JEE-MAIN-2015-SOL
The No. 1 Institute in Eastern India
JEE (Main) 2015
Question & Solution
B
Test Booklet Code :
PART A - CHEMISTRY
1. Which of the following is the energy of a possible excited state of hydrogen ?
(1) – 6.8 eV
(2) – 3.4 eV
(3) + 6.8 eV
(4) + 13.6 eV
Solution : (2)
For hydrogen atom, En
13.6
n2
eV / atom
13.6
 Energy of the excited state =
22
eV / atom = – 3.4 eV/atom
2. In the following sequence of reactions :
KMnO4
Toluene
A
(1) C6H5CH3
SOCl2
B
H2 /Pd
BaSO4
C , the product C is
(2) C6H5CH2OH
(3) C6H5CHO
(4) C6H5COOH
Solution : (3)
Toluene
KMnO4
CO2H
SOCl2
COCl
H2lPd
CHO
BaSO4
3. Which compound would give 5 - keto - 2 - methyl hexanal upon ozonolysis ?
CH3
CH3
(1)
(2)
H3C
(3)
CH3
CH3
CH3
(4)
CH3
CH3
Solution : (1)
O
6
5
4
3
CH3
2
Me
H3C — C — CH2 — CH2 — CH — CHO
1
2
1
5
3
6
Me
4
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4. The ionic radii (in Å) of N3–, O2– and F– are respectively
(1) 1.36, 1.71 and 1.40 (2) 1.71, 1.40 and 1.36 (3) 1.71, 1.36 and 1.40 (4) 1.36, 1.40 and 1.71
Solution : (2)
1
(Z = Nuclear charge)
Z
 decreasing radius order is N3– > O2– > F–
For isoelectronic species r
5. The color of KMnO4 is due to
(1) d – d transition
(3)  – * transition
(2) L  M charge transfer transition
(4) M  L charge transfer transition
Solution : (2)
KMnO4 is coloured due ligand to metal charge transfer.
6. Assertion : Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form
oxides of nitrogen
Reason : The reaction between nitrogen and oxygen requires high temperature
(1) Both assertion and reason are correct, but the reason is not the correct explanation for the assertion
(2) The assertion is incorrect, but the reason is correct
(3) Both the assertion and reason are incorrect
(4) Both assertion and reason are correct, and the reason is the correct explanation for the assertion
Solution : (4)
Nitrogen and Oxygen do not react under normal condition because Ea of the reaction is very high i.e. high
temperature is required.
7. Which of the following compounds is not an antacid ?
(1) Cimetidine
(2) Phenelzine
(3) Ranitidine
(4) Aluminium hydroxide
Solution : (2)
Phenelzine (Nardil) is a tranquiliser which acts as an antidepressant drug.
8. In the context of the Hall-Heroult process for the extraction of Al, which of the following statements is
false ?
(1) Al2O3 mixed with CaF2 which lowers the melting point of the mixture and brings conductivity
(2) Al3+ is reduced at the cathode to form Al
(3) Na3AlF6 serves as the electrolyte
(4) CO and CO2 are produced in this process
Solution : (3)
A mixture of Al2O3, NaAlF6 and CaF2 is the electrolyte.
9. Match the catalysts to the correct processes :
Catalyst
(A) TiCl3
(B) PdCl2
(C) CuCl2
(D) V2O5
(1) (A) - (ii), (B) - (i), (C) - (iv), (D) - (iii)
(3) (A) - (iii), (B) - (i), (C) - (ii), (D) - (iv)
(i)
(ii)
(iii)
(iv)
(2)
(4)
Process
Wacker process
Ziegler - Natta polymerization
Contact process
Deacon’s process
(A) - (ii), (B) - (iii), (C) - (iv), (D) - (i)
(A) - (iii), (B) - (ii), (C) - (iv), (D) - (i)
Solution : (1)
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NH2
NaNO2 /HCl
0 5ºC
10. In the reaction
CuCN/KCN
D
E N2 the product E is
CH3
CN
CH3
(1) H3C
(2)
CH3
COOH
CH3
(3)
(4)
CH3
Solution : (2)
CH3
HN2
NaNO2/HCl
0–5ºC
+
Cl– N
CH3
2
CuCN
KCN
CH3
NC
+ N2
11. Which polymer is used in the manufacture of paints and lacquers ?
(1) Glyptal
(2) Polypropene
(3) Poly vinyl chloride
(4) Bakelite
Solution : (1)
12. The number of geometric isomers that can exist for square planar [Pt (Cl) (py) (NH3) (NH2OH)]+ is
(py = pyridine) :
(1) 3
(2) 4
(3) 6
(4) 2
Solution : (1)
Cl
HOH2N
II
Pt
Py
NH 3
Cl
HOH2N
II
Pt
NH3
Py
Cl
Py
II
Pt
NH3
HOH 2N
13. Higher order (>3) reactions are rare due to :
(1) increase in entropy and activation energy as more molecules are involved
(2) shifting of equilibrium towards reactants due to elastic collisions
(3) loss of active species on collision
(4) low probability of simultaneous collision of all the reacting species
Solution : (4)
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14. Which among the following is the most reactive ?
(1) Br2
(2) I2
(3) ICl
(4) Cl2
Solution : (3)
I – Cl is more reactive due to difference in electronegativity.
15. Two Faraday of electricity is passed through a solution of CuSO4. The mass of copper deposited at the
cathode is (at mass of Cu = 63.5 amu) :
(1) 63.5 g
(2) 2 g
(3) 127 g
(4) 0 g
Solution : (1)
1 eq.  1F
 2F  2 eq.
Cu2+ + 2e  Cu
 Mass = 63.5 g
16. 3 g of activated charcoal was added to 50 mL of acetic acid solution (0.06 N) in a flask. After an hour it was
filtered and the strength of the filtrate was found to be 0.042 N. The amount of acetic acid adsorbed (per gram
of charcoal) is :
(1) 36 mg
(2) 42 mg
(3) 54 mg
(4) 18 mg
Solution : (4)
50 mL 0.06N
 3.00 meq
50 mL 0.042N
 50 × 0.042 meq
 Absorbed CH3COOH = 0.9 meq = 0.9 × 60 mg
= 54.0 mg
 Per gram adsorbed 18 mg.
17. The synthesis of alkyl fluorides is best accomplished by :
(1) Sandmeyer’s reaction
(2) Finkelstein reaction
(3) Swarts reaction
(4) Free radical fluorination
Solution : (3)
18. The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Mol.
wt. 206). What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram
resin ?
(1)
1
206
(2)
2
309
(3)
1
412
(4)
1
103
Solution : (3)
2C8H7SO3Na + Ca
2+
 (C 8H 7SO 3)2Ca + 2Na+
2 × 206gm  1 mole
 412 gm resin requires  1mole Ca2+
1

moles of Ca+2 ions
412
19. Which of the vitamins given below is water soluble ?
(1) Vitamin D
(2) Vitamin E
(3) Vitamin K
(4) Vitamin C
Solution : (4)
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20. The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is:
(1) ion-dipole interaction (2) London force
(3) hydrogen bond
(4) ion-ion interaction
Solution : (3)
Ingeraction hydrogen bond mostly dipole-dipole type & it 
1
r3
21. The following reaction is performed at 298 K.



 2NO2(g)
The standard free energy of formation of NO(g) is 86.6 kJ/mol at 298 K. What is the standard free energy
of formation of NO2(g) at 298 K ? (Kp = 1.6 × 1012)
2NO(g) + O2(g)
(1)
86600 + R(298) ln(1.6 × 1012)
(3) 0.5[2 × 86,600 – R(298) ln(1.6 × 1012)]
l n 1.6 1012
(2) 86600 –
R 298
(4) R(298) ln(1.6 × 1012) – 86600
Solution : (3)
2NO(g) + O2(g)  2NO2(g)
Gº = Gºp – GºR
R × 298 ln (1.6 × 1012) = Gp – 86600

GNO
2
= 0.5 [2 × 86,600 – R(298)ln(1.6 × 1012)]
22. Which of the following compounds is not colored yellow ?
(2) (NH4)3 [As (Mo3O10)4]
(1) K3[Co(NO2)6]
(3) BaCrO4
(4) Zn2[Fe(CN)6]
Solution : (4)
23. In Carius method of estimation of halogens, 250 mg of an organic compound gave 141 mg of AgBr. The
percentage of bromine in the compound is : (at. mass Ag = 108; Br = 80)
(1) 36
(2) 48
(3) 60
(4) 24
Solution : (4)
Br% =
At. wt of Bromine
GMW of AgBr
Br%
80
188
141 10
3
230 10
3
Weight of AgBr
100
Weight of Organic Compound
100
24
24. Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 4.29 Å. The radius of sodium
atom is approximately :
(1) 3.22 Å
(2) 5.72 Å
(3) 0.93 Å
(4) 1.86 Å
Solution : (4)
3a
3
rr 
 4.29 = 1.86Å
4
4
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25. Which of the following compounds will exhibit geometrical isomerism ?
(1) 3-Phenyl-1-butene
(2) 2-Phenyl-1-butene
(3) 1, 1-Diphenyl-1-propane
(4) 1-Phenyl-2-butene
Solution : (4)
4
3
2
1
CH3 — CH  CH — CH2Ph
26. The vapour pressure of acetone at 20ºC is 185 torr. When 1.2 g of non-volatile substance was dissolved in
100 g of acetone at 20ºC its vapour pressure was 183 torr. The molar mass (g mol–1) of the substance is:
(1) 64
(2) 128
(3) 488
(4) 32
Solution : (1)
P
Po
n2
n1 n2
185 183
185
1.2
MB
1.2 100
MB
58
 MB = 64 g mol–1
27. From the following statements regarding H2O2, choose the incorrect statement :
(1) It decomposes on exposure to light
(2) It has to be stored in plastic or wax lined glass bottles in dark
(3) It has to be kept away from dust
(4) It can act only as an oxidizing agent
Solution : (4)
1
2
0
2H2 O2  2H2 O  O2
 H2O2 can act as oxidising as well as reducing agent.
28. Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater than its lattice
enthalpy ?
(1) BeSO4
(2) BaSO4
(3) SrSO4
(4) CaSO4
Solution : (1)
Be+2 is most polarising among the Group II A metal ions, hence hydration energy is maximum.


29. The standard Gibbs energy change at 300 K for the reaction 2A 
 B + C is 2494.2 J. At a given time,
the composition of the reaction mixture is [A] =
= 8.314 J/K/mol, e = 2.718]
(1) reverse direction because Q > Kc
(3) reverse direction because Q < Kc
1
1
, [B] = 2 and [C] = . The reaction proceeds in the : [R
2
2
(2) forward direction because Q < Kc
(4) forward direction because Q > Kc
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Solution : (1)
QC =
2A  B +
C
1
2
1
2
[B][C]
[A]2
2
4
But Gº = – RT ln KC = – 2.303 × 8.314 × 300 log KC
= 2449.2
 KC = 2.716
Q > KC
 reaction goes backward
30. Which one has the highest boiling point ?
(1) Ne
(2) Kr
(3) Xe
(4) He
Solution : (3)
Xe due to more vander Waal’s forces of attraction
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PART B - MATHEMATICS
31. The sum of coefficients of integral powers of x in the binomial expansion of 1 2 x
(1)
1 50
3
2
1 50
3
2
(2)
1
1 50
2
2
(3)
1
(4)
50
is :
1 50
3
2
1
Solution : (4)
1  2 x 
50

 1 50 C1 2 x
Required to find

50

C0  22
1

 50 C2 2 x
50

2

50 C3 2 x
C2  24 50C4  ......  250

3
50

 ..... 50 C50 2 x
C50

50

1  x 50  1 50 C1 x 50 C2 x2  ...... 50 C50 x50 
Put x = 2
1  250  1 50 C1 2 50 C2 22 50 C3 23 50 C4 24  .....
... (i)
Put x = – 2
1 250  150 C1 2 50 C2 22 50 C3 23 50 C4 24  ......
350  1  2 1  22

 1  22
50
50
C2  24
50
... (ii)
C4  .....

C2  24  50 C4  ...... 


1 50
3 1
2
32. Let f (x) be a polynomial of degree four having extreme values at x = 1 and x = 2. If lim 1
x
f (2) is equal to :
(1) – 4
(2) 0
(3) 4
0
f x
x2
3 , then
(4) – 8
Solution : (2)
 f  x 
 lim 1  2   3
x 0 
x 

lim
x 0
f  x
x2
2
Let, f(x) = ax4 + bx3 + cx2 (Since for lim
x 0
f x
x2
0)
 f(x) = 4ax3 + 3bx2 + 2cx
f(x) having frame values at x = 1 and x = 2
 f(1) = 0  4a + 3b + 2c = 0
f(2) = 0  32a + 12b + 4c
to be finite, the coefficient of x and constant term must be
... (i)
... (ii)
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 lim
f x
x 0
lim
x2
f  x
2x
x 0
 2 . By L Hospital’s Rule.
 2  lim

x 4ax 2  3bx  2c
x 0
2x
2c = 4, c = 2
From (i), (ii) and (iii)
32a + 12b = – 8
16a + 12b = – 16
(–)
-----------------------------16a = 8
a
1
,
2
f  x 
 2
... (iii)
b=–2
x4
 2x3  2x 2
2
f(2) = 8 – 16 + 8 = 0
33. The mean of the data set comprising of 16 observations is 16. If one of the observation valued 16 is deleted
and three new observations valued 3, 4 and 5 are added to the data, then the mean of the resultant data,
is :
(1) 16.0
(2) 15.8
(3) 14.0
(4) 16.8
Solution : (3)
16  16  16  12
16
16  15  12
18
16  15  12
 14.
1.8
34. The sum of first 9 terms of the series
(1) 96
(2) 142
13 13  23 13  23  33


 ..... is :
1
1 3
1 3  5
(3) 192
(4) 71
Solution : (1)
13 13  23

 .....
1
1 3
Tn 
13  23  ........  n3
1  3  5  .......   2n  1
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2
2
 n  n  1 
n2  n  1


 2 
4

 
n
n
 2n
1  2n  1
4
2

n  1

n2  n  1
2
4n2
2
4
T1 

22
4
T2 
32
4
T3 
42
4
...................
...................
T9 
102
4
–––––––––––––––––––––––––––––––––––––––––––
S
1 2
1  22  32  ........  102  ........  12 

4

1  10 10  1 2.10  1  1  10.11.26 
 1  
 1

4 
6
6

 4 

1
.384  96
4
35. Let O be the vertex and Q be any point on the parabola, x2 = 8y. If the point P divides the line segment OQ
internally in the ratio 1 : 3, then the locus of P is :
(1) y2 = x
(2) y2 = 2x
(3) x2 = 2y
(4) x2 = y
Solution : (3)
k)
h,
(
P
2
2
Q(2at, at ) = (4t, 2t )
O
h
1 4t  3  0
1 3
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k
1 2t 2  0
t2

2
4
h = t, 2k = t2
 2k = h2
 x2 = 2y
36. Let  and  be the roots of equation x2 – 6x – 2 = 0. If an = n – n, for n  1, then the value of
is equal to :
(1) – 6
(2) 3
(3) – 3
a10  2a8
2a9
(4) 6
Solution : (2)

 
10  10  2  8  8
a10  2a8

2a9
2 9  9



8 ( 2  2)  8 (2  2)
2(9  9 )

8 (6 )  8 (6)
9
9
2(   )

6 9
(   9 )  3
2
37. If 12 identical balls are to be placed in 3 identical boxes, then the probability that one of the boxes contains
exactly 3 balls is :
10
(1)
2
55  
3
12
(2)
 1
220  
3
11
11
(3)
 1
22  
3
(4)
55  2 
3  3 
Solution : ()
* Wrong options given
38. A complex number z is said to be unimodular if |z| = 1. Suppose z1 and z2 are complex numbers such that
z1 2z2
2 z1z2
is unimodular and z2 in not unimodular. Then the point z1 lies on a :
(1) straight line parallel to y-axis
(2) circle of radius 2.
(3) circle of radius 2 .
(4) straight line parallel to x-axis.
Solution : (2)
| z1  2z2 || z1  2z 2 |  | 2  z1z2 |
| z1 |2 4 | z 2 |2 2z 2 z1  2z1 z 2
 4  2 z1 z2  2 z1 z2  | z1 |2 | z 2 |2
a2 + 4b2 – a2b2 = 4
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a2(1–b2) – 4(1– b2)
(a2 – 4)(1 – b2)
|z1|2 = 4
dx
39. The integral
x2 x4
x4
(1)
1
4
1
1
3
equals :
4
x4
(2)
c
1
1
4
x4
(3)
c
x4
1
1
4
c
(4)
x4
x4
1
1
4
c
Solution : (3)
dx
x2 x4
x
1
I
1
1
1
x
dx
3
5
x
1
4
x5
1
x
dx
3
4
4
dz
1
dz
4
dz
z
1
5
z
4
dx
1
4
4
3
1
z
4
4
1
x
1
4
4
3
3
4 dz
1 z
4 3
4
4
1
1
c
z
1
4
c
c
40. The number of points, having both co-ordinates as integers, that lie in the interior of the triangle with vertices
(0,0), (0, 41) and (41, 0), is :
(1) 861
(2) 820
(3) 780
(4) 901
Solution : (3)
(0, 41)
(41, 0)
(0, 0)
x
41
y
1
41
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1 + 2 + 3 + ..... (41 – 2) = 1 + 2 + ..... + 39
39 40
2
= 20 × 39
= 780
=
41. The distance of the point (1, 0, 2) from the point of intersection of the line
x – y + z = 16, is :
(1) 8
(2)
3 21
(3) 13
x
2
3
y 1
4
(4)
z 2
and the plane
12
2 14
Solution : (3)
x
2
3
y 1
4
z 2
12
x = 3 + 2
y = 4 – 1
z = 12 + 2
3 + 2 – 4 + 1 + 12 + 2 = 16
or, 11 = 11
 = 1
x = 5, y = 3, z = 14
 Distance =
= 16
= 13
(1 5)2
9 144
(0 3)2
(2 14)2
169
42. The equation of the plane containing the line 2x – 5y + z = 3 ; x + y + 4z = 5, and parallel to the plane,
x + 3y + 6z = 1, is :
(1) x + 3y + 6z = –7
(2) x + 3y + 6z = 7
(3) 2x + 6y + 12z = –13 (4) 2x + 6y + 12z = 13
Solution : (2)
Putting x = 0
(i)
(ii) × 5
–5y + z = 3
y + 4z = 5
 –5y + z = 3
5y + 20z = 3
21z = 28  z 
4
3
 y = 5 – 4z = 5 – 4.
4
16
1
5

3
3
3
1 4

  0,  ,  is a point on both the lines.
3 3

putting in x + 3y + 6z = 7
4
 1
0  3     6.  1  8  7
3
 3
Which satisfies x + 3y + 6z = 7
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43. The area (in sq. units) of the region described by
{(x, y) : y2  2x and y  4x – 1} is :
5
64
(1)
15
64
(2)
(3)
9
32
(4)
7
32
Solution : (3)
y2  2x, y  4x – 1
1 ,1
2
y = 4x – 1
y – 4x = – 1
x
y
 1
14 1
1
8
1
4
1
2
(4x – 1)2 = 2x
16x2 – 8x + 1 = 2x
16x2 – 8x – 2x + 1 = 0
8x (2x – 1) –1(2x –1) = 0
(2x –1)(8x –1) = 6
1 1
,
2 8
x

 12

1  1 1



A   ydx
     1  


2 2 4

0
 y 2 2x
1

2

0
1 1
2xdx    1 
2 4

 21 
2.

2 
32
3
2
1
 
8
1
8

0
 81 
2
2
1
1

 
3 2 2 8
2xdx 
3
3
2
2

2 
1
8

0

 1  1 1 1
ydx
  2  4  8  2


y 2  2x 

1 1 1
 
2 8 2
1
32
2
1
1


3
32
2 2 2 2 2 2
1 
 1 1  1
  


 3 8   24 32 

5
1
1


24 24 32
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
1 1

4 32

9
32
44. If m is the A.M. of two distinct real numbers l and n (l, n > 1) and G1, G2 and G3 are three geometric means
between l and n, then G14
(1) 4 lm2n
2G24
G34 equals.
(2) 4 lmn2
(3) 4 l2m2n2
(4) 4 l2mn
Solution : (1)
m
l n
2
G1 = lr
G2 = lr2
G3 = lr3
n = lr4
1
 n 4
 l  r
 
l4.r4 + 2l4r8 + l4r12
= l3n + 2l2n2 + ln3
= ln(l2 + 2ln + n2) = ln(l + n)2 = ln(l + n)2 = ln4m 2
45. Locus of the image of the point (2, 3) in the line (2x – 3y + 4) +k (x – 2y + 3) = 0, k R, is a :
(1) straight line parallel to y-axis.
(2) circle of radius
(3) circle of radius
(4) straight line parallel to x-axis
3
2
Solution : (2)
2x – 3y = –4
x – 2y = –3
2x – 4y = – 6
y=2
x=1
(y – 2) = m(x –1)
y = mx + (2 – m)
(2,3)
(h,k)
k 3
h 2
k
3
2
1
m
m
h 2
2
 m
h 2
k
3
2 m
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k
h 2
3
2
k
h2
k 3
2
h 2
2
3
4
2
2 2k 3
2k
h 2
k
3
2h 2
3
(k2 – 9) = [– h2 + 4 + 4k – 12 + 2h – 4]
22
h2 + k2 – 4k – 2h + 3 = 0  r
1 3
46. The area (in sq.units) of the quadrilateral formed by the tangents at the end points of the latera recta to the
ellipse
x2
9
y2
5
1 , is :
(1) 18
(2)
27
2
(3) 27
(4)
27
4
Solution : (3)
One end point of a latus rectum is ae,
b2
a
2,
5
3
Where e
2
3
Equation of T1 :
x
.2
9


2x
9
x
9
2
1 5
.y.
5 3
1
1
y
3
1
y
3
1
1 9
3
2 2
so, area of quadrilateral = 27
 Area of  OAB =
27
4
47. The number of integers greater than 6,000 that can be formed, using the digits 3, 5, 6,7 and 8, without
repetition, is :
(1) 192
(2) 120
(3) 72
(4) 216
Solution : (1)
4 digit nos. = 3 × 4! = 3 × 24 = 72
5 digit nos  5! = 120
72 + 120 = 192
48. Let A and B be two sets containing four and two elements respectively. Then the number of subsets of the
set A × B, each having at least three elements is :
(1) 256
(2) 275
(3) 510
(4) 219
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Solution : (4)
B
1
2
A
a
b
c
d
Required solution
= 28 – 8C0 – 8C1 – 8C2
= 256 – 1 – 8 – 28
= 256 – 37 = 219
2x
49. Let, tan–1y = tan–1x + tan–1
x3
3x
(1)
1 x
(2)
1 3x 2
, where |x| <
2
x3
3x
1
3
. Then a value of y is :
3x
(3)
1 3x 2
x3
1 3x 2
(4)
3x
x3
1 3x 2
Solution : (4)
tan–1 y = tan–1 x + 2 tan–1 x
tan–1 y = 3 tan–1 x
tan–1 y
tan
3x
y
x3
1 3x
1 3x 2
x3
1 3x 2
4
50. The integral
log x2
2
2 log x
x2
log 36 12x
(1) 4
dx is equal to :
(2) 1
(3) 6
(4) 2
Solution : (2)
4
b
log x 2
2 log x
2
4 2
2
log 6
x
2
(1)
s
a
b
f x
a
r s
(~ r
f a b
a
1
51. The negation of ~ s
b
f x dx
dx
f x
f a b
x
x dx
dx
b a
2
s) is equivalent to :
(2)
s
r s
(3)
s r
(4)
s
r
Solution : (3)
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s
r
~s
~r
T
T
T
F
F
F
F
T
F
T
F
T
F
F
T
F
T
T
F
T
F
F
T
T
s r
~ s r
T
F
F
T
F
F
T
T
Option (3) 
~r
s
~s
~r
s
52. If the angles of elevation of the top of a tower from three collinear points A, B and C, on a line leading to the
foot of the tower, are 30º, 45º and 60º respectively, then the ratio, AB : BC, is :
3: 2
(1)
(2) 1: 3
(3) 2 : 3
(4)
3 :1
Solution : (4)
h
[ tan 60º
h
x
h
3 ; x
3
]
60º
A
B
C
h
3
P
h
3h
3 1
1
1
3
53.
lim
x
3 :1
1 cos 2x 3 cos x
x tan 4x
0
(1) 3
is equal to :
(2) 2
(3)
1
2
(4) 4
Solution : (2)
lim
x
1 cos 2x 3 cos x
x tan 4x
0
2 sin2 x 3 cos x
tan 4x
0
x2
4
4x
lim
x
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sin2 x
1
lim
2 x 0
3 cos x
x2
1
1
1 4
2
2
54. Let a, b and c be three non-zero vectors such that no two of them are collinear and a b
c
1
b c a.
3
If  is the angle between vectors b and c , then a value of sin  is :
2
(1)
(2)
3
2
3
(3)
2 3
3
(4)
2 2
3
Solution : (4)
a b
=
c
b.c a
a.c b
1
b c a
3
 b c cos =
sin
1
55. If A
1
9
1
1
b c  cos =
3
3
2 2
3
1 2
2
2 1
2 is a matrix satisfying the equation AAT = 9I, where I is 3 × 3 identity matrix, then the
b
a 2
ordered pair (a, b) is equal to :
(1) (–2, 1)
(2) (2, 1)
(3) (–2, –1)
(4) (2, –1)
Solution : (3)
A
T
1
2
a
2
1
2
2
2 b
1 2
AA
T
2 1
a 2
2
1
2
a
2 2
1
2
0 9 0
2 b
0 0 9
b
2
9 0 0
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a2 + 4 + b2 = 9 .....(I)
2a + 2 – 2b = 0 .....(II)
solving (I) and (II)
2
2b – 2b – 4 = 0
2
b –b–2=0
2
b – 2b + b – 2 = 0
(b – 2) (b + 1) = 0
b = –1
a = –2
b=2
a=1
56. If the function.
k x 1, 0
mx 2 , 3
g x
x
x
3
5
is differentiable, then the value of k + m is :
(1)
16
5
(2)
10
3
(3) 4
(4) 2
Solution : (4)
x
g x
2 x 1
3 < x 5
m
m
k
2 4
0x3
k
4
4m = k .....(1)
2k = 3m + 2
8m = 3m + 2
5m = 2
m=
2
5
8
5
k+m=2
k=
57. The set of all values of  for which the system of linear equations :
2x1 – 2x2+x3 = x1
2x1 – 3x2+2x3 = x2
–x1 + 2x2
= x 3
has a non -trivial solution,
(1) is a singleton.
(3) contains more than two elements.
(2) contains two elements.
(4) is an empty set.
Solution : (2)
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(2 – )x1 – 2x2 + x3 = 0
2x1 – (3 + )x2 + 2x3 = 0
–x1 + 2x2 – x3 = 0
2
2
2
1
3
1
2
0
2
R1  R 1 + R 2 + R 3
3
3
2
1
3
3
2
2
0
C1  C 1 – C 3
0
3
0
1
3
2
3
3
1
1
3
2
3
0
2
1 3
1
2
3
0
0
( – 1)( + 3) (2 – 3 + ) = 0
( – 1)( + 3) ( – 1 + ) = 0
 = 1, 1, – 3
58. The normal to the curve, x2 + 2xy – 3y2 = 0, at (1, 1) :
(1) meets the curve again in the second quadrant (2) meets the curve again in the third quadrant
(3) meets the curve again in the fourth quadrant (4) does not meet the curve again
Solution : (3)
x2 + 2xy – 3y2 = 0
2x
x
2 y
y
dy
dx
dy
dx
x.
x.
dy
dx
dy
dx
3y.
6y.
dy
dx
dy
dx
0
0
x
x
y
y
x.
x
3y
dy
dx
dy
dx
3y.
dy
dx
0
0
3y x
x y
1, 1
3 1
1
1 1
 Equation of normal y – 1 = –1(x – 1)
y–1=–x+1
y=–x+2
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Putting, (2 – x)2 + 2.x.(2 – x) – 3(2 – x)2 = 0
 – 2(2– x)2 + x(2 – x) = 0 – (2 – x)2 + x(2 – x) = 0
 (2 – x)[x – 2 + x] = 0  2(2 – x)(x – 1) = 0
 x = 1, x = 2
Putting x = 1 we get 1 + 2y – 3y2 = 0
So the points are (1, 1), 1,
1
,1
3
1
3
Putting x = 2 we get 4 +4y – 3y2 = 0
So the points are (2, 2), 2,
y
y
2, y
2
3
2
3
59. The number of common tangents to the circles x2 + y2 – 4x – 6x – 12 = 0 and x2 + y2 + 6x + 18y +26 =
0, is :
(1) 2
(2) 3
(3) 4
(4) 1
Solution : (2)
C1 = (2, 3)
r1
4
C2 = (–3, – 9)
9 12
r2
5
52 122
C1C2
9 81 26
64
8
r1 r2
So there are 3 common tangents.
dy
dx
(3) 2e
60. Let y(x) be the solution of the differential equation x log x
(1) 0
(2) 2
y
2x log x, x
1 . Then y(e) is equal to :
(4) e
Solution : (2)
x log x
dy
dx
dy
dx
y.
I.F.
y
2x log x
1
x log x
2
Pdx
e
e
1
dx
x log x
dy
dx
e
dz
z
elog z
z
Py
Q
log x.
 solution is :
y. I.F.
y.log x
Q. I.F. dx
2.log x dx
2 log xdx
2 x log x x
When x = 1,
y e
C
2x logx 1
C
0 = 2.1.(0 – 1) + C  C = 2.
2e loge 1
2
loge
2
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PART C - PHYSICS
61. As an electron makes a transition from an excited state to the ground state of a hydrogen - like
atom/ion :
(1) Kinetic energy, potential energy and total energy decrease
(2) Kinetic energy decreases, potential energy increases but total energy remains same
(3) Kinetic energy and total energy decrease but potential energy increases
(4) its kinetic energy increases but potential energy and total energy decrease
Solution : (4)
KE = +x
PE = – 2x
TE = – x
For inner orbit, x is more
62. The period of oscillation of a simple pendulum is T
2
L
. Measured value of L is 20.0 cm known to 1 mm
g
accuracy and time for 100 oscillations of the pendulum is found to be 90 s using a wrist watch of 1s resolution.
The accuracy in the determination of g is :
(1) 3 %
(2) 1 %
(3) 5 %
(4) 2 %
Solution : (1)
T
2
g
4
g
%
g
L
g
2
L
T2
L
T
100 2
100
L
T
2.7 3%
63. A long cylindrical shell carries positive surface charge  in the upper half and negative surface charge – 
in the lower half. The electric field lines around the cylinder will look like figure given in : (figures are schematic
and not drawn to scale)
(1)
(2)
(3)
(4)
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Solution : (4)
From theory
64. A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz. The frequencies
of the resultant signal is/are :
(1) 2005 kHz, and 1995 kHz
(2) 2005 kHz, 2000 kHz and 1995 kHz
(3) 2000 kHz and 1995 kHz
(4) 2 MHz only
Solution : (2)
Am =(Ac + As sin 2fst) sin 2fct
As
[cos 2(fc – fs)t + cos 2 (fc + fs)t]
2
2000 Hz, 1996 Hz 2005 Hz
Ac sin 2fct

65. Consider a spherical shell of radius R at temperature T. The black body radiation inside it can be considered
U
T 4 and pressure p
V
shell now undergoes an adiabatic expansion the relation between T and R is :
as an ideal gas of photons with internal energy per unit volume u
(1) T  e–3R
(2)
T
1
R
(3)
T
1
R3
(4)
1 U
. If the
3 V
T  e–R
Solution : (2)
dU + pdV = 0
U
dV 0
3V
U3V = const
V3T12V = const
VT 3 = const
R3T3 = const
RT = const
dU
 U = VT4
66. An inductor ( L = 0.03 H) and a resistor (R = 0.15 K) are connected in series to a battery of 15V EMF in
a circuit shown below. The key K1 has been kept closed for a long time. Then at t = 0, K1 is opened and
key K2 is closed simultaneously. At t = 1 ms, the current in the circuit will be : (e5  150)
(1) 67 mA
(2) 6.7 mA
(3) 0.67 mA
(4) 100 mA
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Solution : (3)
15
i0
0.15 10
i
Rt
L
i0e
1
A
10
3
i0
e
1
10
5
1
A
150

0.67 mA
Rt
L
5
67. 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 modulus of the material of the wire is
Y then
1
is equal to : (g = gravitational acceleration)
Y
(1)
TM
T
2
Mg
1
A
(2)
1
TM
2
TM
T
2
A
Mg
(3)
1
T
TM
2
A
Mg
(4)
TM
T
2
1
A
Mg
Solution : (4)
T
L
,
g
2
2
TM
T
1
Y
Mg
A
L
L
1
Y
L
L
A
Mg
L
L
g
L
L
TM
T
2
1
A
Mg
68. A red LED emits light at 0.1 watt uniformly around it. The amplitude of the electric field of the light at a distance
of 1 m from the diode is :
(1) 2.45 V/m
(2) 5.48 V/m
(3) 7.75 V/m
(4) 1.73 V/m
Solution : (1)
I
E
1
2
0E
2
C
2P
4
2
0r C
P
A
P
4 r2
2.45 vm
1
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69. Two coaxial solenoids of different radii carry current I in the same direction. Let F1 be the magnetic force
on the inner solenoid due to the outer one and F2 be the magnetic force on the outer solenoid due to the
inner one. Then :
(1) F1 is radially inwards and F2 is radially outwards
(2) F1 is radially inwards and F2 = 0
(3) F1 is radially outwards and F2 = 0
(4) F1 = F2 = 0
Solution : (4)
From theory
70. Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabatic expansion,
the average time of collision between molecules increases as Vq, where V is the volume of the gas. The value
Cp
of q is :
(1)
Cv
3
5
(2)
6
1
2
(3)
1
2
(4)
3
5
6
Solution : (2)
K1
n
K1V
N
K1V
t
Cm
N
K2V
t
1
K3 V
n
2
K2
3RT
M
V
T
1
K4V
2
TV – 1 = const.,
1
T2
1
K3 V
2
1
2
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71.
An LCR circuit is equivalent to a damped pendulum. In an LCR circuit the capacitor is charged to Q0 and
then connected to the L and R as shown below :
L
R
C
2
If a student plots graphs of the square of maximum charge QMax on the capacitor with time(t) for two
different values L1 and L2 (L1 > L2) of L then which of the following represents this graph correctly ? (plots
are schematic and not drawn to scale)
2
QMax
2
QMax
L1
(1)
L2
L2
(2)
L1
t
t
2
QMax
2
QMax
Qo (For both L1 and L2)
(3)
(4)
t
L2
L1
t
Solution : (4)
q
q0e
A
q2
dA
dt
L
Rt
L
q02e
q20 e
2Rt
L
2Rt
L
2R
L
dA
dt
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R
is removed, as shown in the
2
figure. Taking gravitational potential V = 0 at r = , the potential at the centre of the cavity thus formed
is :
(G = gravitational constant)
72. From a solid sphere of mass M and radius R, a spherical portion of radius
GM
R
(1)
2GM
3R
(2)
(3)
2GM
R
(4)
GM
2R
Solution : (1)
Vhollow = Vsolid – Vcavity
GM
2R
3
2R2
R
2
2
M
8
R
2
2
3G
GM 11
.
2R 4
3GM
8
GM
R
73. A train is moving on a straight track with speed 20 ms –1. It is blowing its whistle at the frequency of
1000 Hz. The percentage change in the frequency heard by a person standing near the track as the train
passes him is (speed of sound = 320 ms–1) close to :
(1) 12 %
(2) 18 %
(3) 24 %
(4) 6 %
Solution : (1)
f
f
f
f
f
f
f
f
f
V
V VS
f 1
VS
V
V
VS
f 1
VS
V
V
2fVS
V
2 20
320
1
8
12.5%
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74.
F
A
B
Given in the figure are two blocks A and B of weight 20 N and 100 N, respectively. These are being pressed
against a wall by a force F as shown. If the coefficient of friction between the blocks is 0.1 and between block
B and the wall is 0.15, the frictional force applied by the wall on block B is :
(1) 80 N
(2) 120 N
(3) 150 N
(4) 100 N
Solution : (2)
F1
F2
F1= 20
F2 = 100 + F1= 120
20
F1
100
75. Distance of the centre of mass of a solid uniform cone from its vertex is z0. If the radius of its base is R and
its height is h then z0 is equal to :
(1)
3h
4
(2)
5h
8
(3)
3h2
8R
(4)
h2
4R
Solution : (1)
3h
4
From theory z0
76. A rectangular loop of sides 10 cm and 5 cm carrying a current I of 12 A is placed in different orientations
as shown in the figures below :
z
z
B
I
B
(a)
(b)
I
I
I
y
x
x
(c)
z
x
B
B
I
I
y
I
y
I
I
z
I
I
I
(d)
I
I
x
I
y
I
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If there is a uniform magnetic field of 0.3 T in the positive z direction, in which orientations the loop would
be in (i) stable equilibrium and (ii) unstable equilibrium ?
(1) (a) and (c), respectively
(3) (b) and (c), respectively
(2) (b) and (d), respectively
(4) (a) and (b) respectively
Solution : (2)
From theory
6V
P 2
1
77.
9V
Q 3
3
In the circuit shown, the current in the 1 resistor is :
(1) 0A
(3) 0.13 A, from P to Q
(2) 0.13 A, from Q to P
(4) 1.3 A, from P to Q
Solution : (2)
6V
i1
P
2
i1 – i2
1
9V
3
Q
i2
3
2i1 + 1(i1 – i2) + 3i1 = 9
6i1 – i2 = 9 ..............(1)
3i2 + (i2 – i1) = 6
– i1 + 4i2 = 6 ..............(2)
Solving i1
42
, i2
23
45
, i2
23
i1
3
23
0.13A along Q to P..
78. A uniformly charged solid sphere of radius R has potential V0 (measured with respect to ) on its surface.
For this sphere the equipotential surfaces with potentials
and R4 respectively. Then
(1) R1  0 and (R2 – R1) > (R4 – R3)
(3) 2R < R4
3V0 5V0 3V0
V0
,
,
and
have radius R1, R2, R3
2
4
4
4
(2) R1 = 0 and R2 < (R4 – R3)
(4) R1 = 0 and R2 > (R4 – R3)
Solution : (2 & 3)
From theory
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79. In the given circuit, charge Q2 on the 2F capacitor changes as C is varied from 1F to 3F. Q2 as a function
of ‘C’ is given properly by : (figures are drawn schematically and are not to scale)
1 F
C
2 F
E
Charge
Charge
Q2
Q2
(1)
(2)
1 F
3 F
C
1 F
3 F
C
Charge
Charge
Q2
Q2
(3)
(4)
1 F
3 F
C
1 F
3 F
C
Solution : (1)
C 3
C 3
Ceq
Q
Q2
dQ2
dC
E
3C
C 3
2
Q
3
2EC
C 3
2E 1
C+3
C 3
+ve
3
2E
3
2
C
dQ2
dC
80. A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v. If the collision is perfectly inelastic, the percentage loss in the energy during
the collision is close to :
(1) 50 %
(2) 56 %
(3) 62 %
(4) 44 %
Solution : (2)
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2m × v = 2mv
1 (2m)v2 = mv 2
2
2mv 2 1
, 3m v
3m
2
v
2
4
mv 2
3
m × 2v = 2mv
1 m(2v) 2 = 2mv 2
2
KE
4
mv 2
3
3
5
3
3
% change
5
mv 2
3
100
500
9
56 %
81. Monochromatic light is incident on a glass prism of angle A. If the refractive index of the material of the prism
is , a ray, incident at an angle , on the face AB would get transmitted through the face AC of the prism
provided :
A
B
(1)
sin
(3)
cos
1
1
sin A
sin
sin A
sin
1
1
1
1
C
(2)
cos
(4)
sin
1
1
sin A
sin
sin A
sin
1
1
1
1
Solution : (4)
r2 < c
r2
sin
1
1
A – r1 < sin
A
sin
sin A
1
1
1
sin
1
r1
1
1
sin r1
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sin A
sin
1
1
sin A
sin
1
1
sin
1
sin A
sin
sin r1
sin
1
1
82. From a solid sphere of mass M and radius R a cube of maximum possible volume is cut. Moment of inertia
of cube about an axis passing through its centre and perpendicular to one of its faces is :
MR2
(1)
(2)
16 2
4MR2
(3)
9 3
4MR2
3 3
(4)
MR2
32 2
Solution : (2)
a 3
a
I
2R
2R
3
ma2
6
a3 a2
6
a5
6
4 3
R
3
3M
M ;
4 R3
]
3M
4 R3
32R5
9 3
[ where
4MR2
M
8 R
3
9 3
83. Match List-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list :
List-I
List-II
(A)
Franck-Hertz
Experiment
(i)
Particle nature
of light
(B)
Photo-electric
Experiment
(ii)
Discrete energy
levels of atom
(C)
Davison-Germer
Experiment
(iii)
Wave nature of
electron
(iv)
Structure of
atom
(1) (A) – (ii) (B) – (iv) (C) – (iii)
(3) (A) – (iv) (B) – (iii) (C) – (ii)
(2) (A) – (ii) (B) – (i) (C) – (iii)
(4) (A) – (i) (B) – (iv) (C) – (iii)
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Solution : (2)
From theory
84. When 5V potential difference is applied across a wire of length 0.1 m, the drift speed of electrons is
2.5 × 10–4 ms–1. If the electron density in the wire is 8 × 1028 m –3, the resistivity of the material is close
to :
(1) 1.6 × 10–7 m
(2) 1.6 × 10–6 m
(3) 1.6 × 10–5 m
(4) 1.6 × 10–8 m
Solution : (3)
l
A
V = iR = Vd e n A 
V
Vd e n l
= 1.6 × 10–5 m
85. For a simple pendulum, a graph is plotted between its kinetic energy (KE) and potential energy (PE) against
its displacement d. Which one of the following represents these correctly?
(graphs are schematic and not drawn to scale)
E
E
PE
KE
(1)
d
KE
d
(2)
PE
E
E
PE
(3)
KE
(4)
KE
PE
d
Solution : (1)
From theory
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86. Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial speed of 10 m/s and
40 m/s respectively. Which of the following graph best represents the time variation of relative position of
the second stone with respect to the first?
(Assume stones do not rebound after hitting the ground and neglect air resistance, take g = 10 m/s2)
(The figures are schematic and not drawn to scale)
(y2 – y1) m
(y2 – y1) m
240
240
(2)
(1)
12
t(s)
8
12
8
12
t(s)
(y2 – y1) m
(y2 – y1) m
240
240
(3)
(4)
8
12
t(s)
t
t(s)
Solution : (2)
When two particles are in air :
S1
10 t
1 2
gt
2
1 2
gt
2
S2 – S1 = 30 t
Time of flight of particle – I is 8 sec.
S2
40 t
S1
S2
240
S2
40 t
S1
240
1 2
gt
2
40 t
1 2
gt
2
Hence option (2)
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87. A solid body of constant heat capacity 1 J/ºC is being heated by keeping it in contact with reservoirs in two
ways :
(i) Seequentially 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 100ºC to final temperature 200ºC. Entropy change
of the body in the two cases respectively is :
(1) ln2, ln2
(2) ln2, 2ln2
(3) 2ln2, 8ln2
(4) ln2, 4ln2
Solution : (1)
From theory
88. Assuming human pupil to have a radius of 0.25 cm and a comfortable viewing distance of 25 cm, the minimum
separation between two objects that human eye can resolve at 500 nm wavelength is :
(1) 30 m
(2) 100 m
(3) 300 m
(4) 1 m
Solution : (1)
x
D
1.22
a
1.22 D
a
x
30 m
89.
L
I
I
Two long current carrying thin wires, both with current I, are held by insulating threads of length L and are
in equilibrium as shown in the figure, with threads making an angle ‘’ with the vertical. If wires have mass
 per unit length then the value of I is :
(g = gravitational acceleration)
(1)
gL
0 cos
2 sin
(2)
2
gL
0
tan
(3)
gL
0
tan
(4)
sin
gL
0 cos
Solution : (1)
0
4
I
2I 2
L
2L sin
L g
2 sin
tan
gL
0 cos
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90. On a hot summer night, the refractive index of air is smallest near the ground and increases with height from
the ground. When a light beam is directed horizontally, the Huygens’ principle leads us to conclude that as
it travels, the light beam :
(1) goes horizontally without any deflection
(2) bends downwards
(3) bends upwards
(4) becomes narrower
Solution : (1)
From theory
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