equal to
28. An electromagnetic wave of frequency 1  1014 hertz is propagating along z-axis. The
amplitude of electric field is 4 V/m. If 0  8.8  1012 C 2 / N  m2 , then average energy
density of electric field will be N  1013 J / m3 .
29. The moment of inertia of a uniform circular disc about its diameter is 200 gm cm 2 . Then
its moment of inertia about an axis passing through its center and perpendicular to its
circular face is ______ gm cm2.
30. A light beam of wavelength 500 nm is incident on a metal having work function of 1.25
eV, placed in a magnetic field of intensity B. The electrons emitted perpendicular to the
magnetic field B, with maximum kinetic energy are bent into circular arc of radius 30
cm. The value of B is _____ 107 T . [Given hc  20  1026 J -m , mass of electron =
9  10 31 kg ]

SECTION – I
31. In which of the following, mixing of 1N strong acid and 1N strongbaseresults in highest
increase in temperature.
A)20 ml acid + 30 ml alkali B)10 ml acid + 40 ml alkali
C)25 ml acid + 25 ml alkali D)35 ml acid + 15 ml alkali
32. Which of the following is an incorrect statement?
A)Half-life of second order reaction decreases with increase in concentration of reactant
B)Half-life of first order is independent of concentration of the reactant
C)The unit of rate constant of zero order is equal to unit of rate
D)The unit of frequency factor A in Arrhenius equation is the unit of half-life of the
reaction
33. A set of solution is prepared by using 180g of water as a solvent and 10g of different
non-volatile solutes A, B and C. The relative lowering of vapour pressure in the presence
of these solutes are in the order [Given, molar mass of A = 100 g mol1 ; B = 200 g mol1 ;
C = 10000 g mol1 ]
A)A > B > C B)A > C > B C)C > B > A D)B > C > A
34. Which of the following is an extensive property?
A)Heat capacity B)Refractive index
C)Specific volume D)Molarity
35. A certain metal when irradiated by light (  3.2  1016 Hz) emits photoelectrons with twice
kinetic energy than the photoelectrons emitted when the same metal is irradiated by
light (  2.0 1016 Hz) . Then 0 of metal is
A) 1.2 1014 Hz B) 8 1015 Hz (c)1.5 1016 Hz D) 4 1012 Hz

36. KMnO4 reacts with KI, in weakly basic medium to form I 2 and MnO2 . When 250 ml of
0.1 M KI solution reacts with 250 mL of 0.02 M KMnO4 in basic medium, what is the
number of moles of I2 formed?
A)0.015 B)0.0075 C)0.005 D)0.01
37. The term invert sugar refers to an equimolar mixture of
A)D-Glucose and D-Galactose B)D-Glucose and D-Fructose
C)D-Glucose and D-Mannose D)D-Glucose and D-ribose
Statement-I: Pure Aniline and other arylamines are usually colourless
Statement-II:Arylamines get coloured on storage due to atmospheric reduction
In the light of the above statements, choose the most appropriate answer from the
options given below
A)Both statement-I and II are incorrect
B)Both statement-I and II are correct
C)Statement-I is correct but statement-II is incorrect
D)Statement-I is incorrect but statement-II is correct
39. Mark out the incorrect option regarding Cannizzaro reaction
A)The aldehyde, which doesn’t have enolisable hydrogen undergo Cannizzaro reaction
B)All types of Cannizzaro reaction e.g. inter, intra and crossed all can be said
disproportionation reaction
OH 
 PhCHD  OH  DCOO 
C)Reaction PhCHO  CD 2  O 
D)Reaction can be first order in base and 2nd order in substrate

40. Identify the alkyne in the following sequence of reactions
H Wacker
2
Alkyne  O3 / Zn
 A   B   H 2C  CH 2
H 2O
Lindlar's Catalyst Process
Only
A) CH3  C  C  CH3 B) CH3  CH 2  C  CH
C) CH2  CH  C  CH D) CH  C  CH2  C  CH
41. The IUPAC name of the following compound is
A)2-nitro-4-hydroxymethyl-5-amino benzaldehyde
B)3-amino-4-hydroxymethyl-5-nitrobenzaldehyde
C)5-amino-4-hydroxymethyl-2-nitrobenzaldehyde
D)4-amino-2-formyl-5-hydroxymethyl nitrobenzene
42. Lassaigne’s test for the detection of nitrogen fails in

A) NH2CONHNH2  HCl B) NH2 NH2  HCl
C) NH2CONH2 D) C6 H5 NHNH 2  HCl
43. Given below are two statements
Assertion (A): Cu 2 in water is more stable than Cu  .
Reason (R): Enthalpy of hydration for Cu 2 is much more negative than that of Cu  .
below
A) Both A and R are correct and R is the correct explanation of A
B) A is correct but R is not correct
C) A is not correct but r is correct
D) Both A and R are correct but R is not the correct explanation of A

44. Which of the following statement is not correct?
A) B(OH)3 is acidic
B)Aqeous solution of Potash alum is acidic in nature
C)The decreasing order of acidic natureof BBr3 , BCl3 and BF3 is BBr3  BCl3  BF3
D) B2H6 contains B – B covalent bonds
45. Among the following, the correct statement is
A)Between NH3 and PH3 , NH3 is a better electron donor because the lone pair of
electrons occupies spherical ‘s’ orbital and is less directional
B)Between NH3 and PH3 , PH3 is a better electron donor because the lone pair of
electrons occupies sp3 orbital and is more directional
C)Between NH3 and PH3 , NH3 is a better electron donor because the lone pair of
electrons occupies sp3 orbital and is more directional
D)Between NH3 and PH3 , PH3 is a better electron donor because the lone pair of
electrons occupies spherical ‘s’ orbital and is less directional
46. Match the list-I with List-II
List-I (Cations) List-II (Group reaction)
(P) Pb 2 , Cu 2 (I) H2S gas in presence of dilute HCl
(Q) Al3 , Fe3 (II) (NH4 )2 CO3 in presence of NH4OH
(R) Co 2 , Ni 2 (III) NH4OH in presence of NH4Cl

(S) Ba 2 , Ca 2 (IV) H2S in presence of NH4OH .
A)P-I, Q-III, R-II, S-IV B)P-IV, Q-II, R-III, S-I
C)P-III, Q-I, R-IV, S-II D)P-I, Q-III, R-IV, S-II
47. Which of the following is correct stability order of oxides of 17th group elements is
A) Cl2  Br2  I2 B) I2  Br2  Cl2
C) I2  Cl2  Br2 (d) Cl2  I2  Br2

48. In the electrolysis of CuCl2 solution using Cu electrodes, the weight of Cu increased by 2
gram at cathode.At the anode:
A)0.2 mole of Cu 2 will go into solution
B)560 ml of O2 liberate
C)No loss in weight
D)2 gram of copper goes into solution as Cu 2
49. The major product of the following reaction is
A) B)
C) D)
50. The conductivity measurement of a coordination compound of Cobalt (III) shows that it
dissociates into 3 ions in solution. The compound is
A)Hexaamminecobalt(III) chloride
B)Pentaamminesulphatocobalt(III) chloride
C)Pentaamminechloridocobalt(III) sulphate
D)Pentaaminechloridocobalt (III) chloride

SECTION-II
51. In the following monobromination reaction, the number of possible chiral product(s)
is/are ____
52. The number of geometrical isomers possible for the complex

[Co L 2 Cl 2 ] (L  H 2 NCH 2 CH 2 O  ) is
53.  5NaCl  A  3H 2O . What is the

Consider the following reaction 6NaOH  3Cl 2 
oxidation number of chlorine in “A”?
54. On complete combustion, 0.492g of an organic compound gave 0.792 g of CO2 . Then
the % of carbon in the organic compound is ____________ (Nearest Integer)
55. The number of d  p bonds in ClO4
56. The value of log10 K for a reaction A   B is (Given:  r H 0298K  54.07 kJ mol1 ,
 rS0298K  10 JK 1 mol1 and R  8.314 JK 1mol1 , 2.303  8.314  298  5705 )
57. A decapeptide (Molecular weight 796) on complete hydrolysis gives glycine (Molecular
weight 75), alanine and phenylalanine. Glycine contributes 47.0% to the total weight of
the hydrolyzed products. The number of glycine units present in the decapeptide is
58. The total number of cyclic isomers possible for a hydrocarbon with the molecular
formula C4 H 6 is
59. Consider the reaction: AB2 (g)  AB(g)  B(g) . If the initial pressure of AB2 is 500 torr
and equilibrium pressure is 600 torr, equilibrium constant K p in terms of torr is
60. Reductive ozonolysis of a terpenoid of following structure gives how many

different products.

SECTION – I
1
x 2 sin    2 x
 x
61. Let f x   1
then lim f x  x 0
 
1  x  x e
2 2 4 4
A) B) C) D)
e e e e
62. In a triangle ABC, BC= 3 and AC= 4 and circle with AB as diameter passes through the
centroid of a triangle, then AB is
5
A) 5 B) 5 C) 3 D)
2

2 x 1  sin x 
63. The value of definite integral 

1  cos 2 x
dx is:
 2
A) B)  C)  2 D)
2 2
64. Let f  x   x3  x  1 and g  x  be its inverse then g   3  ___________.
1 1
A) B)  C) 4 D) –4
4 4
65. Let f  x  be a cubic polynomial on R which increases in the interval  , 0  and in 1,  
and decreases in the interval  0,1 . If f   2   6 and f  2   2 , then the value of
  3 
tan 1  f 1   tan 1  f     tan 1  f  0   is equal to:
  2 
A) tan 1 2 B) cot 1 2 C)  tan 1 2 D)  cot 1 2
66. Let the equations of two adjacent sides of a parallelogram ABCD be 2 x  3 y   23 and
5 x  4 y  23 . If the equation of its one diagonal AC is 3 x  7 y  23 and the distance of A
from the other diagonal is d, then 50d 2 is equal to__________
A) 501 B) 529 C) 468 D) 731

67. A survey of 500 television watchers produced the following information, 285 watch
football, 195 watch hockey, 115 watch basketball, 45 watch football and basketball, 70
watch football and hockey, 50 watch hockey and basketball, 50 do not watch any of the
three games. How many watch all the three games? How many watch exactly one of the
three games?
A) 190 B) 40 C) 25 D) 325
z i 
68. The mirror image of the curve arg  
  , i  1 in the line z  i  1  z 1  i   0 , in
 z 1  4
argand plane , is
z i  z 1 
A) arg  
 B) arg  

 z 1 4  z i  4
z i  z i 
C) arg  
 D) arg  

 z 1 4  z 1  4
69. If x  sin 1  sin10  and y  cos1  cos10  , then y  x is equal to:
A)  B) 0 C) 10 D) 7
70. Statement 1:Length of side of an equilateral triangle is 24. The mid points of the sides
are joined to form another triangle whose midpoints of sides are joined to form another
triangle and continue the process infinite number times. Then sum of perimeters of all
such triangles formed is 144.
Statement 2 : If log 2  a  b   log 2  c  d   4 then the minimum value of a+b+c+d is 8
A) Statements 1 & 2 both are correct
B) Statements 1 & 2 both are false
C) Statement 1 is true, statement 2 is false
D) Statement 1 is false, statement 2 is true
1  x x y
For each real x, -1<x<1. Let A(x) be the matrix 1  x  
1
71.  and z  . Then
 x 1 1  xy
A) A  z   A  x  A  y  B) A  z   A  x   A  y 
1
C) A  z   A  x   A  y  D) A  z   A  x   A  y 

72. If  ,  are the roots of the equation   x 2  x   x  5  0 . If 1 and 2 are two values of
  4  
 for which the roots  ,  are related by   , then the value of 1  2 , is
  5 2 1
A) 254 B) 200 C) 350 D) 154
73. Let A  1, 2,3, 4,5,6, 7 . The number of surjective functions defined from A to A such that
f  i   i for atleast four values of i from i  1, 2,...., 7, is:
A) 7! B) 92 C) 126 D) 407
74. Let a line l pass through the origin and be perpendicular to the lines
 
       
l1 : r  iˆ  11 ˆj  7kˆ   iˆ  2 ˆj  3kˆ ,   R and l2 : r  iˆ  kˆ   2iˆ  2 ˆj  kˆ ,   R .
If P is the point of intersection of l and l1 , and Q  ,  ,   is the foot of perpendicular
from P on l2 , then       is ________
5 9 19 5
A) B) C) D)
9 5 5 3
75. ABCD and EFGC are squares and the curve y   x passes through the origin D and the
FG
points B and F. The ratio is (as shown in the figure).
BC
3 1 3 1 5 1 5 1
A) B) C) D)
4 2 4 2
2 4 
sin 2
sin 2 sin 2
76. The value of the expression 7  7  7 is equal to:
2  2 4
sin sin 2 sin 2
7 7 7
A) 3 B) 4 C) 5 D) 6

77. Two cards are drawn successively with replacement from a well shuffled deck of 52
cards, then the mean of the number of aces is
1 3 2 1
A) B) C) D)
13 13 13 2
STATEMENT-1: The integral part of  8  3 7  is an even integer.

20
78.
STATEMENT–2:  8  3 7    8  3 7  is an even integer.

20 20
A) Statements 1 & 2 both are correct

B) Statements 1 & 2 both are false
79. If an unbiased die, marked with –2,–1,0,1,2,3 on its faces, is thrown five times, then the
probability that the product of the outcomes is positive, is:
881 521 440 27

A) B) C) D)
2592 2592 2592 288
80. The following data gives the distribution of height of students:
Height (in cm) 160 150 152 161 156 154 155
Number of students 12 8 4 4 3 3 7
The median of the distribution is
A) 154 B) 155 C) 160 D) 161
SECTION-II
81. A function f : R  R satisfies the equation f  x  f  y   f  xy   x  y, x, y  R and f 1  0 ,
then f  2  f 1  2   ________
82. For y > 0 and x  R, ydx  y 2dy  xdy where y  f  x  . If f 1  1, then the value of f  3 is

83. Let f  x  be a polynomial of degree 3. If the curve y  f  x  has relative extrema at
2
x and passes through  0, 0  and 1, 2  dividing the circle x2  y 2  4 in two parts,
3
k
then the area bounded by x2  y 2  4 and y  f  x  is . Find the value of k .
2
1  7 cos 2 x g  x
84. Suppose  sin 7 x cos2 xdx  sin 7 x  C , where C is an arbitrary constant of integration. The

find the value of g   0   g    .
4
85. The urns A,B and C contain 4 red, 6 black; 5 red, 5 black and  red, 4 black balls
respectively. One of the urns is selected at random and a ball is drawn. If the ball drawn
is red and the probability that it is drawn from urn C is 0.4 then   ___________.
86. ABCD is a parallelogram. L is point on BC which divides BC in the ratio 1:2 AL
intersects BD at P. M is a point on DC which divides DC in the ratio 1:2 and AM
BD
intersects BD in Q, then  _________.
PQ
87. From a given solid cone of height H, another inverted cone whose axis coincides with
the axis of the given cone, is carved whose height is h, such that it’s volume is
H
maximum, then the ratio = ___________.
h
88. The radius of the circle passing through the points of intersection of the ellipse
x2 y 2
  1 and x 2  y 2  0 is k, then [k] = _____________([.] is G.I.F)
16 9
89. Six X’s have to be placed in the squares of the figure below, such that each row contains
atleast one X. In how many different ways can this be done?
 x 3 2
90. Given the matrix A  1 y 4 . If xyz  60 , 8 x  4 y  3 z  20 and A  adjA is equal to kI ,
 2 2 z 
( k  R , I is identity matrices of order 3) then k = ______________

KEY SHEET
PHYSICS
1 C 2 D 3 B 4 B 5 B
6 C 7 A 8 C 9 D 10 A
11 B 12 C 13 C 14 D 15 C
16 D 17 C 18 B 19 D 20 B
21 -13 22 9 23 8 24 701 25 25
26 10 27 79 28 1 29 20 30 328
CHEMISTRY
31 D 32 C 33 B 34 C 35 B
36 D 37 C 38 C 39 D 40 B
41 C 42 D 43 B 44 B 45 C
46 C 47 C 48 C 49 B 50 B
51 3 52 8 53 4 54 6 55 1
56 2 57 7 58 2 59 5 60 2
MATHEMATICS
61 D 62 C 63 D 64 C 65 D
66 C 67 C 68 D 69 D 70 B
71 C 72 B 73 C 74 B 75 A
76 C 77 B 78 C 79 B 80 B
81 20 82 4 83 10 84 64 85 2
86 3 87 5 88 46 89 3 90 1
SOLUTIONS
PHYSICS
^ ^
 ^ ^

1. Initial velocity of stone w.r.t lift  20sin 300 j  20cos300 i   10 3 i  10 j  m / s
 
 ^ ^

Initial velocity of stone w.r.t ground  10 3 i  12 j  m / s
 
The initial position of stone and lift are same and when they again meet their final positions will also
be same. So both will have same displacement in vertical direction in same time
1 t2
Displacement of lift  2  t    1  t 2  2t 
2 2
1
Displacement of stone  12  t    10  t 2  12t  5t 2
2
2
t
So 2t   12t  5t 2
2
2
11t 20
 10t or t  sec
2 11
20
So time taken by stone to return to the floor of lift is sec
11
dW
2. For W to be maximum;  0;
dx
i.e., F  x   0  x  l , x  0
Clearly for d  l , the work done is maximum.
Alternate Solution:
External force and displacement are in the same direction
 work will be positive continuously so it will be maximum when displacement is maximum.
 I 4  107  18
3. B 0  T  18t
2 r 2  0.2
I I
Now, T  2 and T   2 d
MBH M  BH  B 
T BH T 24
Dividing  or  2
T BH  B T 24  18
T   2  0.1s  0.2 s
4. In the circular motion around the earth, the centripetal force on the satellite is a gravitational force.
Therefore, v 2  GM / R , where M is the mass of the Earth, R is the radius of the orbit of satellite and
G is the universal gravitational constant. Therefore, the kinetic energy increases with the decrease in
the radius of the orbit. The gravitational potential energy is negative and decreases with the decrease
in radius.
5. For an adiabatic process,
0 = dU + PdU
or d(a+bPV) + PdV=0
dV dP
or  b  1 b 0
V P
or (b+1) log V + b log P = constant
V b 1 p b  constant
b 1
or PV b
 constant
b 1
 
b
3RT
6. vrms  . According to problem T will becomes 2T and M will become M/2 so the value of vrms
M
will increase by 4  2 times i.e., new root mean square velocity will be 2v .
7. When sources are coherent, then I R  I1  I 2  2 I1 I 2 cos 
At middle point of the screen,   0 then
I R  I  I  2 II cos 0  4 I
When sources are in coherent, then I R '  I1  I 2  I  I  2I
I R 4I
 =2
I R' 2I
l
8. T 2  4 2
g
4 2l
g
T2
g l T
  100   100  2  100
g l T
g
  y  2x
g
9. Using perpendicular axis theorem I = I1 + I2 and I = I3 + I4 also, I1 = I2 = I3 = I4 hence option 4 is
wrong.
10. We know that
1
PB  PA   2 a 2
2
PD  PA   ga
1 1
PC  PD   2 a 2  PA   ga   2 a 2
2 2
Therefore,
Pc  PA for all the values of  and PB  PD only
2g
If  
a
^
11. For p k it is equatorial point
1 P ^
 E1  k 
4 0 1  
P^
For k it is axial point
2

P   ^
 2 k 
1 2    1 p k^
 E2 
4 0 23 4 0 8
7p ^
 E  E1  E2   k
32 0
mv
12. Radius of circular orbit R 
qB
2mKE 2mT
 
qB qB
2
If T becomes double & ‘B” becomes tripled then radius becomes R
9
13. de  B  x  dx
3L
e  B  x dx
2L
5 B L
2

2
14. I d  1mA  103 A
C  2 F  2  106 F
d dV
I D  I C   CV   V
dt dt
dV I D 103
Therefore,    500 Vs 1
dt C 2  106
Therefore, applying a varying potential difference of 500 Vs-1 would produce a displacement current
of desired value.
15. Radius of circular path described by a charged particle in a magnetic field is given
2mK q 2 B 2 r 2  e  eB 2 r 2
by r  ; Where K = Kinetic energy of electron  K   
qB 2m m 2
2
1  1 
 105   1  8  10 20 J  0.5eV
2
  1.7  1011  1.6 10 19  
2  17 
 12375 
By using  W0  E  K max    eV  0.5eV  4.5eV
 2475 

16.
1. As mutual repulsive force between the particles is internal for the system and as there is no other
external force on the system, linear momentum of the system is conserved in any direction.
2. As the forces on the particles due to one on the other are equal in magnitude. Opposite in direction
and act along the line joining them always, net torque on the system due to these forces about any
point in space in zero. Therefore angular momentum of the system remains constant about any point
in space.
3. As center of mass of the system lies on the line joining the particles always and force on any of
them is passing through C.M always, torque due to this force on any particle about C.M is zero.
Hence angular momentum of any particle about C.M is conserved individually.
4. About any other point except C.M, torque on any individual particle is not zero. Hence angular
momenta of individual particles change but total angular momentum of the system remains constant.
17.
F  mg sin   mg tan   is small  .

dy 2x
i.e, F  mg   mg 
dx 40
x 1
 a    
2 2
18. Potential at any point inside the shell = potential at any point on the surface
 potential at A = potential at C due to ‘q’ and induced charges =
1 q 1 q 1  q  1 q
   
4 0 r 4 0 R 4 0  R  4 0 r
19. For A  B  C   ,  is the positive. This is because Eb for D and E is greater then Eb for F.
20. It is a case of resonance

X L  X C
Z R
Vrms 200
 I rms    2A
Z 100
 Pav  I rms
2
R  4  100  400W
21. Change in linear momentum P   Fdt
 
15
15  v f  u    40cos   t dt
0  10 
40  sin  / 10  t 
15
400
v f  4     4   1  12.5 m / s
15   / 10  0 15

22. ‘O’ is off axis for axis for both the parts. Size of object for upper and lower parts is
2mm and 1 mm respectively.
1 1 1
 
f v u
1 1 1
   v  60Cm
20 v 30
v
m 2
u
 Distance between the two image points is 4 + 2 + 1 + 2 = 9mm
23. Because the collision is perfectly inelastic, the two blocks stick together. By conservation of linear
momentum.
v
2 mV  mv orV 
2
By conservation of energy,
1 v2
2mgh   2mv 2 or h 
2 8.g
 x8
T TL L
24. V & Y &   
 AL L
T Y L Y  
V   
A L 
1.3  1011  1.7  105  20
V  70 m / s
9  103
Cdl
25. dR 
l
l 1
dl dl
0 C l  C l l
1
2 l l0  2 l
l
2 l  22 l
4 l 2
1
 l   0.25m
4
26. The forward biased resistance of a diode is
V 0.7  06
R 
I 15  5   103
01
R  10
10  10 3
27. Energy required to remove first electron is 24.6 eV. After removing first electrons from this atom, it
will become He+
E1   13.6  2 
2
 as EZ 2
and Z  2 
= - 54.4 eV
 Energy required to remove this second electrons will be 54.4 eV.
 Total energy required to remove both electrons
= 24.6 + 54.4
= 79 eV
1 12
28. Req = 0.3   0.3   1.5
1 1 1 10
 
2 4 12
Applying ohm’s law,
V  IR
1.5
I  1A
1.5
29. FBD of the spherical ball
For translation equilibrium

T cos150  Mg   N
T sin150  N
Alsot  L
   NR  t  I0
2
MR 20
t  5
 NR
Solving we get t = 20 s
1
30.  I 0  11
4
2
 I 0  27
4
2  1
 16 cm
4
 1 1 
V     0.64 m
 256 512 
 V  512  0.64 m / s
 328 m / s
CHEMISTRY
31. A) SO3 & CO3 ; Both are sp 2 & planar triangular
B) SO32  & NH 3 ; Both are sp3 & pyramidal

C) PCl5 : sp 3 d & trigonal bipyramidal
POCl5 : sp 3 & tetra hedral
D) XeF2 : sp 3 d & linear
ClF3 : sp 3 d & T–shape
As per M.O.T

32. B2 : By distributing 10 electrons only two electrons in   B .M .O are extra left without cancelling
with A.B.M.O electrons
C2 : By distributing 12 electrons, only 4 electrons in  B.M .O are extra left without cancelling with A
B M O electrons
N2 : By distributing 14 electrons only 4 electrons & 2  electrons in B M O & extra left without
cancelling with A B M O electrons.
 22.44
33. No of moles of M  g  formed =
374
 0.06 moles  0.06  6.023  10 23
 3.613 1022 atoms

At.wt.
 22.44  374
8
 At .Wt .  133.33 g .
34. Due to small B.L of N–N bond lp–lp repulsions on ‘N’ weaken the bond.
35. 0.1 moles of the complex – 28.7 g of AgCl
1 mole gives of complex – 287 g of AgCl
– 2 moles of AgCl
 2 Cl ions should be ionisable.
-
36. The complex cannot show hydration isomerism as no H 2O ligands are present.
37. The colour of KMnO4 is due to charge transfer phenomenon
38. nm.eq NH 3  nm.eq H 2 SO4
 10  1  2  20 meq of NH3 = 20 m mol of NH3
1400  neq NH 3
%N 
wt.of organic compound
1400  20  10 3
  56% .
0.5
39.
40. Reactivity order IV > I > III > II > V on the basis of R and I effect of associated groups.

41.
CH2 CH=CH2 Friedal- craft's aklylation
A=
CH2 CH2 CH2 OH Hydroboration -oxidation

B=
C=
42.
43. Cleavage of the double bond by Ozonolysis, iodoform Rxn, dry distillation of calcium salts to give
cyclopentanone, followed by wolf–kishner reduction to give cyclohexane.
44. Benzyllic oxidation to give potassium salt of Benzoic acid, followed by acidification to give Benzoic
acid.
45. Gabriel pthalamide synthesis
46. Keratin and myosin are fibrous proteins and insoluble in H 2O .

4
1.382  10  1000
47.  61.9  76.3 
S
 S  10 3 M .
48. T f  K f  m
50 1000
9.3  1.86  
62 x
 x  161.29 g
 Amount of ice separated  200  161.29
 38.71g
49. Required energy = I1  I 2
I1  24.6eV
I 2  I H  Z 2  13.6  22  54.4eV
 E  24.6  54.4  79 eV
50. 3A B
t  4 min; a  3 x  x
a
 4x  a  x 
4
 At 4 min 75% of first order is completed.
2t1 t
 t75%   1  2 min .
2 2
51. X  12  Mg  ; Y  15  P 
52. conceptual
53. Greater the stability of carbanion, greater is the rate of decarboxylation.
Except  CH 3 3 C  COOH remaining are more reactive than CH 3COOH .
O
||
54. Except  C  O  R , remaining are ring activating groups.
55.
56. FCH2CHO, O2N CH2CHO are more reactive than acetaldehyde.

57. H 2 g   2 H aq   2e
0.059 2
0.413   log  H  
2
 pH  7 .

58. E 0  E Ag
0

/ Ag
 0.06 log K SP
59. Number of moles of CH 3COOH  0.25  0.3  0.075 moles

Number of moles of CH 3COO   0.56  0.3  0.168
 Number of moles of CH 3COO  left = 0.168  0.006  0.162
Final number of moles of CH 3COOH  0.075  0.006  0.081
0.162
 pH  4.7  log .
0.081
60. NH 4 HS S   NH 3 g   H 2 S g 
x 1  x 
At equilibrium
1  x  x  3  x  1atm
 K P  2  1  2 atm 2 .
MATHS
61. Clearly ( x  1) 2  y 2  ( x  3)  i ( y  3)
 y  3 and ( x  1) 2  9  ( x  3) 2
1
 x
4
1
But    3i does not satisfy the given equation
4
 2k  k 4
62. Let f ( x)  x 2   x   0 by the given data, f (0)  0, f (2)  0, f (3)  0
 k 5  k 5
k 4
 0 ………. (1)
(k  5)
k  24
 0 …….. (2)
(k  5)
4k  49
 0 …….. (3)
(k  5)
 49 
From (1), (2) and (3), k   , 24 
 4 
63. Here A  A  A is an Idempotent matrix
2
 A  A2  A3  ..........  A99
Hence ( I  A)99  I   299  1 A
64. We can take 3 cases namely four odd numbers, two odd numbers and zero odd numbers.
Let X be the number of odd numbers chosen
 P(sum is even)  P ( X  4)  P ( X  2)  P ( X  0)
4 2 2 4
2  2 1 1 41
    4 C2        
3  3   3   3  81
65. Let f ( y ) be the inverse of g ( y )
 f '  g ( y )  g '( y )  1
1 1
 f '  g (2)  g '(2)  1   g 1  y    
y 2 f '  1 14

66. S1 and S2 are skew lines
cos x  x sin x
67. I  dx
x.cos x
Put x.cos x  t 2  I  2 x.cos x  c
 
x2 d x2
0 k  x d
d 0 k  x
d
68. I  lim  lim
 0    sin   0 (1  cos )
(By. L. Hospitals Rule)
  2

 
k  2
 lim  2   1  k  4

 0 2sin  / 2
 k
69. Given equation can be reduced to
dy 1
cos y    sin y   et (1  t )
dt  
1  t
dy dv
Put sin y  v  cos y 
dt dt
sin y
Hence the solution is   et  c 
(t  1)
 cot k  y   tan k  y  2  cot 1 k   0 has real roots
1 2 1 3/ 2 2
70.
 D  0  tan 1 k   / 3  k  3 and sum of roots  0, product of roots  0
 k  0 and k  3
71. Let Tr 1 be the Independent term of x then r  6,
 t61  84 3  6
72. Check for what values of x , g '( x)  0 or does not exist
x x   xi
2
i
2
i 
73. Mean  10 and Variance     4
n n  n 

x 2
i
 104 
x i

4
 102
n n n
 n  20
Hence new Variance = 3.96
1
74. From the graph of the functions, the required area = 2   
x  x 2 dx 
2
3
0
75. f  x   x sin x is differentiable at x  0

76. L.H.L.  lim  f ( )  0 and R.H.L.  lim  f ( )  
   
2 2
 lim f ( ) does not exist
  / 2
77.
x1/ 3  2  g  x 
 f  x  52  x  2  x  2  x 
2 3
 x 3  7 x 2  11x  2  f max  3
Clearly 1 is the root  a + b + c + = 0
x x x
 1   2   n 
     ....   1
 n 1   n 1   n 1 
0
1
78. Clearly k  cot 22  2  1
2
Hence 100 (k  1)   141
79. Use L11 . L22  0 for three sides of the triangle
80. Clearly a  2; b  1 and c  2 and
a1a2  b1b2  c1c2

Use   cos1
a a2
1
2
2
81. p 2  q 2  1 or p 2  q 2  0
1
 z 2018  z   z 2019  1
z
So there are total 2020 solutions
82. The inclination of the line L  x  y  1 is 1350 . So the slopes of the other two sides will be
tan 1350  600 
83. E  a  b  2c. a  c . b 
 a  b  2  c . a  c .b  a .b 
2 2
But  a  c  .  b  c   0
 a .c b . c  a .b 1
 E  8  2  10

84. Let the common ratio of G.P. be ‘ r ’ then x2  x1r , x3  x1r 2 , x4  x1r 3
And x1  x2  1, x1 x2  K , x3  x4  4, x3 . x4  L
 r  2 and x1  1
 K  2, L  32
t t2 1 t3
85. 2t 4t 2 1  8t 3  10
3t 9t 2 1  27t 3
1 1 1 1 1 1
 x 2 4 1  x 2 4 8  10
3 6
1 9 1 3 9 27
 6 x 6  x3  5  0  x 3  5 / 6 or x3  1
dt
86. I  2
dt
 1 
t 2 1  1  
 t
1
Put 1   U
t
 1 2
Then I  2 log 1  1    c
 t 
 1 1 1 
 
 t
2
 f (t ) 
 1
1  1  
 t
35  1
87. The number of un-ordered pairs of subsets of A is  122
2
88. If ‘Q’ is the foot of perpendicular and it divide AB in the ratio  :1 then
 3  4 5  7 3  1 
Q , , 
  1  1  1 
Now PQ perpendicular AB    7 / 4
 5 7 17 
  ,  ,     , , 
3 3 3 
 3  6   9  46
g (2 x) g ( x)
89. Using the derivative from the first principle g 1 ( x)  g (1) 
2x x
g 1 ( x) 1
 
g ( x) x
n (1 y )
90. Write (1  y ) 1/ y
as e y
and evaluate the limit


MISTAKES
MATHS
PHYISCS
CHEMISTRY
SECTION – I
1. A very broad elevator platform is going up vertically with a constant acceleration 1 ms-2.
At the instant when the velocity of the lift is 2 m/s, a stone is projected from the platform
with a speed of 20 m/s relative to the platform at an elevation 300. The time taken by the
stone to return to the floor will be  g  10 m / s 2 
A) 30 sec B) 70 sec C) 20 sec D) 90 sec

11 11 11 11
2. The force exerted by a compression device is given by F  x   kx  x  l  for 0  x  l , where

l is the maximum possible compression, x is the compression and k is a constant. The
work required to compress the body by a distance d will be maximum when:
l
A) d  l B) d  C) d  l D) d  l
4 2 2
3. Figure shows a short magnet executing small oscillations in a uniform magnetic field
directed into page and magnitude 24  T. The period of oscillation is 0.1 s. When the
key K is closed, an upward current of 18A is established as shown. The new time period
is_____(Neglect the effect of earth’s magnetic field)
(Needle oscillates in plane normal to the page)
A) 0.1 s B) 0.2 s C) 0.05 s D) 0.4 s

4. A satellite is moved from one circular orbit around the earth to another of lesser radius.
Which of the following statement is true?
A) The kinetic energy of satellite increases and the gravitational potential energy of
satellite – earth system increases.
B) The kinetic energy of satellite increases and the gravitational potential energy of
satellite – earth system decreases.
C) The kinetic energy of satellite decreases and the gravitational potential energy of
satellite – earth system decreases.
D) The kinetic energy of satellite decreases and the gravitational potential energy of
satellite – earth system increases.
5. The relation between internal energy U, pressure P and volume V of a gas in an

adiabatic process is U = a + bPV.
Where a and b are constant. What is the effective value of adiabatic constant  ?
A) a B) b  1 C) a  1 D) b
b b a a
6. The root mean square speed of the molecules of a diatomic gas is v. when the
temperature is doubled, the molecules dissociate into two atoms. The new root mean
square speed of the individual atom is
A) 2v B) v C) 2v D) 4v
7. In young’s double slit experiment, the two slits are coherent sources of equal amplitude
and wave length  . In another experiment with the same setup, two slits are sources of
equal amplitude ‘A’and wavelength  , but are incoherent. The ratio of intensities of light
at the midpoint of the screen in the first case to that in second case, is
A) 2 : 1 B) 1 : 2 C) 3 : 4 D) 4 : 3
8. The relative error in calculating the value of g from the relation T  2 l is

g
(given the relative errors in calculating T and l are x and  y respectively)

A) x  y B) 2 x  y C) 2 x  y D) x  2 y

9. I1, I2, I3 and I4 are the moment of inertial of square plate about the axis marked 1, 2, 3
and 4 respectively and I is the moment of inertia about an axis passing through O and
perpendicular to the plate, Relation between them is given as
i) I = I1 + I2 ii) I = I1 + I3
iii) I = I2 + I4 iv) I = I1 + I2 + I3 + I4
Which of ṭhe following options is incorrect?
A) (i) and (iii) B) (ii), (iii) and (iv) C) Only (ii) D) Only (iv)
10. A cylindrical container filled with a liquid is being rotated about its central axis at a
constant angular velocity  . Four points A, B, C and D are chosen in the same plane
such that ABCD is a square of side length a and AB is horizontal while BC is vertical.
A and D lie on the axis of rotation. Let the pressure at A, B, C and D be denoted by
PA, PB, PC and PD respectively. Now, consider the following two statements.
2g
(i) PC> PA for all values of  (ii) PB> PD only if  
a
Which of these options is correct?
A) Both (i) and (ii) are correct B) (i) is correct and (ii) is incorrect
C) (ii) is correct and (i) is incorrect D) Both (i) and (ii) are incorrect

Two point dipoles p k and p k are located at (0, 0, 0) and (1 m 0, 2m) respectively. The
^ ^
11.
2
resultant electric field due to the two dipole at the point (1m, 0, 0) is…
9p ^ 7p ^ 7p ^ 11 p
A) k B)  k C) k D)
32 0 32 0 32 0 32 0
12. A electron having kinetic energy T is moving in a circular orbit of radius R


perpendicular to a uniform magnetic induction B .If kinetic energy is doubled and
magnetic induction tripled, the radius will become
3R 3 2 4
A) B) R C) R D) R
2 2 9 3
13. A metallic rod of length ' l ' is tied an insulating string of length 2l and made to rotate
with angular speed  on a horizontal table with one end of the string fixed. If there is a
vertical magnetic field ‘B’ in the region, the e.m.f. induced across the ends of the rod is:
3Bl 2 4Bl 2 5Bl 2 2Bl 2

A) B) C) D)
2 2 2 2
14. In order to establish an instantaneous displacement current of 1 mA in the space between
the plates of 2  F parallel plate capacitor, the time varying potential difference need to
apply is
A) 100 Vs-1 B) 200 Vs-1 C) 300 Vs-1 D) 500 Vs-1
0
15. Light of wavelength 2475 A is incident on barium. Photoelectrons emitted describe a
1
circle of maximum radius 100 cm by a magnetic field of flux density 105 Tesla.
17
Work function of the barium is (nearly (Given e  1.7 1011 ), hc  12375  eV  A0 
m
A) 1.8 eV B) 2.1 eV C) 4.5 eV D) 3.3 eV

16. Two positively charged particles are projected along two parallel lines on a smooth
horizontal surface as shown. Which of the following statement is incorrect
corresponding to their subsequent motion? [Before any collision (except between the
particles) takes place]
A) The linear momentum of the system of particles is conserved in any direction

B) The angular momentum of the system of particles is conserved about any point in
space
C) The angular momentum of each particle is individually conserved about their center
of mass
D) The angular momentum of each particle is individually conserved about any point in
space
17. A particle of mass 5 x 10-5 kg is placed at lowest point of smooth parabola
x2  40 y  x and y in m  . If it is constrained to move along parabola, angular frequency of
small oscillations (in rad/s) will be approximately (g=10 m/s2)
1
A) 2 B) 10 C) D) 5
2
18. A point charge q is placed at a distance r from the center of a thin metallic neutral
spherical shell of radius R as shown in fig. electric potential at point A is
1 q 1 q 1 q q 1 1 
A) B) C) D)   
4 0 R 4 0 r 4 0 R r
2 2 4 0  r R 

19. There 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
(i) A  B  C   (ii) C  A  B   (iii) D  E  F   and (iv) F  D  E  
Where  is the energy released? In which reactions is  positive?
A) (i) and (iii) B) (ii) and (iv) C) (ii) and (iii) D) (i) and (iv)
20. An R-L-C series circuit with 100  resistance is connected to an AC source of 200 V
and  = 300 rad/s. When only capacitor is removed, the current lags behind voltage by
600. When only inductor is removed, the current leads voltage by 600. The power
dissipated in the R-L-C circuit is
A) 200 W B) 400 W C) 200 3 W D) 100 W
SECTION-II
21. A 15 kg block is initially moving along a smooth horizontal surface with a speed of
v  4 m / s to the left. It is acted by a force F, which varies in the manner shown.
If the velocity of the block at t = 15 seconds is ‘X’. Then the value of  X  =__
([ ] – greatest integer function)
 
Given that, F  40cos  t
 10 
22. A convex lens of focal length f = 20 cm is cut into two equal pieces and the pieces are
separated by 3mm as shown in the figure. A point object O is placed at a distance of 30
cm. The distance between the two image points formed will be (in mm)
23. A ball of mass m moving horizontally with a velocity  strikes the bob of a pendulum at
rest. The mass of the bob is also m. If the collision is perfectly inelastic, the height to
v2
which the system will rise is given by h  , then the value of x is
x.g
24. A copper wire is held at the two ends by rigid supports. At 300 C, the wire is just taut,
with negligible tension. Find the speed of transverse waves (in m/s) in this wire at 100 C
in decimeter/second [Given, for copper: Young’s modulus = 1.3 x 1011 N/m2, coefficient
of linear expansion = 1.7 x 10-5C-1, density = 9 x 103 kg/m3.]
25. In a meter bridge, the wire of length 1 m has a non - uniform cross section such that, the
dR
variation of its resistance R with length l is dR  1 . Two equal resistances are
dl dl l
connected as shown in the figure. The galvanometer has zero deflection when the jockey
is at point P. The length AP is X (in m), then 100X = …
26. The forward – bias voltage of a diode is changed from 0.6 V to 0.7 V, the current
changes from 5 mA to 15 mA. What is the value (in  ) of the forward bias resistance?

27. The binding energy of an electron in the ground state of the He atom is equal to
E0  24.6eV . Find the energy required (in eV) to remove both electrons from the He atom.
28. In the circuit shown in the adjoining figure, the reading of ideal ammeter A (in ampere)
is:
5
29. A solid spherical ball of radius m is connected to a point A on the wall with the help
9
of a string which makes an angle   450 with the vertical. The sphere can rotate freely
about its central axis and it is set into rotational motion against the vertical face of the
wall with an angular velocity 100 rad s-1. In how much time (in s) will it come to rest?
   0.1& g  10 m / s 2 
30. A resonance tube is old and has jagged end. It is still used in the laboratory to
determine velocity of sound in air. A tuning fork of frequency 512 Hz produces first
resonance when the tube is filled with water to a mark 11cm below a reference mark,
near the open end of the tube. The experiment is repeated with another fork of
frequency 256 Hz which produces first resonance when water reaches a mark 27 cm
below the reference mark. The velocity of sound in air (in m/s), obtained in the
experiment, is close to_____

SECTION – I
31. In which of the following pairs, the hybridization of central atom is same, but shape is
not the same?
A) SO3 , CO32 B) SO32 , NH 3 C) PCl5 , POCl3 D) XeF2 , ClF3
32. Number of  –bonds in B2 , C2 , N2 respectively as per molecular orbital theory is :
A) 1,2,3 B) 0,1,2 C) 1,2,2 D) 1,1,2
33. 22.44 kJ of energy is required to convert 8 g of gaseous metal, M to M+(g). If the first
ionisation energy of the metal is 374 kJ/mol, select the incorrect statement from the
following.
A) 0.06 moles of gaseous M+ are formed
B) Same energy can convert all the M+ to M2+
C) Gram atomic mass of the metal is 133.33 g
D) 3.613  1022 atoms of M are converted to M+
34. Assertion(A):The single N–N bond is weaker than the single P–P bond
Reason(R):High inter electronic repulsion of the non–bonding electrons due to the small
N–N bond length
below:
A) Both (A) and (R) true but (R) is not the correct explanation of (A)
B) (A) is false but (R) is true.
C) Both (A) and (R) true and (R) is correct explanation of (A).
D) (A) is true but (R) is false.

35. A coordinate complex of formula CrCl3 .6H 2O has green colour. 1L of 0.1 M solution of
the complex when treated with excess of AgNO3 gave 28.7g of white precipitate. The
formula of the complex would be: [At.wt. Ag : 108 amu; At.wt. of Cl = 35.5 amu]
A) Cr  H 2O 6 Cl3  B) CrCl  H 2O 5  Cl2 .H 2O
C) CrCl2  H 2O 4  Cl.2 H 2O D) Cr  H 2O 3 Cl3 
36. Which type of Isomerism cannot be shown by Co  en 2  NO2 2  Cl ?
A) Geometrical B) Ionisation C) linkage D) hydrate

37. Which of the following does not involve d  d transition for causing colour ?
4
A)  Fe  CN 6  B) CuSO4 .5H2O C) KMnO4 D) CrCl3 .4 NH 3
38. The NH3 evolved from 0.5 gm of the organic compound in KJeldhal’s estimation of
Nitrogen neutralizes 10 ml of 1M H2 SO 4 . Identify the incorrect statement out of the
following.
A) Percentage of nitrogen in the organic compound is 56%
B) 20 milli moles of NH3 is produced
C) 10 milli moles of NH3 is produced
D) if the evolved NH3 were neutralized by 10 ml of 1M HCl, the % of nitrogen, would
have been 28%.
39.
CH3
KMnO4 H HBr (C)

CH3 CH (A) (B)
  H 2O  ROOR (Major)
CH3
The product “C” in the above reaction formed is
A) B) C) D)

Br Br Br Br Br
40. O NH
O
O
(I) (II) (III) (IV) (V)
Ease of SN1 reactions among these compounds upon treatment with aqueous NaOH will
be in the order as:
A) (I) > (II) > (III) > (IV) > (V) B) (IV) > (I) > (III) > (II) > (V)
C) (I) > (IV) > (III) > (II) > (V) D) (V) > (IV) > (III) > (II) > (I)
41.
 A 
 i  BH /THF
3
 ii  H O / OH
2 2
  B
 B  
HF
C 
The compound(C) is
A) B)
C) D)
42. When neopentyl alcohol is heated with an acid, it slowly converted into an 85 : 15
mixture of alkenes A and B, respectively. Then, the ratio between number of hyper
conjugated structures for A and B is :
A) 5 : 9 B) 5 : 1 C) 1 : 5 D) 9 : 5

43. End product in the following sequence of reactions is :
A) B) C) D)
44.
A) B)
C) D)
45. Given below are two statements, one is labelled as Assertion (A) and other is labelled
as Reason(R):
Assertion (A): Gabriel phthalimide synthesis cannot be used to prepare aromatic
primary amines.
Reason (R): Aryl halides do not undergo nucleophilic substitution reaction at room
temperature.
below:

A) Both (A) and (R) true but (R) is not the correct explanation of (A)
B) (A) is false but (R) is true.

C) Both (A) and (R) true and (R) is correct explanation of (A).
D) (A) is true but (R) is false.
46. Which of the following statement is false?
A) In fibrous proteins, poly peptide chains are held by hydrogen & disulphide bonds.
B) In Globular proteins, chains of polypeptides coil around to give a spherical shape
C) Keratin & myosin are fibrous proteins and soluble in water
D) Insulin & albumin are globular proteins and soluble in water
47. The conductivity of a saturated aq.solution of AgCl at 298 K is found to be
1.382  10 2  1 m 1 . The ionic conductance of Ag  and Cl  at infinite dilution are
61.9  1cm 2 mol 1 and 76.3 1cm 2 mol 1 respectively. The solubility of AgCl is
A) 1101 mol L1 B) 1102 mol L1 C) 1103 mol L1 D) 1.9 105 mol L1
48. 50 g of antifreeze (ethylene glycol) is added to 200 g water. What amount of ice will
separate out at 9.30 C ?  K f  1.86 K kg mol 1  .
A) 38.71 mg B) 42 g C) 38.71 g D) 42 mg
49. An energy of 24.6eV is required to remove the first electron from helium atom. The
energy required to remove both electrons from helium atom is
A) 54.4eV B) 79 eV C) 49.2 eV D) 51.8 eV
50. For the first order reaction 3 A  B concentration varies with time as shown in the
adjacent graph. The half – life of the reaction would be
A) 4 minutes B) 2 minutes C) 6 minutes D) 8 minutes

SECTION-II
51. When the ionization energy Vs atomic number is plotted for the elements, of atomic
number 11 to 18, two peaks are observed for the element X and Y in between the curve.
What is the difference between atomic number of element X and element Y?
52. For a low spin , Cr 2  complex, the value of spin only magnetic moment is x B.M in an
octahedral field. Then the value of x2 is_____
53. The number of compounds among the following more reactive than acetic acid towards
decarboxylation by soda-lime is
HC  CCH 2COOH , H 2C  CH  COOH , PhCH 2COOH , CCl3COOH , Me3CCOOH
54. How many of the following groups activates benzene ring towards electrophilic aromatic
substitution?
O
||
 NHCOR,  OCOR,  C  O  R,  NR 2 ,  NH 2 ,  OH,  OR
55. .
The number of iodoform molecules produced per molecule of the reactant in above
reaction is ______
56. How many of the following are more reactive than acetaldehyde towards nucleophilic
addition?
FCH2CHO, O2N CH2CHO , CH3CH2CHO, CH3COCH3, PhCHO, PhCOCH3
57. At what pH the oxidation potential of hydrogen electrode will be 0.413 V?
 2.303RT 
  0.059V  (Given: PH 2  1 atm )
F 
58. If E Ag
0

/ Ag
 0.8V ; K sp  AgCl   10 10 M 2 and ECl0  / AgCl / Ag is ‘x’ volts, then the value of 10x is:
 2.303RT 
  0.06V 
F 
59. A buffer is 0.25 M in CH 3 COOH and 0.56 M in CH 3 COONa . What is the value of pH if
0.006 mol of HCl is added to 0.300 L of a buffer solution?
pK a  CH3COOH  4.7  log10 2  0.3 
0
60. NH 4 HS S  is added to a closed vessel containing H 2 S g  at 1 atm and 27 C. If the total
pressure at equilibrium is 3 atm at 270C, then, the value of K P [in (atm)2] is :

SECTION – I
61. If ‘  ’ is a complex number satisfying the relation   1    3(1  i) . Then  equals to
1 1
A)   3i B)  3i
4 4
1
C)  3i D) No such complex number exists
4
62. The greatest integral value of k such that both the roots of the equation
 k  5 x 2  2kx  (k  4)  0 are positive, one root is less than 2 and the other root lies
between 2 and 3 is
A) 13 B) 14 C) 23 D) 24
 3 1
 then  I  A  (where I is a unit matrix of order 2) equals to
99
63. Let matrix A  
 6 2 
A) I  298 A B) I  299 A C) I   299  1 A D) I   299  1 A
64. In a bag there are three tickets with number 1, 2, 3. A ticket is drawn at random and the
number is noted and put back in the bag, this is repeated for four times. Then the chance
that the sum of those numbers is even is
39 40 41 42
A) B) C) D)
81 81 81 81
d
65. If g ( y )  y 5  2 y 3  3 y  4, then the value of 28
dy
 g 1 ( y)  at y  2 is
1
A) 2 B) 1 C) D) 2
14
x  4 y  5 z 1 x  2 y  1 z
66. Given lines   ,  
2 4 3 1 3 2
S1: The lines are intersecting
S2: The lines are not parallel
A) Both S1 and S2 are true B) S1 is true and S2 are false
C) S1 is false and S2 is true D) Both S1 and S2 are false

 cos   
67.  
 
cos 
. sin   d 
 
A)  .sin   c B)   .cos   c C) 2  .cos   c D) c  2  .cos 


x2
0 k  x dx
68. lim
  0   sin  
 1 where k  0 then the value of ‘k’ is
1 1
A) B) C) 2 D) 4
2 4
dy tan y
69. The solution of the differential equation   (1  t ) et sec y is, where ‘ c ’ is an
dt (1  t )
arbitrary constant
A) cos y   et  c   t  1 B) cos y   et  c   x  1
C) sin y   et  c   t  1 D) sin y   et  c   t  1
70. The minimum value of k for which the quadratic equation
 cot k  y   tan k  y  2  cot 1 k   0 has both positive roots is:  k  I 

1 2 1 3/ 2 2
A) 1 B) 2 C) 3 D) 4
71. If  3   6  2 then the maximum value of the independent term of x in the expansion of
 x   x 1/6    0,   0  is
1/3 9
A) 42 B) 68 C) 84 D) 148
72. The set of critical points of the function g ( x)  ( x  2)2 / 3 (2 x  1) is
1
A) {1} B) {1, 2} C) {1, 2} D)  , 1
2 
73. The mean and variance of the marks obtained by the students in a test are 10 and 4
respectively. It is known that one of the students got ‘12’ instead of 8. If the new mean
of the marks is 10.2 then the new variance is equal to
A) 4.04 B) 4.08 C) 3.96 D) 3.92

74. The area enclosed between the curves y  x 2 and y  x is
1 2 4
A) B) C) D) 2
3 3 3
75. Statement 1: f  x   x sin x is differentiable at x  0
Statement 2: If f  x  is not differentiable and g  x  is differentiable at x  a , then

f  x  .g  x  can still be differentiable at x  a
A) Statement 1 is true, Statement 2 is true

B) Statement 1 is true, Statement 2 is false
C) Statement 1 is false and Statement 2 is true
D) Statement 1 is false and Statement 2 is false
 1 ;   /2

76. If f ( )   sin cos   , where {.} represents fractional part function, then
 (   / 2) ;    / 2

A) f ( ) is continuous at    / 2
B) lim f ( ) exists but not continuous at    / 2
 / 2
C) lim f ( ) does not exist

 / 2
D) lim f ( )  1
 / 2

A) g(x) = 2 – x1/3 and f(g(x)) = - x + 5x1/3 – x2/3, the local
P) 0
maximum value of f(x) is
z 3 
B) No. of points of intersection of the curves arg    and
 z 1  4 Q) 1
z 1  i   z 1  i   4  0
C) If f(x) = ax3 + bx2 + cx + d, (a, b, c, d  Q) and two roots of

f(x)=0 are eccentricities of a parabola and a rectangular hyperbola, R) 2
then a + b + c + d =
D) Number of solution of equation 1x + 2x + 3x …. + nx = (n + 1)x
S) 3
are
A) A – Q ; B – S ; C – R ; D – P B) A – S ; B – Q ; C – P ; D – Q
C) A – S ; B – Q ; C – R ; D – Q D) A – S ; B – Q ; C – P ; D – R

440
 cos r o
The value of 100 (k  1)  where [ x] represents the G.I.F. and k 

r 1
78. 440 is
 sin r
r 1
o
A) 144 B) 142 C) 141 D) 140
79. The sum of possible integral values of k for which the point P(0, k) lies on or inside the
triangle formed by the lines y  3x  2  0, 3 y  2 x  5  0 and 4 y  x  14  0 is
A) 4 B) 5 C) 6 D) 7
x 1 y  1 z x  1 y  3 z 1
80. The acute angle between the lines   and   where a  b  c
a b c b c a
and a, b, c are the roots of the equation t 3  t 2  4t  4  0 is
 63   3 
C) tan 1  
4 2
A) sin 1   B) cos 1 D) cos 1  
 9  9 3  13 
SECTION-II
81. If ( p  iq) 2018  p  iq . Then the number of real ordered pairs ( p, q) that satisfy the given
equation is N. Then   where [ x] represents G.I.F. equals to _________

N
100 
82. An equilateral triangle has its centroid at the origin and one side is x  y  1 , then the sum
of the slopes of the other two sides is
If a  b  2, c  1,  a  c  .  b  c   0 . Then the value of a  b  2c .  a  b  is equal to

2
83.
84. If x1 , x2 are the roots of x 2  x  K  0 and x3 , x4 are the roots of x 2  4 x  L  0 such that
x1 , x2 , x3 , x4 are in G.P. Then the product of the integral values of K and L is ______

t t2 1 t3
85. The total number of distinct real values of ‘t’ for which 2t 4t 2 1  8t 3  10 is _____
3t 9t 2 1  27t 3
86. If 
dt  1
 
 2 log 1  1    f (t )  c then 3 lim f (t )  _________
 
2
t  t (1  t )  t t 
87. Let A  1, 2, 3, 4, 5 . The number of unordered pairs of subsets P and Q of A such that
P  Q   is ‘n’. Then the sum of digits of n is __________
88. If  ,  ,   is the foot of perpendicular drawn from the point P(1, 0, 3) to the line joining
the points A(4, 7, 1) and B(3, 5, 3) . Then the value of 3  6  9  ______
If g   
xy g ( x) . g ( y ) g (3)
89. ; x, y  R, g (1)  g 1 (1) then 1  ________
 2  2 g (3)
(1  y )1/ y  e1
90. The value of real number ‘k’ for which the right hand limit lim is a non-
y  0 yk
zero real number is _________

KEY SHEET
PHYSICS
1 C 2 C 3 B 4 B 5 A
6 C 7 B 8 C 9 B 10 B
11 A 12 C 13 C 14 C 15 A
16 A 17 C 18 A 19 A 20 A
21 54 22 2 23 13 24 7 25 2
26 30 27 4 28 8 29 1410 30 34
CHEMISTRY
31 D 32 B 33 A 34 D 35 C
36 C 37 C 38 C 39 D 40 D
41 A 42 D 43 A 44 A 45 D
46 B 47 C 48 C 49 A 50 C
51 8 52 125 53 60 54 231 55 1565
56 6 57 6 58 6 59 3 60 3
MATHEMATICS
61 D 62 A 63 A 64 D 65 B
66 D 67 B 68 A 69 C 70 B
71 D 72 A 73 A 74 B 75 A
76 B 77 D 78 C 79 A 80 A
81 3 82 6 83 0 84 141 85 2
86 3 87 1 88 2 89 13 90 1
SOLUTIONS
PHYSICS
v '2 v2
1. L'  ,L 
a' a
2
L'  v'  a  2  1
      3 / 3
L  v   a '     
m' F ' a 1 1 1
    2 2
m F a '    
Time =Velocity /Acceleration , i.e.,
Momentum =Mass X Velocity
V2 dv
2. ac  & at  , a Net  a c 2  a t 2
r dt
3. Vx  4 sin 30 and Vy  e U y  0.5  4 cos 300 
0
 /2
L ml 2 6v m
4. mv 0  w  w  0 & F   . w 2 x dx
2 12  0

5. Conceptual
6. Conceptual
2mVN cos 
7. 
A
4 2
ml
I 3 8l
8. T  2  2  2
mgd l 9g
2mg .3
4
9. E is uniform and conservative, hence total energy ‘T’ is constant, K increases, U decreases.
 A
10. C 0 & C  2C1  2C2  C3
d
 V
11. i 
R  r 2R
 M
12. B 0 3
4 d
 I 
13. d  0 .a d x & q 
2 x R
V  VR 2   VL  VC  & for LR circuit Z  R 2  X L 2
2
14.
1
15. I 0 E 2 C & Sˆ  Eˆ  B
ˆ
2
1 1 1
16.  
f v u
d d
17.  n
 
 L    
e e nh
18. m
2m 2m  2 
E
19. i
R1  R 3  R d

20.
12   3  6  2 
21. a  0.2 & for 2kg
5
T  5  0.4  T  5.4N

22. W – E theorem , W   f.dx
0
I
23. K
M
Fs  Fm  M e a1  2G  Mm 
24.  a1  a 2 
Fs  Fm  Mea 2  r2
 dh   hdgr
4
25. A   
 dt  8 
 I  p 2
26. dB  10 log10   & I 
 I0  2pV
 i
27. B1  B2  0 & B  B12  B2 2
4 d
h h
28.    x 8
mV 2mqV
NE mN 0 E
29.   .
t M t
30. d  P.S.R.   C.S.C  L.C. & S.A.  2rL

CHEMISTRY
31. P: Markonikoff product with rearrangement
Q: Antimarkonikoff product
R: Markonikoff product without rearrangement
32. Conceptual
33. Conceptual
34.
CH3 CH CH2 H2SO 4 CH3 CH CH3 CH3

CH
CH3
(P)
O2 /h
CH3
OH +
CH3 C O H2O/H C O OH
CH3 CH3
(S) (R) (Q)
35. Lassaigne’s test for nitrogen is given by those compounds in which N is bonded to carbon.
36. CH2Cl2, NF3 and ClO2 have non–zero dipole moment.
PCl3F2 has zero dipole moment
F
Cl
Cl
P
Cl
Obviouslly res.  0
+4
37. Even though Ce is favoured by its noble gas configuration, it is strong oxidant, reverting to
common oxidation state of +3. E0 of Ce4+/Ce+3= 1.74V suggests that Ce4+ can oxidize even water(but
reaction is slow)
 
38. Co NH3   NO2   and Co  NH3 4 NO2  ONO   are linkage isomers.
 2 2

Co NH  NO   exhibits geometrical isomerism but both the geometrical isomers are optically
 3 4 2 2

inactive Co  NH3 4 NO2 2  Cl and Co  NH3 4 NO2  Cl NO2 are ionisation isomers.
39. 

Sb2S3  3
 2Sb  3s
2
Eq. Conc. 2S 3S (Suppose solubility of Sb2S3 is S moles L–1 )

 3s 3  108s5  108  105 
5
K sp   2s 
2
 Ksp = 108  10–25
40. For isotonic sol  1   2

8.21 = C x 0.0821 x 310 .
C = 0.323 mol/lit ; wt of glucose = 0.323 x 180 = 58.14

41.
42. Conceptual
43.
Br Br Br
Br Br Br
HNO3 H2SO 4
NO2
NO2
44. Conceptual
45.  NH 2 acts as both ortho, para and meta directing group in the presence of acid due to salt formation.
46. O22  BO  1 O2  BO  1.5 O2  BO  2 O2  BO  2.5
47. Conceptual
48. No. of angular nodes = 

No. of radial (spherical) nodes = n    1
The no of peaks in radial probability distribution curves= n  
49. Conceptual
50. Inert pair effect
51. Conceptual
52. NaHC2 O4 & KMnO4
M1 100 0  0.1

5 2
NaHC2O 4 & NaOH
M1 100 0.1 V

1 1
53 - 180 = - RE – 240
RE = 180 – 240 = - 60Kcal mol 1
Rate 0.693
54. k  3 103 
 A t1/ 2
2
0.06  Fe 2 
55. E cell  E cell 
0
log
 PO2  H  
4
4
56. Conceptual
57. Conceptual
58.   n  n  2  and n = 5 or Mn 2
 3x 
3
59. Kc   2  103
 2  x  4  2x 
2
x : no. of moles ‘A’ dissociated.

60. Conceptual

MATHS
61. Given equation can be written as
x 3  x 2   a 4  4a 2  1 x  a 2  0
, ,  are roots     1,    a 4  4a 2  1,   a 2
Now,
    3                    3    
 1 1 1              
 , ,   
                
     1  1   a 4  4a 2  1 1
        3  a2  2 1 3
     
2
a a
62.
63. Statement-2 is true.

Consider Statement-1.
Let  and  denote the roots of the quadratic x 2  5 x  9  0 .
Then,    , but f    f     0
 f  x  is not one one
 Statement-1 is true.

64. Given
 sin  x  1  a 1  x  
lim  
x 1
  x  1  sin  x  1 
1  x 1  x   1
1 x 4
1 x
 sin  x  1 
 x 1  a 
    1
 lim   
 1  sin  x  1  4
x1
  x  1 
2
1 a  1
  
 2  4
  a  1  1
2
 a  2or 0
Hence, the maximum value of a is 2.
 x 2 1  x 7  
5
65. Required no.of ways = coeff of x in x  x  ......  x

30
 2 3

8 5
 coeff of x in 
30
 1 x 

 
 coeff of x 20 in 1  x 7  1  x   24C 20  5  17C13  10  10C6  826
5 5
66. 1,0,1
R   x, y  : x, y  z , x 2  3 y 2  8
1
For domain of R
Collection of all integral of y ' s
For x  0,3 y  8
2
 y  1,0,1
sin A.sin B.cos C
67.  tan A tan B   cos A.cos B.cos C
1
  sin A.sin B.cosC  cos A.sin B.sin C  sin A.cos B.sin C
cos A.cos B.cos C

1

cos A.cos B.cos C
 
sin B. sin  A  C    sin A.cos B.sin C
1

cos A.cos B.cos C
1  cos 2 B  cos Bsin A.sin C 
1 1  cos A.cos B.cos C

cos A.cos B.cos C
1  cos b  sin Asin C  cos B   
cos A.cos B.cos C

8
3 1

1 4 3 1 1 
68. We have f  x   ax  cos 2x  sin x  cos x
As f   x   0 for any real number x  a  2sin 2x  sin x  cos x ….
 
Let t  sin x  cos x  2 sin  t     2  t  2 .
 4
So the inequality can be written as a  2t  t  2
2
2
 1  17
Let g  t   2t  t  2  2  t   
2
 4 8
The range of g  t  for  2  t    1

2 is g  2  g  t   g    2  2  g  t  
4
17
8
17 17 
So, the range of a can be a  max t  2  a   a   ,   Hence,
8 8 
 m  n least  17  8  25
69. Let E1 ,E 2 ,E 3 be respectively Events that P,Q,R ride the horse A.
A = Event that horse A win the race
1 1 1
P  E1   ;P  E 2   ;P  E 3  
2 3 6
1 2 3
P  A   ;P  A   ;P  A  
 E1  8  E 2  8  E 3  8
3 A 5
P  A    P  E i P   
i 1
 E i  24
70.
xi x i2
1 1
2 4
2 4
3 9
  x i  8;  x i2  18
x   x i  18  8  2 9
2 2
1 1
Variance        4  1
i
log1/2
n  n  4  4 2 2 2

71. We can observe that 3  3i  A but  B
Line
(3,3)
(0,0)
 n  A  B  0
72.
Solving the variable line y  mx  1 with x  2 y , we get

2
x1  1
2m  1
Solving with y  2 x , we get
1
x2   2
m2
Now, y1  y2  m  x1  x2   2
Let the centroid of triangle OAB be  h, k  . Then,
x1  x2
h
3
y  y2 m  x1  x2   2
and k  1 
3 3
3k  2
or m 
3h
2 1
So, 3h  x1 x2  
 3k  2   3k  2 
2  1  2
 3h   3h 
[Using (1) and (2) ]
2 1
or  2
6k  3h  4 6h  3k  2
Simplifying, we get the final locus as 6 x  9 xy  6 y  3x  4 y  0 which is a hyperbola
2 2
passing through the origin, as h  ab and   0 .

2

73.
74. 4km  5m 2  16  k 2 k  5m 2  4km   k 2  16   0 ;m  R

12
  0  16k 2  20  k 2  16   0  4k 2  320  0  k 2  80  k  8  m  & m  4 for
5
12 128
m = 4 ; common difference = 0 & for m = ; common difference   25.60
5 5
75. The family of parabolas is
a3 x2 a 2 x
y   2a  Ax 2  Bx  C
3 2
and the vertex is P   B / 2 A,  D / 4 A    h, k  . Therefore,
a2 / 2 3
h 
2  a / 3
3
4a
a / 2   4a 3  2a  / 3
2 2
and k  
4  a 3 / 3
3 35a
or h   and k  
4a 16
Eliminating a , we have hk  105 / 64 .
Hence, the required locus is xy  105 / 64 .
1 1
log  e  x    log    x 
76. f  x   x ex
 log  e  x  
2
log  e  x    e  x      x  log    x 

   x  e  x   log  e  x  
2
Since log function is an increasing function and e   ,

log  e  x  ,log    x 
Thus,  e  x  log  e  x    e  x  log    x      x  log    x  for all x  0 .
Thus, f   x   0 .
Therefore, f  x  decrease on  0, 
n!
 a  b  c 
n
77. .a p b q c r , p  q  r  n
p!.q!.r!
In statement – 1 p + q + r exceeds n

78.
y  tan 1 x

y
2
At x  

tan 1 x 
2
and
f1x  0
lim
x 
 f  x   f 1
 x   L
79. Let point of intersection be (h,k)
h k a b
   1and ah  kb  1and   1
a b b a
h k
    ah  bk   1
a b
b a
h 2  k 2  hk     1
a b
(A)  I  A   8Co I  8C1IA  8C2 IA 2  .........  8C8IA8
8
80.
 8C0 I  8C1A  8C2 A  ......  8C8A8
= IA  8
C1  8C2  ......  8C8 
8 8
= I + A(2 – 1)  = 2 – 1
 
(B) adj A 1 | A 1 |2 
1
| A |2
  
1 1
adj A 1  1
| A |2  22  4
| adjA |
a11 a12 a13
(C) | A | a 21 a 22 a 23
a 31 a 32 a 33
a11  1a12  2a13  2a11 a12 a13
1
 | B | a 21 a 22  1a 23 = 3
 2 a 21 a 22 a 23 | A |
2
 2
 a 31 a 32 a 33  a 31 a 32 a 33
Hence, |A| = |B|  = 1.
(D) A diagonal matrix is commutative with every square matrix, if it is a scalar matrix.
So every diagonal element is 4.
 |A| = 64.

81.  
10  zz   3i z 2   z   6  0
2
or 5  x 2  y 2   6 xy  8  0 ....1
Let  r cos , r sin   be a point on (1), then
8
5r 2  6r 2 sin  cos   8  0  r 2 
5  3sin 2
Clearly1  r 2  4  r  2
r1 r max  2 and
r1 r max  1  r1  r2  3

1  x  x  1 x2  x  1 
2
1
82. f   1 x  2x  3x 2 3  2x  dx
cot  ....1

1 1
x  dx   2 dt
t t
1


 t 2  t  1 t 2  t  1   1  1 1  t 2  t  1 t 2  t  1 
f      t cot  1
 2    dt   cot   2  dt
  2t  3 3t  2t   t 2  1 t  2t  3 3t  2t 

1 t 2  t  1 

1  t  t  1
2
     cot     dt ... 2 
1 t  3  2t 2t  3t 2  

Equation (1) + (2)

   1 
2f         ln   ln     2 ln   f      ln 
1 t    

Now
 
 x 2  3x  2   x  1 x  2 
ln   ln 
1
gx    
 1 dx   1.dx  ln   ln    2ln 
1 x 1  x 1  
ln

   
1
ln  

 
 Odd function i.e f   x  f  x  
 
f  200   g  50    ln  200    ln  50    ln 4  3  ln 4  a  3,b  4 .
2 3
83. Clearly, g ( x)  0x  R
 f ( x )  2
 f ' ( x)  0

9
84. Total no.of ways 
333
   36
 
1 1
Favourable cases    3    2

  2 3

 
 only one way
 6  2  1  12
12 12  6  6  6 1
Pr obability     p  q  141
9 9  8  7  6  120 140
333
85. 1  cos 2 x  2 sin 1 (sin x)
 2 cos x  2 sin 1 (sin x)
 cos x  2 sin 1 (sin x)
When we draw the graph both functions (shown below) we can actually see that they intersect only at
two points x    x  
3
cos x 2
sin 1  sin x 
0
-4 -3 -2 -1 0 1 2 3 4
 
-1
-2
86. Let the side length of cube be unity, vertices of cube are as shown in figure. The direction ratios of its
four body diagonals OR,PB,QC and SA respectively, are
(1,1,1), (1,1,-1), (1,-1,1),(-1,1,1).

Let l , m, n be the DC’s of any line, which makes angles
, ,  and  with the body diagonals of cube respectively. Then, we have

l mn l mn
cos   ,cos  
3 3
l mn l  m  n
cos   ,cos  
3 3
We have
cos 2   cos 2   cos 2   cos 2 
4 4
 l 2  m2  n2  
3 3
18 r  1
r r 3r
 1  18 18
Tr 1   1 C r  9x  
r 18 18 r
87.   C r  9  r
 x 2
 3 x  3
3r 96
18   0  r  12  C12  12    18C12    1
18
2 3
88. Since B  C  75
BAC  30, BOC  60
OBC is equilateral with BC = OB = 3
M is the mid point of BC.

9 3 3
 OM  9  
4 2
 3 3 3  3 3 3 9
 B  ,  and C   ,    y1 y2 
 2 2  2 2 4
1 1
89. f x  
 4 16   2 4  12
 x  4   2 x  2   4
 x   x 
 2 4 16 
 x  2  4and x  4  8
4
 x x 
m  1,n  12

90. We have


MISTAKES
MATHS
PHYISCS
CHEMISTRY
SECTION – I
1. The velocity, acceleration, and force in two systems of units are related as under
2  1 
'
i) v  v ii) a'    a iii) F '   F
   
All the primed symbols belong to one system and unprimed ones belong to the other
system.  and  are
dimensionless constants. Which of the following is incorrect?
3 
A) Length standards of the systems are related by L '   3 
L
 
 1
B) Mass standards of the two systems are related by M '   M.
  
2 2
 
C) Time standards of the two systems are related by T '    T
  
 1 
D) Momentum standards of the systems are related by P '   3 
P.
 
2. A particle is moving along a circle with velocityV=kt, here k=0.5 SI units. The
th
acceleration of the particle at the moment when it covered   of circle after beginning
1
 10 
of motion is ______ (nearly)
A) 1ms 2 B) 1.2 ms 2 C) 0.8 ms 2 D) 1.4 ms 2
3. A ball with velocity of 4ms1 impinges at 300 with vertical on a smooth horizontal fixed
plane. If the coefficient of restitution is 0.5, the velocity and direction of motion with
vertical after impact is _______
A) 3 ms 1 , 600 B) 7 ms 1 , Tan 1  2 / 3 
C) 2 ms 1 , 300 D) 1 ms1 , Tan 1  3 / 2 

4. A uniform rod of mass m and length L rests on a smooth horizontal surface. One end of
the rod is struck by a small ball of same mass in a horizontal direction at right angles to
the rod with ‘ V0 ’elastically. The force act on one half of the rod by the other half is…
9mV0 2 9mV0 2 3mV0 2 3mV0 2

A) B) C) D)
4L 2L 4L 2L
5. Statement 1 :When there is a thin layer of water between two glass plates there is a
strong attraction between them
Statement 2 :The pressure between the plates become less than atmospheric pressure as
pressure difference is created due to surface tension.
6. Statement 1 :In an adiabatic process the change in internal energy of a gas is equal to
negative of the work done by the gas
Statement 2 :Temperature of the gas remains constant during an adiabatic process

7. The mass of a hydrogen molecule is 3.23  1027 kg . If 1023 hydrogen molecules strike on
2 cm 2 area of a wall per second at an angle 450 with normal to the wall with a speed
105 cm s 1 , the pressure they exert on the wall is _____ Pa. (Take 2  1.4 )
A) 3.32  103 B) 2.30  103 C) 1.27  103 D) 1.67  103

8. A point mass m is suspended from free end of rod of length  , mass m. Then the time
period for small amplitude of oscillations will be:
 4 8 8
A) 2π B) 2π C) 2π D) 2π
g 3g 9g 15 g
9. A particle of charge –q, mass m moves in a region of space between two plates of a
capacitor from a plate at potential –V to the plate at potential +V. The plate separation is
d. If K, U, T and E be the respective kinetic energy potential energy, total mechanical
energy of the particle and E be the electric field between the plates, then match the facts
in Column-I with those in Column-II
Column - I Column – II
(A) K (P) constant
first increases and then
(B) U (Q)
decreases
(C) T (R) increases
(D) E (S) decreases
Other than those in (p), (q),
(T)
(r) or (s)
A) A – S; B – R; C – P; D – P B) A – R; B – S; C – P; D – P
C) A – R; B – S; C – Q; D – P D) A – S; B – R; C – P; D – T
10. The upper plate of parallel plate capacitor of plate area A is modified into 5 equal
segments as shown. The equivalent capacitance between the terminals is _____
10 0 A 2 0 A 3 0 A 3 0 A
A) B) C) D)
3d 3d 10d 2d

11. A voltage V is applied to a d.c.electric motor of resistance R. The current flowing in the
motor to get maximum power produced by the motor is …
V V V 4V
A) B) C) D)
2R 4R R R
12. A bar magnet of length 6 cm has a magnetic moment of 4 JT-1. Find the strength of
magnetic field at a distance of 200 cm from the center of the magnet along its equatorial
line.
A) 4 108 T B) 3.5 108 T C) 5 108 T D) 3 108 T
13. A square loop of a side a and straight infinite conductor carrying current I are in the
same plane as shown, The Resistance of the loop is "R". The frame is turned through
1800 about the axis oo1 . Find the electric charge that flows in the square loop. (Ignore
inductance)
o
I
a o1
b
μ 0 Ia  ab  μ 0 Ia  a  2b  μ 0 Ia  2a  b  μ 0 Ia  2a  b 
A) log   B) log   C) log   D) log  
2 πR  2a  b  4πR  a  2πR  b  4πR  a 
14. In a series LCR circuit the voltages across resistance, capacitance, inductance are 20V
each. If the capacitance short-circuited, the voltage across inductance will be ______
20
A) 20V B) 20 2V C) V D) 10V
2
15. A plane electromagnetic wave of wavelength λ has an intensity I. It is propagating along

the positive Y-direction. The allowed expressions for the electric and magnetic fields are
given by
 2I  2   1  2I  2   1
A) E  cos   y  ct   kˆ ; B   Eiˆ B) E  cos   y  ct   kˆ ; B  E î
0 c   c 0 c   c
 I  2   1  I  2   1
C) E  cos   y  ct   kˆ ; B  E î D) E  cos   y  ct   î ; B  E kˆ
0 c   c 0 c   c

16. A converging lens and a diverging mirror are placed at a separation of 15 cm. The focal
length of the lens is 25 cm and that of mirror is 40cm. At what distance from mirror a
point source of light placed between two so that, a parallel beam of light comes out from
the lens after getting reflected from mirror.
A) 13.3cm B) 6.66 cm C) 20cm D) 4.44 cm
17. A parallel beam of microwaves of wave length 0.5 mm falls normally on Young’s
double slit apparatus. The separation between the slits is 1.5 mm and the screen is placed
at a distance 1.0 m from the slits. Find the number of maxima in the interference pattern
observed on the screen.
(Excluding maxima formed at infinity)

A) 8 B) 9 C) 5 D) 11
18. An orbital electron in the ground state of hydrogen has magnetic moment 1 . This orbital
electron is excited to 3rd excited state by some energy transfer to the hydrogen atom. The
new magnetic moment of the electron is  2 , then
A) 1  42 B) 21  2 C) 161  2 D) 41  2
19. Figure shows a circuit in which three identical diodes are used. Each diode has forward
resistance 20 and infinite backward resistance. Resistors R1  R2  R3  50 . Battery
voltage is 6 V. The current
through R3 is :
A) 50 mA B) 100 mA C) 60 mA D) 25 mA
20. In an experiment for measurement of Young`s modulus, following readings are taken :
Load = 3.00 kg, length = 2.820 m, diameter = 0.041 cm and extension = 0.87 mm. The
percentage error in measurement of Y is around
A) 6% B) 8% C) 1% D) 3%

SECTION-II
21. The figure shows two blocks placed on a rough horizontal surface, under the action of
x
two forces F1  3N and F2  12N . The tension in the string is N . Find the value of
10
‘ x ’(take g=10m/s2)
F1 2 kg 3 kg F2
1  0.1  2  0.2
22. A Particle moving along the x-axis is acted upon by a single force F  F0 e kx , here F0 and
k are constants. The particle is released from rest at x  0 . It will attain a maximum
2F0
kinetic energy of , find the value of N.
NK
R
23. A circular hole of radius is cut from a circular disc of radius R. The radius of gyration
2
of this disc about an axis passing through its original centre and normal to its plane is
N
, find the value of N.
24
24. If the change in the acceleration of the earth when the position of the moon changes from
solar eclipse position to on exactly other side of the earth is N x 10-5 ms2 , find the value
of N. Ignore the effect of other planets (mass of the moon = 7.36 x 1022 kg, radius of
Lunar orbit = 3.8 x 108 m , distance between the sun and the earth is 150 million
kilometers, take G  6.7  1011 S.I.units ) (Mark the nearest integer only)

25. A cylindrical vessel of area of cross-section A and filled with liquid to a height of h1 has
a capillary tube of length l and radius r protruding horizontally at its bottom. If the
viscosity of liquid is  and density  . Find the time in which the level of water in the
X  lA h1
vessel falls to h2 is ln , find the value of X/Y.
 gr Y h2
26. Due to a point source of sound, loudness at a point is 40dB. The speed of sound is
15
330 ms 1 and air density is kgm 3 , If the pressure amplitude at this point is x  104 Pa.
11
find the value of X.
27. A very long wire carrying a current 10A is bent at right angles at O. If the magnetic
induction at a point "P" perpendicular to the plane of the wire which is at a distance d
from O is X  10 6 T , find the value of X .(Here d = 35cm, take 2 = 1.4).
28. A proton and an  - Particle are accelerated through a potential difference of 100 V. The
ratio of wavelength associated with the proton to that of  -particle is x , find the value
of x.
29. A reactor is developing nuclear energy at a rate of 32 MW. To run this reactor for 1000
hr of continuous operation, the mass of U 235 will be required ______gram. (Average
energy per fission of U 235 is 200 Mev) (Molar mass of U 235 is 235grmas) (Avagadro’s
number: 6  1023 )
30. A Screw gauge of pitch 0.5 mm is used to measure the diameter of uniform wire of
length 6.8cm. The main scale reading is 1.5mm and circular scale reading is 7. The
curved surface area of wire is x  105 m 2 .Find ‘ x ’ [Screw gauge has to 50 div on its
circular sale] (take   3.14&  2  10 )

SECTION – I
31. Which is correct for product, P, Q and R (P, Q, R are major product)
A) Product P & R are identical

B) Product Q & R are identical
C) Product P & Q are functional group isomers
D) Product P, Q & R are different
32. Which of the following statement is incorrect?
A) SRP values of halogens X 2  g  / X   aq  F2  Cl2  Br2  I2
B) Bond dissociation enthalpy of Cl2  F2  Br2  I2
C) Boiling points of I2  Br2  Cl2  F2

D) Reducing power of I  Br   Cl  F 
33. An electron in an atom jumps to the higher energy level in such a way that its kinetic
y
energy changes from ‘y’ to . Then change in its potential energy will be
2
y y
A) y B) – y C)  D) 
2 2
34.
H2SO4 O2 H+ / H2O
+ H3C CH CH2 P Q R + S
h
If R is aromatic and S is aliphatic, then:

A) Rate of EAS of (R) is more than that in benzene.
B) enol content of “S” is more than the enol content of acetaldehyde
C) R is more acidic than C2 H 5OH
A) A and C are correct B) B and C are correct
C) Only C is correct D) All A, B and C are correct

35. Which of the following will not give Lassaigne’s test for N in sodium extract?
A) C6H5NHNH2 B) NH2CONH2
NH2
C) NH2  NH2 D) SO 3H
36. The compound which has zero dipole moment is

A) CH 2Cl2 B) cis But – 2 – ene
C) PCl3 F2 D) ortho - Dichlorobenzene
37. Cerium  Z  58  is an important member of lanthanoids, which of the following
statements about cerium is incorrect?
A) The common oxidation states for cerium are + 3 and +4
B) The +3 oxidation state of cerium is more stable than +4
C) The +4 oxidation state of cerium is not known in solutions
D) Cerium (IV) acts as an oxidizing agent.
38. The complex Co NH3 4 NO 2 2  Cl , exhibits
A) Ionization isomerism, geometrical isomerism and optical isomerism
B) Linkage isomerism, geometrical isomerism and optical isomerism
C) Linkage isomerism, geometrical isomerism and ionization isomerism
D) Optical isomerism, ionization isomerism and Linkage isomerism
39. Choose the correct statement(s)

A) If the solubility of Sb2 S3 is 1.0  10 5 mol / L at 298 K, its solubility product will be
108  1025
B) pH of an aqueous solution having  H    108 M is 8.
C) Ammonia is a leveling solvent for stronger acids like HCl, HBr, HI while glacial
acetic acid is differentiating solvent.
A) A and B B) B and C C) A and C D) A, B and C

40. The osmotic pressure of blood at 370 C is 8.21 atm. The amount of glucose (in gm) that
should be added per litre for an intravenous injection so that it is isotonic with blood is
( GMW of glucose = 180g and R= 0.082 L atm mol1K 1
A) 20 gm B) 36 gm C) 42 gm D) 58 gm
41. Which of the following statement is correct for an aqueous solution of CH3COOH with
concentration 5  10 2 M and having Ka  2  10 5 ( log 2  0.3 )
A) Its pH = 3.0
B) If equal moles of NaOH are added then pH =7
C) It acts as acidic buffer if NaCl is added
D) It acts as basic buffer on adding NaOH
42. The product P in the following reaction is

O
OH (i) NaBH4
P
O (ii) H3O+
O O
A) B)
O O
C) D)
O O O O
43. A compound having the molecular formula C6H4Br2 when heated with nitration mixture
gave two mono nitro derivatives. The compound is
A) 1, 2–Dibromobenzene B) 1, 4–Dibromobenzene
C) Either 1, 2 or 1, 4–dibromobenzene D) 1,3-di tert butyl benzene

44. The compound of xenon that has the same number of lone pairs as in I3 is
(on central atom)
A) XeF2 B) XeO3 C) XeF4 D) XeO4
45. Assertion (A): Aniline on nitration gives meta nitro aniline in maximum yield.

Reason (R) :  N H 3 acts as meta directing group.
46. According to MO theory which of the list ranks the oxygen species in terms of
decreasing Bond order O2 , O2 , O2 , O22 
A) O22 , O2 , O2 , O2 B) O2 , O2 , O2 , O22
C) O2 , O2 , O2 , O22 D) O22 , O2 , O2 , O2

47. Which of the following statement is incorrect?
2.303RT
A) E 0AgCl / Ag / Cl  0.24 if E 0Ag
   0.84V and K sp AgCl  1010 (use  0.06 )
/ Ag
F
B)  0M for H aq  is highest is aqueous solution
C) In the electrolysis of aqueous Na 2SO4 if 11.2L of H 2 g  is liberated at cathode, then at
the anode the volume of O2g  liberated is 22.4 L at STP
D) In lead – acid battery the equivalent weight of H 2SO4  98 .
48. In an atom, for a 3 p  orbital there exist
A) Two spherical nodes
B) Two nonspherical nodes
C) One spherical and one nonspherical nodes
D) One spherical and two nonspherical nodes

49. Which one of the following is incorrect?
4 3
A)  Fe  CN 6  and  Fe  H 2 O 6  have same number of unpaired e in central metal ion.
2 2
B) A solution of  Ni  H 2 O 6  is green but a solution of  Ni  CN 4  is colourless
3 2
C) Cr  NH 3 6  is paramagnetic while  Ni  CN 4  is diamagnetic
D) d – orbital occupation of the central metal ion in the complex  CoF6  is t 52g eg2
4
50. Statement-I: Among 13th group elements, Gallium has maximum liquid range.
Statement-II:Oxidation state of Tl in TlI3 is +3
Choose the correct option.
A) Both Statement-I and Statement-II are correct
B) Both Statement-I and Statement-II are incorrect
C) Statement-I is correct but Statement-II is incorrect
D) Statement-I is incorrect but Statement-II is correct
SECTION-II
51. Number of OH groups in one molecule of sucrose is….
52. 100mL of NaHC2O4 requires 50 mL of 0.1 M KMnO4 solution in acidic medium for its
complete oxidation. Volume of 0.1 M NaOH required by 100 mL of same NaHC2O4 for
its complete neutralization is.
53. Given that
Compound ‘A’
What is the resonance energy of‘A’ (in magnitude) is…

54. For a first order reaction A  B the reaction rate at reactant concentration of 0.01 M
is found to be 3.0  105 mol L1 s 1 . The half-life period of this reaction in seconds is
55. Consider the following cell reaction
2Fe  s   O2  g   4H  2Fe2  aq   2H2O  l  ,E0  1.67 V
at Fe 2    10 3 M,P  O2   0.1 atm and pH  3 , the cell potential (Volts) at 250 C is V  103 .
2.303RT
The value of ‘V’ is…. (  0.06 )
F
56. Find the number of basic radicals among the following, which can form complex on
adding excess of KCN
Pb 2 , Ag  , Fe 2 , Fe3 , Cu 2 , Ni 2 
57. Identify the no. of molecules with 'sp3d ' hybridization on central atoms among the
following.
PCl5 , PO 34 ,SF4 , XeF2 , I3 , XeO 2 F2 , PCl3 F2 ,SO 2 Cl2
Spin – only magnetic moment of  MnBr4  is….BM (to its nearest integer)
2
58.
59. For the given reaction A g   2B g   3C  g  ; K c  2  10

3
1
Find the equilibrium concentration of B  mol L1  in L cylinder if 2.0 moles ‘A’ and 4.0
8
mole ‘B’ are taken initially
60. The number of moles of ozone required for the complete ozonolysis of one mole of the
given compound is
CH3 H CH3
CH3 CH CH C CH CH C CH CH
H CH3

SECTION – I
61. If , ,  be the roots of x 3   a 4  4a 2  1 x  x 2  a 2 ( where a  R ), then minimum
     1 
value of       is
     
A) 6 B) 8 C) 4 D) 3
62. The area enclosed by y  g  x  , x  axis, x  1 and x  37 , where g  x  is inverse of
f  x   x 3  3 x  1 is 297/m. Then value of ‘m’ will be
A) 4 B) 6 C) 8 D) 2
x2  5x  9
63. Statement-1: f  x   , x  R is not a one-one function.
3x 2  2 x  7
Statement-2: f  x  is not one-one, if for any x1, x2  domain of f  x  where x1  x2 ,
f  x1   f  x2  .
A) Statement-1is True, Statement-2 is True; Statement-2 is a correctExplanation

forStatement-1
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
D) Statement-1 is False, Statement-2 is True
64. The largest value of the non-negative integer ‘ a ’ for which

1 x
  ax  sin  x  1  a 1 x
1
lim    is
x 1
 x  sin  x  1  1  4
A) -2 B) 0 C) 2 D) 2

65. In paper of English there are 5 questions such that the sum of marks is 30 and the marks
for any question is not less than 2 and not more than 8. If the number of ways in which
2
marks can be awarded is a 3 digit number xyz then the value of  x  y  z  is equal to
5
? (Given that marks can be allotted in integers only)
A) 5.4 B) 6.4 C) 7.4 D) 8.4
66. If R   x, y  : x, y  Z, x 2  3y 2  8 is a relation on the set of integers Z, then the domain
of R 1 is :
A) 2, 1,1, 2 B) 0,1 C) 2, 1,0,1,2 D) 1,0,1
3 1 3 3
67. In a ABC if cos A.cos B.cosC  and sin A.sin B.sin C  , then the value
8 8
of tan A.tan B  tan B.tan C  tan C.tan A is equal to
A) 5  4 3 B) 5  4 3 C) 6  3 D) 6  3
68. For constant number ‘a’, consider the function f  x   ax  cos 2x  sin x  cos x on R
(the set of real numbers) such that f  u   f    for all u   . If the range of ‘a’ is
m 
,   , then the minimum value of  m  n  is.
 n 
A) 25 B) 35 C) 45 D) 15
69. A is one among the 8 horses in a race. A is to be ridden by one of the 3 jockeys P,Q,R. if
P rides A all the horses are equally likely to win, if Q rides A his chances are doubled
and if R rides A his chances are tripled. A die is thrown if 1 or 2 or 3 appears then P
rides A, if 4 or 5 appears then Q rides A other-wise R rides A. Then the probability that
A wins is
1 3 5 7
A) B) C) D)
12 16 24 48
70. If the variance of 1,2,2,3 is  , then the value of log1/2 
A) 8 B) 1 C) –1 D) –2

   2 
71. Let z  C and if A   z,arg  z    and B  z,arg  z  3  3i    . Then n  A  B 
 4  3
is equal to
A) 1 B) 2 C) 3 D) 0
72. A variable line y  mx  1 cuts the lines x  2 y and y  2 x at points A and B. Then
locus of centroid of triangle OAB (O being origin) is a curve passing through origin will
be
A) 6 x 2  9 xy  6 y 2  3x  4 y  0 B) 6 x 2  9 xy  6 y 2  4 x  3 y  0
C) 4 x 2  8 xy  4 y 2  2 x  3 y  0 D) 4 x 2  8 xy  4 y 2  3x  2 y  0
 x   x 4  x 2 
6 3/2
73. If x  x4  x 2  2 x4  3x 2  6 dx   C where C is constant then,

6
18
1
the value of        is equal to
4
A) 1.75 B) 2.75 C) 0.75 D) 3.75
74. Let k be the greatest integer for which 5m2  16,2km,k 2 are distinct consecutive terms
of an A.P. (arithmetic progression) where m  R . The common difference of the A.P. is
equal to :
A) 25.40 B) 25.60 C) 25.80 D) 25.20
a3x2 a2 x
75. The locus of the vertex of the family of parabolas y    2a is
3 2
A) xy  105 / 64 B) xy  3 / 4 C) xy  35 / 16 D) xy  64 / 105
ln    x 
76. The function f  x   is
ln  e  x 
A) increasing in  0, 
B) decreasing in  0, 
C) increasing in  0,  / e  , decreasing in   / e,  
D) decreasing in  0,  / e  increasing in   / e,  

8!
77. Statement – 1: Coefficient of a 2 b 3c 4 in the expansion of  a  b  c  is
8
2!3!4!
Statement – 2: Coefficient of a  b c  , where    n , in the expansion of  a  b  c n is

n!
.
 !! !
A) Statement-1is True, Statement-2 is True; Statement-2 is a correct Explanation for

Statement-1

for Statement-1
78. Let f be a differentiable function on  0,  and suppose that
lim  f  x   f   x    L where L is a finite quantity, then which of the following must be

x
true ?
A) lim f  x   0 and lim f   x   L

x  x 
L L
B) lim f  x   and lim f   x  
x 2 x  2
C) lim f  x   L and lim f   x   0

x  x 
D) Nothing definite can be said
x y
79. Given   1 and ax + by = 1 are two variable lines, ‘a’ and ‘b’ being the parameters
a b
connected by the relation a 2  b 2  ab . The locus of the point of intersection has the
equation
A) x 2  y 2  xy  1  0 B) x 2  y 2  xy  1  0
C) x 2  y 2  xy  1  0 D) x 2  y 2  xy  1  0

80.
Column I Column II
(A) A is a matrix such that A2 = A. If (I + A)8 = I + A, (P) 64
then  + 1 is equal to
(B) If A is a square matrix of order 3 such that |A| = 2, (Q) 1
 adjA 
1
1
then is equal to
(C) Let | A | a ij  0 . Each element aij is multiplied by (R) 256

33
i  j . Let | B | the resulting determinant, where

| A |  | B | , then  is equal to
(D) If A is a diagonal matrix of order 3  3 is commutative (S) 4

with every square matrix of order 3  3 under
multiplication and trace (A) = 12, then |A| =
A) A – R, B – S, C – Q, D – P B) A – P, B – S, C – Q, D – R
C) A – P, B – S, C – R, D – Q D) A – R, B – P, C – Q, D – S
SECTION-II
81. If r1 and r2 are the maximum and minimum distance of a points on the curve
 
10  zz   3i z 2   z   16  0 from origin, then value of ( r1  r2 ) will be
2
 2 1 3 4  3 4 
82. Consider three matrices A    , B  2 3 , and C   2 3  . Then the value of
 4 1    
 ABC   A  BC 2   A  BC 3 
the sum tr  A   tr    tr    tr    ...   is
 2  4   8 
   
83. If f ( x)  g ( x) |( x  1)( x  2)..........( x  10) | 2 is derivable for all x  R , where
g ( x)  ax 9  bx 6  cx3  d , a, b, c, d  R , then value of f ( 1) is

84. The numbers 1,1,1,2,2,2,3,3,3 are placed randomly in a 3  3 matrix. The probability that
p
each row and each column contain all three different numbers is given by , where p
q
and q are coprime then value of (p+q) is :
85. The number of real solutions of the equation 1  cos 2x  2 sin 1  sin x  in   x  
is
86. A line makes angles , , ,  with the four body diagonals of a cube. If the value of
cos 2   cos 2   cos 2   cos 2   4 / k . Then ‘k’ will be
18
 1 
87. The term independent of ‘x’ in the expansion of  9x   , x  0 is  times the
 3 x
corresponding binomial coefficient. Then the value of  is :
88. An isosceles triangle ABC is inscribed in the circle whose equation is x 2  y 2  9 with
base angles B and C each equal to 75 , then the absolute value of product of the
ordinates of B and C is
89. If maximum value of function
x4 m
f x  8 is (where m and n are coprime natural number)
x  2x  4x  8x  16
6 4 2
n
then (m+n) is equal to
90. Let f be a function defined on the interval  0,2 such that
x 0
  f   t   sin 2t  dt   f  t  tan tdt and f  0   1. Then the maximum value of f  x  is

0 x
___.

KEY SHEET
PHYSICS
1 D 2 C 3 A 4 A 5 A
6 A 7 C 8 A 9 A 10 C
11 D 12 B 13 C 14 A 15 B
16 A 17 B 18 A 19 A 20 A
21 5 22 5 23 2 24 2 25 3
26 1 27 3 28 120 29 6 30 3
CHEMISTRY
31 D 32 D 33 D 34 C 35 B
36 A 37 A 38 D 39 A 40 D
41 B 42 B 43 C 44 A 45 D
46 C 47 C 48 D 49 B 50 A
51 3 52 5 53 4 54 3 55 4
56 3 57 4 58 5 59 27 60 5
MATHEMATICS
61 B 62 D 63 B 64 B 65 B
66 D 67 B 68 B 69 C 70 D
71 C 72 A 73 A 74 C 75 C
76 D 77 C 78 C 79 C 80 A
81 2 82 6 83 3 84 2 85 1
86 18 87 101 88 4 89 4 90 2
SOLUTIONS
PHYSICS
1.
F  2T  2ma 1
3T  4ma 2
2a 1  3a 2
solving we get
3F
a2 
17m
A 0 A0

cc x (a  b  x)
2. c 1 2 
c1  c2 A0  A 0
x (a  b  x)
A0
c independent of x
(a  b)
1 PV
3. Wcyclic  V0P0  0 0
2 2
For theprocess AB,P  KV  PV 1  constant
Molar heat capacity of the gas in the process AB,
R 3R R
C  Cv     2R
(1  x) 2 2
Q AB  nCT  n2R(4T0  T0 )  6nRT0  6P0 V0
Q BC  0and Q CA  0
Wcyclic P0 V0
The efficiency of the cyclic process,    100   100
Qsup plied 2  6P0 V0
25
  8.33%
3 
4. By the symmetry, B total  0
5. Let us observe the motion of A and B relative to C.
AA BB

2d d

1 2
AA vt  at
BB   2
2 2
Vt 1 2
BB'   at
2 4
V
is initially speed of B w.r.t C as well as ground.
2
a
is acceleration of B w.r.t C
2
  
aB/ g  aB/C  aC/ g
a 1
 a  a
2 2
6. Conceptual
7. v cos   u cos 
u cos 
v
cos 
v sin   u sin   gt
u cos  sin 
 u sin    gt
cos 
u cos  sin   u sin  cos 
  gt
cos 
u sin    
t
g cos 
8. m1s1T1  m 2s 2 T2
Since m1s1  m 2s 2
T1  T2
9.
3
tan 2 
D
10. Q  (80  7)  (120  8)  (200  6.5) MeV  220 MeV
11. By the property of full wave rectifier
n
 V0 
12. Pfinal  Pinitial  
 V0  V 
w 
13. 15Vg  Vg  w    V  a
g 
 14Vg  w 
a  g
 w  Vg 
14. Time period of a spring mass system will remain constant when fluid is non-viscous.
15. v c  2v 0  1.414 v 0 % increase in orbital
v0  v0
velocity   100  41.4%
v0
16.
A) phase difference between current and voltage in a purely resistive AC circuit is zero.

B) phase difference between current and voltage in a pure inductive AC circuit is ; current lags
2
voltage.

C) phase difference between current and voltage in a pure capacitive AC circuit is ; current leads
2
voltage.
 X  XL 
D) phase difference between current and voltage in an LCR series circuit is tan 1  C .
 R 
1 1
17.    R 2B1  B2 3R 2
2 2
d R  dB1 3R 2 dB2
2
  
dt 2 dt 2 dt
R 2 3R 2
 .2K 0   4K 0
2 2
 emf  5R 2 K 0
18. At node, energy is maximum when all particle reach to there extreme position.
19. Vs  tan .   C
4V  tan . 2 0  C
10V  tan . 4 0  C
3V
6V  20 tan   tan  
0
Wc
4V  6V 
e
Wc  (2V)e
(3V)e
h
0

20.
v2 V 2
w1   P
R v2
P
2
 V  4V 2 .P 4P
w2    .4R  
 5R  25V 2 25
21.
a 0  a cm  R........1
a1  a cm  R....... 2 
Solivng equation 1 &  2 
a 1  a 0  2R   4a 0
a 1  5a 0
K  5.00
22. Tension of rope is maximum at lowest point
mv 2
Tmax  mg  ........(1)

By energy conservation,
 1
mg  mv2  v  g
2 2
From (1),
m(g)
Tmax  mg   Tmax  2mg

For 8 kg block, Tmax  f L
2mg   (8g)
4g   (8g)
  0.5
23.
Phasor diagram

I8  I C  I L  5  3  2A
24. E  E0 cos  kz   t  i
 6  108
k  2
Vwave 3  108
P02 V
25. I
2B
P02  340
102 
2  1.6  105
320 160
P0  
34 17
P0  3
p 2 hc hc h2
26.     
2m   2m 2
h 6.60  1034 11
  31
  1011 m  1.2 pm
2mc 2  9.1  10  3  10 91
8
27. Consider prism of mass 4 m by joining 4 prism given in question.Total MOI of this system will be
 4m   
2
2a ma 2
I   4I prism    I prism 
6 3
28.   R Th. C, R Th is the Thevenin’s resistance at the capacitor terminals.
R Th  8  (20 (9  (70 30))  20 k
  0.12 s
2T cos 
29. h  59.6mm
 rg
Here h is greater than protruding part of tube hence water will rise to maximum length of tube such
hr
that radius of meniscus is given by R 
l
30. White spot on screen would be central maxima
Where
d d 3d
x  0 y   
2 8 8

CHEMISTRY
31. As 'q' is noble gas, p, r and s having atomic number Z-1, Z+1 and Z+2 should belong to halogen,
alkali metal and alkaline earth metal respectively. As halogen has one electron less than stable noble
gas configuration it has greater tendency to accept an additional electron forming anion. Alkaline
earth metal having valence shell configuration ns2 exists in +2 oxidation state.
32. VA group hydrides: (a) reducing properties increases down the group.
(b) Basic nature decreases down the group.
(c) Thermal stability decreases down the group.
(d) Bond angles decreases down the group.
34. Mn2O7: multiple bonds

AgNO 3   Ag   NO 2   O 2 
 5  6   7 
V O  Cr2 O  Mn O 4

2
2
7
When I is oxidized by MnO4 in alkaline medium, I converts into IO3
37. Conceptual
38. Yellow coloured solution of metal ion belonging to basic radicals of group-III → Fe³ is metalion
Fe+3 + SCN– → [Fe(SCN)2]+ (Red solution) + other species
Fe+3 + C2O42– → [Fe(C2O4)3] (stable complex)
Fe3+ + F– → [FeF6]3- (stable complex)
F- and C2O42– are stronger ligand than SCN–.
Hg2+ forms stable complex with SCN–
39. Conceptual
41. X = 1-Bromo 1,2-dimethylcyclopentane, Y= 1,2 -dimethylcyclopentene,
Z = Heptane -2,6 –dione
43. Decarboxylation of betaketo acid after ester hydrolysis
44.
CH 3
N
3 2
N
N (CH 3 )2 CH 3
N
1 2
46. Initial volume of gas  1000  VO 2

x
Final volume of gas  1000  VO2 
2
 vol of CO  20ml
47. Freezing is exothermic process. The heat released increases the entropy of surrounding.
48. Psolution  160
Solution have positive deviation from Raoult’s law.
Ptotal  200  0.5  100  0.5  150
G m ix  0 Vmix  0 H mix  0
Ssurr  0
49. Justification: K increases with increase in temperature.
Q > K, Therefore, reaction proceeds in the backward direction.
∆n > 0, Therefore, ∆S > 0.
50. Conceptual
51. Two cis, one trans
52. Conceptual
53. Conceptual
55.
56. III, V & VI are less reactive

 B  3k1  3   at any time
57.  
C 8k1 8
C  8k1  8   at any time
 
 D 7.5k 2 7.5
   0.4
n1 3n 2 n
58.  0.8   2
2000 8000 4000
n  0.8 5n 2
 1 
2000 8000
n 5
 1 
n 2 4  0.8
n 5 5
 1  
4n 2 16  0.8 12.8
P1V1 P2 V2 T 6
59.   2 
T1 T2 T1 3
 T P
S  2.303  n CP log 2  R log 1 
 T1 P2 
S  27.22 J / K / mole
60. pH of HOCl  2.85
But,  pH  log  H  
2.85  log  H  
3.15  log  H  
For weak mono basic acid  H +   K a  C
1.413  103 
2 2
 H  
Ka  
C 0.08
 24.957  10  2.4957  105
6

MATHS
61. Since  0, 0   R, R is not reflexive, we have
z1  z2
 z1 , z2   R  is real
z1  z2
z2  z1
 is real   z2 , z1   R
z1  z2
Therefore R is symmetric,
Since  0, z   R and  z , 0   R , but  0, 0   R there fore R is not transitive. Hence R is not an
equivalence relation.
63 Suppose that   2    2  x 2     2  x  1  0
For all real x,
 2    2  0 and    2   4   2    2   0
2
   2    1  0 and 5 2  8  4  0
2    1 and    2  5  2   0
2
2    1 and 2   
5
These inequalities imply
 2
   2, 
 5
64. Note that every solution of f  x   x is also a
solution of f  f  x    x
f  x   x  x 2  4 x  3  0  x  3 or 1
Therefore, 3 and 1 are roots of f  x   x , also
f  f  x    x   x 2  3 x  3  3  x 2  3 x  3  3  x
2
 x 4  6 x3  12 x 2  10 x  3  0
Since 3 and 1 are roots of f  x   x , then are roots of
f  f  x    x also and therefore.
f  f  x    x   x  3 x  1  x 2  2 x  1   x  3 x  1
3
Therefore 3, 1, 1, 1 are solutions of f  f  x    x . Hence

The number of arrangements of the solutions is
4!
4
3!
3 n 2n  4
66. e  1 
1 n n 1
Put n  48
10
e is a rational number
7
x2 y 2
 1
49 51
2b 2 102
l 
a 7

69. Let
tan x cot x
t dt
F  x   dt  
1/ e
1 t 2
1/ e t 1  t
2
 
Then
 tan x  2 1
F ' x    sec x 
 1  tan x 
2
cot x 1  cot x 
2   cos ec 2 x 
 tan x  1 / cot x  =0
Therefore F is a constant function. Now
 
1 1
t 1
F   dt   dt
 4  1/ e 1  t 1/ e t 1  t 
2 2
1
t2 1
 dt   log e t 1/ e
1
1/ e t 1  t 
2
= 0   0  log e e  =1
Hence F  x   1
70. The two curves intersect at  4, 4  which is a vertex of the given square. Therefore
4
Required area (Shaded portion ) =  2 x
1
4
x2 2 1
  dx   2  1  1 = 2   x 3/ 2    x3   1
4 4
1
4 3 1 12 2
Y
(1,4)
y2  4x
(4,4)
x2  4 y
(1,1)
(2,1)
O 1 2 4 X
4 1 28 56 112  56  12
  8  1   64  8   1 =   1 
3 12 3 12 12
44 11
 
12 3
71. The given equation is
dy
sec 2 y  x  2 tan y   x3
dx
Put tan y  z , Therefore
dz
  2 x  z  x3 (Linear in z)
dx
The integrating factor is
I.F = e 
2 xdx 2
 ex
Therefore
ze x   x 3e x dx  c
2 2

1 2 x2
  x 2 e x xdx  c  x e  2 x  dx  c
2
2
1 1
  tet dt  c where t  x 2  et  t  1  c
2 2
1 x2 2
 e  x  1  c
2
So
1
tan y   x 2  1  ce x
2
2
The curve passes through  0,  / 4  . This implies
1 3
1  c  c 
2 2
1 2 3  x2
Therefore tan y   x  1  e
2 2
72. 3  a 2  b 2  c 2  1  2  a  b  c  ab  bc  ca   0
 a  1   b  1   c  1   a  b    b  c    c  a   0
2 2 2 2 2 2

 a=b=c=1
73. Let a1 , a2 , a3 , a 4 be the coefficients of rth,  r  1 th ,  r  2  th and  r  3 th terms, respectively.
Then
a1  n Cr 1 , a2 n Cr , a3  n Cr 1 , a4 n Cr  2
We know that
n
CK n  K 1
n

CK 1 K
Therefore
a2 n  r  1 a n 1
  1 2 
a1 r a1 r
a3 n  r a n 1
  1 3 
a2 r  1 a2 r  1
a4 n  r  1 a n 1
  1 4 
a3 r 1 a3 r  2
And hence
a1 a3 r r2  r 1   a2 
    2   2 
a1  a2 a3  a4 n  1 n  1  n 1   a2  a3 
74.  x   x 
2 2
 2 x1 x j  300;
x 2
1
 30
1 i
10
2
x12  x1 
   ;   30  25  5
10  10 
75. (A) tan 2   sin  cos   sin   cos 3 
 (1  sin 2  )  (1  3sin 2  )  3sin 4   sin 6 
 cos 2   (1  sin 2  )3  cos 2   cos 6   cos 2   sin 2   1
(B) sin 400  sin(600  200 )
3 1
2sin 200 cos 200  cos 200  sin 200
2 2

4cos 20  3 cot 20  1
0 0
3  cot 760 cot160 3sin 760 sin160  cos 760 cos160

(C) 
cot 760  cot160 sin(760  160 )
2sin 760 sin160  cos(760  160 ) cos 600  cos 920  cos 600 1  cos 920
    tan 460  cot 440
sin(76  16 )
0 0
sin 92 0
sin 92 0
   2  2  
(D) sin 2    sin 2    ....  sin   = 5
 18   18  2
76. We have
k 3  1  k  1  k  k  1  k  1   
2
k 2  k 1
     
k 3  1  k  1  k 2  k  1  k  1    k  12   k  1  1 
For k  2, 3,.............n Therefore
 2 1 3 1 4 1 n  2 n 1   7 13 21 n2  n  1 
Pn   , , ..... .    . . ......... 
 2 1 3 1 4 1 n n 1  n  1   n  1  1 
2
 3 7 13
1 2 3 n  2 n  1   7 13 21 n2  n  1 
=  . . ..... .   . . ..... 
3 4 5 n n  1   3 7 13  n  1   n  1  1 
2
 2   n2  n  1  2  1 
      1  
 n  n  1   3  3  n  n  1 
2 2
Therefore lim Pn  1  0  
n  3 3
3
77. We have seen that x is not differentiable at x  0 , Whereas x is differentiable at x  0 . Also
2
x  x 2 is differentiable for all real x. If a2  0 , then
f  x   a0 x  a1 x  a3
3 2
Is differentiable at x  0 . Conversely, if f  x  is differentiable at x  0 , then

a2 x  f  x   a0 x  a1 x  a3
3 2
Is differentiable at x  0 which is possible when a2  0

78. Let n  X  denote the number of elements in X
Then,
n  A  B  C   n  A  n  B   n  C   n  A  B 
n  B  C   n  C  A  n  A  B  C 
 n  A   n  A  B 
(since A  B  C   )
Now
AB   A  B    B  A    A  B    A  B 
Therefore
n  AB   n  A  B   n  A  B 
n  A   n  B   2n  A  B 
And
300  n  AB     n  A   n  B   2n  A  B    2 n  A   n  A  B  
Therefore n  A  B  C   n  A   n  A  B   300 / 2  150

79.
80.
81.
B  0, 2 
A  0,1
yx
A ' 1, 0
PA  PB will be minimum
Where A and A ' are mirror image
A ' , P, B are collinear equation of line A ' B : 2x+y=2 Solve A ' B with y =x
2 2
x  , y 
3 3
2
 P  1  i 
3
k  2

82. Number of ways to distribute at least one toy to each 141 C31 13 C2  78
If toys are distributed in the following way then two will get equal number of toys
No.of ways
3!
1 1 12  3ways
2!
2 2 10 3 ways
338 3 ways
446 3 ways
554 3 ways
662 3 ways
 Required number of ways = 78-18= 60
83.
84. Differentiating the given function we have

2
f '  x   x 2/3   x  1 x 1/3
3
3x  2  x  1 5 x  2
 
3 x1/3 3 x1/3
Now f '  x   0  x  2 / 5 . Also f  x  is defined and continuous at x  0 , f is not differentiable
at x  0 . Thus, zero is a critical point. Therefore, 0 and 2/5 are critical points of f  x 
(i) x  0  f '  x   0 and x  0  f '  x   0 . Therefore at x  0 , f is maximum and the
maximum value = f  0   0
ii) x  2 / 5  f '  x   0 and x  2 / 5  f '  x   0 . Thus f is minimum at x  2 / 5 and the
minimum value
1/3
2 3 4 
f    
5 5  25 
86. Let the equation of the line L, by hypothesis, be y  2  m  x  8 
Where m  0 . Therefore
 2 
p   8  , 0  and Q   0, 2  8m 
 m 
Now,

 2
OP  OQ   8     2  8m  m  0 
 m
2  2
10    8m   10  2   8m 
m  m
 10  8  AM  GM 
And equality occurs if and only if
2 1
  8m or m  
m 2
Hence, the absolute minimum of OP  OQ is 12 + 6 = 18
87.
88.

5x  7
8 6
5 x 6  7 x 8
89.  14  1 1  dx    1 1 2 dx
x  5  7  2 2 5  7 
x x   x x 
1 1
Put 2  5  7  t
x x
 5 x  7 x  dx  dt
6 8
 dt 1
  c
t2 t
x7
f  x  7 c
2 x  x2  1
1
f  0   0  C  0  f 1  k 4
4
90. 6  x 2  3ax  2a 2 
= 6  x  a  x  2a  ; a  0
x = a is point of maxima
x = 2a is point of minima
 a 2  2a
 a  0 or a  2
But a  0  a  2


MISTAKES
MATHS
PHYISCS
CHEMISTRY
SECTION – I
1. The acceleration of the block B shown in the figure will be (Assuming the surfaces and
the light pulleys P1 and P2 all are smooth)
F F F 3F
A) B) C) D)
4m 6m 2m 17m
2. The distance between two parallel plates of a capacitor is a. A conductor of thickness b
(b < a) is inserted between the plates as shown in the figure. The variation of effective
capacitance between the plates of the capacitors as a function of the distance (x) is best
represented by
A)
B)
D)
C)

3. One mole of an ideal monoatomic gas is taken through a cyclic process ABCA as shown
in the P-V diagram. The efficiency (in percentage) of the cyclic process is
A) 8.33 B) 12.33 C)16.33 D) 20.33
4. Two infinitely long conductors carrying equal currents are shaped as shown . The short
sections are all of equal lengths. The point P is located symmetrically with respect to the
two conductors. The magnetic field at point P due to any one conductor is B. The total
magnetic field at point P is
A) zero B) B C) 2B D) 2B
5. At the initial moment three point A,B and C are on a horizontal plane along a straight
line such that AB = BC . Point A begins to move upward with a constant velocity ‘v’ and
point C downward without any initial velocity at a constant acceleration ‘a’. If the point
begin to move simultaneously, then the initial velocity and acceleration of point B for all
the three particles to be constantly on same straight line must be:
v a v a
A) upwards, downwards B) upwards, upwards
2 2 2 2
v a v a
C) downwards, downwards D) downwards, upwards
2 2 2 2

6. Assertion (A): The magnetic field at the center of the current carrying circular coil
shown in the fig. is zero.
Reason (R): The magnitudes of magnetic fields are equal and the directions of magnetic
fields due to both the semicircles are opposite.
A) Both A and R are true but R is the correct explanation of A
D) A is false but R is also false
7. A particle projected at an angle  grazes the inclined surface BC at point P as shown.
Find the time required to reach P from O.
2u sin  u sin  u sin      2u sin     

A) B) C) D)
g cos  g cos  g cos  g cos 
8. Objects A and B that are initially separated from each other and well isolated from their
surroundings are then brought into thermal contact . Initially temperature of A and B are
00 C and 100 0 C respectively. The specific heat of A is less than the specific heat of B.
After some time, the system comes to an equilibrium state. The final temperatures are
A) TA  TB  500 C B) TA  TB  500 C
C) TA  TB  500 C D) TB  TA  500 C

9. A large plane mirror with its bottom on the floor is tilted at an angle 30° to the vertical.
A boy whose eyes are at height √3 m above the floor is standing in front of the mirror.
At what maximum distance (in m) from mirror should the boy be to see his full image in
mirror?
A) 1m B) 2m C) 2.3m D) 1.5m
10. A heavy nucleus X having mass number 200 gets disintegrated into two small fragments
Y and Z of mass numbers 80 and 120 respectively. If binding energy per nucleon for the
parent atom X is 6.5 MeV and for daughter nuclei Y and Z are 7 MeV and 8 MeV
respectively. Energy released in the decay will be
A) 200 MeV B) 240 MeV C) 220 MeV D) 180 MeV
11. Which of the following circuits will provide a full wave rectification of an AC input?
A) B)
C) D)

12. A vessel of volume of V0 is evacuated by means of a piston air pump. One piston stroke
captures the volume V  0.2V0 . If process is assumed to be isothermal then find the
 1 
minimum number of strokes after which pressure in the vessel becomes    Pinitial  .
 1.728 
A) 2 B) 3 C) 5 D) 7
13. A balloon of volume V, contains a gas whose density is  and the density of the air at
the earth’s surface is 15 . If the envelope of the balloon be of weight w but of negligible
volume. Find the acceleration with which it will begin to ascend.
 7Vg  w   2Vg  w 
A) 
Vg  w g B) 
Vg  w g
   
14Vg  w   7Vg  w 
C)  g D)  g
 Vg  w   Vg  w 
14. Assertion (A): A small body suspended by a light spring, perform SHM. When the entire
system is immersed in a non-viscous liquid, the period of oscillation does not change.
Reason (R): The angular frequency of oscillation of the particle does not change.
A) Both A and R are true but R is the correct explanation of A
D) A is false but R is also false
15. A satellite is revolving around the earth in an orbit such that it time period of revolution
as same as that of earth and it revolve in same sense as of earth. To make it escape from
gravitational field of earth, its velocity must be increased by
A) 100% B) 41.4% C) 50% D) 59.6%

16. Match the List-I with the List-II.
List-I List-II

Phase difference between current and ; current leads
A. 1. 2
voltage in a purely resistive AC circuit
voltage
Phase difference between current and

B. 2. Zero
voltage in a pure inductive AC circuit

Phase difference between current and ; current lags
C. 3. 2
voltage in a pure capacitive AC circuit
voltage
Phase difference between current and  X  XL 

D. 4. tan 1  C 
voltage in an LCR series circuit  R 
Choose the most appropriate answer from the options given below:
A) A-2, B-3, C-1, D-4 B) A-1, B-3, C-4, D-2

C) A-2, B-3, C-4, D-1 D) A-2, B-4, C-3, D-1
17. In the given figure two concentric cylindrical region in which time varying magnetic
field is present as shown. From the centre to radius R magnetic field is perpendicular
dB
into the plane varying as  2k 0 and in a region from R to 2R magnetic field is
dt
dB
perpendicular out of the plane varying as  4k 0 . Find the induced emf across an arc
dt
AB of radius 3R.
A) 6R 2 k 0 B) 5R 2 k 0 C) 7R 2 k 0 D) None of these

18. If a string of length  fixed at both ends vibrates with a standing wave
 2 
y  Asin  x  sin 2t in resonance. Then the minimum time (from t=0) after which
  
energy is maximum at mid-point of string will be –
1 1 1 1
A) sec B) sec C) sec D) sec
4 5 8 6
19. Figure shows the graph of stopping potential versus the frequency of a photosensitive
metal. The plank’s constant and work function of the metal are (V and  0 are two
different constant.)
(3V)e (2V)e
A) Wc  (2V)e; h  B) Wc  (2V)e; h 
0 0
(3V)e (2V)e
C) Wc  (3V)e;h  D) Wc  (3V)e; h 
0 0
20. There are two bulbs B1 (P,V),B2 (P,2V) their rated power and voltages are mentioned
W1
with them. Calculate the ratio of consumed power ?
W2
25 4 10 4
A) B) C) D)
4 25 4 10

SECTION-II
21. A system of two planks and a sphere of radius R is in motion as shown in figure. Radius
of the sphere is R and there is no slipping anywhere. It is given that R   2a 0 where 
is angular acceleration of sphere and acceleration of upper block is a 1  ka 0 where k is a
+ve constant then the value of k will be
22. In the system shown, the mass m  2 kg oscillates in a circular arc of amplitude 600 .
The minimum value of coefficient of friction between mass = 8 kg and surface of table
to avoid slipping is  . Then find 10 .
23. Consider a circuit with an alternating source and contains inductor and capacitor. Given
reading of A 1 and A 2 as 3 ampere and 5 ampere respectively. Find the magnitude of
reading of A in ampere.
24. The electric field associated with e.m. waves in vacuum is given by

 
E  î40cos kz  6  108 t , where E, z and t are in volt/m, meter and seconds
respectively. The value of wave vector k is _______per m

25. A point like sound source emits sound in all direction uniformly. An observer who is at a
distance of 50 m from the source detects sound of intensity 10-2 watt/m². If the bulk
modulus of air is 1.6 x 105 N/m² and velocity of sound is 340 m/s. Find pressure
amplitude (in N/m²) at the position of observer in nearest integer.
26. An x-ray beam of monochromatic photon are incident on a metallic surface having a
negligible work function  take   0  . It is seen that the wavelength of most energetic
photoelectrons is equal to the wavelength of x-ray photons. Find the wavelength (in pm).
Round off to nearest integer.

27. Figure shows a right angle solid prism ( O  ) and mass of prism is m. Moment of
2
ma 2
inertia of prism about axis perpendicular to plane passing through O is , find value
k
of k.
28. Calculate time constant (in milli-sec) of the circuit.
29. A vertical capillary tube with inside radius 0.25 mm is submerged into water so that the
length of its part protruding over the water surface is equal to 25 mm. Surface tension of
water is 73 x 10-3 N/m and angle of contact is zero degree for glass and water,
acceleration due to gravity is 9.8 m/s². Then value of 10R approximately (in mm) is
(where R is radius of meniscus and h is height of water in capillary tube)
30. In the figure, if a parallel beam of white light is incident on the plane of the slits
kd
S1 & S2 then the distance of the central maxima on the screen from O is . Find the
8
value of k.  Assume D  d,d    .

SECTION – I
31. There are four elements 'p', 'q', 'r' and 's' having atomic numbers Z-1, Z, Z+1 and Z+2
respectively. If the element 'q' is an inert gas, select the correct answers from the
following statements.
(i) 'p' has most negative electron gain enthalpy in the respective period.
(ii) 'r' is an alkali metal.
(iii) 's' exists in +2 oxidation state.
A) (i) and (ii) only B) (ii) and (iii) only
C) (i) and (iii) only D) (i), (ii) and (iii)
32. The following are some statements related to VA group hydrides. INCORRECT
statement is:
A) Reducing property increases from NH 3 to BiH3 .
B) Tendency to donate lone pair decreases from NH 3 to BiH3 .
C) Thermal stability of hydrides decreases from NH 3 to BiH 3 .
D) Bond angle of hydrides increases from NH 3 to BiH 3 .
33. The incorrect statement is
A) The first ionization enthalpy of K is less than that of Na and Li
B) Xe does not have the lowest first ionization enthalpy in its group
C) The first ionization enthalpy of element with atomic number 37 is lower than that of
the element with atomic number 38.
D) The first ionization enthalpy of Ga is higher than that of the d-block element with
atomic number 30.

34. Which one of the following statements is incorrect?
A) Highest oxidation state of manganese in fluorides is +4 (MnF4) but highest oxidation

state in oxides is 7 (Mn2O7).
B) VO2+< Cr2O7 2  < MnO4 is the correct order of oxidizing power.
C) When I  is oxidized by MnO4 in alkaline medium, I  converts into I2.
D) Silver nitrate on heating decomposes to give two types of paramagnetic gases along
with a residue.
35. Consider the following co-ordination compounds.

(i) Ni(CO)4 (ii) [Co(CO)4]- (iii) [Fe(CO)4]2-
The M-C bond strength follows the order :
(A) (ii)>(iii)>(i) (B) (iii)>(ii)>(i) (C) (i)>(ii)>(iii) (D) (i)>(iii)>(ii)

3 4
36. Assertion (A):  Fe  CN 6  is more stable than  Fe  CN 6 
Reason (R): Complexes where the iron is in the (III) oxidation state are generally more
stable than those in (II) oxidation state.
A) Assertion is True, Reason is True; Reason is correct explanation for Assertion
B) Assertion is True, Reason is True; Reason is NOT a correct explanation for Assertion
C) Assertion is True, Reason is False
D) Assertion is False, Reason is True
37. Which of the following statements is CORRECT?
A) In the formation of dioxygen from oxygen atoms, 10 molecular orbitals will be
formed.
B) All the molecular orbitals in the dioxygen will be completely filled.
C) Total number of bonding molecular orbitals will not be same as total number of anti-
bonding orbitals in dioxygen.
D) Number of filled bonding orbitals will be same as number of filled anti bonding
orbitals.

38. Yellow coloured solution of metal ion [which belongs to group-III of basic radicals] +
SCN   Blood red colour solution.
The intensity of red colour will not change on addition of:
A) Hg(NO3)2 B) F- ions C) C2O42- ions D) Br- ions
39. The IUPAC name of the following compound is

CN
OHC COOH
A) 5-cyano-3-formylcyclohex-3-en-1-carboxylic acid
B) 3-cyano-5-formylcyclohex-4-ene-1-carboxylic acid
C) 5-cyano-3-oxocyclohex-3-ene-1-carboxylic acid
D) 5-carboxy-3-formylcyclohex-2-ene-1-carbonitrile
Br2 , NaHCO3
40    A , Major organic product
The CORRECT structure of A is

OH Br
OH OH O
O O
Br
O O
A) Br B)Br C) D) Br
41. Compound X(C7H13Br) is a tertiary bromide. On treatment with potassium ethoxide in

ethanol, X is converted into Y(C7H12). Ozonolysis of Y gives Z as the only product. Z
can also be obtained by the oxidation of heptane-2,6-diol.
Select the INCORRECT statement from the following:
A) Z can also be obtained by oxidation of Y with conc. hot KMnO4 solution.
B) Z gives negative 2,4-DNP test.
C) Z gives positive iodoform test.
D) Z can be reduced to a diol with NaBH4.

42. Find out correct product of reaction:
HBr
(Excess)
O
Br
O OH Br
A) B)
Br Br Br OH
C) D)
43. Analyse the following reaction sequence:
O
H3C i) Me3COK/THF i) dil. KOH

X Y
ii) CH 3COCl +
COOC 2H5 ii) H 2O/H , 
The compound Y is the above sequence of reaction is

O O
O O O
COOC 2H5 COOH
A) B) C) D) COOH
H3C CH3
COCH 3 COcH 3
44. The nitrogen atom in each of the following tertiary amines may be removed as trimethyl
amine by repeated Hoffmann eliminations (exhaustive methylation followed by heating
with moist Ag2O). Which of the amines requires the greater number of such Hoffmann
sequences to accomplish complete removal of nitrogen.
A) B) C) D)
45. Which of the following statement is correct?
A) α-D-glucose and β -D-glucose are enantiomers.
B) Glycine is optically active.
C) Cellulose on hydrolysis yields only -D-glucose.
D) Arginine is most basic amino acid

46. One litre of a mixture of CO & CO2, is burnt in excess of oxygen. The reduction in
volume after combustion and cooling was found to be 100 ml at same temperature and
pressure. Calculate the volume ratio of CO and CO2 in initial mixture.
A) 1 : 9 B) 9 : 1 C) 1 : 4 D) 4 : 1
47. The entropy change can be calculated by using the expression S  qrev / T . When water
freezes in a glass beaker, choose the correct statement amongst the following.
A) ∆S (system) decreases but ∆S (surroundings) remains the same.
B) ∆S (system) increases but ∆S (surroundings) decreases.
C) ∆S (system) decreases but ∆S (surroundings) increases.
D) ∆S (system) decreases and ∆S (surroundings) also decreases.
48. A liquid mixture contain 10 moles of A(PA = 200 mmHg) and 10 moles of
B(PB = 100 mmHg). The vapour pressure over liquid mixture is 160 mm Hg. Which is
correct statement?
A) ∆Gmix = +ve B) ∆Vmix= -ve
C) ∆Ssurrounding= +ve D) ∆Ssurrounding = -ve
49.  2NO2  g  , the value of K is 50 at 400 K and 1700 at 500

For the reaction N2 O4  g  
K. Which of the following options is INCORRECT?
A) The reaction is endothermic.
B) The reaction is exothermic.
C) If NO2 (g) and N2O4 (g) are mixed at 400 K at partial pressures 20 bar and 2 bar
respectively, more N2O4 (g) will be formed.
D) The entropy of the system increases in the forward direction.
50. Match the items of Column I and Column II.
Column I Column II
(i) Leclanche cell (a) cell reaction 2H2 + O2→ 2H2O
(ii) Ni–Cd cell (b) does not involve any ion in solution and is used in hearing aids.
(iii) Fuel cell (c) rechargeable
(iv) Mercury cell (d) reaction at anode, Zn → Zn2+ + 2e–

(e) converts energy of combustion into electrical energy

A) (i) →(d); (ii) →(c); (iii) →(a), (e); (iv) →(b)
B) (i) →(c); (ii) →(d); (iii) →(e); (iv) →(b)
C) (i) → (c); (ii) →(e); (iii) →(a), (c); (iv) →(b)
D) (i) → (d); (ii) →(d); (iii) →(a); (iv) →(c)
SECTION-II
51. Total number of geometrical isomers for the square planar complex [RhCl(CO)
(PPh3)(NH3)] is /are _______ .
52. The total number of compounds having pp bonding among the molecules given
below are ________ .
SO 2 , SO3 , CO2 , C3O2 , N 2 O5 , Cl2 O7 and Cl2 O6
53. Find the number of resonating structures of the given carbanion where negative charge is
on 2 carbon.
N
–
CH2
54. From which position does NO2+ replace a hydrogen from the following compound
predominantly?
H
N
1 O
2
3 4
5 8
6 7
55.
The degree of unsaturation in major product (B) will be:

56. How many of the following diazonium salts are less reactive than (I) towards diazo
coupling reaction?
+N  + + +
2 Cl N2 Cl  N 2 Cl
 N Cl 
2
NO2 O 2N NO2
I. II. III. IV.
NO2 NO2 NO2
+
N  +N Cl 
2 Cl 2
V. VI.
CH3
CH3 CH3
57. Analyse and complete the following reactions

(i) CH2CH=CH2 + H2 –(Ni) →
(ii) CH3CH2CH2Cl+ H2 –(Zn/H+) →
(iii) CH2CH2CH2COONa + NaOH –(CuO/∆) →
Find the sum of molar masses (in g) of byproducts (if any) formed in above reactions
[Take atomic masses – H = 1, C = 12, O = 16, Na = 23, Cl = 35, Cu = 63, Zn =65
58. A light of wavelength 200nm falls upon a surface and two different wavelength photons
λ = 800nm and λ =400nm are emitted from the surface. 80% of the energy absorbed is
re-emitted in the form of photon. Number of photons emitted as λ = 800nm is 3 times
that of number of photons emitted as λ = 400nm. If the ratio of total absorbed photon to
total emitted photon is x. then find the numerical value of (12.8x)
59. When 1 mole of an ideal gas at 20 atm and 15 L volume expands such that the final
pressure becomes 10 atm and volume is 60 L. The entropy change of the process in
J/K/mole is (Round off your answer to nearest double digit integer)
Given that: Cp,m = 30.96 J/mole/K, R = 8.314 J/mol/K, ln 2 = 0.69
60. pH of 0.08 moldm–3HOCl solution is 2.85. If its ionization constant is A × 10–B
Find the value of B. [antilog (0.15) = 1.413] (Given : 1  A  10, log 2  0.3 )

SECTION – I
61. Let C be the set of all complex numbers and
 z z 
R   z1 , z2   C  C : 1 2 is real 
 z1  z2 
Then, on C, R is a
A) reflexive relation B) symmetric relation
C) transitive relation D) equivalence relation
62. Let A be a 3  3 real matrix such that A  I  2 BBT , where BT is transpose of column
matrix B, whose sum of the squares of elements is unity. Given the statements
I) Tr  A   1( Tr  A  is sum of the elements in the principal diagonal of matrix A)
II) A is a symmetric matrix
III) A is an orthogonal matrix
Then the number of the correct statements among the above is
A) 0 B) 1 C) 2 D) 3
63. If  is real and   2    2  x 2     2  x  1 for all real x, then  belongs to the interval
A)  2,1 B)  2, 2 / 5 C)  2 / 5,1 D) 1, 2 
64. Let f  x   x 2  3 x  3 and x1 , x2 , x3 and x4 be solutions of the equation f  f  x    x . Then

the number of arrangements of x1 , x2 , x3 and x4 taken all at a time, is
A) 24 B) 4 C) 6 D) 1

65. Let S  {1, 2,3, 4,5,6,7} .A subset of A is selected from ‘S’ and now keeping back
elements in S, again a subset of S,B is selected. Let E1  event that
A  B  {1,2,3,4,5}, E2  event that A  B  {1,2}, if
E  a 1 b 
P  2    a, b  N   G.C.D of  a, b   1 ,then equals___
 E1  b 10  a 
([.] represents G.I.F)
A) 2 B) 3 C) 4 D) 5
x2 y2
66. Let H n :   1 , n  N . Let k be the smallest even value of n such that the
1 n 3  n
eccentricity of H k is a rational number. If l is the length of the latus rectum of H k , then
21l is equal to
A) 101 B) 204 C) 102 D) 306
67. Two adjacent sides of parallelogram ABCD are given by AB  2iˆ  10ˆj  11kˆ and
AD  î  2ˆj  2kˆ . The side AD is rotated by an acute angle  in the plane of the
parallelogram so that AD becomes AD ' . If AD ' makes a right angle with the side AB,
then the cosine of the angle  is given by
8 17 1 4 5
A) B) C) D)
9 9 9 9
68. Consider a complex number z on the argand plane satisfying

i 2
  
arg  z      arg  z       e 3  . If minimum value of z  2  2i z  2  2i is
2 2 2
2  
a b ab
 a, b  N  ,then the value of 
2 52
A) 2 B) 5 C) 10 D) 20
tan x cot x
t dt
69. 
1/ e
1 t 2
dt  
1/ e t 1  t 
2

A) 2  tan e  1 B) 2 tan e C) 1 D) tan e  cot e

70. The area enclosed between the curves y 2  4 x and x 2  4 y inside the square formed by
the lines x  1, y  1, x  4, y  4 is
8 16 13 11
A) B) C) D)
3 3 3 3
71. The curve passing through the point  0,  / 4  satisfying the differential equation
dy
 x sin 2 y  x 3 cos 2 y is
dx
1 2 3 2 1 2 3 2
A) tan y 
2
 x  1  e x / 2
2
B) tan y 
2
 x  1  e  x / 2
2
1 2 3 2 1 2 3 2
C) tan y 
2
 x  1  e x
2
D) tan y 
2
 x  1  e  x
2
72. Statement – 1: If a,b,c are non zero real numbers such that
3  a 2  b 2  c 2  1  2  a  b  c  ab  bc  ca  , then a,b,c are in A.P. as well as in
G.P.
Statement – 2: A series is in A.P. as well as in G.P. if all the terms in the series are equal
and non zero.
A) Both Statement - 1 and Statement - 2 are true
B) Both Statement - 1 and Statement - 2 are false
C) Statement - 1 is true, Statement - 2 is false
D) Statement - 1 is false, Statement - 2 is true
73. If a1 , a2 , a3 and a4 are the coefficients of any four consecutive terms in the expansion of
a1 a2 a3
1  x  , then
n
, , are in
a1  a2 a2  a3 a3  a4
A) AP B) GP C) HP D) AGP
74. x1 , x2 ,...........x10 are ten observations such that x i  50 and x x i j  1100 1  i  j  10 , then
standard deviation of x1 , x2 ........x10 is equal to
A) 5 B) 10 C) 5 D) 10

Column I Column II
If tan  is the G.M. between sin  and cos
A) P) 1
then 2  4 sin 2   3sin 4   sin 6  can be
B) 3 cot 200  4 cos 200  Q) 0
C) cot160 cot 44 0  cot 440 cot 760  cot 760 cot160  R) 3

9
 r 
D)  sin
r 1
2
 
 18 
S) 5
A) A – S; B – P; C – R; D – P B) A – P; B – R; C – P; D – S
C) A – P; B – P; C – R; D – S D) A – R; B – D; C – P; D – S
76. Let
23  1 33  1 n3  1
Pn  . ........... ; n  2, 3, 4......
23  1 33  1 n3  1
Then lim
n 
Pn is equal to
1 7 3 2
A) B) C) D)
2 11 4 3
Let f  x   a0 x  a1 x  a2 x  a3 . Then
3 2
77.
A) f is differentiable at x = 0 if a2  0
B) f is not differentiable at x=0, whatever be a0 , a1 , a2 , a3
C) f is differentiable at x  0 if and only if a2  0
D) If f is differentiable at x  0 , then a0  0 and a2  0
78. Let A, B and C be finite sets such that A  B  C   and each one of the sets AB, BC
and C A has 100 elements. The number of elements in A  B  C is
A) 250 B) 200 C) 150 D) 300

79. Let Z be the set of all integers,
A  x, y   Z  Z :  x  2  2

 y2  4
B   x, y   Z  Z : x 2  y 2  4 and
C  x, y   Z  Z :  x  2 2
  y  2  4
2

If the total number of relations from A  B to A  C is 2p , then the value of p is
A) 16 B) 49 C) 25 D) 9
80. Let A,B and C be three events such that the probability that exactly one of A and B
occurs is 1  k  , the probability that exactly one of B and C occurs is 1  2k  , the
probability that exactly one of C and A occurs is 1  k  and the probability of all A, B
and C occur simultaneously is k 2 , where 0  k  1 . Then the probability that at least one
of A, B and C occur is
1 1
A) Greater than B) Exactly equal to
2 2
1 1 1 1
C) Greater than but less than D) Greater than but less than
8 4 4 2
SECTION-II
81. P is a point satisfying arg z=  / 4 , such that sum of its distances from two given points
k
 0,1 and  0, 2  is minimum, then P must be 1  i  . Then numerical value of k is _____
3
82. The number of ways can 14 identical toys distributed among three boys so that each one
n
gets atleast one toy and no two boys get equal number of toys is n, then is equal
10
to____


83. Let P be an interior point of a triangle ABC such that PA  2PB  3PC  0 . Then the ratio
Area of triangle ABC

is equal to ______
Area of triangle APC
84. The number of values of x at which the function f  x    x  1 x 2/3 has extremum values is
If the complete solution set of the inequality  cosec1 x   2  cosec 1 x   cosec 1 x 

2 π π
85. is
6 3
(, a]  [b, ) , then (a  b) is equal to
86. A straight line L with negative slope passes through the point (8, 2) and cuts the positive
coordinate axes at points P and Q. Then the absolute minimum value of OP+OQ as L
varies (Where O is the origin) is
87. The sum of first 100 terms of the series is
 1   1   1   1 
tan 1  2 
 tan 1  2 
 tan 1  2 
 tan 1  2 
 .... (if x>0) is
 1 x  x   3  3x  x   7  5x  x   13  7x  x 
 100 
tan 1   , then a  b  ____
 1  ax  bx 
2
88. Let the curve C be the mirror image of the parabola y 2  4x with respect to the line
x  y  4  0 . If A and B are the points of intersection of C with the line y  5 , then the
distance between A and B is
5 x8  7 x 6 1
89. If f  x    dx,  x  0  f  0   0 and f 1  , then the value of K is ______
x 2
 1  2x 
7 2 K
90. Suppose x1 and x2 are the point of maximum and the point of minimum respectively of
the function f  x   2 x 3  9ax 2  12a 2 x  1 respectively,  a  0  then for the equality x12  x2
to be true the value of ‘a’ must be

KEY SHEET
PHYSICS
1 B 2 B 3 C 4 D 5 A
6 B 7 C 8 A 9 B 10 A
11 B 12 B 13 A 14 A 15 A
16 C 17 A 18 D 19 B 20 B
21 1 22 30 23 3 24 780 25 16
26 2000 27 8 28 3 29 3 30 20
CHEMISTRY
31 B 32 D 33 A 34 B 35 A
36 D 37 C 38 C 39 D 40 A
41 D 42 B 43 A 44 A 45 A
46 A 47 C 48 D 49 B 50 C
51 7 52 11 53 3 54 2 55 4
56 5 57 2 58 50 59 2 60 3
MATHEMATICS
61 C 62 C 63 C 64 D 65 C
66 C 67 B 68 B 69 B 70 C
71 D 72 B 73 A 74 A 75 D
76 D 77 B 78 B 79 A 80 D
81 1777 82 9090 83 249 84 3 85 64
86 28 87 30 88 31 89 1049 90 6
SOLUTIONS
PHYSICS
1.  = Q A
A 0
Q
=
0
d 1 dQ I
= × =
dt 0 dt 0
 
2. Angle between a and p is :

p 

a

–1 a.p
 = cos  
| a || p |
 
32  18
= cos-  

 (16  9) (64  36) 
cos–1  
14
=
 50 
 = 73.73°
Since 0° < 90°, the motion is an acceleration one.
3. Retardation = g (sin  +  cos ) = 5 ( 3  )
Now v = u – at
u
 a= as v = 0
t
 5 ( 3  ) = 10  = 0.27
4 m/s v
4.
1 m/s 1 m/s
Before collision After collision
Let v be the velocity of ball after collision, collision is elastic

 e =1
or
relative velocity of separation = relative velocity of approach
 v – 1 = 4 + 1
or v = 6 m/s (away from the wall)
GH
5.  g=
(R  h ) 2
GM GM
 2
=
9R (R  h ) 2
 3R = R + h
 h = 2R
So option (1) is correct.
6. Dq = – Du
–R R P dV
C = –CV = = +
 –1  –1 n dT
P dV 2R
– 
n dT  – 1
T5V = const.
const.
V=
T5
dV const
=–5 6
dT T
PV = Nrt
P/n = RT/V
RT 5  const  2R
+ T × –5 6  =
const.  T   –1
5 1
=  – 1 = 2/5
2  –1
 = 7/5
adiabatic compressibility
1 5
= =
P 7P
1
7. PV = m o Nv 2rms
3
1
(2P) (2V) = m o Nv 2rms
3
rms = 2vrms = 2v
8. 4 trips means 32 m
v
4m
d d 32
t=  v= = = 40 m/s
v t 0 .8
T
v=

T v2
0 .2 2 16  10
T= × (40)2 =
4 4
T = 80 N

9. Time period of both A and B T = 2
g
After first collision, B acquires amplitude of A and after second collision it acquires its own
amplitude in this process time taken is
T T T T 
= + + + = T = 2
4 4 4 4 g
10. The distribution of charge on the outer surface, depends only on the charges outside, and it distributes
itself such that the net electric field inside the outer surface due to the charge on outer surface and all
the outer charges is zero. Similarly the distribution of charge on the inner surface, depends only on the
charges inside the inner surface, and it distributes itself such that the net, electric field outside the inner
surface due to the charge on inner surface and all the inner charges is zero.
Also the force on charge inside the cavity is due to the charge on the inner surface. Hence answer is
option (A).
11. Conceptual

12. Point A shall record zero magnetic field (due to -particle) is at position P and Q as shown in figure.
The time taken by -particle to go from P to Q is –
P

O 60º
A
60º
12 2
t= or  =
3 3t
13. Curie temperature is temperature above which Ferromagnetic materials obey Curie’s law.
1 2
14. U= LI
2
= LI  
dU dI
Rate =
dt  dt 
At t = 0 , I = 0
 Rate = 0
dI
At t = , I = I0 but = 0, therefore, rate = 0
dt
15. The given lens is a convex lens. Let the magnification be m, then for real image
1 1 1
+ = ... (i)
mx x f
1 1 1
and for virtual image  = ... (ii)
 my y f
From Eq. (i) and Eq. (ii), we get
xy
f=
2
16. e = 4 × LC = 4 × 0.01 cm = 0.04 cm
c = –0.04 cm
17. Zener diode is in parallel to load resistance and
is connected in reverse bias.
R
18. R = R0 A1/3  = A1/3
R0
 R  1
log   = log A
 R0  3
1
or y= x
3
a straight line passing through origin with slope 1/3.
19. Conceptual
20. S1
S2
y 2 2y
x =  = × x =
  
Inet = I + I + 2I cos 

  2y    y 
= 2I 1  cos   = 4I cos2  
      
y  
= I0 cos2   [ I0 = 4I]
  
k 10 2
21. F=  F = 
2x 2 2x 2
F 10 2 1 dv
a= = 2 2  a = 2 =v
m (10 )2 x 2x dx
0.5 v 0.5
dx 1 v2
– =  vdv  –  x  2 dx =
1
2x 2 0
2
1
2
0.5 2
 1   1 v2
  1 =
v 1 1
 =
2  x 1 2 2  0 .5  2
 v = 1 m/s
22.
R
y
d
2
y2 = R2 –  
d
2
vCM/g
× y
vCM/g = × y
= 30 cm/s
23. v = Rcos
v 2
 cos= = cm
R 5
 h = R(1 – cos) = 3 cm.

24. If m is mass of single drop then as it drops

mg = 2 Rt
If number of drops in M = 10 grams is N then,
M Mg Mg
N= = =  779.86  780
m mg 2rT
25. For sonometer wire
F
n

n =
2
 n F = constant
 [are constant for two cases
 of comparison]

n12
 F2 = . F1
n 22
 m2 = 25 kg
 Additional mass = 16 kg
26.
Potential to centre is same as potential at the inner surface of the spherical shell.
27. At resonance reactance = 0
V 60 1
I= = = Amp.
R 120 2
VL = I × XL = I × L
VL
L= ………(i)
I
1
0 =
LC
1
C= ……….(ii)
L 0 2
Calculate L & C from (1) & (2) current will lag
the applied voltage by 45°
1
L –
if tan 45° = c
R
Solve for   = 8 × 105 rad/sec
28. Energy present in 3  108 m is 10 Mj, hence energy in 90 cm is
90  102  10  103
E  3  10 11 J
3 10 8
29.
4º
Aº
deviation = 0
(µ1 – 1) 4 – (µ2 – 1)A = 0
(1.54 – 1) 4 – (1.72 – 1) A = 0
0.54  4
A= = 3º
0.72
1
30. From graph  0.2cm 1
f
 f  5cm
 Power of lens equal to 20 D.

CHEMISTRY
31.  Xe 4f 14 1 2
5d 6s ----------- Lu
32. INERT PAIR effect NCERT (Physical properties of group 13 elements )
33.  K 2SO 4  Cr2  SO 4 3  H 2O
K 2 Cr2O 7  SO 2  H 2SO 4 
34. X : NaC O
Y : NaC O3 .
35. By using Lanthanoid contraction concept we get correct answer.
36. Th  Z  90  : 86  Rn  6d 2 7s 2
37. y  4, x  6
38. PC 5  sp3  TBP  ; AsH 4  sp 3  Tetrahedral  ; XeF4  sp 3d 2 -Square planar;
XeO 2 F2  sp3d  Seesaw 
39. synergic bond concept in complex compounds
40. Conceptual
41. Product is
OH
Br Br
HC  CH  COOH
| |
Br Br
43.
  
44. Hinsberg’s test is used for separation of 1 , 2 and 3 Amines.
46. 2  n  3  2   0 n  4
 y y
C x H y  g   x   O 2  g  
 xCO 2  g   H 2O   
 4 2
10 70 - -
 y 5
- 70  10  x   10x -70  10x  y  10x  65
 4 2
5y  5  2 y  210 x  20 x2
C x H y  C 2 H 2 Molar mass  26g / mol
47. At r  a 0   0  4r 2 2  0
48. In liquid phase
PT  PB0 . B  PT0 .T  120  0.6  50  0.4
= 72 + 20 = 92 mmHg
PToluene 20
In vapour phase  T   0.22
PTotal 92
49. C3H5COOH  NaOH 
 C6 H5COONa  H 2O
 WA  SB Salt 
D
Conductance
A C
B
VNaOH
50.
51. NCERT data
52. a  3;b  35;c  3;d  3
2
53. Option i) Manganate  MnO 4
Permanganate  MnO 4
O O
Mn Mn
O O O O
O O
hybridisation hybridisation
of Mn  d s 3
of Mn  d 3s
After excitation
2 × 2p   3d 

2 × 2p   3d 
1 × 2p   4p 
2
(ii) MnO 4  green
MnO 4  purple / violet
(iii) Manganate contains I unpaired electron hence it is paramagnetic Where as permanganetic
contains no unpaired electrons hence it is diamagnetic.
3
(iv) Both have d s hybridization hence both have tetrahedral geometry.
54.
55.

56. (i), (ii), (vi), (vii),(ix) can evolve H 2 on reaction with Na metal.
57. X:3
Y:3
Z:3
58.  M  50  70  120 1 cm 2 mol1
C.C 0.2 1 1 K  1000 0.2  1000
K   cm   M  0.4975mol / t
R 33.5 M 120  33.5
M  49.75milli mol / t
60. Average of pK1 and pK 2
pH = 2.95 is isoelectric point
MATHS
61.

62.
63.

64.
65.
66.
67.

69. CONCEPTUAL
70.
71.
72.CONCEPTUAL
73.
74.

75.
76.
77.
78.

79.
80. conceptual
81.
82.
83.

84.
85.


MISTAKES
MATHS
PHYISCS
CHEMISTRY
SECTION – I
1. A circular parallel plate capacitor of radius R and distance d between the plate is given.
A capacitor is being charged with a current I flowing through the wire. Neglect fringing
effect.
I I
What is the rate of change of electric flux through plane in middle of capacitor with
d
respect to time (i.e. )–
dt
2I I 4I 6I
A) B) C) D)
0 0 0 0
2. A particle is moving in a circular path. The acceleration and momentum of the particle at
 2 
a certain moment are a  (4î  3 ĵ) m/s and p  (8î  6 ĵ) kg-m/s. The motion of the particle at
that instant is
A) Uniform circular motion
B) accelerated circular motion
C) decelerated circular motion
D) We cannot say anything with 

a and p only
3. A block starts moving up a fixed inclined plane of inclination 60° with a velocity of
20 m/s and stops after 2 sec. The approximate value of coefficient of friction is
(g = 10 m/s2)
A) 3 B) 3.3 C) 0.27 D) 0.33

4. A ball of mass m approaches a wall of mass M (>> m) with speed 4 m/s along the normal to
the wall. The speed of wall is 1m/s towards the ball. The speed of the ball after an elastic
collision with the wall is -
A) 5 m/s away from the wall B) 9 m/s away from the wall
C) 3 m/s away from the wall D) 6 m/s away from the wall
g
5. The height at which the acceleration due to gravity becomes (where g = the
9
acceleration due to gravity on the surface of the earth) in terms of R, the radius of the
earth, is
R
A) 2R B) C) R / 2 D) 2 R
2
6. An ideal gas is expanded so that amount of heat given is equal to the decrease in internal
energy. The gas undergoes the process TV1/5 = constant. The adiabatic compressibility of
gas when pressure is P, is –
7 5 2 7
A) B) C) D)
5P 7P 5P 3P
7. An ideal gas is held in a container of volume V at pressure P. The average speed of a gas
molecule under these conditions is v. If now the volume and pressure are changed to 2V
and 2P, the average speed of a molecule will be
A) 1/2 v B) v C) 2v D) 4v
8. A wire is 4 m long and has a mass 0.2 kg. The wire is kept horizontally. A transverse
pulse is generated by plucking one end of the taut (tight) wire. The pulse makes four
trips back and forth along the cord in 0.8 sec. The tension in the cord will be -
A) 80 N B) 160 N C) 240 N D) 320 N
9. Two identical simple pendulums A and B are fixed at same point. They are displaced by
very small angles  and  (    ) and released from rest. Find the time after which B
reaches its initial position for the first time. Collisions are elastic and length of strings is
.
 
A B
    2 
A)  B) 2 C) D)
g g  g  g

10. The figure shows a charge q placed inside a cavity C in an uncharged conductor. Now if
an external electric field is switched on then :
C
q
A) only induced charge on outer surface will redistribute.
B) only induced charge on inner surface will redistribute
C) Both induced charge on outer and inner surface will redistribute.
D) force on charge q placed inside the cavity will change
11. STATEMENT – 1
If potential difference between two points is zero and the resistance between the same
two points is also zero, then current may flow between those two points
STATEMENT – 2
Kirchhoff’s 1st law is conservation of charge.
A) Statement – 1 is True. Statement – 2 is True; Statement – 2 is a correct explanation


C) Statement – 1 is True, Statement is False.
D) Statement – 1 is False, Statement is True.

12. An  particle is moving along a circle of radius R with a constant angular velocity  .
Point A lies in the same plane at a distance 2R from the centre. Point A records magnetic
field produced by  particle. If the minimum time interval between two successive
times at which A records zero magnetic field is 't', then the angular speed  is :
2 2  
A) B) C) D)
t 3t 3t t
13. Curie temperature is the temperature above which -
A) a ferromagnetic material behaves like paramagnetic
B) a paramagnetic material behaves like diamagnetic
C) a ferromagnetic material behaves like diamagnetic
D) a paramagnetic material behaves like ferromagnetic

14. In an LR circuit connected to a battery, the time rate of change of energy stored in the
inductor is plotted against time during the growth of current in the circuit. Which of the
following best represents the resulting curve?
Rate Rate
A) B)
Time Time
Rate Rate
C) D)
Time Time
15. When an object is at distance x and y from a lens, a real image and a virtual image is
formed respectively having same magnification. The focal length of the lens is –
xy
A) B) x – y C) xy D) x + y
2
16. If the zero of the Vernier lies on the right hand side of zero of the main scale and fourth
division coincide with the main scale division when the jaws are in contact, the zero
correction will be __ (assume standard Vernier Calipers)
A) + 0.04 cm B) + 0.06 cm C) –0.04 cm D) –0.06 cm

17. A zener diode is to be used as a voltage regulator. Identify the correct set up –
Rs Rs
+ +
A) RL B) RL
– –
Rs Rs
+ +
C) RL D) RL
– –
 R 
18. The graph of log  
 versus log A (R = radius of a nucleus and A = mass number) is -
 R0 
A) a circle B) an ellipse C) a parabola D) a straight line
19. Match the entries of Column I with those in Column II
Column I Column II
Mass of products less than mass

(A) (P) Most stable
of reactants
Total internal reflection

(B) Iron nucleus (Q)
take place
(C) Angle of contact is obtuse (R) Nuclear fission
Incidence angle more than critical

(D) (S) Stronger cohesive forces
angle
A) A – R; B – P; C – Q ; D – S B) A – R; B – P; C – S ; D – Q
C) A – Q; B – S; C – R ; D – P D) A – Q; B – S; C – P ; D – R

20. In a YDSE experiment, I0 is given to be the intensity of the central bright fringe &  is the
fringe width. Then, at a distance y from central bright fringe, the intensity will be -
(y is very small)
 y   y   2y   y 
A) I0 cos   B) I 0 cos 2   C) I 0 cos  D) I 0 cos 2  
          2 
SECTION-II
21. A particle of mass 10–2 kg is moving along the positive x-axis under the influence of a
force F(x) = – K/(2x2) where K = 10–2 Nm2. At time t = 0 it is at x = 1.0 m and its
velocity is v = 0. Find its velocity when it reaches x = 0.50 m. (in m/s)
22. A uniform ball of radius R = 10 cm rolls without slipping between two rails such that the
horizontal distance is d = 16 cm between two contact points of the rail to the ball. If the
angular velocity is
5 rad/s, then find the velocity of centre of mass of the ball in cm/s.
23. A disc of radius '5cm' rolls on a horizontal surface with linear velocity v = 1 î m/s and
angular velocity 50 rad/sec. Height of particle from ground on rim of disc which has
velocity in vertical direction is (in cm) -
y

v x
24. A liquid flows out drop by drop from a vessel through a vertical tube with an internal
diameter of 2 mm, then the total number of drops that flows out during 10 grams of the
liquid flow out: [Assume that the diameter of the neck of a drop at the moment it breaks
away is equal to the internal diameter of tube and surface tension is 0.02 N/m,
g = 9.8 m/s2]

25. A tuning fork is in unison with a sonometer wire vibrating in its fourth overtone. Mass
hanged with wire is 9 kg. When additional mass is hanged wire vibrates in unison with
tuning fork in its 3rd harmonic. Additional mass hanged in kg is.
26. Two conducting concentric spherical shells are present. If the electric potential at the
centre is 2000 V and the electric potential of the outer shell is 1500 V. then the potential
of the inner shell in volts is
27. A series LCR circuit containing a resistance of 120 has angular resonance frequency
4 × 105rads–1. At resonance the voltage across resistance and inductance are 60 V & 40
V respectively. At what frequency the current in the circuit lags the voltage by 45°. Give
answer in -------×105 rad/sec.
28. The energy stored in a 90 cm length of laser beam operating at 10 mW is X × 10–11 J.
Find the value of X.
29. A thin prism P1 with angle 4º and made from glass of refractive index 1.54 is combined
with another thin prism P2 made from glass of refractive index 1.72 to produce
dispersion without deviation. The angle of prism in degrees is
30. An experiment with convex lens gives certain result which is represented by a student in
the shown graph. The power of the lens in diopters is
0.2
1 –1
cm
0.1 v
0.2 0.1
1
cm–1
u

SECTION – I
31. The element with  Xe  4f 14 5d1 6s 2 belongs to
A) 7th period and Group 4 of the modern periodic table

B) 6th period and Group 3 of the modern periodic table
C) 5th period and Group 3 of the modern periodic table
D) 6th period and Group 5 of the modern periodic table
32. The relative stability of +1 oxidation state of group 13 elements follows the order:
A) A  Ga  T  In B) Ga  A  In  T
C) T  In  Ga  A D) A  Ga  In  T
33. Assertion (A) : SO2 turns acidified K 2Cr2O7 green.
Reason (R) : K 2Cr2O7 is reduced to Cr 3 (green) salt by SO2 .

34. Product (X and Y) of the following reactions (1 and 2) are
1) 2NaOH  C 2  NaC  X  H 2O
( Cold and dilute )
2) 6NaOH  3C 2  NaC  Y  H 2O

( Hot and conc.)
A) X  NaCO3 and Y  NaOC B) X  NaOC and Y  NaCO3
C) X  NaHCO3 and Y  NaOC D) X  NaC O3 and Y  NaHC O3
35. Arrange Ce3 , La 3 , Pm3 , Yb 3 in increasing order of their ionic radii
A) Yb 3  Pm 3  Ce 3  La 3 B) Ce3  Yb 3  Pm 3  La 3
C) Yb 3  Pm 3  La 3  Ce 3 D) Pm3  La 3  Ce 3  Yb 3

36. Which one of the following does not contain electrons in 5f subshell?
A) Lr  Z  103 B) U  Z  92  C) Cm  Z  96 D) Th  Z  90 
37. Consider octahedral complex of Ca 2 with EDTA 4 .

x = co-ordination number of complex
y = Total number of O  donors . Find (x+y)?
A) 13 B) 14 C) 10 D) 11
38. In which of the following species maximum atoms can lie in same plane?
A) PC 5 B) AsH4 C) XeF4 D) XeO 2 F2
39. Which of following involves maximum C  O bond length:
A) Ni (CO ) 4 B) [ Fe(CO )5 ] C) [ Mn(CO )6 ] D) [V (CO)6 ]
40. Assertion (A) : Coloured cations can be identified by borax – bead test
Reason (R) : Transparent bead  NaBO2  B2O3  forms coloured bead with coloured
cation.
41. Para-coumaric acid, an antioxidant in coffee, is treated with aqueous solution of

bromine. What is the maximum number of bromine atoms that can be incorporated into a
molecule of para-coumaric acid?
O
HO
OH
para  coumaricacid
A) 1 B) 2 C) 3 D) 4

42. Which of the following reaction produces anisole as a major product?
A) B)
C) D) All
43. The oxidation of toluene with hot KMnO 4 gives
A) Benzoic acid B) Benzaldehyde
C) Benzene D) Benzyl alcohol
44. Primary, secondary and tertiary amines can be separated by
A) Hinsberg’s test B) HNO 2 test
C) Carbylamine test D) Lucas reagent
45. Identify the CORRECT option(s) among the following ?
A) [R] & [S] on treatment with excess of dil NaOH gives same compound
B) Position of double bond in [P] & [Q] in same as per IUPAC
C) [U] & [V] on treatment with excess of dilute NaOH gives different product as [R] &
[S] on same treatment
D) [U] & [R] are constitutional isomers

46. A 10cm sample of an unknown gaseous hydrocarbon was mixed with 70cm3 of oxygen
3
and the mixture was set on fire by means of an electric spark. When the reaction was
over and water vapours were liquefied, the final volume of gases decreased to 65cm3 .
This mixture then reacted with a potassium hydroxide solution and the volume of gases
decreased to 45cm3 . Find the molar mass of hydrocarbon in gm/mol, if volumes of
gases were measured at standard temperature and pressure (STP) conditions ?
A) 26 B) 36 C) 16 D) 46
47. Radial wave function for 2s orbital of hydrogen – like atoms is given as
3
zr
 z 2  zr 
3
  2 2    2   e 2a 0
 a0   a0 
The radial probability  4r 2  2  for an electron in He  at r  a 0 is

3 3
3
 1 2 3  1 2 1
A)   B) 2 2   C) Zero D)  
 a0   a0   a0 
48. Benzene and toluene are nearly ideal solutions. If at a temperature
0
PBenzene  120mm Hg and PToluene
0
 50mm Hg . Then what will be the mole fraction of
Toluene in vapour phase of this solution if mole fraction of Benzene in liquid phase of
solution is 0.6.
A) 0.70 B) 0.6 C) 0.4 D) 0.22
49. Choose the correct representation of conductometric titration of benzoic acid vs sodium
hydroxide.
Conductance
Conductance
VNaOH  VNaOH 
A) B)
Conductance
Conductance
VNaOH  VNaOH 
C) D)

50. Consider the reaction 2 H 2  g   2 NO  g   N 2  g   2 H 2O  g  . The rate law for this
reaction is: rate = k  H 2  NO  under what conditions could these steps represents the
2
mechanism?
Step: 1 2 NO  N 2O2
Step: 2 N 2O2  H 2  N 2O  H 2O
Step: 3 N 2O  H 2  N 2  H 2O
A) These steps cannot be the mechanism under any circumstances
B) These steps cannot be the mechanism if Step 1 is sloe step
C) These steps cannot be the mechanism if Step 2 is sloe step
D) These steps cannot be the mechanism if Step 3 is sloe step
SECTION-II
51. How many of these elements have lower electron affinity than fluorine ?
Cl , S , O, N , P, Br , I , C
52.
Complex Spin-only magnetic moment
 Fe  CO 4  C 2 O 4  
 a
 FeCl4  b

 Fe  CN 6 
3
c
 Cu  NH 3 4 
2
d
Then (a+b+c+d) = _______

(Given : 3  1.7; 2  1.4; 5  2.2; 15  3.8, 35  5.9; 6  2.4 )

53. The number of correct statements is /are :
(i) In manganate and permanganate ions. The  - bonding takes place by overlap of p-
orbitals of oxygen and d-orbitals of manganese
(ii) Manganate ion is green in colour and permanganate ion in purple in colour
(iii) Manganate and permanganate ions are paramagnetic
(iv) Manganate and permanganate ions are tetrahedral
54. The total number of chiral compound/s from the following is____________
OH
OH
OH OH CH2
HO OH
OH
55. How many of the following reactions will give an alcohol as the major product ?
(i) CH 3CH 2C 

NaOH
aq

(ii) CH3CH  CH 2 

aq Na CO2
 3
(iii) CH3CH 2  C 

CH OH
3


 i  mCPBA

 ii  LiAlH  4
 iii  H 
(iv)
Br
aq 
AgNO 3
(v)
OH
NH 2 
HC 
NaNO 2
(vi)
OH OH
 i  H SO
CH3  C  C  CH 3 
 ii  NaBH 
2 4
4
 ii  H
CH 3 CH 3
(vii)
56. Number of compounds which can evolve H 2 on reaction with Na metal.
OH OH O  CH2  H
(i) (ii) (iii) (iv)
(v) H 3C  C  C  CH 3 (vi) Ph  SO 3H
CH 3
H  C  OH H 3C  C  H
O CH 3
(vii) (viii) (ix) H 3C  C  CH

57. Among the following oxides
Cr2O3 , V2 O5 , FeO, Mn 2O 7 , ZnO,Cu 2 O, MnO 2 ,CrO3 , NiO
If x = number of basic oxides
y = number of acidic oxides
z = number of amphoteric oxides
then the value of

 x  y is
z
58. Find the solubility of Co  NH 3 4 C 2  CO 4 .
 
If the  0m Co  NH3 4 C 2   50 1cm2mol1   0m  CO 4  70  1cm 2 mol1 and the

measured resistance was 33.5 in a cell with cell constant of 0.20cm 1 ?
Represent your answer in milli mol/litre WITH NEARST INTEGER
59. How many of following ethers are difficult to make or cannot be synthesized as a major
product by S N 2 reaction ?
60. At what pH given molecules doesnot migrate towards any electrode when electric field
is applied. (give answer in nearest integer. )
p ka1
 2;p ka  3.90;p ka  10.0 
2 3
NH 3
HOOC  CH 2  CH  COOH

SECTION – I
61. Let z1 , z2 , z3 be three distinct complex numbers lying in a circle with a centre at the origin
such that z1  z2 z3 , z2  z3 z1 and z3  z1 z2 are real numbers, then z1 z2 z3 equals to
A) -1 B) 0 C) 1 D) None of these
62. The number of irrational solutions of the equation x 2  x 2  11  x 2  x 2  11  4 is
A) 0 B) 2 C) 4 D) 11
63. If H1 , H 2 ,.....H n are n harmonic means between a and b,  b  a  , then the value
H1  a H n  b
 is equal to
H1  a H n  b
A) n  1 B) n  1 C ) 2n D) 2n  3
1 1 12 12  22 12  2 2  32
64. Let H n  1     then the sum to n terms of the series    .... is
2 n 13 13  23 13  23  33
4 4 1 4 4 2 n
A) Hn 1 B) Hn  C) Hn D) Hn 
3 3 n 3 3 3 n 1
65. If a  log12 18, b  log 24 54 then the value of ab  5  a  b  is
A) 0 B) 4 C) 1 D) 6
66. The number of ordered pairs  m, n  , m, n {1, 2....100} such that 7m  7n is divided by 5 is
A) 1250 B) 2000 C) 2500 D) 5000
The number of irrational terms in the expansion of  51/6  21/8  is

100
67.
A) 96 B) 97 C) 98 D) 99
68. If A, B and C are three sets such that A  B  A  C and A  B  A  C , then
A) A=C B) B=C C) A  B   D) A=B

69. Two n  n square matrices A and B are said to be similar if there exists a non-singular
matrix Psuch that P 1 AP  B . If A and B are similar and B and C are similar, then
A) AB and BC are similar B) A and C are similar
C) A+C and B are similar D) None of these
17. A letter is taken at random from the letters of the word ‘STATICSTICS’ and another
letter is taken at random from the letters of the word ‘ASSISTANT’. The probability that
they are the same letter is
1 13 19 3
A) B) C) D)
45 90 90 16
71. sin 470  sin 610  sin110  sin 250 is equal to
A) sin 360 B) cos 360 C) sin 70 D) cos 70
72. A1 , A2 , A3 ,... An are n points in a plane whose coordinates are
 x1 , y1  ,  x2 , y2  ,  x3 , y3  ...,  xn , yn  respectively. A1 A2 is bisected at the point G1 , G1 A3 is divided
in the ratio 1:2 at G2 , G2 A4 is divided in the ratio 1:3 at G3 , G3 A5 is divided in the ratio 1:4
at G4 , and so on until all n points are exhausted. The coordinates of the final point G so
obtained are
A)  n  x1  x2  ...  xn  , n  y1  y2  .... yn  
x  x  ...  xn y1  y2  ....  yn 
B)  1 2 , 
 n n 
x  2 x2  3 x3 ...  nxn y1  2 y2  3 y3 ....  nyn 

C)  1 , 
 n n 
D) None of these
73. Let PQ and RS be tangents at the extremities of a diameter PR of a circle of radius r.

Such that PSand RQ intersect at a point X on the circumference of the circle, then
diameter of the circle equals to
PQ  RS 2 PQ  RS PQ 2  RS 2
A) PQ.RS B) C) D)
2 PQ  RS 2

a3 x 2 a 2 x
74. The locus of the vertices of the family of parabolas y    2a is
3 2
105 3 35 64
A) xy  B) xy  C) xy  D) xy 
64 4 16 105
75. A triangle ABC is placed so that the mid-point of its sides are on the x, y and z axes
respectively. Lengths of the intercepts made by the plane containing the triangle on these
axes are respectively  ,  ,  , then the coordinates of the centroid of the triangle ABC are
A)   / 3,  / 3,  / 3 B)  / 3,  / 3,  / 3
C)  / 3,  / 3,  / 3 D)  / 3,  / 3,  / 3
e| x|  e  x
76. Let f : R  R be a function defined by f  x  = .Then
e x  e x
A) f is both one-one and onto B) f is one-one but not onto

C) f is onto but not one-one D) f is neither one-one nor onto
cos  sin x   cos x
77. The value of f  0  so that the function f  x   is continuous at each point
x4
in its domain, is equal to
A) 2 B) 1/6 C) 2/3 D) -1/3
f ''  0  f '''  0  f n  0
If f  x   1  x   n  N, x  R  , then the value of f  0   f '  0  
n
78.     is
2! 2! n!
(where f n  0  represents nth derivates of f at 0)
A) n B) 2n C) 2n1 D) None of these
dx
79. x 1/2
 x1/3

A) 2 x  3 x1/3  6 x1/6  log 1  x1/6   C B) 3x1/3  6 x1/6  log 1  x1/6   C
C) x  3x1/3  6 x1/6  log 1  x1/6   C D) 3x1/3  x  6 x1/6  log 1  x1/6   C
80. The solution of the differential equation y ' Ay   By 2 ( A, B  0) is (C is arbitrary constant)
A) y    B / A e Ax  C 
1
B) y   B / A   Ce Ax
C) y  e Ax   B / A  e Ax  C  D) y  e Ax   B / A  e Ax  C 
1 1

SECTION-II
81. If tan 1  x    tan 1  x    tan 1 then the value of 250 x4 +320 x2 +137 is equal to
2 2 4
 x  x x
82. If O is the origin and the coordinates of A and B are (51, 65) and (75, 81) respectively.
Then OA  OB cos AOB is equal to
k
k k
83. Let 0   r  1  r  1, 2,  and  cos  r = for any k  1 and A   (  r ) r ,then
r 1 2 r 1
lim
498.
1  x  2 1/3
 1  2 x 
1/4
x A
x  x2
7 1
dx  e1   e2  1 is
1
 e  x  1
2 x 1 n
84. A positive integer n  5 such that 0 4 16
85. Let a, b, c are given vectors,  ,  , v are scalars If

d    a  b     b  c   v  c  a  ,  abc   1/ 8 and d   a  b  c   8 then     v is equal to
   
2 3
86. The maximum value of f  x   4  x2  3  4  x 2  1 is
50 n
87. Let U X i  U Yi  T , where each X i contains 10 elements and each Yi contains 5 elements. If
i 1 i 1
each elements of the set Tis an element of exactly 20 of sets X i ' s and exactly 6 of sets
Yi ' s then n is equal to
88. If A  4, 3,5 , B  0, 6, 0  , C  8,1, 4  are three consecutive vertices of parallelogram ABCD . If

 is the angle between AC and BD the sum of digits of 3025 sec 2  is
89. Let x1 , x2 ...., x100 be in an arithmetic progression with x1  2 and their mean equal to 200.If
yi  i  xi  i  ,1  i  100. Let y be the mean of y1 , y2 ,.... y100 . Then y  9000.5
90. Let A  1, 2, 3, 4 , R be a relation on the set A  A defined by
R   a, b  ,  c, d   :2a  3b  4c  5d  . Then the number of elements in R is

KEY SHEET
PHYSICS
1 B 2 D 3 A 4 A 5 B
6 D 7 A 8 A 9 C 10 C
11 C 12 B 13 A 14 A 15 C
16 B 17 C 18 A 19 D 20 D
21 400 22 7 23 2 24 0 25 4
26 2 27 -182 28 3 29 5 30 45
CHEMISTRY
31 A 32 B 33 C 34 D 35 D
36 D 37 B 38 B 39 B 40 B
41 C 42 B 43 D 44 D 45 B
46 C 47 B 48 D 49 C 50 B
51 3 52 3 53 4 54 5 55 3
56 0 57 4 58 4 59 4 60 3
MATHEMATICS
61 A 62 C 63 A 64 A 65 C
66 D 67 B 68 C 69 A 70 D
71 A 72 A 73 B 74 B 75 A
76 B 77 A 78 B 79 C 80 D
81 48 82 42 83 3000 84 75 85 34
86 75 87 6 88 8 89 1680 90 5
SOLUTIONS
PHYSICS
1. Here, S  13.8  0.2  m
and t   4.0  0.3 sec
Expressing it in percentage error, we have,
0.2
S  13.8   100%  13.8  1.4%
13.8
0.3
and t  4.0   100%  4  7.5%
4
s 13.8  1.4
V     3.45  0.3 m / s
t 4  7.5
2. Conceptual
Q2 Q2
3.  8R S  U 
2
 8R 2S  U
2  4 0 R 8 0 R
dU
0
dR
Q2
 R3 
8 0 16 S
4. Conceptual
T2 / T  100
5.  1 ,  1 2  / 
T1 T1  100
6.
PV P 1 h V0
0 0
 0 V 
T0 T0 1 h 1  h
1 1
dv  V0   dh
2 1 h3/2
h h
v0 1
 dw   P dv   P 1 h
0
2 1 h3/2
dh
0 0
 1 h1/2 
h
0 0
PV 1 0 0 
PV 

2  1 h
1/2
dh 
2  1 / 2   

 0
 PV
0 0  1  h 
3.310  3.119 0.191
7.    2.01
3.310  3.215 0.095
8. Consider two small elements of ring having charges +dq symmetrically located about y-axis

The potential due to this pair at any point on y-axis is zero. The sum of potential due to all such
 R 
possible pairs is zero at all points on y-axis. Hence potential at P 0,  is zero
 2 

d d d 0 A 
9. I Q   CV   V  
dt dt dt  d 0  a cos t 
V 0 A  a sin t  V 0 A  a sin t 
I 
 d 0  a cos t   d 0  a cos t 
2 2
When sin t  1, cos t  0 'I' becomes maximum

V 0 Aa
 I0 
d 02
I0 d 02
a
V 0 A
1 1 1  1 1 1
10.  R 2  2   /  R 2  2 
 1 2   2 3 
1 1 

  12 22  3 / 4 3 36 27
/
    
  1  1  5 / 36 4 5 5
 2 2 
2 3 
11. Theorical concept
12. Conceptual
13. Initially when key is closed, the capacitor acts as short-circuit, so bulb will light up. But finally the
capacitor becomes fully charged, so it will act as open circuit, so bulb will not glow
14.
15. The magnetic induction due to both semicircular parts will be in the same direction perpendicular to
the paper inwards
 0i  0i  0i  r1  r2 
 B  B1  B2     
4r1 4r2 4  r1r2 
16. Diamagnetic material shows weak repulsion towards any magnetic pole
17.
18. At angular frequency  , the current in RC circuit is given by

Vrms
irms   i 
2
 1 
R2   
 C 
irms Vrms Vrms
Also   ... ii 
2  
2
9
R  2 2
2
 1  C
R2   C 

 
3 
From equation (i) and (ii), we get
1
5 3 X 3
3 R 2  2 2  C   C 
C R 5 R 5
19. Conceptual
1 1 1
20.  
u v f
1 1 1
 
f  x1 f  x2 f
f  x2  f  x1 1
or 
 f  x1  f  x2  f
or f  fx2  fx1  x1 x2  2 f  f  x1  x2 
2 2
or f
2
 x1 x2 or f  x1 x2
21. v  1.5t 2  2t
dv
a  3t  2
dt
a 3t  2 8
    400
r 0.02 0.02
22. T3  mg
T2   mg  2T3  3mg
F   mg  2T2  7  mg
23. Ist case
30 l 30 37.5
  
P  Q 100  l  P  Q 100  37.5 
30 37.5 30  62.5
  PQ 
P  Q 62.5 37.5
P  Q  50 ... i 
IInd case
30 l

PQ 100  l 
PQ
30  P  Q  71.4

PQ 100  71.4 

30  50 71.4 30  50  28.6
  PQ 
PQ 28.6 71.4
P  600 ... ii 
So, from Eqs. (i) and (ii)
P  30  and Q  20 
24. Conceptual
25. Conceptual
26.
As springs and supports  M 1 and M 2  are having negligible mass. Whenever springs pull the
massless supports, springs will be in natural length. At maximum compression, velocity of will be
zero
And by energy conservation

1 1 y 1
 4 K  y 2  Kx 2  
2 2 x 2
For adiabatic process TV  constant
1
27.
1 1
T2  V1   V1 
     T2     T1
T1  V2   V2 
1.251 0.25
1 1
 T2     273     273
 81   81 
273
  91 K  1820 C
3
r 22 r  2v 
2
22 r 2 v 2
2
28. h  
2g 2g g
2  2   0.05   2 2
2
 = 0.02 M = 2 cm
9.8
dx
29. v  3  8t  3t 2
dt
 v0  3 m / s and v4  19 m / s
1
W  m  v42  v02  [According to work energy theorem]
2
1
  0.03  192  32   5.28 J
2
a
30.   tan 1    tan 1 (1)  450
g
CHEMISTRY
32. The density of diethyl ether is less than water
33. 
2 MnO2  2 K 2CO3  O2   2 K 2 MnO4  2CO2  g 
X   air  green
 y
2 K 2 MnO4  Cl2 
 2 KMnO4  2 KCl
Y   y Pink
35. Oxidative ozonolysis
36. eg3t2g3
38. Reactivity directly proportional to electron density
39. b)HEH bond angle NH3 (107.8)  PH3 (93.6)  AsH3 (91.8)  SbH3 (91.3) (E= central atom)
40. Na gives Golden yellow
K gives Lilac
Ba Green
41.
COOH
COOH
COOH
43. Phenyl alkyl ether become phenol
45.
Column-I Column-II Column-III
a) Bromine iii) Liquid non-metal q) 4s 2 4 p5
b) Gold i) Noble metal r) Transition metal
c) Mercury iv) Liquid metal p) Amalgam
Crystalline non-
d) Iodine ii) s) Violet
metal
46. All bond angles are same
47. Benzaldehyde cannot give Fehiling’s test
48. Refer NCERT
49. PCC cannot oxidise ter. alcohols
50. Secondary amine
52. Acetyloide anion acts as donor
56. All carbons are Sp3
59. Two for self and two for crossed
50. Complex is square planar
MATHS
61. (i) y = f(x) is symmetric about y = x  x = f(y)
 f(f(x)) = f(y) = x
 statement 1 is true
 x , x is rational
(ii) f x   is
1  x , x is irrational
Symmetric about y = x
 f(f(x)) = x
62. Given that PQ  kI
P . Q  k3
 P  2k  0  P is an invertible matrix
 PQ  kI  Q  kP 1I  P 1P  I 
adj.P k
Q   q 23  
2 8

 3  4  k  k  12  16 ….(i)
2 8
 P  2k  k  10  6 ….(ii)
From (i) and (ii) we get   1, k  4  2  k 2  17
63.  is 7th root of unity

 1     2  ....   6  0, p  q  1
pq   4   6   5   7   8   7   9   10  3     2   3  ....   6   3  1  2
 x2  x  2  0
Both I and II are true and II is the correct explanation
64.
M A N K I N D
 4  6!   4! 3 
    5! 0       3! 2    2! 1  1! 1   0! 0   1  1492
 2!   2! 
 1440  36  12  4  1492
65. (A) No. of such triangles = 10 6C1 = 60

(B) No. of such triangles = 10
(C) Number of such quadrilaterals = 10 5C1 + = 75.
(D) Number of such quadrilaterals = 10 (when four consecutive points are taken)
66. Sol: Let p  E1   x, p  E 2   y and p  E 3   z
     
  p E1  E 2  E 3  p  E1  .p E 2 .p E 3
   x 1  y 1  z  …(i)
Similarly,
  1  x  .y 1  z  ..(ii)
  1  x 1  y  .z ..(iii)
p  1  x 1  y 1  z  …(iv)
From (i) and (iv),
x  
 x
1 x p p
From (iii) and (iv),
z  
 z
1 z p p

  p
1 
p  E1  x p  
    …(v)
p  E2  z  1 p 
  p   
Given that
   2  p    p     2p   …(vi)
  3  p  2  3p   p  2   …(vii)
From (vi) and (vii),
   2p
  p  6p  5 …(viii)
3 p  2 
p 6p p p 
   5   1  6   1
    
From (v) and (viii),
p  E1 
6
p  E3 
67.
Slope of AC = 
Slope of PD = 0
 a a b3  b 3
D ,   D  a, 
 2 2   2 
b3
 1  0;b  3  2  0  b  1
2
b  1
 b  a 5  b   a 1 
E ,  ,2
 2 2   2 
Slope of BC  Slope of EP  1
 
5b  2 1 
 
    1
 b  a   a 1 1 
 2 
 6   2 
    1  12  1  a  a  3
 1  a   a  3 
 12  a 2  3a  a  3  a 2  2a  15  0
 a  5  a  3  0
ab  0  a  1  0; a  0;a  0

a  3 Accept
Equation of AP A  3,3 , P 1,1
 3 1 
y 1     x  1
 3  1 
2y  2  x  1
 x  2y  3
Equation of BC B  1,5  , C  3, 1
 5 1 
y 1     x  3  3x  9
 1  3 
3x  y  8  0
 13 17 
Solving AP and BC Q   , 
 7 7
68. Given equation is
e 4 x  4e3x  58e 2 x  4e x  1  0
 1  1  
Take, f  x    e2 x  2 x  4  e x  x   58 
 e  e  
1
Let e x  x  p   0  ……(i)
e
p 2  4 p  60  0
p  6 or  10
Only p  6 is allowed
1
ex  6
ex
Two real and distinct values of x
d 2
69. Since, given  1   2  20   1
d 1
   
2 2
Now, A1   1  and A 2    2 
4  2 
 3
2 2
Let S  2A1  3A 2  1  2
8 4
For max or min
ds 2 6 d  6 
0 1  2. 2 0 1  2  1 6
d 1 8 4 d 1 4 4 2
70. Let f  x   4x  11x  8x  5 x  R
3 2
 f '  x   12x 2  22x  8

f ' x   0
 2  6x 2  11x  4   0  6x 2  8x  3x  4  0
  2x  1 3x  4   0
1 4
 function is decreasing in  , 
2 3
71. Since, we know AB  BC  CA  0
   2,   4,     3
1
Now, AB  AC  5 6
2
  9    2  12   100  600    5,   8
2 2
Hence, CB.CA  60
72. Given points and direction ratios are shnown below.

a 1  1, 2,3 ,a 2   2, 4,5  , b 2  2iˆ  3jˆ  kˆ

b 2  î  4ˆj  5kˆ
Apply shortest distance formula,
a 2
 a 1  .  b1  b 2 
Shortest distance =
b1  b 2
 
S.D.=
  2  1 î   4  2 ˆj  5  3 kˆ . b  b  1 2
…..(i)
b1  b 2
î ˆj kˆ
 
Take, b1  b 2  2 3 
1 4 5
 î 15  4   ˆj    10   kˆ  5  15  4  î     10  ˆj  5kˆ
 
b1  b 2  15  4      10   25
2 2
From equation (i),
S.D. 
 î  2ˆj  2kˆ  . 15  4  î     10  ˆj  5k 
 
15  4      10 
2 2
 25
15  4  2  20  10 1

15  4      10  3
2 2
 25
Take square both sides,
3  5  2   225  16 2  120   2  100  20  25
2
12 2  75  60  17 2  140  350

5 2  80  275  0   2  16  55  0
   5    11  0    5,11
Sum of values of   5  11  16
15sin 4   10 1  sin 2    6  25sin 4   20sin 2   4  0
2
73.
 25sin 4   10sin 2   10sin 2   4  0
2
  5sin 2   2   0  sin 2  
2
5
4 3
 cos 2   1  
25 5
3 3
5 5
Now, 27sec   8cos ec   27    8    125  125  250.
6 6
3 2

74. For, S1 we have
 x  2   x 2  3x  5
 0
x 2  3x  2
 x   , 2  1, 2 
For S2 , we have
 3x  3x  3  32  3x  3  0
For S2 , x  1, 2
  , 2  1, 2
x2
75.   x sin x  cos x 2 dx
d
  x sin x  cos x   x cos x
dx
x cos x  x  x  1 
  x sin x  cos x 2  cos x dx  cos x  x sin x  cos x 
 I
x sin x  cos x  1 
  dx
cos x  x sin x  cos x 
2
x  1 
     sec2 xdx
cos x  x sin x  cos x 
 x sec x
  tan x  C
x sin x  cos x
76. Given equations are y 2  8x and y  2x
Put the value of y in other equation

 8x  2x 2 , 2x 2  8x  0
2x  x  4   0  x  0 & 4
4
 x 32  4
   
4
x 2
Area : 2 2 x  2x dx  2 2    2
 2
3/ 2  1
 1
1
Apply the limit,

4 2 2 28 2 15 2 11 2
  8  1  16  1   
3 2 3 2 6
77. Given, 1  x  dy  y  x  y  dx
2

where, y  0   1, y 2 2   
 yx  y 2 
dy   2 
dx
 1 x 

dy   x   1  2
 y 2 
 2 
y
dx 1 x  1 x 
Divide y 2 both side
1 dy 1   x  1
  2 

y dx y  1  x  1  x 2
2
1 1 dy dt
Put  t then 2 
y y dx dx
dt x 1
 t 
dx 1  x 1  x 2 2
 1 x dx
x
 e In 1 x  1  x 2
2
I.F. = e
2
 1 
 t 1 x2    . 1  x 2  dx
1 x 
2
1  x2
y

 In x  x 2  1  c 
At, y  0   1  c  1
 1  x 2  y In e x  x 2  1 
  In  e  3  2 2  
3 3


IN e 3  2 2  
 
3
e  e 3 2 2

78. First common term of both the series is 23 and common difference is 7  4  28
 Last term  407
 23   n  1  28  407   n  1  28  384
384
n  1  n  14.71
28
Hence, number of terms common are 14
79.
C.I. fi xi fi x i C.F.
0-6 a 3 3a a
6-12 b 9 9b a+b
12-18 12 15 180 a+b+12
18-24 9 21 189 a+b+21
24-30 5 27 135 a+b+26
N=(26+a+b)  504  3a  9b 
504  3a  9b 309
Mean = 
26  a  b 22
 243a  111b  3054
 81a  37b  1018 …..(i)
Median class is 12 – 18

a  b  26
 a  b
Now, Median =  12  2  6  14
12
a  b  26  2a  2b
  4  a  b  18 …(ii)
2
on solving eqs. (i) and (ii), we get
a  8, b  10
  a  b    8  10   4
2 2
50 n
80.  X i   Yi  T;  n  X i   10, n  Yi   5
i 1 i 1
50 n 500 5n
So,  X i  500,  Yi  5n    n  30
i 1 i 1 20 6
81. z 1  i   z 1  i   10
  
 z  z  i z  z  10  x  y  5  0 
And z  5  4 is interior of a circle with centre  5,0  and radius 4.
 z  1 represents the distance of z from -1.
z  1 is maximum at A.
On solving equation of circle and line we get

A 2 2  5, 2 2 
   
2 2
2
z  1  AB2  2 2  4  2 2
   2  32  16 2
So,     32  16  48 .
  
82. Given matrix is A    2
2 2
    
Applying, R 3  R 3  R 1
  
 A        2 2  2
1 1 1
 A                   
 adjA  A
n 1
adj  adjA   A
 n 1 2

adj adj  adj  adjA    A   n 14
A
24
 A
16
         232.316
16

          22.3  12 
16 16 16
       12. Hence, , ,   N
    1     1     1  9
Number all tuples  , ,   11 C 2  55
1 case for     
And 12 case when any two of these are equal
So, No. of distinct tuples  , ,    55  1  12  42
83. Since 54  33  2
Given that number whose G.C.D with 54 is 2.
 Numbers should be divisible by 2 but not by 3
N = (Numbers divisible by 2) – (Number divisible by 6)
9000 9000
N   4500  1500  3000
2 6
1 d
0 1  3  d 1 ..(i)
84. 4
1  2d 1 3
0 1    d  …(ii)
4 2 2
1  4d 3 1
0 1    d  …(iii)
4 4 4
1  3d 1
0 1    d 1 …(iv)
4 3
From (i), (ii), (iii) and (iv)
1 1 1
  d  Minimum value of d  
3 4 3
1  2d 2 1  4d  3 1  3d 
Mean = 0   
4 4 4
6  3d 1  1 5 5
X   6  3     60X  60   75
4 4 3 4 4
85. Given system of equations are x  y  z  6
2x  5y  z  
x  2y  3z  14
From the given equations.
x y6z ….(i)
x  2y  14  3z ….(ii)
Subtract (i) from (ii),
 y  8  2z, then x = z-2.
Now, put the values x & y
In eq. zx  5y  z   .
2  z  2   5  8  2z   z  
   8 z    36
For having infinite solutions
  8  0&   36  0
  8,   36
Required sum =     44

86. We have
x 2  4y 2  2x  8y    0
 x  1  y  1
2 2
2b 2
  1  4
5 5 a
4
   5  4
2
4
 5 
On solving    59
  5
1  2a  2   5  2 65  16
     59  16  75
87. (d) Given function is

x3  x
f x  dx
e xx
 1
Apply property
f  x  dx   f  a  b  c  dx. &  f  x  dx  2 f  x  dx
b b a a
a a a 0
2
x x3
I dx …(i)
2 e xx
 1
2
x3  x
I dx ….(ii)
2 e x x
 1
Add (i) & (ii)
 x3  x
2
x3  x 
2I  2   x x   dx .
  e  1  e x  x  1 
0
 
2 
x3  x x3  x  2 
x3  x x3  x 
I   xx  x x  dx  I    x x  x x  dx
0

  e  1  e  1  
 0

  e  1  e  1 

2 
 x3  x x 3  x 
I  x dx
0  e

x

1 e 1 
2
   2

 x3  x e  x  x  
2
2 x 3 2
I  

  dx =    x 3  x  dx
 1 e 1 e
2 2
x x
0
 0
2
 x4 x2 
     42  6
 4 2 0
88.
x  1 y  3 z 1
  
Given line is 2 3 1

x  2  1, y  3  3, z    1.
 2  1  a  2   3  1 3     1 1  0
 4  2  2a  9  3    1  0
 14  4  2a  0  7  2  a  0and,
  5  1   3  1     1  24
2 2 2
 35 2  14  21  0     1 35  21  0

For,   1  a  5
Let  1 ,  2 , 3  be reflection of point P
1  5  2  2  4  12 3  2  0
1  3 2  8 3  2
a  1   2  3  8
 z 2  8iz  15 
89.  2 R
 z  3iz  2 
 1 2
11iz  13  R
 z  3iz  2 
13
Put z    i Given 
11
  z 2  3i z  2  is imaginary
Put z = x + iy
  x 2  y 2  2xyi  3ix  3y  2   Imaginary
 Re  x 2  y 2  3y  2   2xy  3x  i   0
 x 2  y 2  3y  2  0
 x 2  y 2  3y  2  x 2   y  1 y  2 
13 13
As z    i;Put x  , y   , we get
11 11
 13  13 
2    1  2
 11  11 
 24  35 
 2     242  45  35  1680
2
 121 
90. Given function is
 In 1  5x   In 1  x 
 : x0
f x   x
 10 : x 0

In 1  5x   In 1  x 
lim  10
x 0 x
Apply expansion of In (1 + x).
lim
 5x  .....   ax  ....  10
x 0 x
lim  5     10
x 0
5    10    5


MISTAKES
MATHS
PHYISCS
CHEMISTRY
SECTION – I
1. A body travels uniformly a distance of 13.8 0.2  m in a time ( 4.0  0.3) s. The
velocity of the body within error limits is
A)  3.45  0.2  ms 1 B)  3.45  0.3 ms 1
C)  3.45  0.4  ms 1 D)  3.45  0.5  ms 1
2. Statement 1 : In a cyclic process initial and final states are not same
Statement 2 : Initial and final temperatures are same, therefore the change in internal
energy is zero
3. A soap bubble of radius R has uniformly distributed charge Q on it's surface. It's energy
is the self-energy of charges & surface energy due to surface tension. In equilibrium, this
energy is minimum. Surface tension is S. At equilibrium radius of bubble is R &
pressure inside is P. Pressure outside is P0 . Then
1/3 1/3
 Q2   Q2  2S 2S
A) R    B) R    C) P  P0  D) P  P0 
 128 0 S   64 0 S 
2 2
R R
4. Statement 1 : If a boy standing on a frictionless horizontal surface throws a ball, then he

will move backward.
Statement 2 : In the absence of external force the linear momentum of the system
remains same

5. The work of 146 kJ is performed in order to compress one-kilo mole of gas adiabatically
and in this process the temperature of the gas increases by 7C . The gas is
(R  8.3Jmol1K 1 )
A) monoatomic B) diatomic
C) triatomic D) a mixture of monoatomic and diatomic
6. A light string is tied at one end to a fixed support and to a heavy string of equal length L
at the other end A as shown in the figure (Total length of both strings combined is 2L ).
A block of mass M is tied to the free end of heavy string. Mass per unit length of the
strings are  and 16  and tension is T . Find lowest positive value of frequency such
that junction point A is a node.
1 T 5 T 3 T 1 T
A) B) C) D)
L  2L  2L  2L 
7. In the determination of refractive index of material of a parallel sided slab using a
travelling microscope the following observations are made. Given least count of
microscope is 0.001cm. Then the value of refractive index of material of slab is
Reading of the microscope when focused on
Sl. Mark Mark on paper Particles on
no Made on paper Through the slab top of the glass surface
M.S.R. V.S.R. M.S.R. V.S.R. M.S.R. V.S.R.
a1=M+N  L.C. a2=M+N  L.C. a3=M+N  L.C.
M N M N M N
(cm) (cm) (cm)
(cm) (cm) (cm) (cm) (cm) (cm)
3.10 19 3.20 15 3.30 10

A) 2.01 B) 1.50 C) 1.25 D) 2.25

8. A thin ring of radius R metres is placed in x-y plane such that its centre lies at the
origin. The half ring in region x<0 carries uniform linear charge density  C / m and
the remaining half ring in region x>0 carries uniform linear charge density  C / m .
 R 
The electric potential (in volts) at point P whose coordinates are 0m,  m is
 2 
1  1  1
A) Zero B) C) D) 
40 2 40 4 40
9. A parallel plate capacitor (plate Area: A) connected to battery of emf 'V' and negligible
internal resistance, so that one of the plate is made to oscillate and distance between
plate varies as d  d0  a cos  t  , a  d0 . If maximum current observed in circuit is I0 ,
then the corresponding amplitude of vibration (a) is
a 2I0 I0d 0 I0d 02 I0d 0
A) B) C) D)
VA0 V A0 VA0 VA0
10. Wavelength of first line in Lyman series is  . The wavelength of first line in Balmer series is
5 36 27 5
A)  B)  C)  D) 
27 5 5 36
11. Pick out the correct statements of the following
A: If a rigid body is in translational equilibrium, it should be in rotational equilibrium
also
B: If a rigid body is in rotational equilibrium, it should be in translational equilibrium
Also
C: A body in mechanical equilibrium should be in both translational and rotational
equilibrium
D: When a force acting on a body produces turning effect, the force should be a skew
vector with respect to the axis of rotation
A) A, B and C are only correct B) B and Care only correct
C) C and D are only correct D) Only C is correct

12. Match the entries in the Column I with those in Column II
Column I Column II
Binding energy per nucleon
(A) (P) Photoelectric effect
of a nucleus
(B) Particle nature of light (Q) Nuclear fission and fusion
Binding energy of products
(C) is greater than that of (R) Uncontrolled chain reaction
reactants
(D) Atom bomb (S) Measure of stability
A) A – S; B – P; C – R; D – Q B) A – S; B – P; C – Q; D - R
C) A – P; B – S; C – Q; D – R D) A – S; B – Q; C – P; D – R
13. A light bulb, a capacitor and a battery are connected together as shown here, with switch
initially open. When the switch is closed, which one of the following is true
A) The bulb will light up for a short interval of time and then puts off
B) The bulb will light up when the capacitor is fully charged
C) The bulb will not light up at all
D) The bulb will light up and go off at regular intervals
14. In a cylindrical magnetic field B is changing as B  B0   t . The value of emf induced
in the loop as shown in the figure is_____(AOBA is semi circular and ,B0 are
constant values )
 
   
0  
 
A 2a B
 
 a 2  
A) B) a 2   1 C) a 2 D)  a 2
2 2 

15. In the figure shown there are two semicircles of radii r1 and r2 in which a current i is
flowing. The magnetic induction at the centre O will be
 0i  0i  0i  r1  r2   0i  r2  r1 
A)  r1  r2  B)  r1  r2  C)   D)  
r 4 4  r1r2  4  r1r2 
16. If a diamagnetic substance is brought near north or south pole of a bar magnet, it is
A) Attracted by the poles
B) Repelled by the poles
C) Repelled by the north pole and attracted by the south pole
D) Attracted by the north pole and repelled by the south pole
17. The potential energy of a particle of mass is given by

 E ; 0  x  1
Ux   0
 0; x  1 
1 and  2 are the de-Broglie wavelengths of the particle, when 0  x  1 and

1
x  1 respectively. If the total energy of particle is 2E0 the ratio will be
2
1
A) 2 B) 1 C) 2 D)
2
18. An AC source of angular frequency  is applied across a resistor R and a capacitor C in
series. The current registered is I. If the frequency of source is changed to  / 3 .
(maintaining the same voltage), the current in the circuit is found to be halved. The ratio
of reactance to resistance at the original frequency  is
3 2 1 4
A) B) C) D)
5 5 5 5

19. A Zener diode of Zener break – down voltage 10 v is connected as shown in the figure.
Current through Zener diode is
A) 4mA B) 0.6mA C) 6mA D) Zero
20. With a concave mirror, an object is placed at a distance x1 from the principal focus, on
the principal axis. The image is formed at a distance x2 from the principal focus. The
focal length of the mirror is
x1  x2 x1
A) x1 x2 B) C) D) x1x2
2 x2
SECTION-II
21. A particle is moving in a circular path with velocity ( in ms-1) varying with time as
v  1.5t 2  2t . The radius of the circular path is 2cm. Then the angular acceleration of
the particle at t  2sec is ______(in rad/sec2)
22. On a rough table, three blocks (including the first block) are placed as shown in the
figure. Mass of each block is m and coefficient of friction for each block is  . A force F
is applied on the first block so as to move the system. If the minimum value of F
required is nmg , find n.
T2 T3
F m m m
  

23. In a meter bridge a 30 resistance is connected in the left gap and a pair of resistances P
and Q in the right gap. Measured from the left, the balance point is 37.5 cm, when P and
2
P
Q are in series and 71.4 cm when they are parallel. Find the value of   ?
Q
24. A ball is projected with speed 20 2 m/s at an angle of 45 with horizontal. It collides
first with the right wall A (e = 1/2) and then with the left wall B, and finally returns to
the projection point. Then find the coefficient of restitution between ball and wall B.
(g = 10 m/s2)
25. The ratio of amplitudes of two coherent waves in Young’s double – slit experiment is
A1 1
 . What is the ratio of maximum and minimum intensities of fringes?
A2 3
26. A block (B) is attached to two unstretched springs S1 and S2 with spring constants k
and 4 k , respectively (see figure I). The other ends are attached to identical supports M 1
and M 2 not attached to the walls. The springs and supports have negligible mass. There
is no friction anywhere. The block B is displaced towards wall 1 by a small distance
x (figure II) and released. The block returns and moves a maximum distance y towards
wall 2. Displacements x and y are measured with respect to the equilibrium position of
the block B. Find the ratio ‘ x / y ’ ?

27. A certain mass of an ideal gas at 273K is expanded to 81 times its volume under
adiabatic condition. If   1.25 for the gas, then find the final temperature of the gas?
28. A uniform rod AB is hinged at its end A and the other end of the rod is connected to a
block through a massless string as shown in the figure. The pulley is smooth and
massless. Masses of the block and the rod are same and are equal to m .The acceleration
ng
of the block just after the release from this position is found to be . Find the value of
8
‘n’?
29. A force acts on a 30 g particle in such a way that the position of the particle as a function
of time is given by x  3t  4t 2  t 3 , where x in meters and t is in seconds. Find the
work done during the first 4 seconds in Joules?
30. An open vessel containing water is given a constant acceleration ‘g’ in the horizontal
direction. Then the free surface of water gets sloped with the horizontal at an angle  in
degrees is

SECTION – I
31. One litre of mixture of oxygen and ozone at STP (1atm, 273K) was allowed to react with
excess of acidified solution of KI. Released iodine required 80 mL of M/20 sodium
thiosulphate for titration. Volume of oxygen in original mixture at STP will be
2 KI  H 2O  O3  2 KOH  I 2  O2
I 2  2 Na2 S 2O3  Na2 S 4O6  2 NaI
A) 955.2 mL B) 800 mL C) 0.056 L D) 955.2 L
32. Consider the Assertion and Reason given below

Assertion: If diethyl ether and water are used for differential extraction, diethyl ether
would stay as an upper layer and water would stays as lower layer in separating funnel.
Reason: The density of diethyl ether is less than water
Choose the correct answer from the following
A) Both (A) and (R) are correct but (R) is not the correct explanation of (A)
B) Both (A) and (R) are correct and (R) is the correct explanation of (A)
C) (A) is correct but (R) is wrong
D) (A) and (R) both are wrong
33.  X   K 2CO3  Air 

heat
 Y   CO2 
Y   Cl2 
  Z  Purple
Which of the following is correct?
A) X= black, MnO2 , Y=Blue, K 2CrO4 ,Z= KMnO4
B) X= green, Cr2O3 , Y= yellow, K 2CrO4 , Z  K 2Cr2O7
C) X= black, MnO2 , Y= green, K 2 MnO4 , Z= KMnO4
D) X= black, Bi2O3 , Y= colourless KBiO2 , Z  KBiO3

34. Maximum number of different spectral lines observed when a sample contains 1000
hydrogen atoms in third excited state undergoes transition of electrons to ground state
are
A) 3 B) 4 C) 5 D) 6
35. The major end products of the following reaction are

CH3
i) (CH3)3COK/
CH3
H3C
ii) O3
Cl
iii) H2O2
CH3 CH3
+ CH 3CHO + HCHO
CH 3 O CH 3 CHO
A) B)
CH3 CH3
+ CH 3COOH + HCOOH
CH 3 O CH 3 COOH
C) D)
36. Consider that a d6 metal ion (M2+) forms a complex with aqua ligands, and the spin only
magnetic momentum of the complex is 4.90 BM. The geometry and the crystal field
stabilization energy of the complex is
A) Tetrahedral and -1.6  t +1 P B) Octahedral and -2.4  0 +1 P
C) Octahedral and 1.6  0 D) Tetrahedral and 0.6  t
37. Consider the following equilibria:
A( s )  B( g )  C( g ) K   x
p 1
B( g )  D( g ) K   2
p 2
Initially only A(s) was present and final total pressure at equilibrium is 20 atm.
Equilibrium constant ( x ) is found to be
A) 100 B) 33.33 C) 50 D) 24

38. The correct order of reactivity towards nitration using HNO3/H2SO4 mixture for the
given compounds is
Cl Cl
CH3 Cl
Cl
I) II) III) IV)
A) I>II>IV>III B)I>IV>II>III C)II>I>IV>III D)II>IV>I>III
39. Which of the following order is Correctly matched
A) Melting point NH3  PH3  AsH3  SbH3
B) HEH bond angle NH3  PH3  AsH3  SbH3 (E= central atom)
C) Boiling point NH3  PH3  AsH3  SbH3
D) E –H Distance /pm NH3  PH3  AsH3  SbH3 (E= central atom)
40. Which of the following metals can’t be detected in the laboratory through flame test
A) Na B) Mg C) K D) Ba
41. Which two isomeric compounds give same tri carboxylic acid after completion of the
below scheme of reactions?
i) Mg / Dry ether
ii) Anhy. CO2
Isomeric compound Tri carboxylic acid
iii) H+
iv) KMnO4
CH3 Br
Br CH3 H3C Br Br
CH3 H3C
CH3 CH3 CH3
I II III IV
A) I and II B) III and IV C) I and IV D) II and III

42. Consider the Assertion and Reason given below
Assertion: C-O bond order in the complexes is

Cr  CO 5 PF3   Cr  CO 5 PCl3   Cr  CO 5 PMe3 
Reason: Order of Electrons in the ABMO of CO

is Cr  CO 5 PF3   Cr  CO 5 PCl3   Cr  CO 5 PMe3 
Choose the correct answer from the following
A) Both (A) and (R) are correct but (R) is not the correct explanation of (A)
B) Both (A) and (R) are correct and (R) is the correct explanation of (A)
C) (A) is correct but (R) is wrong
D) (A) and (R) both are wrong
43. Consider a chemical reaction nA  products for which following graphical

representation is given:
At is concentration of A at time t in mol L–1.

Which of the following the most appropriate graph indicating relation between half life
and initial concentration of A.

44. The major product formed in the following reaction is
O HI
?
Heat
O
OH I I
OH
A) OH B) I C) I D) OH
45. Match the columns I,II and III and mark the appropriate choice
Column-I Column-II Column-III
a) Bromine i) Noble metal p) Amalgam
b) Gold ii) Crystalline non-metal q) 4s 2 4 p5
c) Mercury iii) Liquid non-metal r) Transition metal
d) Iodine iv) Liquid metal s) Violet

A) a – ii, p, b – i, s c – iii, q d – iv, r
B) a – iii, q b – i,r c – iv, p d – ii, s
C) a – i, s b – ii, p c – iv, r d – iii, q
D) a – iv, q b – i,r c – iv, p d – ii, s
46. Which of the following is correct order for the mentioned property
i) SH4=SiF4 =SiCl4 (Bond angle)
ii)Vinyl chloride >Allyl Chloride ( C-Cl Bond length)
iii) Na2O2  KO2  O3 (O – O Bond order)
1) i,ii,iii 2) i,ii only 3) i only 4) ii,iii only
47. Benzaldehyde shows positive
(i) Tollen’s test (ii) Fehiling’s test (iii) lodoform test (iv) DNP test
A) iii B) i and iv C) ii and iv D) ii and iii

48. Choose the correct statement(s), regarding molecular orbitals
i)If inter nuclear axis is z-axis, bonding molecular orbital obtained from 2px and 2px
orbitals are not symmetrical around the bond axis.
ii)The combining atomic orbitals must have the same symmetry about the molecular
axes.
iii)Sigma bonding molecular orbital are symmetrical around bond axis where as pi
bonding molecular orbitals are unsymmetrical.
iv)Stability order of orbitals is Molecular orbital > Atomic orbital > Anti bonding
Molecular orbital
A) i,ii,iii only B) i and iv only C) ii and iv only D) i,ii,iii and iv
49. The major product of following reaction is
i) CH3MgBr (excess)
O Major Product
ii) H3O+
iii) PCC in CH2Cl2
A) B) C) D)
50. Which amine gives solid with benzenesulphonylchloride and that solid is insoluble in
KOH ?
A) CH 3CH 2 NH 2 B) CH 3 NHCH 3 C)  CH 3 3 N D) CH 3CH ( NH 2 )CH 3
SECTION-II
51. For a unielectronic species, the radial component of Schrodinger wave equation for ns
3
2  Z 2 
a
Zr
orbital is given as    (27  18  2 ) e [  ] , value of principle
2 3
81 3  ao  ao
quantum number(n) will be

52.
Cl
Cl
i) ''X'' moles of NaNH2
C C CH3
ii) CH3Br
One mole
One mole
The value of [X] is
53. Following reversible processes are performed for one mole a monoatomic ideal gas :
Select the number of CORRECT statements among the following:

P :  E  A B  0,  H  A B  0 Q :  w  B C  0
R :  S C  D  0 S :  q  A B  0
54. 5 mol of KI is mixed with 1 mol of HgCl2 in 3.5 L of water. Magnitude of freezing point
of the solution will be nearest to ( KI  HgCl2  K 2 HgI 4  KCl )
( k f  1.86 K .kg.mol 1 )
55. Neutralisation curve for orthophosphoric acid and NaOH is observed as

How many of the following statements are CORRECT ?
P : At points A, C and E, pH  pK a1 , pH  pK a2 , pH  pK a3 respectively.
pK a1  pK a2 pK a2  pK a3
Q : At point B and D, pH  and pH  respectively.
2 2
R : At point “ C ”, H 2 PO4 and HPO42 are in equal amount.
56. The number of sp2 carbon atom(s) present in gammaxane is (are)
57. Thermodynamic efficiency of a cell with cell reaction A( s )  B(2aq )  A(2aq )  B( s ) is 80%.
If H  965kJ . Ecell
o
will be ___ V.
58. How many of the following statements are CORRECT ?
P : A zero order reaction must be a complex reaction.
Q : Molecularity of a complex reaction is undefined.
R : Molecularity is independent of external conditions, however order of a reaction may

be affected by external variables such as high concentration or pressure.
S : Ea> 0 indicates that effective collisions are LESS than available collisions.
59. Find the number (excluding stereo isomers) of aldols (Both self and crossed) formed in
the given reaction is
O dil. NaOH
H
+ H3C
H3C H
O
60. The number of geometrical isomers possible for the complex

[ Pt ( NH 3 )( NH 2OH )( Py )Cl ] is ___________

SECTION – I
61. Statement – 1: If a function y = f(x) is symmetric about y = x, then f(f(x)) = x
 x : x is rational
Statement – 2: If f  x    , then f(f(x)) = x
1  x : x isirrational
A) Statement-1is True, Statement-2 is True
B) Statement-1 is False, Statement-2 is False
 3 1 2 
62. Let P   2 0   , where R . Suppose Q  q ij  is a matrix satisfying PQ  kI3 for some
 3 5 0 
k k2
non-zero kR , If q 23   and Q , then 2  k 2 is equal to
8 2
A) 13 B) 15 C) 17 D) 21
2  2
63. Statement – I: If   cos  
  i sin 

 , p       , q       , then the equation
2 4 3 5 6
 7   7 
where roots are p and q is x2 + x + 2.
Statement – II : If  is a root of Z 7  1 , then 1     2  .....   6  0
A) Statement 1 is true, Statement – 2 is true
B) Statement 1 is false, Statement 2 is false
C) Statement 1 is true; Statement 2 is false
D) Statement 1 is false; Statement 2 is true

64. The letters of the word ‘MANKIND’ are written in all possible orders and arranged in
serial order as in an English dictionary. Then the serial number of the word
‘MANKIND’ is
A) 1492 B) 1493 C) 1490 D) 1491

65. Match the following:
Column I Column II
(A) Number of triangle that can be made using the (p) 75

vertices of a polygon of 10 sides as their vertices and
having exactly one side common with the polygon is
(B) Number of triangles that can be made using the (q) 110
having exactly 2 sides common with the polygon is
(C) Number of quadrilaterals that can be made using the (r) 60

having exactly 2 sides common with the polygon
(D) Number of quadrilaterals that can be made using the (s) 10

vertices of a polygon of 10 sides as their vertices had
having 3 sides common with the polygon is
A) A-r, B-s, C-p. D-q B) A-s. B-r, C-p, D-s
C) A-r, B-s, C-p, D-s D) None of these
66. Let there be three independent events E1 , E 2 and E 3 . The probability that only E1 occurs
is  , only E 2 occurs is  and only E 3 occurs is  . Let ‘p’ denote the probability of
none of events occurs that satisfies the equations   2  p   and    3  p  2 .
All the given probabilities are assumed to lie in the interval  0,1 .
Probability of occurrence of E1
Then is equal to
Probability of occurrence of E 3
A) 9 B) 3 C) 7 D) 6
67. In Let the circumcentre of a triangle with vertices A(a, 3), B(b, 5) and C(a, b), ab > 0 be
P(1, 1). If the line AP intersects the line BC at the point Q  k1 , k 2  , then k1  k 2 is equal to
4 2
A) 2 B) C) D) 4
7 7

68. The number of real solutions of the equation e4x  4e3x  58e2 x  4e x  1  0 is
A) 4 B) 6 C) 2 D) 8
69. A wire of length 20 m is to be cut into two pieces. A piece of length  1 is bent to make a
square of area A1 and the other piece of length  2 is made into a circle of area A 2 . If
2A1  3A 2 is minimum then   1  :  2 is equal to
A) 6 : 1 B) 3 : 1 C) 1 : 6 D) 4 : 1
70. Let f  x   4x 3  11x 2  8x  5, x  R . Then f:
1 3
A) has a local minima at x  . B) has a local minima at x 
2 4
1 3 1 4
C) is increasing in  ,  D) is decreasing in  , 
2 4 2 3
71. Let for a triangle ABC,
AB  2i  j  3k
CB   i   j   k
CA  4i  3j   k
If   0 and the area of the triangle ABC is 5 6 , then CB.CA is equal to
A) 60 B) 120 C) 108 D) 54
x 1 y  2 z  3 x 2 y4 z5 1
72. If the shortest distance between the lines   and   is ,
2 3  1 4 5 3
then the sum of all possible values of  is :
A) 16 B) 6 C) 12 D) 15
73. If 15sin 4   10cos4   6, for R , then the value of 27 sec6   8cos ec6  is equal to:
A) 350 B) 250 C) 400 D) 500

  x  2   x  3x  5 
2
74. Let S1  x  R  1, 2 :  0 and S2  x  R : 32 x  3x 1  3x  2  27  0 . Then, S1  S2

 2  3x  x 2

is equal to:
A)  , 2  1, 2  B)  , 2  1, 2 C)  2,1   2,   D)  , 2

2
75. The integral   x 

 dx is equal to (where C is a constant of integration):
 x sin x  cos x 
x sec x x tan x
A) tan x  C B) sec x  C
x sin x  cos x x sin x  cos x
x tan x x sec x
C) sec x  C D) tan x  C
x sin x  cos x x sin x  cos x
76. 
The area of the region S   x, y  : y2  8x, y  2x, x  1 is 
13 2 11 2 5 2 19 2
A) B) C) D)
6 6 6 6
77. Let y  y  x  , y  0 be a solution curve of the differential equation
1  x  dy  y  x  y  dx . If y  0  1 and
2
 
y 2 2   , then
1

A) e3   e 3  2 2  1
B) e   e2 5  2  
1

C) e   e2 3  2 2  1
D) e3   e 5  2  
78. The number of terms common to the two A.P,’s 3, 7, 11, …., 407 and 2, 9, 16, …., 709
is __________.
A) 7 B) 14 C) 21 D) 28
79. Consider the following frequency distribution :
Class: 0-6 6-12 12-18 18-24 24-30
Frequency : a b 12 9 5
If mean = 309 and median = 14, then the value  a  b 2 is equal to ________.
22
A) 2 B) 6 C) 4 D) 8

50 n
80. Let U Xi  U Yi  T ,
i 1 i 1
where each Xi contains 10 elements and each Yi contains 5 elements.
If each element of the set T is an element of exactly 20 of sets Xi 's and exactly 6 of sets
Yi 's , then n is equal to
A) 15 B) 50 C) 45 D) 30
SECTION-II
81. Let z be the complex numbers which satisfy z5  4 and z 1  i   z 1  i   10,i  1 . If the
maximum value of 2
z  1 is    2, then the value of      is __________.
    
82. Consider a matrix 
A  2
 2
 2  where , ,  are three distinct natural numbers.
         
det (adj(adj(adj(adjA)))
If  232  316 , then the number of such 3-triples  , ,   is _________.
            
16 16 16
83. The total number of 4-digit numbers whose greatest common division with 54 is 2, is
84. The probability distribution of X is :
X 0 1 2 3
P X 1 d 1  2d 1  4d 1  3d
4 4 4 4
For the minimum possible value of d, sixty times the mean of X is equal to ______
85. If the system of equations.
x  y  z  16
2x  5y  z  
x  2y  3z  14
Has infinitely many solutions, then    is equal to

86. If the length of the latus rectum of the ellipse x 2  4y 2  2x  8y    0 is 4, and l is the
length of its major axis, then λ +l is equal to __________
2
x3  x
87. The value of the integral  dx is equal to :
2 e xx
1 
88. If the length of the perpendicular drawn from the point P  a, 4, 2  ,a  0 on the line
x 1 y  3 z 1
  is 2 6 units and Q 1 ,  2 , 3  is the image of the point P on this line,
2 3 1
3
then a    i is equal to
i 1
 z 2  8iz  15  13
89. Let S  z  C  i, 2i : 2  R  . If   i  S,     0 , then 242 2 is equal to
 z  3iz  2  11
90. Let the function
 log e 1  5x   log e 1  x 
 ; if x  0
f x   x
 10 ; if x  0

Be continuous at x = 0. Then absolute value of ‘  ’ is equal to

KEY SHEET
PHYSICS
1 2 2 1 3 4 4 3 5 1
6 4 7 2 8 1 9 4 10 3
11 3 12 1 13 3 14 1 15 1
16 3 17 2 18 4 19 4 20 1
21 8 22 0 23 25 24 13 25 1
26 180 27 1 28 5 29 1 30 0
CHEMISTRY
31 4 32 2 33 3 34 1 35 1
36 3 37 2 38 1 39 4 40 1
41 2 42 1 43 2 44 3 45 3
46 4 47 3 48 3 49 4 50 1
51 2 52 138 53 7 54 9 55 158
56 6 57 7 58 144 59 26 60 12
MATHEMATICS
61 B 62 B 63 A 64 A 65 B
66 B 67 A 68 A 69 B 70 C
71 C 72 D 73 D 74 D 75 B
76 D 77 A 78 A 79 B 80 D
81 5 82 7 83 0 84 2 85 2
86 6 87 5 88 3 89 4 90 0
SOLUTIONS
PHYSICS
1. v cos  300   v0 cos 600
v0
v
3
v0
1
e 2 3 
v0 3 3
2
2. P  2 Pl  Pm
3. Conceptual
4. Applying conservation of energy
2
1 1 1  5v  9 MR 2
Mgh  Mv02  I 02  M  0   I 
2 2 2  4  16
3R
Mx 2  I  x 
4
n3
5. For conservative forces, dU   dW
U f  U i  Wi  f
Or Wi  f  U i  U f  q Vi  V f 
5  105  2  106  2a  0.1  0
2
 
Or a  1.25  10 V / m  1250 V / m 2
3 2
6. Magnetic moment M  NIA

3
  3
2
M  I0 2a  I0a2
4 2
7.
Mass of coal
P 1
8. S av   0 cE02
4 R 2
2
P
 E0 
2 R 2 0 c
3

2  3.14  100  8.85  1012  3  108
=1.34 V/m

4
9. Mass of the sphere is given by M   y 3 
3
4 
G   y3  m
Gravitational force F   
3
2
y
F
a .
m
a0 g
10.  tan   a0 
g 3
11. v cos 370  20cos 530
3
v cos 37 0  20   v  15m / s
5
m
12.  
l r 2
 m 2r l
  
 m r l
0.003 2  0.005 0.06 4
     4%
0.3 0.5 6 100
D   2D 
13.  n  1 t  
d d
 
 n  1 t  5892 A
2
3
14.  2v0   e  v0  3e
2
Here, v0  gR
kT
15.    p  constant
2 D 2 p
16. Conceptual
17. Since ML = Pt
Pt
L
M
18. When one e is removed from neutral helium atom, it
For one e species we know
13.6Z 2
En  eV/atom
n2
For helium ion, Z=2 and for first orbit n=1
13.6 2
 E1   2  54.4eV
1
2
 Energy required to removed this e =+54.4eV

 total energy required =54.4+24.6=79eV
103
19. n
200  106  1.6  10 19
20. use wheatstone bridge condition
21. Wave velocity as a function of distance (x) from top is

T
v  g 9  x 

2h
g
h
dx  2h 
   dt  time takenby stone  
0 g 9  x  0  g 
 h  8m
22. At steady state 10 resistor will be short circuited
u2
23.  100
g
  450
u 2 sin 2 
 25m
2g
24. Heat is extracted from the source in path DA and AB is
3  PV  5  2 PV  13
Q  R  0 0   R  0 0   PV 0 0
2  R  2  R  2
  
25. B  B1  B2
  
B  0  B1   B2
0 I 0 0 I1
 0
2 d 2 R
I R
I1  0
d
1
s1   60 18 
26. 2
s2  40  18
27. Conceptual
20   3
t

28. I1  1  e 5104
   1.5 A
10   2
4
20V
L  5mH
I1
6
5
5
20V I2
C  0.1mF

t
20 103
I2  e  1.0 A
10
From superposition I  I1  I 2  2.5 A
dl
29.  dR  k  l
R1  k 2 l

R2  k 2  2 l 
R1  R2
30. No current flow through capacitor, it remains un charged
CHEMISTRY
31. Hybradisation
7.8
32. S  10 4 mol / l  10 5 mol / l
78
K SP  4 S 3  4  10 15
log10 K SP  log 4  15  14.4
33. Friedal Crafts alkylation, rearrangement
34.
35. Conceptual
36. H  H
C   A
0.06 0.001
Ecell   log  0.06 log 2  102  0.06  0.3  2   0.06  2.3  138V
1 0.2

37. A  I 2 , B  IO3
38. Conceptual
39. Ncert Points
40. Conceptual
41. BCD are correct
42. H  120  350  380  610 Kj / mol
43. Conceptual
44. Conceptual
45. Sulphur
46. Conceptual
47. Conceptual
48. Conceptual
49. T f  K f mi  5  0.4  3  6
50. Conceptual
51. 2 r  n , 2  0.53  2  
OH
COOH
52.
53. Conceptual
54. Conceptual
55.

OH OH
Br Br
S= =331 T= =173
Br Br
56. Conceptual
57. Conceptual
58. x  6, y  0, z  4, p  2
0.693 2.303  0.3010
59. t1/ 2    0.3010
k 2.303
t75%  2t1/2  0.6020
2.303 100
t99%   log 2
2.303 1
 2  0.6020  2.6020
60. DU=12
MATHS
 8 
61. 61   cis  
 11 
1
Re     2   3   4   5   
2
1          ....    0
2 3 4 10
 2 Re     2  ....   5   1
62. Use the theory of combined mean and combined variance formulae
 1 1 1 1 1
63.  1  3    3  3    3  3       1
 2  2 3  3 4 
64. x 3  6x 2  3px  2p  0
x 4  x 2  x3  2
  
 0 
2 4 4
1 x
x2 x3
65. f  x   4e 2  1  x  
2 3
 7 1 1
g '    
 6  f ' 1 5

66. g  t    2 cot 1  3 t 
2
 
g   t    2 tan 1  3 t     2 cot 1  3 x 
2 2
 odd
1
g '  t   2. 2t
.3 t   log 3
1 3
 decreasing

dy  1  cos x  
67.     sin x , f    1
dx  y  2
dy  sin x.y
 
dx 1  cos x
1 x
 dy   tan dx
y 2
x
log y  2 log sec    c
2
0   log 2  c
x
 y  2 cos 2  
2
y  0  2
x y
68.  1
a b
1 4
 1
a b
b
Sab  b
b4
4
 1 b 
b4
ds 4
 1
 b  4
2
db
b  4  2, 2
b  6, 2
b6;a 3
A9
69.  a  b  .c  1
c  a b
1
a  2, b  3, c 
3
1
   2  3  sin  
3
1
 sin  
3 2
  a  b  .c    2  3  sin 2    1
6sin   3 2
1
sin  
2


4

x 1 y z  2
70. Any point on the line   is B  t  1, t , 2t  2  , t  R
1 1 2
Also AB is perpendicular true the line where A is (1,2,-4)
That is 1 t    t  2   2   2t  2   0
 t  1
 B=  0,1, 4 
 AB  2
71. Standard problem
x 1 x2  2 x2  x
72. ax 2  bx 6  cx 5  ....  h  x 2  x x  1 x2  1
x2  2 x2  x x  1
1 2 0
 h  0 1 1  1  2  2   5
2 0 1
73. sin 4 x   k  2  sin 2 x   k  3  0
k  2  k  3 1
 sin 2 x   k  3  sin 2 x  1  0
 0  k 3 1
 3  k  2 
74. 112012  232014  32012  1
1 1
I dx
0 1
1  x2  x 
x
1
x 
 dx 
0 1 x  x 1 x
2 2 4
2
75. A) Area  2  ydy
0
B) Standard formula
 1
C) Area  ab  ab .
4 2
2
D) Area   1dx
1
77. S  1, 2,3,........,50

Ways  250  225
 225  225  1
78. 1, 2,3,........,50
6m  9n multiple of 5
6m  6
9n  9 / 1
ways  50  25
63
79. P E  1
127

tan 1  x  sin x cos x
80. Lim
x 0 x 1  x 
x  0 , n  0  x  x
x  0 , n  1  x  x  1
81. V  ai  bj  ck ; a, b, c  1,1
ways 8 C3  6 4 C3  8  7  6  4  25
82. 0  a, b, c, d  
2cos a  6cos b  7 cos c  9 cos d  0
2cos a  6sin b  7 cos c  9sin d  0
 4  81  36 cos  a  d   36  4  84 cos  b  c 
cos  a  d  21 7
 
cos  b  c  9 3
a3 a7 a13
83. a4 a6 a12 
a5 a4 a11
16
 1  x  x 2    a r x16  r
8
r 0
So,   0
n  2  2r  1 
84. Sn   tan 1  
 4  r 2  r 2  2r  1 
r 1
 
n 
2
 tan 1  
2 
n
   lim   cot Sn 1  cot Sn 
n 
x2
  cot S1  cot Sn 
I    x 2  1   x  1 e x  dx
2
85.
x 2
 1 e x  t
  x  1 e x dx  dx
2
1

 x 2  1 e x 
2
I c
2
1
 f  x   c
2

2
2A  f  0   1  1  2
86. Basic type
A  95 
87. P     
 B   25  9  5 
 1 2 
 23  1
88. P E   
 1 1  2  2
 2 3

89. ways  D 4  D5
x  y
2
90. min
 x  2y   2
 x  y   2 1 y  y 
2 2
y
1 y  
2
3y y
 1, 1 
2 2
2
y , y2
3
y  2, x  2
2 2
y , x
3 3


MISTAKES
MATHS
PHYISCS
CHEMISTRY
SECTION – I
1. A billiard ball of mass m moving with speed v0strikes a smooth floor at an angle of
0
300 with normal to floor. If ball rebounds at an angle of 60 with vertical, then coefficient
of restitution is
1) 1/2 2) 1/3 3) 1 / 2 3 4) 1/ 3 2
2. A plano-convex lens of refractive index 1.5 and radius of curvature 30 cm is silvered at

the curved surface. Now this lens has been used to form the image of an object. At what
distance from this lens an object is placed in order to have a real image of the size of the
object ?
1)20 cm 2)80cm 3)60 cm 4)30 cm
3. STATEMENT – 1: Kinetic energy of a particle is conserved if it is acted upon by a

conservative force only.
STATEMENT – 2: Work done by a conservative force in a closed path is zero.
(A) Statement – 1 is True, Statement – 2 is True; Statement – 2 is a correct explanation

(B) Statement – 1 is True, Statement – 2 is True; Statement – 2 is NOT a correct

(C) Statement – 1 is True, Statement – 2 is False.
(D) Statement – 1 is False, Statement – 2 is True.

4. A thick-walled hollow sphere has outer radius R. It rolls down on an inclined plane
without slipping and its speed at bottom is v0 . Now the incline is waxed so that the
friction becomes zero. The sphere is observed to slide down without rolling and the
speed now is  5 v0 / 4  . The radius of gyration of the hollow sphere about the axis through
nR
center is . Then the value of n is
4
1)1 2)4 3)3 4)2

5. Electric potential V in volt in a region is given by V  ax 2  ay 2  2az 2 , where a is a
constant. Work done by the field when a 2 c charge moves from point (0,0,0.1m) to
origin is 5 105 J . Find a ? ( in V / m 2 )
1)1250 2)1520 3)1750 4)1500
6. Calculate the magnetic moment associated with the loop carrying current I 0 as shown in
figure is
3 3 2 3 2 3
1) I0a2 2) I0a2 3) I0a2 4) I0a2
2 2 3 2
7. What mass (approximately) of coal with calorific value of 30 kJ/g is thermally
equivalent to the heat liberated during the formation of one gram of He4 from deuterium
H2 ?
m  H 2   2.01410 amu, m  He 4   4.002603amu
1) 1  104 kg 2) 2  104 kg 3) 3  104 kg 4) 4  104 kg

8. A lamp emits monochromatic green light uniformly in all directions. The lamp is 3%
efficient in converting electrical power to electromagnetic waves and consumes 100 W
of power. The amplitude of the electric field associated with the electromagnetic
radiation at a distance of 10 m from the lamp will be
1)1.34 V/m 2)2.68 V/m 3)5.36 V/m 4)9.37 V/m

9. Suppose a vertical tunnel is dug along the diameter of the earth assumed to be a sphere
of uniform mass having density  . If a body of mass m is thrown in this tunnel, its
acceleration at a distance y from a centre is given by
4 3 4 4
1) g y 2)  G  y 3)   y 4)  G  y
3 4 3 3
10. A vessel contains liquid of density  up to height h. Vessel is given horizontal

acceleration a0 such that free surface of liquid makes an angle of 30° with horizontal.
Value of a0 is
1) g/2 2) g/3 3) g / 3 4) 3 g / 2
11. A projectile is projected with velocity 20 m/s at an angle of 53° with horizontal. Speed
of the projectile when its velocity is perpendicular to its initial velocity, is
1) 16 m/s 2) 12 m/s 3) 15 m/s 4) 18 m/s
12. A wire of length l  6  0.06 cm and radius of cross-section r  0.5  0.005 cm and mass
m  0.3  0.003 gm . Maximum percentage error in density is
1) 4 2) 2 3) 1 4) 6.8
13. In Young’s double slit experiment using monochromatic light, the fringe patterns shifts
by a certain distance on the screen when a mica sheet of refractive index 1.6 and
thickness 1.964 micron is introduced in the path of one of the interfering waves. The
mica sheet is then removed and the distance between the plane of slits and the screen is
doubled. It is found that the distance between successive maxima now is the same as the
observed fringe shift upon the introduction of the mica sheet. The wavelength of light is
0 0 0 0
1) 5762 A 2) 5825 A 3) 5892 A 4) 6500 A

14. Suppose a narrow, and straight tunnel passes through the center of earth as shown in the
figure. Two small balls A and B are released simultaneously at the two ends of the
tunnel. The balls collide with each other and after collision one ball comes to rest and the
other ball just manages to escape the earth’s gravitational field. If the coefficient of
restitution between the balls is e , then select the correct option.
3 1
1) e  2) e 
2 2
1 3
3) e 4) information is insufficient to predict
2 2
15. An ideal gas undergoes an isothermal process. The pressure (P) of the gas is plotted
against the mean free path  of the molecules. Select the correct graph.
1) 2)
3) 4)

16. If two soap bubbles of different radii are in communication with each other, then
1) air flows from the larger bubble into the smaller one until their sizes become equal
2) the sizes of the bubbles remain unchanged
3) air flows from the smaller bubbles into the larger one and the larger bubble grows at
the expense of the smaller one
4) air flows from the larger into the smaller bubble until their radii interchange
17. A copper wire and a steel wire of the same diameter and length are connected end to end.
A force is applied which stretches their combined length by 1 cm. Then the two wires
have
1) the same stress and strain
2) the same stress but different strains
3) the same strain but different stresses
4) different stresses and strains
18. An energy of 24.6eV is required to remove one of the electrons from a helium atom. The
energy (in eV) required to remove both the electrons from a neutral helium atom is
1) 38.2 2) 49.2 3) 51.8 4) 79.0
19. If 200 MeV energy is released per fission of 92U235, how many fissions must occur per
second to produce a power of 1 mW?
1) 6.25  107 2) 12.25  107 3) 25  107 4) 3.125  107
20. Find the equivalent resistance between points A and B (in  )_
1) 4 2) 6 3) 11.2 4) 10.4

SECTION-II
21. A uniform string of length 9m and mass 4.5 kg is fixed at one end and hangs freely
under its own weight. A wave pulse is generated at top end which runs down to other
end. At the same moment a stone is released from rest and falls freely from the top of
string. How far from the top does stone pass the wave ?
22. Two resistors of 10 and 20 and an ideal inductor of 5 H are connected to a 2 V
battery as shown in the below figure. The key is plugged in at t  0 the value of the
current in the 10 resistor at steady state is
23. A person can throw a ball upto a maximum range of 100m. The maximum height of this
projectile is
24. The p-v diagram represents the thermodynamic cycle of an engine, operating with an
ideal monoatomic gas. The amount of heat extracted from the source in a single cycle is
x
  p0v0 . Find the value of x.
2

25. Current, I0 flows in long straight conductor as shown. If magnetic field at center of
n I0 R
circular loop in the same plane is zero, then current in the circular loop is , then the
d
value of n is
I0
R O
26. At the instant a motor bike starts from rest in a given direction, a car overtakes the motor
bike, both moving in the same direction. The speed time graphs for motor bike and car
are represented by OAB and CD respectively. Then at t = 18s find the distance between
then motor bike and car.
27. A solid disc of uniform thickness has density that varies by quadrants as shown, with
number indicating relative densities. If x-y axes are as indicated with centre of disc at
origin, then the equation of straight line drawn through origin and centre of mass of the
disc is y  nx .Find the value of n
y
3 1
x
4 2

28. In the circuit shown, the key (K) is closed at t=0, the current through the key at the
x
t  103 ln 2 s is ampere . Find the value of x
2
4 5
20V
L  5mH
5
6 C  0.1mF
29. In a meter bridge, the wire of length 1m has a non – uniform cross-section such that, the
dR dR 1
variation of its resistance R with length l is  . Two equal resistances are
dl dl l
connected as shown in the figure. The galvanometer has zero deflection when the jockey
x
is at point P. The length of AP is m . Find the value of x
4
30. In the circuit shown in figure the switch S is closed at t =0, find the energy stored on the
capacitor at steady state.
4
10V 10 10

3
C  2 F 60V
S
3 10V 10

SECTION – I
31. Match the Column-I with Column-II
Species Shape
A) S2O32 1) pyramidal
B) ClO3 2) linear
C) C3O2 3) squarepalanar
D) Ni  CO 4 4) tetrahedral
1)A-3,B-1,C-2,D-4 2) A-4,B-1,C-2,D-3
3)A-2,B-1,C-3,D-4 4)A-4,B-1,C-2,D-4
32. The Solubility of Calcium fluoride in water is 7.8 104 g / L . The value of log10 K sp of
calcium fluoride is
1) 4 1015 2) -14.4 3)14.4 4) 4 1015
33.
AlCl3
X (major product)
+
Cl
1) 2)
3) 4)

34. In which of following pairs “A’” is less stable than “B”?
(A) (B)
1) 2) Ph  CH 2 C6 H 5
( A) (B)
O
O
(A) (B) (A) (B)

3) 4)
35.
NH 2
NaNO2  HCl NaNO2
NH 2 Q;
P;
05 C 0
HCl , 0  5 c0
.
P and Q are
OH OH N 2Cl
N 2Cl
1) 2)
N 2Cl
N 2Cl OH
N 2Cl
3) 4)
36. Electromotive force of the following cell at 298K
 2.303RT 
Pt , H 2 / H  / / H  / H 2 , Pt   0.06 
1atm   0.001 M   0.2 M  1atm  P 
1)120mv 2) 0.12V 3) 138mv 4) 0.138mv
37. 2 MnO4  10 I   16 H   2 Mn 2  8 H 2O  A ,
2 MnO4  H 2O  I   2 MnO2  2OH   B .
A & B respectively are

1) IO3 , I 2 2) I 2 , IO3 3) I 2 , I 2 4) IO3 , IO3

38. The number of geometric isomers possible for square planar complex
 Pt  Cl  py  NH 3  NO2   is  py  pyridine 
1) 6 2) 8 3) 3 4) 12
39. Correct statements among the following are
A) At isoelectric point amino acids having least solubility
B) All natural amino acids are optically active
C) Globular protiens have coiled (spherical) like structure and are water soluble
D) Fibrous protiens have sheet like (run in parallel) structure and are water soluble
1)A, B only 2) B, C only 3)C, Donly 4) A, C only
40. The radius of La 3  z  57  is 108 Pm . The radius of Lu 3  z  71 will be closest to
1) 85Pm 2)108 Pm 3)180 Pm 4)160 Pm
41. Consider following statements about complexes
A) FeSO4 .  NH 4 2 SO4 .6 H 2O and FeSO4 .4 KCN are complex compounds
B) All Zn 2 complexes (C.N=4 and 6) are diamagnetic

C) In K 4  Fe  CN 6  , iron undergoes d 2 sp3 hybradisation
D) Cu  NH 3 4  SO4 is square planar complex
Select the correct statement(s) from the above?

1)A, B, C only 2)B, C, D only 3) A, C, D only 4) A, B, C, D
1
42. Enthalpy change for the conversion of Cl2  g  to Cl 1  aq  is _______ KJ/mole
2
(Given that  dis H  Cl2   240KJ / mol ,  eg H  Clg   350 KJ / mol and
 hyd H  Cl g   380 KJ / mol )
1)-610 2)610 3)-490 4)490
43. The species given that doesn’t undergoes disproportionation reaction is
1) Cl2 2) MnO4 3) BrO2 4) H2O2

44. Match List-I & List-II
List-I List-II
(Atomic Number) (Block of periodic table)
A) 56 P) d-block
B) 49 Q) f-block
C) 79 R) p-block
D) 64 S) s-block
1)A-R,B-P,C-S,D-Q 2)A-S,B-P,C-Q,D-P
3)A-S,B-R,C-P,D-Q 4)A-S,B-R,C-Q,D-P
45. The chalcogen with highest negative electron gain enthalpy is
1) O 2) Se 3) S 4) Te
46. The nitrogen of following compound doesn’t converted into ammonium sulphate, in
estimation of Nitrogen by Kjeldhal’s method.
1) propanamine 2) urea 3) Aniline 4) Nitro benzene
47. Identify the INCORRECT combination
1) Oxidising power -F2  Cl2  Br2  I 2 2) Bond energy  Cl2  Br2  F2  I 2
3) S.R.P  F2  Cl2  Br2  I 2 4) Water solubility at 250C  Ne  Ar  Kr  Xe
48. In which of following pairs of halogen compounds first one undergoes SN2 reaction
faster.
Cl ,
Cl Cl , I
1) 2)
Ph Ph
Br
Br , H Br
Br
3) , 4) Ph Ph

49. Molal depression constant for a solvent 5.0kkgmol-1. The depression in freezing point of
solvent for 0.4 mole kg-1 solution of Na2 SO4 is (undergoes 100%ionisation ).
1) 3 2) 4 3) 2 4) 6
Cr2O3 CrO2Cl2 , CS2 Conc.KOH , 

A B X+Y

773K ,10  20atm H 3O 
H
50.
X gives “B” when treated with MnO2 . Then Y is
O
H
1) O 2) OH 3) CHO 4)
SECTION-II
51. The de-Broglie wave length of electron present in second Bohr orbit of H-atom is __
 A0 (Round off to nearest integer)
52.
NH 2
NaNO2  HCl H 2O CHCl3 1) AgNO3  NH 4OH

A B C D
0  50 C Steam KOH , H  2) H 
Molecular weight of D is______ g/mol.

(Atomic weight of H=1, C=12,N=14, O=16 g/mol).
53. At 250C, the pH of 1.0 108 M HCl aqueous solution is ______
(Round off to nearest integer).
54. The number of molecules or ions from the following having non-planar structure is ___
BF3 , H 2O2 , SO3 , CH 4 , CH 3 , NO3 , SF6 , BF4 , PH 4 , PCl5 , SF4 , PCl3 , B2 H 6

55.
OH
Br2 Br2
T S
CS2 Water
.
The difference between molecular weights of “S” and “T” is _____ (g)
(Atomic weight H=1, C=12, O=16, Br=80 g/mol).
56. Number of alkenes when treated with HCl gives rearranged product
, , , , , ,
57. How many of following reduces ammonical silvernitrate solution?
Glucose, Fructose, Galactose, Sucrose, Maltose, Lactose, Ribose, De-oxy ribose, Starch,
Cellulose.
58. The d-electronic configuration of  Ru  en 3  Cl2 is t2xg eg y and that of  Fe  H 2O 6  Cl2 is
t2z g eg p . Then 12  x  z  p  y  is
59. Rate constant of a reaction is 2.303sec1 . Calculate t75%  t99% value of same reaction is
______ 101 sec
60.
O
OH
NH 2 Br2  NaOH
X Y
OH  (Major)
(major organic)
.
Degree of unsaturation in “Y” is ____

SECTION – I
8 8
61. If   cos    i sin   , then Re     2   3   4   5  is equal to
 11   11 
1 1
1) 2)  3) 0 4) 11
2 2
62. The mean of two samples of sizes 200 and 300 were found to be 25, 10 respectively.
Their standard deviations were 3 and 4 respectively. The variance of combined sample
of size 500 is
1) 6.72 2) 67.2 3) 672 4) 0.672
7 19 37 61
63. The sum of the series    
23 63 123 203
1) 1 2) 2 3) 3 4) 4
64. If x1 , x 2 and x 3 are the positive roots of the equation x 3  6x 2  3px  2p  0, p  R  0 then
1 1   1 1   1 1
the value of sin 1     cos 1     tan 1    is equal to
 x1 x 2   x 2 x3   x 3 x1 
  3
1) 2) 3) 4) 
4 2 4
1 x
x2 x3
65. Let f  x   4e  1  x  
2
for any real number x , and let g be the inverse function
2 3
of f. Then the value of g '    is

7
 6
1 1
1) 5 2) 3) 3 4)
5 3
  
66. Let the function g : R    ,  be given by g  t    2 cot 1  3 t  .
 2 2 2
P : g is an odd function
Q : g is strictly increasing in  ,   .
1) P is true ; Q is true 2) P is true ; Q is false

3) P is false ; Q is true 4) P is false ; Q is false

dy  1  cos x  
67. y  f (x) is a solution of     sin x and f    1 then f  0  is
dx  y  2
1) 2 2) 1 3) 3 4) 4
68. Let a straight line passing through P(1, 4) with negative slope cuts the coordinate axes at
A, B then the area of the triangle OAB when OA + OB is minimum is __________
1) 9 2) 18 3) 4 4) 14
If a, b & c are such that  a  b  .c  1, c   a  b , a  2 , b  3 & c 

1
69. , then the angle
3
between a & b is
   
1) 2) 3) 4)
6 4 3 2
x 1 y z  2
A) Assertion: There exists two points on the line   which are at a distance
1 1 2
of 2 units from the point (1,2,-4)
x 1 y z  2
(R) Reason : Perpendicular distance of point (1,2,-4) from the line   is 1
1 1 2
unit,
1) A,R are true and R is correct explanation of A
2) A,R are true but R is not correct explanation of A
4) R is true and A is false
71. If ABCD is a square of unit side, 4-circles of unit radius are described with centres at
A,B,C,D then area common to 4 – circles is
  3  
1)1   3 2)1   3) 1   3 4) 1   3
4 4 2 3 3
x 1 x2  2 x2  x
72. Let ax 7  bx 6  cx 5  dx 4  ex 3  fx 2  gx  h  x 2  x x  1 x 2  1 then
x2  2 x2  x x 1
1) g = 3 and h = -5 2) g = -3 and h = -5
3) g = 3 and h = 9 4) g = -2 and h = 5

73. If the equation sin x   k  2  sin 2 x   k  3  0 has a solution then k must lie in the interval:
4
1)  4, 3 2)  2, 0  3)  2, 2  4)  3, 2
a
dx
74. If ‘a’ be the digit at unit`s place in 112012  232012  32012 , then  1

a 1 1 x  x 
2
   
1) 2) 3) 4)
6 3 2 4
75.
Column-I Column-II
A Area bounded by y  x and y=2 is p 4
Area bounded
B x y q
  2  ab
by   1, when a, b  0 is 4
a b
x2 y2
Area between the ellipse 2  2  1
a b
C r 1
x y
and the chord   1 (a,b>0) is
a b
Area bounded by y   x  , the x-axis
D and x  1, x  2 is [.] denotes greatest s 2ab
integer function
1) A-S,B-P,C-Q,D-R 2) A-P,B-S,C-Q,D-R
3)A-S,B-P,C-R,D-Q 4) A-Q,B-S,C-R,D-P
K
76. If the probability that the random variable X takes values x is given by P(X  x) 
3x
x  0,1,2,3,... where K is constant then P(x  2) 
1 1 1 1
1) 2) 3) 4)
3 27 18 9
77. Let S  1, 2,3,.........50 . The number of non empty subsets A of S such that the product of
elements of A is even
1) 225  225  1 2) 225  1 3) 225  1 4) 250  1

78. The number of ordered pairs (m, n) where m, n  1, 2,3,......., 50 , such that 6m  9n is a
multiple of 5 is
1) 1250 2) 2500 3) 625 4) 500
79. A set S contains 7 elements. A non-empty subset A of S and an element x of S are

chosen at random. Then the probability that x  A is:
1 64 63 31
1) 2) 3) 4)
2 127 128 128
tan 1  x  sin x cos x

80. Lim
x 0 x 1  x 
(where x denotes fractional part and  x  denotes greatest integer)
1) 1 2) tan 1 3) cos 1 4) does not exist
SECTION-II

81. Consider the set of eight vectors V  aiˆ  bjˆ  ckˆ ; a, b, c  1,1 . Three non-coplanar
vectors can be chosen from V in 2 p ways. Then p is
82. Let 0  a, b, c, d   , where b and c are not complementary, such that

2cos a  6cos b  7 cos c  9cos d  0 and such that
cos  a  d 
2sin a  6sin b  7sin c  9sin d  0 , then the value of 3 is _____
cos  b  c 
 a3 a7 a13 
16
 
83. Given 1  x  x 
2 8
  a k x then the value of  a 4
k
a6 a12   .........
k 0 a a4 a11 
 5
n  2  2r  1  n
84. If Sn   tan 1  
 4  r 2  r 2  2r  1 
then find the value of lim
n 
  cot  Sk 1   cot  Sk   .
r 1
  k 2

If I    x 2  1   x  1 e 
x 2
dx  A  f  x    C , where C is constant of integration and
2
85.
2
f  1  , then 2 A  f  0  is
e
 2 1 3 4  3  4 
86. Let three matrices A    ; B   2 3 and C   2 3  then
 4 1    
 ABC   A  BC  
Tr  A   Tr    Tr    ........  (where Tr denotes trace of matrix)
 2   4 
87. Mr.A has two fair cubic dice one with faces numbered from 2 to 7 and the second with
faces numbered from 4 to 9. Twice, he randomly picks one of the dice (selection of each
dice is equally likely) and rolls. If it is known that the sum of the resulting two rolls is 10
m
then the probability he rolled the same dice twice is  m, n  N  , (G.C.D. (m,n)=1), find
n
the least value of  n  m  .
88. A is targeting to B. B and C are targeting to A. The probability of hitting the targets by
2 1 1
A, B, C are , and respectively. If A is hit, then the probability that B hits the target
3 2 3
and C does not is P then value of 6P is ________
89. 6 letters L1 , L 2 ......L6 be inserted into 6 addressed envelopes E1 , E 2 ......E 6 one letter each
into one envelope such that no letter goes into its corresponding envelop. If the number
of ways in which letter. L 2 is placed in envelop E 3 is N then value of   is (where .

N
13 
is GIF).
If x, y  R and log 4  x  2y   log 4  x  2y   1 . Then, the minimum value of  x  y  is

2
90.
……..


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PHYISCS
CHEMISTRY
SECTION – I
MgL3
1. A student determined young’s modulus of elasticity using the formula Y  . The
4bd 3
value of g is taken to be 9.8 m / s 2 , without any significant error, his observations are as
following.
Then, the fractional error in the measurement of Y is

A) 0.0083 B) 0.0155 C) 0.155 D) 0.083
2. A box weighs 196 N on a spring balance at the north pole. Its weight recorded on the
same balance if it is shifted to the equator is close to (Take g  10 ms 2 at the north pole
and the radius of the earth = 6400 km)
A) 195.66 N B) 194.66 N C) 194.32 N D) 195.32 N

3. Two identical vessels A and B contain one mole of O2 each. The pressure in vessel A is
‘ P0 ’ and that in B is ‘P’. The r.m.s velocity of molecules in both the vessels are equal.
The vessel A is at rest and B is moving with constant speed V0 , then which of the
following is correct.
Vo
1 mole 1 mole
of O2 of O2
Vessel A Vessel B
Vrms
2
 V02 
A) P  P0 B) P  P0 C) P  P0 D) P  P0  2 
 Vrms 

4. In the given circuit, the potential difference across the capacitor in steady state is 12V.
Each resistance is of 3 . The cell is ideal. The emf of the cell as
A) 15V B) 9V C) 12V D) 24V
5. The potential difference (V) across the 2 F capacitor increases with time, and
dV d 2V
 1V / s and  2V / s 2 at particular instant. The potential difference across the 3H
dt dt 2
inductor is___(Assume current through resistor is constant)
A) 4V B) 12V C) 6V D) 8V
6. When voltage vs  200 2 sin t  150  is applied to an AC circuit the current in the circuit

is found to be i  2sin  t   then average power consumed in the circuit is
 4
A) 200 watt B) 400 2 watt C) 100 6 watt D) 200 2 watt
7. In the young’s double-slit experiment, when a glass – plate (refractive index 1.5) of
thickness t is introduced in the path of one of the interfering beams (wavelength  ), the
intensity at the position where the central maximum occurred previously remains
unchanged. The minimum thickness of the glass-plate is
2 
A) 2 B) C) D) 
3 3

8. A sonometer wire resonates with a given tuning fork forming standing waves with five
antinodes between the two bridges when a mass of 9kg is suspended from the wire.
When this mass is replaced by mass M, the wire resonates with the same tuning fork
forming three antinodes for the same positions of the bridges. The value of M is
A) 25 kg B) 5 kg C) 12.5 kg D) 1/25 kg
9. The circuit contains two diodes each with a forward resistance of 50 and with infinite
reverse resistance. If the battery voltage is 6V, the current through the 120 resistance
is…..mA
A) 5 B) 10 C) 15 D) 20
10. A spherical glass vessel filled with liquid is kept in uniform gravity. Horizontal surface
represents meniscus of liquid. Now complete system is taken to gravity free space. C is
the center of sphere.
A) Vessel and liquid is wetting combination
B) Finally Liquid forms a drop
C) No effect on the liquid system
D) Finally liquid spreads over entire surface of vessel

11. A plane electromagnetic wave of frequency 500 MHz is travelling in vacuum along y-
direction. At a particular point in space and time, B  8.0 108 zT
ˆ . The value of electric
field at this point is (speed of light  3 108 ms 1 xˆ, yˆ , zˆ are unit vectors along x, y and z-
direction)
V V V V
A) 24 xˆ B) 2.6 xˆ C) 24 xˆ D) 2.6 xˆ
m m m m
12. A particle is taken from point A to point B under the influence of a force field. Now it is
taken back from B to A and it is observed that the work done in taking the particle from
A to B is not equal to the work done in taking it from B to A. If Wnc and Wc is the work
done by non – conservative forces and conservative forces present in the system
respectively, U is the change in potential energy, k is the change in kinetic energy,
then choose the correct option
A) WC  U  K B) WC  U C) Wnc  Wc  K D) Wnc  U  K


13. The following set of figures show two cases of collision between two balls. Let u1 and
  
u2 be the velocities before collision and v1 , v2 be the velocities after collision.
u2
 
v2  v1
The coefficient of restitution, e   
u1  u2
Student-A: Equation for ‘e’ holds for both cases as e is property of material of the
colliding bodies.
Student-B: Equation for ‘e’ holds for case II only.
A) Student A is incorrect, student B is correct
B) Student A is correct, student B is incorrect
C) Both are incorrect
D) Both are correct

14. A uniform wooden plank floating on water is tied to the bottom of the pool such that in
static equilibrium the diagonal plane of symmetry coincides with water surface, as
shown in figure. The specific gravity of the wood is_______
1 1 2 3
A) B) C) D)
2 3 3 4
15. A body of mass 10 kg placed on a rough surface is pushed by force F making an angle of
300 to the horizontal. If the angle of repose (between the block and the surface) is also
300 , then the magnitude of minimum force F required to move the body is equal to ___
(g = 10 m/s2)
A) 100 N B) 50 2N C) 100 2N D) 50 N
16. A container is filled partially with a non-viscous fluid of density '  ' upto a height ‘h’.
Initially, the entire system is at rest. Now, the container starts rotating with a constant
angular velocity o about its vertical axis. Choose the appropriate plot from below which
depicts the shape of free surface of the fluid after a long time.

17. One mole of an ideal monoatomic gas is expanded till the temperature of gas is doubled
under the process TV 2  const (T is temperature and V is volume of gas). The initial
temperature of gas is 400 K, the total work done in the process is
A) 400R B) 600R C) 300R D) 200R
18. In a reactor, 2 kg of 92U 235 fuel is fully used up in 30 days. The energy released per
fission is 200 MeV. Given that the Avogadro number, N  6.023 1026 per kilo mole and
1eV  1.6  1019 J . The power output of the reactor is close to
A) 125 MW B) 35 MW C) 63 MW D) 54 MW
19. A closed organ pipe has a fundamental frequency of 1.5 kHz. The number of overtones
that can be distinctly heard by a person with this organ pipe will be. (Assume that the
highest frequency a person can hear is 20 kHz)
A) 7 B) 5 C) 6 D) 4
20. Two magnetic dipoles X and Y are placed at a separation d, with their axes
perpendicular to each other. The dipole moment of Y is twice that of X. A particle of
charge q is passing through their midpoint P, at angle   450 with the horizontal line, as
shown in figure. What would be the magnitude of force on the particle at that instant? (d
is much larger than the dimensions of the dipole)
   2M
A)  0  C) 2  0  D)  0 
M M
 qv B) 0  qv  qv
 4   d   4   d   4   d 
3 3
     
2 2 2

SECTION-II
21. A stone is dropped from the top of a building. When it crosses a point 5m below the top,
another stone starts to fall from a point 25 m below the top. Both stones reach the bottom
of building simultaneously. The height of the building is (in meters)  g  10 m / s 2 
22. A certain mass of a solid exists at its melting temperature of 200 C . When a heat Q is
4
added of the material melts. When an additional Q amount of heat is added the
5
material transforms to its liquid state at 500 C . Find the ratio of specific latent heat of
fusion (in J/g) to the specific heat capacity of the liquid (in J g 1 0C 1 ) for the material
23. A positive point charge +q is placed at the origin. There is an electric field
 x x2 
E x   E0  2  3 2  , that accelerates the point charge along the x-axis. The kinetic energy
 d d 
of the charge when it reaches the position x = 2d is XqdE0 then find the value of X .
24. In a meter bridge, gaps are closed by two resistances P and Q and the balance point is
obtained at 40cm. When Q is shunted by a resistance of 10 , the balance point shifts to
50 cm. Find the value of Q? in 
25. A particle of charge q and mass m starts moving from the origin under the action of an
  
electric field E  E0iˆ and magnetic field B  B0iˆ with a velocity v  v0 ˆj . The speed of the
PmV0
particle will become 2v0 after a time t  find the value P.
qE0
26. A monochromatic light ray is incident making an angle " " with the axis of a transparent
5
cylindrical fiber of refractive index placed in vacuum as shown. Find the maximum
4
value of  (in degrees) so that the light entering the cylinder does not come out of the
curved surface.

27. The total energy of an electron in the ground state of hydrogen atom is -13.6 eV. The
magnitude potential energy (in eV) of an electron is the ground state of Li 2 ion will be
28. A rod of mass m and length L is pivoted at its centre and can rotate in a vertical plane.
Two springs each of force constant k are connected at its ends as shown in the figure.
m
The time period of SHM of rod is T  2 for small oscillations. Find the value of P.
Pk
L

O
29. A rod of mass m and length  rests on a smooth horizontal ground and is hinged at one
of its ends. At the other end, a horizontal force F is applied whose magnitude is constant
and the direction is always perpendicular the rod. When the rod rotates by 900 angle,
xF 3 l
power supplied by this force at that instant is . Find the value of x .
m
30. The energy flux of sunlight reaching the surface of the earth is 1.33 103 W / m 2 . How
many photons (nearly) per square metre are incident on the earth per second? Assume
that the photons in the sunlight have an average wavelength of 550 nm (express the
answer in multiples of 1021 )

SECTION – I
31. Read the following Statements
1) First IP of Mg is less than that of Al
2) O has lower electronegativity than that of O
3) Electron affinity of Cl is lesser than that of F
4) Second IP of oxygen is greater than that of N
Among the following, the correct option is
A) All are correct B) 1 only correct C) 1 and 2 only correct D) only 4 is correct
32. The correct statement regarding ClOn molecular ion is:
A) On decreasing value of ‘n’, 'Cl  O' bond order increases
B) On increasing value of ‘n’, 'Cl  O' bond length increases
C) On increasing value of ‘n’, oxidation number of central atom increase
D) On increasing value of ‘n’, hybrid orbitals on central atom increase
33. Which one of the following is a correct statement?

A) Cr 2 is oxidising and Mn 3 is reducing when both have d 4 configuration
B) Many copper(I) compounds are unstable in aqueous solution and undergo
disproportionation as 2Cu   Cu 2  Cu
C) Oxidation state of iron in Fe3O4 is fractional.
D) Actinoid contraction is lesser from element to element than lanthanoid contraction
34. Increasing value of magnetic moment of following species is
4 3 3 2
I .  Fe  CN 6  II .  Fe  CN  6  III . Cr  NH 3  6  IV .  Ni  H 2O  6 
A) I  II  III  IV B) IV  III  II  I C) II  III  I  IV D) I  II  IV  III

35. The longest CO bond length will be with
A) [Mn(CO)6]+ B) [V(CO)6]– C) Cr(CO)6 D) [Ti(CO)6]2–

36. Assertion (A) : Boron always forms Covalent bond.
Reason (R) : The small size of B3 favours formation of covalent bond
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
37. Match the Column – I with Column-II
COLUMN - I COLUMN – II
A NH4Cl P Covalent bond
B HNC Q Ionic bond
C Liquid H2O2 R Hydrogen bond
D CuSO4 .5H2O S Co-ordinate bond
A) A  p, q, s; B  p, s; C  p, r; D  p, q, r , s
B) A  p, r , s; B  p, q; C  q, r; D  p, q, r , s
C) A  p, r , s; B  p, q; C  q; D  r , s
D) A  p, r , s; B  p, q, s; C  q, s; D  q, r , s
38. Match List – I with List – II and select the correct answer using the code given below
lists
Molecular weight When CrI3 oxidizes into Cr2O72
(P) Eq.wt.  (1)
33 and IO4
(Q Molecular weight When Fe SCN2 oxidizes into

Eq.wt.  (2)
27 Fe3 , SO24 , CO32 and NO3
Molecular weight When NH4SCN oxidizes into

(R) Eq.wt.  (3)
28 SO24 , CO32 and NO3
Molecular weight When As 2S3 oxidizes into AsO3

(S) Eq.wt.  (4)
24 and SO24
A) P  1; Q  2; R  4; S  3 B) P  2; Q  1; R  3; S  4
C) P  2; Q  1; R  4; S  3 D) P  1; Q  2; R  3; S  4

 2
18  Z 
39. Based on equation E  2.17810 J  2  , certain conclusions are written. Which of
 n 
them is not correct?
A) The negative sign in equation simply means that the energy of electron bounded to
the nucleus is lower than it would be if the electrons were at the infinite distance from
the nucleus.
B) Larger the value of n, the larger is the orbit radius.
C) Equation can be used to calculate the change in energy when the electron changes
orbit.
D) For n = 1, the electron has a more energy than it does for n = 6
40. For the reaction: 2A(g)  B(g) 

 2D(g)
U o298  2.5kcal and So298  10.5cal / K . Calculate approximate G o298 for the reaction, and
predict whether the reaction may occur spontaneously.
A) 0.061 k cal, spontaneous B) -0.033 k cal, spontaneous
C) 0.061 k cal, non spontaneous D) 0.033 k cal, non-spontaneous
41. The equivalent conductivities of K  , Al 3 and SO42 ions x,y and Z S cm 2 Eq 1 respectively.
The  0 eq for K 2 SO4 Al2 SO4 3 .24 H 2O (Potash Alum)
A)    z  Scm2 Eq 1
x 3y
B) x  3 y  z Scm2 Eq 1
4 4 
x yz
C) Scm2 Eq 1 D) 2 x  3 y  4 z Scm 2 Eq 1
8
42. Consider the following carbocations.

   
I. C6 H 5 C H 2 II. C6 H 5CH 2 C H 2 III. C6 H 5 C H CH 3 IV. C6 H 5 C  CH 3 2
The correct sequence of the stability of these carbocation is
A) II  I  III  IV B) II  III  I  IV C) III  I  II  IV D) IV  III  I  II

43. A mixture of ethyl iodide and n-propyl iodide is subjected to Wurtz reaction. The
hydrocarbon which will not be formed is (exclude side reaction products)
A) Butane B) Propane C) Pentane D) Hexane
44. The number of optically active products obtained from the complete ozonolysis of the
given compound is
A) 0 B) 1 C) 2 D) 4
45.
From the above compounds correct order of reactivity in electrophilic aromatic

substitution reactions will be
A) II  I  III  IV B) IV  III  II  I C) I  II  III  IV D) II  III  I  IV
46. The major product in the following reaction.

47. The order of reactivity of phenyl magnesium bromide with the following compounds is
A) II  III  I B) I  III  II
C) II  I  III D) All react with the same rate
48. The compound that undergoes decarboxylation most readily just on heating is…
A) B) C) D)
49. In the reaction,
The product E is
A) B) C) D)

50. Which of the following compounds will behave as a reducing sugar in an aqueous KOH
solution?
A) B)
C) D)
SECTION-II
51. How many of the following reagents convert isopentyl alcohol into alkyl halide without
any rearrangement?
a) HCl , ZnCl2 ,  b) SOCl2 c) PCl5 d) PCl3 e) PBr3 f) HI
52. How many of the following substances can act both as oxidizing and reducing agents
H3PO2, H3PO3, H3PO4, HNO2, SO2, NO, N2O3, NO2, SeO2, TeO2
53. How many of the following molecules are non-polar?

NO2 COOH
OH
Cl
Cl
i) OH ii) Cl iii) NO2 iv) COOH
Cl
v) Cl  CH 2  Cl vi) Cl Cl vii) HO  CH  CH  OH
2 2

54. What is the molecular weight of the naturally occurring optically inactive   amino
acid?
55. The number of revolutions made by an electron in one second in H – atom 2nd orbit is
eight times of numbers of revolution made by electron in one second in nth orbit of H –
atom, then n is ……….
56. Calculate the percentage dissociation of H 2 S g  If 0.1 mole of H 2 S is kept in 0.5 L vessel
at 1000K. The value of K c for the reaction 2 H 2 S g   2 H 2 g   S2 g  is 1.0 107
57. A 4.0 molar aqueous solution of NaCl is prepared and 500 mL of this solution is
electrolysed. This leads to the evolution of chlorine gas at one of the electrodes. If the
cathode is a Hg electrode, the maximum weight (g) of amalgam (Na – Hg) formed from
this solution is : (atomic mass: Na=23, Hg=200)
58. Two substances A and B are present such that  A  4  B  and half-life of A is 5 minute
and of B is 15 minute. If they start decaying at the same time following first order, how
much time (minutes) later will the concentration of both of them would be same?
59. The total number of monochloro products obtained on chlorination of 3-Methylpentane
in presence of sunlight is_____ (including stereo isomers).
60. How many of the following can give aldol condensation reaction?
a ) CH 3  CHO b) CH 3  C  CH 3 c)CH 3  CH 2  CHO d ) CH 3  COOH ,
O
e) C6 H 5  CHO f ) CH 3  COOCH 3 g )C6 H 5  CH 2  CHO h)  CH 3 3 C  CHO

SECTION – I
 /3 8sin x  sin 2 x 
61. Let I   / 4  dx . Then
 x 
2 3  5 5 2 3
A) I B) I C) I  D)  I 
3 4 5 12 12 3 4
62.  
General solution of x 2 ydx  x3  y 3 dy, is (where c being arbitrary constant)
x2 x2 x3 x
A) 2  l n x  c B) 2  l n y  c C) 3  l n y  c D)  2l n x  3 l n y  c
2y 2y 3y y
 k 1 2
63. The set of values of k  k  R  such that det  adj  adjA  =16, where A  0  1 1  , is
 4 1 1
equal to
A) 5,7 B) 2,7 C) 5, 2 D) 5,3
f 1  h   f 1
64. Let f  x   3 x10  7 x8  5 x6  21x3  3 x 2  7, then the value of Lim is equal
h 0 h3  3h
to
53 22 22 53
A) B) C) D)
3 3 3 3
z  2 1  i 
65. The complex number z which satisfies the equations z  1 and  1 is
z
1 i 1  i
A) 1 B) 1+i C) D)
2 2
66. Let A and B be two sets each containing three elements then number of subsets of A  B ,
each having at least two and at most 7 elements, is equal to
A) 502 B) 492 C) 456 D) 1002
       
67. If V1  i  j  k and V2  ai  b j  c k where a, b, c  2, 1, 0,1,2 , then number of possible
  
non-zero vectors V2 such that V2 is perpendicular to V1 is
A) 10 B) 13 C) 15 D) 18

68. Let y  y ( x ) be the solution curve of the differential equation
    
   
sin 2 x 2 log e tan x 2 dy   4 xy  4 2 x sin  x 2    dx  0, 0  x 
 4  2
, which passes through the

     
point  ,1 . Then y   is equal to
 6   3
A) 1 B) 3 C) 2 D) 3
69. If f  x  is a differentiable function and satisfies f  x  y   f  x  . f  y  x, y  R and
f 1  2, then area enclosed by 3 x  2 y  8 is (in sq.units)
1 1 1
A) f  4  B) f  6 C) f 6 D) f  5
2 3 3
70. If roots of the equation x 2  ax  b  0 are ‘c’ and ‘d’ then one of the roots of the equation
 
x 2   2c  a  x  c 2  ac  b  0 is always equal to  c  d 
A) c B) d  c C) 2c D) 2d
2n iAi  Ai 1 
71. If A1 , A2 , A3 ......... are in A.P then   1   is equal to
i 1 A
 i  Ai 1 
A) 2n-1 B) n-1 C) -2n D) n+2

72. Let e be the eccentricity of hyperbola and f(e) be the eccentricity of its conjugate
1
hyperbola then   f  e   f  f  e   de  g  e  and g 1  , then g  e  =
2
1 2 e2 1 2 e2 e2 e2
A) e 1  B) e 1  C) e 1 
2
D) e 1 
2
2 2 2 2 2 2
73. Nine balls of the same size and colour, numbered 1, 2, ..9 , were put into a packet. Now
A draws a ball from packet, noted that it is of number a, and puts it back. Then B also
draws a ball from the pocket and noted that it is of number b . Then probability for the
inequality a  2b  10  0 to hold is
52 59 60 61
A) B) C) D)
81 81 81 81
74. A scientist is weighing each of 30 fishes. Their mean weight worked out is 30 gm and a
standard deviation of 2 gm. Later it was found that the measuring scale was misaligned
and always under reported every fish weight by 2 gms (2gms less than the original
weight of the each fish). The ratio of correct mean and standard deviation (in gm) of
fishes are respectively
A) 16 B) 18 C) 22 D) 16.5

75.  
Let z1 , z2 , z3  C such that z1  1, z2  1  3 i  3 and z3  3  3 3 i  7. If
z3  3z1
3z1  2 z2  4 z3 is maximum then the value of z3  z1  is equal to
z2
A) 8 B) 12 C) 16 D) 18
76. The number of equivalence relations in set A = {a, b, c, d} is equal to
A) 3 B) 8 C) 12 D) 15
77. Each of three identical jewellery boxes has two drawers. In each drawer of the first box
there is a gold watch. In each drawer of the second box there is a silver watch. In one
drawer of the third box there is a gold watch while in the other there is a silver watch.
The probability of selecting any drawer after selecting a box is half. If we select a box at
random, open one of the drawers and find it to contain a silver watch, then the
probability that the other drawer has the gold watch in it, is
1 2 1 1
A) B) C) D)
3 3 2 4
78. Let ABC be a triangle with vertices at points A  2,3,5  , B  1,3, 2  and C      in three
dimensional space. If the median through A is equally inclined with the axes, then
     is equal to
A)  7,5  B) 10, 7  C)  5, 7  D)  7,10 
If the function f  x   axebx have the local maximum value 1 at x  2 , then
2
79.
e 1  e 1
A) a  and b  B) a  and b 
2 8 2 8
1 e  e 1
C) a  and b  D) a  and b 
8 2 2 8
80. Water is pouring into a conical vessel with vertex upwards and closed base at the rate of
3 cubic meters per minute through a tiny hole at the vertex. The radius of the base is 5m
and the height is 10m. When the level is 7m from the base, the rate at which water level
increases is (in m/min)
18 9 4 9
A) B) C) D)
7 7 3 7
SECTION-II
 a
 
1
81. The value of definite integral  e x 1
e 3 x
dx is  a, b, c  N  then the value of
1 bec
a  b  c is

 
x 2
1
 tan t dt
2
82. If Lim 0
 , then k is
x 
1  x2 k
A A2 B
sin  t cos ec 1
83. If A   dt , B   dt , then e A B B 2  1 equals
1 1 t2 1 
t 1 t 2
 1 A2  B 2  1
84. The number of values of x  0, n  , n  I (the set of integers) that satisfy
log|sin x| 1  cos x   2 is_____
85. For a biased die the probabilities for the different faces to turn up are given below
Face 1 2 3 4 5 6
Probability 0.1 0.32 0.21 0.15 0.05 K
This die is tossed and you are told that either face 1 or face 2 has turned up. If the
a
probability that it is face 1, is where a and b  a, b  N  are coprime to each other then
b
ab 
If for positive integers r  1, n  2, the coefficients of the  3r  and  r  2  powers of x in
th th
86.
 nr 5 
the expansion of 1  x  are equal, then 10 
2n
  .… (where nr  k means value of n when
 nr  2 
rk)
87. If the area bounded by the parabolas y 2  4  x    and y 2  4  x    , where   0 is 48
square units then  is equal to
 

The value of tan   tan 1  2   is equal to
4
88.
4r  3
 r 1  
x2 y 2 x2 y 2
89. Let the equations of two ellipses be E1 :   1 and E2 :  2  1 . If the product of
3 2 16 b
1
their eccentricities is , then the product of all possible lengths of the minor axis for all
2
possible positions of ellipse E2 is
90. Let f  x  be a differentiable function satisfying
 x  y  f  x  y    x  y  f  x  y   4 xy  x 2  y 2  for all x, y  R . If f 1  1 , and the area of the
region bounded by the curves y  f  x  and y  x 2 is
a
b
 
a,b  N,G.C.Dof  a,b  is1 ,then
b
the value of is equal to
a

KEY SHEET
PHYSICS
1 D 2 D 3 B 4 B 5 C
6 A 7 A 8 D 9 A 10 B
11 C 12 B 13 A 14 D 15 B
16 A 17 C 18 D 19 B 20 D
21 2 22 5 23 3 24 8 25 246
26 5 27 27 28 4 29 3 30 50
CHEMISTRY
31 A 32 B 33 A 34 D 35 D
36 A 37 D 38 A 39 C 40 D
41 B 42 A 43 D 44 B 45 B
46 B 47 B 48 A 49 A 50 B
51 1758 52 3 53 25 54 6 55 1
56 2 57 9 58 6 59 2 60 8
MATHEMATICS
61 C 62 D 63 A 64 A 65 D
66 A 67 B 68 A 69 B 70 A
71 D 72 C 73 D 74 D 75 B
76 C 77 C 78 A 79 B 80 B
81 3 82 30 83 7 84 2 85 7
86 1 87 2 88 8 89 4 90 4
SOLUTIONS
PHYSICS
1
1. T  2
mgd
2. Angular momentum is defined by the equation L=MVR
3. Let height of liquid above second hole (2) be ‘H’
 v1  2 g  H  h 
Thus the force experienced by the tank at point
d
‘A’ is F1   AV12
dt
Similarly F2  AV22
V2 is the velocity of liquid coming out at second hole (2).
At point B V2  2 gH
  
 The net force on tank is, F  F2  F1
 A V22  V12   A 2 gh  A  2 g  h
Fnet h
4.
YA
K'
L
KYA
KK ' L  KYL
 K eq  
K  K ' K  YA KL  YA
L
m
 T  2
K eq
m  KL  YA 
 2
KYA
5. The given graph represent isothermal process and for isothermal process internal energy is
constant.
6.
+q -q
A
B
C
D
Kq KQ Kq
  0
3a 3a 4 a
Q  q
4
q
Kq K 4 Kq Kq
VA    
2a 3a 4a 6a
Kq
VA  VC 
6a
1 8
7. sin   , cos  
3 3
According to conservation of momentum mu  2mvcos 
8
u  2v
3
v 9
e 
u cos  16
8. As the branch of the circuit containing 3 resistor is open so no current flows through it.
9. In equilibrium electrostatic attraction between the plates = spring force
q2
  kx
2 0 A
 CE 
2
  k  d  0.8d 
2 0 A
2
 0 A  2
  E
  0.8d   0.2dk
2 0 A
 0 AE 2 4 0 AE 2
 k 
0.256d 3 d3
10. As magnetic flux is same through inductors, L1i1  L2i2
 i1  2i2 …..(i)

 10V
Also, i1  i2    1A …..(ii)
R 10Ω
(i) and (ii)
2
 i1  A
3
R1R2
11. Time constants of both the circuits   C but that does not ensure same charge at any time on
R1  R2
capacitors or same current in branches of the capacitor.
mv 2
12. At the bottom, N  mg 
r
13. r2  c ; A  r1   c
r1  A  c
Sin r1  sin  A   c 
sin i
 sin  A   c 

sin i    sin A cos c  cos A sin  c 
7 3 3 3 1 1 1
 1    1  
3 2 7 7 2 2 2
1
sin i  or i  300
2
14. According to Lenz’s law e.m.fs of the same magnitude in the clockwise direction are induced in the
two loops into which the figure is divided. So, current is induced in the clockwise direction in the
outer boundary but no current in wire AB.
1D
15.  4.84mm
d
D
Let required wavelength is 2 the according to given information 2  4.84mm
2d
1
  1  2  1200mm
2 / 2
16.

17. The field due to current (either conventional or displacement) is normal to the direction of current.
18. The first photon will excite the hydrogen atom ( in ground state) in first excited state
 E2  E1  10.2 eV  . Hence, during de-excitation a photon of 10.2eV will be released. The second
photon of energy 15eV can ionize the atom. Hence the balance energy 1.4eV is retained by the
electron.
 2M
19. The magnetic fields at P due to horizontal and vertical magnets are respectively 0 3 towards
4 d
 M
right and 0 3 upwards.
4 d
 M  M 5
Their resultant is 0 3 2 2  12  0
4 d 4 d 3
20.
21.
 2R 
PEmax  mgR   mg  R 
   
 mgR  mgR  mg  
2R
  
 PEmax  mg  
2R
  

F F
22. Horizontal acceleration of the system is a   .
2 m  m  2 m 5m
Let N be the normal reaction between B and C.
Free body diagram of C gives
Now B will not slide downwards if  N  mB g
    F   mg
2
or
5 
5
or F mg
2
5
So, minimum value of F is mg
2
ml 2
23. Angular momentum L  I   .
3
24. For all point out the sphere we can treat it as point mass at its centre so, effectively it will be the
force between ring and point mass.
Gravitation field at x  3a on axis of ring is
E
Gmx

Gm  3a 
R  x2  a  3a 2 
3 3
2 2 2 2
3Gm

8a 2
3GMm
F  ME 
8a 2
26. Let V is total volume of iceberg & n is the fraction of the volume of iceberg that appears above the
surface of sea water. According to principle of floatation
V  0.9  103  g  1  n  V  1.125  103 g
 n  0.2
0i 0iR 2
27. Bcentre  & Bdis tan ce  28. 8  1014 h  0  0.5
 
3
2R

2 2
2 R2  2 2R
12 1014 h  0  2
12 0  2
Dividing, we get 
8 0  0.5
3 0  2

2 0  0.5
30  1.5  20  4 or 0  2.5eV
29. m1v1  m 2 v2

m1 v 2

m 2 v1
m1 27

m2 8
4
p  r13
3 27

4
p  r13 8
3
r1 3

r2 2
3
x
2
y3
R1 l1 l1
30.  
R2 l2 100  l1
x 20 1 4x l
   1 and    2
y 80 4 y 100  l
From (1) and (2) we get l  50 cm
CHEMISTRY
31. CONCEPTUAL
32. CONCEPTUAL
33. M.O configuration of N 2 is
 
* *
1s 1s 2 s 2 s  22 px   22p y 2p
2 2 2 2 2
z
1 i
34.  and  f  ik f m
1 1
n
35. TLC is a technique used to isolate non-volatile mixtures.
X  2Y Z  PQ
36. 1 0 1 0 0
1   2 1  
4 2 P1  2 P2
KP  K P2 
1 1  1 2
2
K P1 1 4 P1 P 1
   1 
K P2 9 P2 P2 36
37. 5d series member has more  0 than 3d and 4d series.

38. -I is a good leaving group and also bonded to benzylic carbon.
39. Cu 2 is reduced to Cu 
CH3 CH3 CH3
| | |
CH 3  C  Cl  CH 3  C  O  Na   CH 3  C  CH 2
40. | |
CH 3 CH 3
 major 

41. CONCEPTUAL
42. CONCEPTUAL
43. CONCEPTUAL
44. The charge on the complex ion is not always equal to the oxidation state of the metal atom. It
actually
depends on the nature of ligands and oxidation state of the metal atom.
45. CONCEPTUAL
46. The purple colour of KMnO 4 is due to charge transfer transition
The intense colour, in most of the transition metal complexes is due to d – d transition
47. CONCEPTUAL
48. CONCEPTUAL
49. CONCEPTUAL
50. CONCEPTUAL
h h
51. e  and n 
me ve mn vn
52.
H 2 SO4 KOH
53.
N1V1  N 2V2
0.3  V1  0.5 15
0.5 15
V1   25ml
0.3

CH 3  CH 3 
CH 3 
CH  NH 2 , CH 3 
CH  NH 3 Cl 
54. , wt % of Cl  37.2%
A
55. Tripeptide  2 H 2O  3 moles of Amino Acid
 NH| 2 
 
  CH 2   COOH  3  189  36
 
 n
 n 1
NH 2
|
X is CH 2  COOH
56. H 2O and XeF4
57. Tf  K f  m
58. ALL
59.


60. CONCEPTUAL
MATHS
61. a  n!
62.
is continuous at
1 0 0 x  a2 ab ac x  a2 ab ac
63. f '  x   ab xb 2
bc  0 1 0  ab xb 2
bc
ac bc x  c2 ac bc x  c2 0 0 1
64. a  A  R  p 1log a  log A   p  1 log R
65.
xi  observation  0 2 22 2n
f i  frequency  n
C0 n
C1 n
C2 n
Cn
x
fxi i
f i
0  C0  2 n C1  22 n C2 ......2n n Cn 3n  1 728
n
 n  n
n
C0  n C1  n C1.......n Cn 2 2
 3n  36
n6
66. a 2  a  2 x 2  1  x 4  x  0
2 x 2  1   2 x  1
a
2
ax x
2
a  x2  x  1
a1
4
a3
4  x  R 
 
 1 0 x
2

 0  3
x
 2 4
 3
 1  x 
     4
67. sin x  cos x    2 sin  x     
  4   3
2  x
 2
 3 7
 1 x
 2 2
 7
0  x  2
 4
68. Statement I : Req. number of solutions

10
  21 x  2 1 C21  132.
x0
Statement II : Is true by the definition of n!.
69. n  s   54
n  A   2.5 C4  10
10 2
P  A  
625 125
a
70.  a  ar  26 .....  i 
r

f  5 2  7 
19
71. Let
x  f  an integer   x    x  f  an integer
  x  f  an integer,but -1<   x  f  1  f
So, x  x  x. f  119  1
x 2dt
72. If t  tan , then dx 
2 1 t2
1
dt
A) a  1, I1  
0 2 1  t   2t  1  t 2 
2
1 1
dt dt 1
I1  2  2  2  log 3
0 4   t  1
0
t  2t  3 2
2
Similarly, for others
73. xy  x   x 2 y '  x   2 xy  x  , xy  x   x 2 y '  x   0

dy
x  y  0 , xy  c
dx
 
tan    
74. Taking first two  4 5
  3
tan    
 4 

75.
y
x
db
b  5 ft / sec , when b  12 then l  5
dt
h2  l 2  b2
b  h cos 
db d
  h sin 
dt dt
d
5  l
dt
d
 1 rad / sec
dt
76. Required probability = P  I ' P  II   P  I ' P  II '  P  I ' P  II   .......

 0.7  0.2  0.7  0.8  0.2  ......
0.14
  0.312
1  0.56
77.

y
Q
P
y  sec 1   sin 2 x   
a
 16  x 2  1
A  a  4
    dx where a  2  4    2
e0h  3
78. lim f  x   lim  2
x 0 x 0 0  x  1
e 1 h  3 e1 h  3
lim f  x   lim  lim  
x 0 h  0 1  h  1 h 0 h
79. f ''  x   0
 f ' is inc.fn
To find : where g is nec.Inc
G is inc  g '  0
1 1
 . f '  2 x 2  1  4 x   f ' 1  x 2   2 x   0
4 2

 x f '  2 x  1  f ' 1  x 2   0
2

  
Case I: x  0  1 f ' 2 x 2  1  f ' 1  x 2 
 2 x2  1  1  x2
 2  2 
 x   ,   ,     2
 3   3 
 2 
1   2   x   ,   .........  3
 3 
Case II: x  0   3 f '  2 x 2  1  f ' 1  x 2 
 2x2  1  1  x2
 2 2
 x    ,    4 
 3 3 
 2 
 3   4     , 0    6 
 3 
 g is inc in x   5    6 

 2   2 
 x    , 0    ,  
 3   3 
80. tan 1  x   t
t 2  4t  3  0
t   ,1   3,  

tan 1  x     ,1
 2 
x   , tan1
Largest integral x 1
81.
82.
83. Equating the components
3 x  2 y  3 z   x , 2 x  2 z   y and 4 x  2 y  3 z   z

Hence,
3    x  2 y  4z  0
2x   y  2z  0
4x  2 y  3    z  0
3  2 4
For non-trivial solution 2  2 0
4 2 3 
Hence   1 or 8  sum  7
84. Curve is rectangular hyperbola
 
P n, n 2  1  d n 
n  n2  1
2
1
 lt n.d n  K 2
n 2 2
85. f ' x  0
x
f '  x    3t  3t  4  dt  x.3x  3x  4   3x.x  3x  4 
0
1
f ' x 
2 ln 3
 32 x  8.3x  7 
1
f ' x 
2 ln 3
 3x  1 3x  7 
x  log3 7 is the point of minima

 3a  3log3 7  7
86. 4x  2 y  6  0
4x  2 y  8  0
86 2 1
  
16  4 20 5
6x  3 y  9  0
6 x  3 y  15  0
15  9 6 6 2
   
36  9 45 3 5 5
1 4
 2   2    1
5 5
88. There are three choices for the first letter and two choices for each subsequent letters. Hence using
fundamental principal.
No. of good words  3.2 n 1  384  2n 1  128  n  8
89. 2ab 2  4ab  6b 2  ab  2a  3b   a  3 b  2a
2a 2a 2
b  ab  K
a 3 a 3
dk 2  a  6a 
2
  0 a  0, 6
 a  3
2
da
dk 2
At a  6,  Ve
da 2
When a  6 & b  4, ab is min imum
6 4

ab
6
|min 
6
4
1 4 2 4  4
90.   4 x2 dx ; x  x  t  1  x2  dx  dt
 x   8
 x
1  t  1  x  4 
2
dt 1 1
2  tan    c  tan  c
 
2
t  2 2 2 2 2 2 2 2  2x 2 
a  2, b  2  a  b  4


MISTAKES
MATHS
PHYISCS
CHEMISTRY
SECTION – I
1. A wire frame in the shape of an equilateral triangle is hinged at one vertex so that it can
swing freely in a vertical plane, with the plane of the triangle always remaining vertical.
1
The side of the frame is m . The time period in seconds of small oscillations of the
3
frame will be
  
A) B)  2 C) D)
2 6 5
2. The dimensions of angular momentum are
A) M 0 L1T 0 B) M 0 L2T 0 C) M 1LT 2 D) M 1L2T 1
3. There are two identical small holes on the opposite sides of a tank containing liquid. The
tank is open at the top. The difference in height between the two holes is h. As the liquid
comes out of the two holes, the tank will experience a net horizontal force proportional
to.
3
A) h B) h C) h 2
D) h 2
4. One end of a long metallic wire of length L is tied to the ceiling. The other end is tied to
a mass less spring of force constant K. A mass m hangs freely from the free end of the
spring. The area of cross-section and Young’s modulus of the wire are A and Y
respectively. If the mass is slightly pulled down and released, it will oscillate with a time
period T equal to
m YA  KL  mYA mL
A) 2 m / K B) 2 C) 2 D) 2
YAK KL YA
5. Internal energy of the gas as it expands according to the graph AB which is a
recatangular hyperbola
A) Increasing continuously B) decreasing continuously
C) Always constant D) Initially Increasing then decreasing.
6. There are four concentric shells A,B,C and D of radii a,2a,3a and 4a respectively. Shells B
& D are given charges +q & -q respectively. Shell C is now earthed. The potential
1
difference VA  VC is___(take K)
4 0
Kq Kq Kq Kq
A) B) C) D)
6a 2a 3a 4a
7. Two Identical discs initially at rest are in contact on a table. A third disc of same mass
but of double radius strikes them symmetrically and itself comes to rest after impact.
The co-efficient of restitution is:
9 3 1 1
A) B) C) D)
16 4 2 16
8. Find out the value of current through 3 resistance for the given circuit
A) 10 amp B) 6 amp C) 4 amp D) zero

9. One plate of a capacitor is connected to a spring as shown in figure. Area of both the
plates is A. In steady state separation between the plates is 0.8d (spring was unstretched
and the distance between the plates was d when the capacitor was uncharged). The force
constant of the spring is approximately
4 0 AE 2 2 0 AE 6 0 E  0 AE 3
A) B) C) D)
d3 d2 Ad 3 2d 3
10. The current through 3 mH inductor in steady state after closing switch S is
1 2 3
A) ampere B) ampere C) 1 ampere D) ampere
3 3 2
11. STATEMENT-1 : Time constants of the circuits shown in the figure are same.
AND
STATEMENT-2 : Instantaneous current through the capacitor branch is same at any
instant for both the circuits, if batteries are inserted in the circuits at t=0.
A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for
Statement-1
for Statement-1
12. A small sphere A of mass m and radius r rolls without slipping inside a large fixed
hemispherical bowl of radius R(>> r) as shown in figure. If the sphere starts from rest at
the top point of the hemisphere. Find the normal force exerted by the small sphere on the
hemisphere when it is at the bottom B of the hemisphere.
10 17 5 7
A) mg B) mg C) mg D) mg
7 7 7 5
7
13. Refractive index of a prism is and the angle of prism is 600 . The minimum angle of
3
incidence of a ray that will be transmitted through the prism is ______
A) 300 B) 400 C) 600 D) 900
14. The radius of the conducting loop shown in figure is R. Magnetic field is decreasing at a
constant rate  . Resistance per unit length of the loop is  . Then current in wire AB is (
AB is one of the diameters)
R R 2R
A) from Ato B B) from B to A C) from Ato B D) zero
2 2 
15. In YDSE, coherent monochromatic light having wavelength 600 nm has fallen on slits.
First order bright fringe is at 4.84 mm from central maxima. Determine the wavelength
for which the first order dark fringe will be observed at same location on screen ? Take
D  3m
A) 600 nm B) 1200 nm C) 300 nm D) 900 nm

16. Two capacitors having capacitance C1 & C2 are connected in series and a potential
difference V is applied across them. Then:

V1 , V2 & U1 , U 2 be the potentials drop & energy store in C1 & C2 respectively.
Match the entries in Column-I with Column-II :
Column-I Column-II
a) V1  V2 p) C1  C2
b) U1  U 2 q) C1  C2
c) V1 r) C1V /  C1  C2 
d) V2 s) C2V /  C1  C2 
A)  a  q  ;  b  q  ;  c  s  ;  d  r  B)  a  r  ;  b  q  ;  c  p  ;  d  s 
C)  a  q  ;  b  p  ;  c  r  ;  d  s  D)  a  s  ;  b  r  ;  c  p  ;  d  q 
17. A changing electric field produces magnetic field. The direction of this magnetic field is
A) In the direction of electric field
B) In the direction opposite to the electric field
C) Perpendicular to the direction of electric field
D) Independent of the direction of electric field
18. A photon collides with a stationary hydrogen atom in ground state ineleastically. Energy
of the colliding photon is 10.2 eV. After a time interval of the order of microsecond
another photon collides with same hydrogen atom inelastically with an energy of 15 eV.
What will be observed by the detector?
A) 2 photons of energy 10.2 eV
B) 2 photons of energy 1.4 eV
C) one photon of energy 0.2 eV and an electron of energy 1.4 eV
D) one photon of energy 10.2 eV and an electron of energy 1.4 eV
19. Two short bar magnets of magnetic moment M each are placed at a distance 2 d apart.
The magnetic field. Midway between them at P is
 0 3M 0 M 5 0 2M 0 M
A) B) C) D)
4 d 3 4 d 3 4 d 3 4 d 3

20. Graph shows a hypothetical speed distribution for a ample of N gas particles (for
dN
V  V0 ,  0 ). If Vave and Vrms are the average speed and rms speed of the gas
dV
molecules, then
V0 V0
A) Vave  B) Vrms  C) Vave : Vrms  3: 2 D) Vave : Vrms  3 : 2
3 2
SECTION-II
21. A half section of thin uniform pipe of mass m and radius r is released from rest. Pipe
nmgR
rolls without slipping. The change in PE of pipe when it has rolled through 900 is .

Then the value of n is_
22. The system is pushed by a force F as shown in figure. All surfaces are smooth except
between B and C friction coefficient between B and C is  .Minimum value of F to
 n 
prevent block B from downward slipping is   mg . Then the value of n is _____.
 2 

23. A uniform stick of length l and mass m lies on a smooth table. It rotates with angular
velocity  about an axis perpendicular to the table and though one end of the stick. The
ml 2
angular momentum of the stick about the end is .Find the value of n_______.
n
24. A uniform ring of mass m is lying at a distance 3a from centre of a sphere of mass M
just over the sphere where a is the radius of ring as well as that of sphere. Then,
3GMm
gravitational force exerted is then the value of n is ____
na 2
25. Two tuning forks A & B when sounded together produces 4 beats/s. If B is loaded with
wax then also beat frequency remains same. Frequency of A is 242 Hz, find frequency
of B?
26. The relative density of ice is 0.9 and that of sea water is 1.125. The fraction of the whole
1
volume of an iceberg appears above the surface of the sea is then the value of x is ___
x
27. An electric current is flowing through a circular coil of radius R. The ratio of the
magnetic field at the centre of the coil and that at a distance 2 2 R from the centre of the
coil and on its axis is
28. When a metallic surface is illuminated by a light of frequency 8 1014 Hz , photoelectron of

maximum energy 0.5 eV is emitted. When the same surface is illuminated by light of
frequency12 1014 Hz , photoelectron of maximum energy 2 eV is emitted. The work
10
function is  ev  .Then the value of n _____
n
29. An unstable heavy nucleus at rest breaks in to two nuclei which moves away with
velocities in the ratio of 8:27. The ratio of the radii of the nuclei (assume to be spherical)
is x. Then y = 2x is______
30. In a meter bridge experiment null point is obtained at 20 cm from one end of the wire
when resistance ‘x’ is balanced against another resistance y. If x  y , then where will be
the new position of the null point from the same end, if one decides to balance a
resistance of 4 x against ' y ' in ___cm.

SECTION – I
31. Which of the following is not the name of an element with atomic number 104?
A) Nobelium B) Rutherfordium C) Kurchatovium D) Unnilqauadium
32. During halogen test, sodium fusion is boiled with con.HNO3 to
A) remove unreacted sodium
B) decompose cyanide or sulphide of sodium
C) extract halogen from organic compound
D) Maintain the p H of extract
33. In the ionization of N 2 to N 2 , the electron is lost from a

* *
A)   orbital B)   orbial C)   orbial D)   orbial
34. A certain substance ‘A’ tetramerises in water to the extent of 80%. A solution of 2.5g of
‘A’ in 100g of water lowers the freezing point by 0.30 C . The m.wt of ‘A’ is
K f  1.86 k.kg.mol 1 
A) 122 B) 31 C) 344 D) 62
35. Which technique among the following is most appropriate in separation of a mixture of
100 mg of
p-nitrophenol and picric acid?
A) Steam distillation B) Distillation under reduced pressure.
C) Sublimation D) Thin layer chromatography.
36. The equilibrium constants K p and K p for the reactions X  2Y and Z  P  Q
1 2
respectively are in the ratio of 1: 9 . If the degree of dissociation of X and Z be equal then
the ratio of total pressure at these equilibria is
A) 1: 36 B) 1:1 C) 1: 3 D) 1: 9

37. Transition metal complex with the highest value of crystal field splitting energy   0 
will be
3 3 3 3
A) Cr  H 2O6   B)  Mo  H 2O6  C)  Fe  H 2O6   D) Os  H 2O6 
38. The major product of the reaction is
I I
Br SPh PhS F
Cl Cl
A) B) NO2 C) D) NO2
39. When an aldehyde is heated with Fehilings solution, a reddish brown precipitate is
formed which is
A) CuO B) Cu C) Cu2O D) Cu  C  C  Cu
40. Which of the following cannot be prepared by using willamson’s synthesis?
A) Methoxy benzene B) Benzyl p-nitro phenyl

ether
C) t-butyl methyl ether D) Ditertiarybutyl ether
41. Among the following the metal with the highest melting point will be
A) Hg B) Ag C) Ga D) Cs
42. The element of group 15 which can form a strong bond with hydrogen is
A) Nitrogen B) Phosphorous C) Arsenic D) Antimony

43. Which of the following is not a mineral of fluorine?
A) Fluorspar B) Cryolite C) Fluoroapetite D) Carnallite
44. Consider the following statements:
According to Werner’s theory
i) Secondary valencies are non-ionisable and directional
ii) Secondary valencies are satisfied by neutral molecules or negative ions .
iii) Ligands form coordinate bonds with metal ions or atoms.
Iv) The charge on the complex ion is always equal to the oxidation state of the metal
atom.
A) all are correct B) (i), (ii) and (iii) are correct
C) (i), (ii) and (iv) are correct D) (iii) and (iv) are correct
45. Zinc and mercury do not show variable valency like other d-block elements because
A) They are soft
B) Their (n-1) d-shells are completely filled
C) They have only two electrons in the outermost shell
D) Their d-shells are incompletely filled.
46. Assertion (A): The purple colour of KMnO 4 is due to charge transfer transition
Reason (R): The intense colour, in most of the transition metal complexes is due to d – d
transition.
47. Which of the following doesn’t undergo Friedal-craft’s reaction?
A) Xylene B) Nitrobenzene C) Cumene D) Toluene

48. Match the transition element ions given in Column I with the characteristic (s) of
products given in Column II.
Column –I Column – II
a) Cu 2 P) Form amphoteric oxide
b) Zn 2 Q) Diamagnetic and colourless compounds
c) Cr 3 R) Coloured hydrated transition metal ion
d) Ni 2 S) Paramagnetic
A) a – RS, b – PQ, c – PRS, d – QRS
B) a – PQ, b – RS, c – PRS, d – QRS
C) a – PRS, b – RS, c – PQ, d – QRS
D) a – RS, b – PQ, c – QRS, d – PRS
49. Oxidation state of iodine in H 4 IO6 is
A) +7 B) +5 C) +1 D) -1
50. Which of the following does not give a white precipitate when dilute hydrochloric acid
is added?
A) Ag  B) Ba 2 C) Pb 2 D) Hg 22
SECTION-II
51. The wave length of an electron and a neutron will become equal when the velocity of
electron is x times the velocity of neutron. The value of x is (nearest integer)
 Mass of electron  9.110 31

kg  ,  Mass of neutron  1.6  1027 kg 
52. The major product in the following reaction is
The degree of unsaturation of product P is _______

53. The volume of 0.15M H 2 SO4 solution required to neutralise 15 ml of 0.5M KOH in
presence of phenolphthalein indicator is ___ ml.
54. An known amine A with benzene sulphonyl chloride yields a derivative which dissolves
in aq.KOH solution. Compound A on reaction with nitrous acid gives an alcohol B
which responds to iodoform test. If hydrochloride of A contains 37.2% of chlorine by
weight, the total number of carbon atoms present in compounds A and B is ___.
55.
If the molecular weight of tripeptide formed by X is 189, the value of n is ________

56. Amongst the following, the number of species with only two lone pairs of electrons on
central atom is ______.
SF6 , XeF4 , CF4 and H 2O
57. 50 g of ethylene glycol is dissolved in 170.3 g of water and is cooled to 9.30 C . The
amount of water in g, separated as ice  K f  1.86 K / molal 
58. How many of the following can exhibit geometrical isomerism?

 3  
Co  en 2 Cl2  CrCl2  C2O4 2 
,   2  4  2  ,   3 2   4  ,
 Fe H O OH   Fe NH CN 
,
I II III IV
2 2
Co  en 2  NH 3  Cl  
,   3  4  2   
Co NH H O Cl 
V VI
59. 10 ml of a gaseous organic compound ( vapour density = 23 ) containing C,H and O only
was mixed with 100 ml of oxygen and exploded under conditions which allowed the
water formed to condense. The volume of gas after explosion was 90 ml. On treatment
with potash solution, a further contraction of 20 ml in volume was observed. The number
of moles of CO2 formed by 1 mole of organic compound is ____
60. How many moles of  CH3CO 2 O are required to react with 1 mole of sucrose?

SECTION – I
61. If a denotes the number of permutation of n different things taken all at a time, b the
number of permutation of n-2 different things taken 10 at a time and c, the number of
permutations of n-12 different things taken all at a time such that a  182bc , then value of
n is:
A) 10 B) 12 C) 14 D) 18
62. STATEMENT – 1: f  x   x  x  is discontinuous at all Integers , where [.] denotes G.I.F
STATEMENT – 2: If a function is non-differentiable at a point then it may be

continuous at that point
A) Statement 1 is true, statement 2 is true
B) Statement 1 is false, statement 2 is false
x  a2 ab ac
63. If a,b,c are real, then f  x   ab xb 2
bc is decreasing in:
ac bc x  c2
  a 2  b2  c2  
A)    a 2  b 2  c 2  , 0  B)  2 2 2 
2
 0,  a  b  c   C)  0,  D) Never decreases
2
 3   3   3 
 
64. If pth , q th , r th , terms of a G.P are the positive numbers a,b,c respectively then angle
between the vectors log a3 i  log b3 j  log c3 k and  q  r  i   r  p  j   p  q  k is:
   1 
A) B) C) 0 D) sin 1  
2 3  p 2  q 2  r 2 


65. Consider the data on X taking the values 0, 2, 4, 8, …. , 2n with frequencies
728
n
C0 ,n C1 ,n C2 ,..., n Cn respectively. If the mean of this data is , then n is equal to ___.
2n
A) 15 B) 8 C) 4 D) 6
66. Complete set of real values of ‘a’ for which the equation x 4  2ax 2  x  a 2  a  0 has all its
roots real
A)  ,  
3
B) 1,   C)  2,   D)  0,  
4 
20
67. The value of  sin x  cos x  dx is: (where . denotes greatest integer function)
20
A) 10 B) 20 C) 20 D) 10
68. Statement-1: The number of non-negative integral solutions of 2x + y + z = 21 is 132
Statement-2: For n  N ,  n 2 ! is divisible by (n!)n .
A) Statement 1 is true, statement 2 is true
B) Statement 1 is false, statement 2 is false
69. Of all the functions that can be defined from the set A : 1, 2,3, 4  B : 5, 6, 7,8,9 a
mapping is randomly selected. The chance that the selected mapping is strictly
monotonic is
2 4 5
A) 1105 B) C) D)
125 4096 2048
70. Three distinct numbers a1 , a2 , a3 are in increasing G.P a12  a22  a32  364 and a1  a2  a3  26 ,
then the value of a10 if an is the nth term of the given G.P is:
A) 2.39 B) 39 C) 2.310 D) 312

Let x   5 2  7  , then x  x  x denotes the fractional part of x  is equal to:
19
71.
A) 219 B) 319 C) 0 D) 1
 /2
dx
72. If I a  
0
2 cos x  sin x  a
, then the value of I a for
3
A) a = 1 P) ln  
2
1
B) a = 3 Q) log 3
2
2  1 3 1 
C) a = 2 R)  tan  tan 1 
11  11 11 
S) tan 1  
1
D) a = 4
3
A) A – Q ; B – P ; C – S ; D – R B) A – Q ; B – S ; C – R ; D – P
C) A – Q ; B – S ; C – P ; D – R D) A – Q ; B – P ; C – Q ; D – S
73. A curve passing through  2,3 and satisfying the differential equation
x
 ty  t  dt  x y  x  ,  x  0  is :
2
9 x2 y 2
A) x 2  y 2  13 B) y 2  x C)  1 D) xy  6
2 8 18
     
tan     tan     tan    
4  4  4  . Then 12sin 2     15sin 2     7 sin 2   
74. Let      
5 3 2
is equal to
1 1
A)  B) C) 1 D) 0
2 2

75. A 13 ft. ladder is leaning against a wall when its base starts to slide away. At the instant
when the base is 12 ft. away from the wall, the base is moving away from the wall at the
rate of 5 ft/sec. The rate at which the angle  between the ladder and the ground is
changing is
12 13 10
A)  rad / sec B) 1 rad / sec C)  rad / sec D)  rad / sec
13 12 13
76. Two aero planes I and II bomb a target in succession. The probabilities of I and II
scoring a hit correctly are 0.3 and 0.2, respectively. The second plane will bomb only if
the first misses the target. The probability that the target is hit by the second plane is
A) 0.06 B) 0.14 C) 0.312 D) 0.70
16  x 2
77. The area of the region bounded by the curve y  and y  sec 1   sin 2 x 
4
( where . denotes greatest integer function is:)
1 3 3 8 3 8 1
A)  4    2
B) 8  4    2
C)  4    2
D)  4    2
3 3 3
e   3
x x
78. If f  x   , then: (where . represents greatest integer function)
 x  x  1
A) lim f  x   2 B) lim f  x   0 C) lim f  x   2 D) lim f  x  exist

x0 x0 x0 x0
1 1
79. Let g  x   f  2 x 2  1  f 1  x 2  x  R , where f ''  x   0x  R, g  x  is necessarily
4 2
increasing in the interval
 2 2  2   2 
A)   ,  B)   ,0  ,
 3 3   3   3 
C)  1,1 D) None of these
The largest integral value of x satisfying the inequality  tan 1  x    4  tan 1  x    3  0 is:
2
80.
A) 0 B) 1 C) 2 D) 3

SECTION-II
   
81. Let O be an interior point of ABC such that OA  2OB  3OC  0 then ratio of the area of
ABC to the area of AOC is _______.
82. Find number of integral values of k for which the line 3x  4 y  k  0 , lies between the
circles x 2  y 2  2 x  2 y  1  0 and x 2  y 2  18 x  12 y  113  0, without cutting a chord on
either of circle.
  
83. Let a  3i  2 j  4k , b  2  i  k  and c  4i  2 j  3k . If the equation
  
 
xa  yb  zc   xi  y j  zk has a non-trivial solution, then find the sum of all distinct
possible values of  .
84. For each positive integer n, consider the point P with abscissa n on the curve y 2  x 2  1 . If
d n represents the shortest distance from the point P to the line y  x then lt  n.d n  has
n 
1
value , then K is
K K
x
85. Let f  x    3t  3t  4   x  t  dt  x  0  . If x  a is the point where f  x  attains it’s local
0
minimum value then value of 3a is ;

y 3 y 3
86. The distance between x    0 & 2 x  y  4  0 is  , and distance between x    0 &
2 2 2 2
3 y 15
3x    0 is  , then find  2   2
2 2
87. The number of points of intersection of Z   4  3i   2 and Z  Z  4  6, Z  C is

88. Defined a “good word” as a sequence of letters that consists only of the letters A, B and
C and in which A never immediately followed by B, B is never immediately followed by
C, and C is never immediately followed by A. If the number of n-letter “good words” is
384, find n.
ab
89. If 2ab 2  4ab  6b 2 , a  0, b  0 then find the minimum value of .
6
x2  4 1  x  4 
2
1
90. If  4 dx  tan    c then a  b is __.
x  16 a b  ax b 

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Sri Chaitanya IIT Academy.,India.

 A.P  T.S  KARNATAKA  TAMILNADU  MAHARASTRA  DELHI  RANCHI

Name of the Candidate (in Capital): ________________________________________________

09‐01‐24_ Sr.Super60_Elite, Target & LIIT-BTs _ Jee‐Main_GTM‐10_Test Syllabus

Sec: Sr.Super60_Elite, Target & LIIT-BTs Page 2

1) M 2 L1T 0 2) M 1L2T 0 3) M 1L3T 1 4) M 1L2T 3

and b is given by:

Sec: Sr.Super60_Elite, Target & LIIT-BTs Page 4

Sec: Sr.Super60_Elite, Target & LIIT-BTs Page 5

1) Will not be there

1) He AND Ar 2) He AND O2 3) O2 AND N 2 4) O2 AND He

Sec: Sr.Super60_Elite, Target & LIIT-BTs Page 7

Sec: Sr.Super60_Elite, Target & LIIT-BTs Page 8

surface of the earth is 

Sec: Sr.Super60_Elite, Target & LIIT-BTs Page 9

energy is 4 K . Find the value of K .

in s. the particle velocity at y = 25 cm in time t = 1/100s will be 10 n m / s. What is the

Sec: Sr.Super60_Elite, Target & LIIT-BTs Page 10

1 x 108 M? ( K b for C6 H 5 NH 3  2.4  10 5 M )

1) 102 M 2) 101 M 3) 103 M 4) 104 M

250 C . The equilibrium constant of the reaction at 250 C will be:

1) 1 1010 2) 10 3) 29.5 102 4) 2.95  1010

Sec: Sr.Super60_Elite, Target & LIIT-BTs Page 11

3) [Co( NH 3 )4 Cl2 ]Cl 4) [Co( NH 3 )5 .H 2O]Cl3

1) [Cr ( H 2O ) 6 ]2  ,[CoCl4 ]2  2) [Cr ( H 2 O ) 6 ]2  ,[ Fe( H 2 O )6 ]2 

3) [ Mn ( H 2 O ) 6 ]2  ,[Cr ( H 2 O )6 ]2  4) [CoCl4 ]2  ,[ Fe( H 2 O )6 ]2 

37. Identify the product “Z” in the following series of reactions:

Sec: Sr.Super60_Elite, Target & LIIT-BTs Page 12

 -Si bond angle in O( SiH ) is greater than Cl - O

3) The nature of back bond in both molecules is (2 p  3d )

45. Consider the following conversions:

3) H1, H 2 and H 3 are negative whereas H 4 is positive

4) H1 and H 3 are negative whereas H 2 and H 4 are positive

46. What is the geometry of the IBr2 ion?

47. The formation of PH 4 is difficult compared to NH 4 because:

Sec: Sr.Super60_Elite, Target & LIIT-BTs Page 14

E  ( X / X  )  0.33V . From this data one can deduce that:

Sec: Sr.Super60_Elite, Target & LIIT-BTs Page 15

solution of 0.816 g of compound A when dissolved in 7.5 g of benzene freezes at 1.59o C .

of C2O42 in acidic medium____

Sec: Sr.Super60_Elite, Target & LIIT-BTs Page 16

1) 1 2) zero 3) n 4) n(n –1)

66. The number of point(s) on the line 3x  4 y  5, which are at a distance of

sec 2  2cosec 2 ,   R, from the point 1,3 is

Sec: Sr.Super60_Elite, Target & LIIT-BTs Page 17

1) n1 Cn2 2) n1 Cn2  n1 ! n2 ! 3) n1 Cn2 1  n1 ! n2 ! 4) n1 1

Sec: Sr.Super60_Elite, Target & LIIT-BTs Page 18

If coefficient of x y z in  x  y  z  is A, (where A  0 ) then coefficient x 4 y 4 z is

Sec: Sr.Super60_Elite, Target & LIIT-BTs Page 19

perpendicular then the value of r 2 – h2 is ___

84. If minimum distance between the curve y 2  4 x and x 2  y 2  12 x  31  0 is equal to  ,

equal to  k1  k2  sq.unit , then value of  k1  k2  is where k1 , k2  N .

Sec: Sr.Super60_Elite, Target & LIIT-BTs Page 20

      

a  b where a, b  N , then the value  a  b    a  b  is

90. Consider a system of equations x 4  y 4  z 4  1, x 4   y 4  z 4    2 and

x 4  y 4   z 4   , where  is a real parameter. If the number of real solutions of the system

Sec: Sr.Super60_Elite, Target & LIIT-BTs Page 21

Sec: Sr.Super60_Elite, Target & LIIT-BTs Page 22