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JEE ADVANCED
EXAMINATION 2014
QUESTIONS WITH SOLUTIONS
PAPER - 2 [CODE - 4]
394,50 - Rajeev Gandhi Nagar Kota, Ph. No. : 93141-87482, 0744-2209671
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IIT-JEE 2014 Solutions by Motion Edu. Pvt. Ltd. Kota
PHYSICS
1. A metal surface is illuminated by light of two
different wavelengths 248 nm and 310 nm. The maximum speeds of the photoelectrons corresponding to
these wavelengths are u1 and u2, respectively. If
the ratio u1 : u2 = 2 : 1 and and hc= 1240 eV nm,
the work function of the metal is nearly
(A) 3.7 eV
(B) 3.2 eV
(C) 2.8 eV
(D) 2.5 eV
Sol.
A
E1 =
12400
=5eV
2480
12400
= 4eV
3100
 5 =  + 4k
4 = + k
On solving
Sol.
A
912×R2 = 5.7×10–18×1×[T4–(300)4] × 4 R2
 108 (121) = T4
 T = 330 K
4.
During an experiment with a metre bridge,
the galvanometer shows a null point when the jockey
is pressed at 40.0 cm using a standard resistance of
90 , as shown in the figure. The least count of the
scale used in the metre bridge is 1 mm. The unknown resistance is
E2 =
=
.....(1)
.....(2)
11
= 3.66 = 3.7 eV
3
2.
If Cu is the wavelength of K X-ray line of
copper (atomic number 29) and M0 is the wavelength iof the K X-ray line of molybdenum (atomic
number 42), then the ratio Cu/Mo is close to
(A) 1.99
(B) 2.14
(C) 0.50
(D) 0.48
Sol.
B
By Mosley’s Law :
Sol.
2
R = 90 × 
1
n R = n 90 + n 2 + n 1
d 2
d 1
dR
=  + 
R
2
1
f = a (z – 1)
c

 Inverse relation and hence :
As v = f  f =
dR
0.1
0.1
=
+
60
60
40
dR = 0.1 + 0.15
= 0.25
 R = (60 ± 0.25) 
1
  (Z  1)2
2

 Z2  1 
1
(42  1)2

  1 =
=
= 2.14
 0
2
(24  1)2
 Z1  1 
3.
Parallel rays of light of intensity I=912
Wm-2 are incident on a spherical black body kept in
surroundings of temperature 300 K. Take StefanBoltzmann constant =5.7×10-8Wm-2K-4 and assume
that the energy exchange with the surroundings is
only through radiation. The final steady state
temeprature of the black body is close to
(A) 330 K
(B) 660 K
(C) 990 K
(D) 1550 K
Page 2
(A) 60  0.15
(B) 135  0.56
(C) 60  0.25
(D) 135  0.23
C
R × 60 = 90 × 40
1
×(radius of Earth)
10
has the same mass density as Earth. Scientists dig
5.
A planet of radius R 
R
on it and lower a wire of the same
5
length and of linear mass density 10-3 kgm-1 into it.
If the wire is not touching anywhere, the force applied at the top of the wire by a person holding it in
place is (take the radius oif Earth=6×106m and the
acceleration dur to gravity on Earth is 10 ms-2)
(A) 96 N
(B) 108 N
(C) 120 N
(D) 150 N
a well of depth
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Sol.
B
Sol.
R=

1
R
10 e
dF =
= 10–3
= 10–3
9 
 .dx. 10 1  R
x 


e / 10 
2
R
 Re  
10
e
× 1  50   50   × R  2

 

e
1 
1
× 6 × 106  50  900 


h = T cos ( + /2)
h=
= 108
6. A wire, which passes through the hole in a small
bead, is bent in the form of quarter of a circle. The
wire is fixed vertically on ground as shown in the
figure. The bead is released from near the top of
the wire and it slides along the wire without friction.
As the bead moves from A to B, the force it applies
on the wire is
Sol.
D
(A) always radially outwards.
(B) always radially inwards.
(C) radially outwards initially and radially in
wards later.
(D) radially inwards intially and radially out
wards later.
D
By theory
7. A glass capillary tube is of the shape of a truncated cone with an apex angle  so that its two
ends have cross sections of different radii. when
dipped in water vertically, water rises in it to a height
h, where the radius of its cross section is b. If the
surface tension of water is S, its density is , and
its contact angle with glass is , the value of h will
be (g is the acceleration due to gravity)
(A)
2s
cos     
bg
(C)
2s
2s
cos     / 2  (D)
cos     / 2 
bg
bg
(B)
2s
cos ( + /2)
bg
8. Charges Q, 2Q and 4Q are uniformly distributed in
three dielectric solid spheres 1,2 and 3 of radii R/2,
R and 2R respectively, as shown in figure. If magnitudes of the electric fields at point P at a distance R
from the centre of spheres 1,2 and 3 are E1, E2 and
E3 respectively, the
Sol.
(A) E1>E2>E3
(C) E2>E1>E3
C
KQ
E1 = 2
R
4KQ.R
KQ
E3 = (2R )3 =
2R 2
(B) E3>E1>E2
(D) E3>E2>E1
E2 =
2KQ
R2
 E2 > E1 > E3
9. A tennis ball is dropped on a horizontal smooth
surface. It bounces back to its original position after hitting the surface. The force on the ball during
the collision is proportional to the length of compression of the ball. Which one of the following
sketches describes the variation of its kinetic energy K with time t most appropriately? The figures
are only illustrative and not to the scale.
(A)
(B)
(C)
(D)
2s
cos     
bg
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Sol.
Sol.
B
V = gt
C
 0ia2
1
 KE =
m (gt)2
2
 Relation is parabolic

2
2a h
10. A point source S is placed at the bottom of a
transparent block of height 10 mm and refractive
index 2.72. It is immersed in a lower refractive index
liquid as swhon in the figure. It is found that the
light emerging from the block to the liquid forms a
circular bright spot of diameter 11.54 mm on the top
of the block. The refractive index of the liquid is
 oi


2 3/2
0i

2 a2  h2
a
a
2
h

2 3/2
a
a
2
Sol.
(A) 1.21
(C) 1.36
C
a
2
133.3
n = 1.36
h
a2
0.6a2
2




cos
2

2
 a  h2

2

4
2
= 0.4a2 + 0.4h2
= 0.4h2
2
 11.54 

  100
 2 
Also sin C =
5.77

2 1/ 2
a2
(B) 1.30
(D) 1.42
11.54
2
sin C =
h
cos
2 a2  h2
=
h2 =
n
2.72
M
2.72
Paragraph 11 & 12
The figure shows a circular loop of radius a with two
long parallel wires (numbered 1 and 2) all in the
plane of the paper. The distance of each wire from
the centre of the loop is d. The loop and the wires
are carrying the same current I. The current in the
loop is in the counterclockwise direction if seen from
above.
3 2
a
2
h  1.2 a
12. consider d>>a, and the loop is rotated about its
diameter parallel to the wires by 30° from the position shown inthe figure. If the currents in the wires
are in the opposite directions, the torque on the
loop at its new position will be (assume that the net
field due to the wires is constant over the loop)
(A)
(C)
Sol.
0l2a2
d
30l2a2
d
(B)
(D)
0l2a2
2d
30l2a2
2d
B
11. When d  a but wires are not touching the loop,
it is found that the net magnetic field on the axis of
the loop is zero at a height h above the loop. In
that case
(A) current in wire 1 and wire 2 is the direction PQ
 = MB sin 
and RS, respectively and h  a
2 oi
(B) current in wire 1 and wire 2 is the direction PQ
= ia2×
sin 300
2 d
and SR, respective and h  a
(C) current in wire 1 and wire 2 is the direction PQ
 i2 a2 1  oi2 a2
and SR, respectively and h  1.2a
 
= o
d
2
2d
(D) current in wire 1 and wire 2 is the direction PQ
and RS, respectively and h  1.2a
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Paragraph 13 & 14
In the figure a containner is shown to have a movalble
(without friction) piston on top. The container and
the piston are all made of perfectly insulating material allowing no heat transfer between outside and
inside the container. The container is divided into
two compartments by a rigid partition made of a
thermally conducting material that allows slow transfer of heat. The lower compartment of the container is filled with 2 moles of an ideal monatomic
gas at 700 K and the upper compartment is filled
with 2 moles of an ideal diatomic gas at 400 K. The
heat capacities per mole of an ideal monatomic gas
3
5
R, CP  R , and those for an ideal diatomis
2
2
5
7
gas are CV  R, CP  R .
2
2
are CV 
13. Consider the partition to be rigidly fixed so that
it does not move. When equilibrium is achieved, the
final temperature to the gases will be
(A) 550 K
(B) 525 K
(C) 513 K
(D) 490 K
Sol.
D
3
2×
R (700 – T)
2
= 2R (T – 400) +
Paragraph 15 & 16
A spray gun is shown in the figure where a piston
pushes air out of a nozzle. A thin tube of uniform
cross section is connected to the nozzle. The other
end of the tube is in a small liquid container. As the
piston pushes air through the nozzle, the liquid from
the containwer rises into the nozzle and is sdprayed
out. For the spray gun shown, the radii of the piston
and the nozzle are 20 mm and 1 mm respectively.
The upper end of the container is open to the
atmosphere.
15. If the piston is pushed at a speed of 5 mms-1, the
air comes out of the nozzle with a speed of
(A) 0.1 ms-1
(B) 1 ms-1
-1
(C) 2 ms
(D) 8 ms-1
Sol.
C
From equation of continuity
A1V1 = A2V2
(20)2 × 5 × t = (1)2 × v × t
v = 400 × 5 = 2000 mm/s
= 2 m/s
16. If the density of air is a and that of the liquid
l, then for a given piston speed the rate (volume
per unit time) at which the liquid is sprayed will be
proportional to
Sol.
(A)
a
l
(B)
(C)
l
a
(D) l
a l
A
5
R (T – 400) × 2
2
T = 490 K
14. Now consider the partition to be free to move
without friction so that the pressure of gases in
both compartments is the same. Then total work
done by the gases till the time they achieve equilibrium will be
(A) 250 R
(B) 200 R
(C) 100 R
(D) -100 R
Sol.
D
|Q1| = |Q2|
5
7
× 2 × R (700–T) =
× 2 × R (T – 400)
2
2
 T = 525 K
W.D. by lower gas = 2 × R × 175 = 350 R
W.D. upper gas = 2 × R × 125 = 250 R
 net W.D. = –100 R
Pressure at point in nozzle  P
P = P0 – 1 a V12
.....(1)
1
 V 2 – lgh
2 l 2
equation (1) and (2)
P = P0 –
......(2)
1
1
a V12 =
 V 2 + lgh
2
2 l 2
Neglecting the term regh
V2 =
a
V1
l
Rate of liquid flow = AV2
= AV1
a
l
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a
l
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17. A person in a lift is holding a water jar, which
has a small hole at the lower end of its side. When
the lift is at rest, the water jet coming out of the
hole hits the floor of the lift at a distance d of 1.2 m
from the person. In the following, state of the lift’s
motion is given in list I and the distance where the
water jet hits the floor of the lift is given in List II.
Match the statements from List I with those is List
II and select the correct answer using the code
given belsow the lists.
List I
List II
P. Lift is acceleration vertically up. 1. d = 1.2 m
Q. Lift is accelerating vertically
2. d> 1.2 m
with an acceleration less than
the gravitational acceleration.
R. Lift is moving vertically up with 3. d < 1.2 m
constant speed.
S. Lift is falling freely.
4. No water
leaks out of the
jar
Code :
(A) P-2, Q-3, R-2, S-4
(B) P-2, Q-3, R-1, S-4
(C) P-1, Q-1, R-1, S-4
(D) P-2, Q-3, R-1, S-1
Sol.
C
List I
List II
P. Q1, Q2, Q3, Q4 all positive
1. +x
Q. Q1, Q2 positive; Q3, Q4 negative
2. -x
R. Q1, Q4 positive; Q2, Q3 negative 3. +y
S. Q1, Q3 positive; Q2, Q4 negative
4. -y
Code :
(A) P-3, Q-1, R-4, S-2
(B) P-4, Q-2, R-3, S-1
(C) P-3, Q-1, R-2, S-4
(D) P-4, Q-2, R-1, S-3
Sol.
A
(P) All +ve 
(Q) Q1, Q2 +ve 
Q3, Q4 –ve 
d = vt = 2gh
2h
= 2h which is independent
g
of g.
But when the lift falls freely no water leaks
out of t he jar as g eff = 0.
18. For charges Q1, Q2, Q3 and Q4 of same magnitude are fixed along the x axis at x = -2a, -a, +a
and +2a, respectively. A positive charge q is placed
on the positive y axis at a distance b > 0. Four
potions of the signs of these charges are given in
List I. The direction of the forces on the charge q is
given in List II. Match List I with List II and select
the correct answer using the code given below the
lists.
(R) Q1, Q3 –ve 
Q2, Q3 –ve
(S) Q1, Q3 +ve 
Q2, Q4 –ve
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19.
Four combinations of two thin lenses are
given in List I. The radius of curvature of all curved
surface is r and the refractive index of all the lenses
is 1.5. Match lens combinations in List I with their
focal length in List II and select the correct answer
using the code given below the lists.
List I
List II
feq = -r
P.
1. 2r
=
=
2. r/2
1
3
  1 1
 1   
f2 =  2
  r
1
1 1
 
feq r 2r
feq = 2r
Q.
R.
3. -r
S.
1
3
 1 1
  1   
=
f1
2
 r r
1
r
P-2
Q-4 R-3
1
2r
S-1
20.
A block of mass m1 = 1 kg another mass
m2=2kg are placed together (see figure) on an
inclined plane with angle of inclination . Various
values of  are given in list I. The coefficient of
static and dinamic friction between the block m2
and the plane are equal to =0.3. In List II expressions for the friction on block m2 are given. Match
the correct expression of the friction in List II with
the angles given in List I, and choose the correct
option. The acceleration due to gravity is denoted by g.
4. r
Code :
Sol.
1
1 1

feq
2r 2r
(A) P-1, Q-2, R-3, S-4
(B) P-2, Q-4, R-3, S-1
(C) P-4, Q-1, R-2, S-3
(D) P-2, Q-1, R-3, S-4
D
 L
  1
1 
1
=    1  R  R 
f
2 
 m
  1
1 3
 1 1 
   1   
f1  2
 r r 
1 1

f1 r
1 1 2

f1  2  r
1
1 1
 
feq r r
1 3
1
   1
f1  2
r
1
1

f1
2r
1
1
1


feq
f1 f2
feq = r
1
1

feq r
feq 
r
2
List I
List II
P.  = 5°
1. m2g sin 
Q.  = 10°
2. (m1+m2) g sin 
R.  = 15°
3. m2 g cos 
S.  = 20°
4. (m1+m2) g cos q
Code :
(A) P-1, Q-1, R-1, S-3
(B) P-2, Q-2, R-2, S-3
(C) P-2, Q-2, R-2, S-4
(D) P-2, Q-2, R-3, S-3
Sol.
D
There will be no slipping if friction
balances the net force acting downwards
along the incline
fsmax =  m2 g cos 
Force in t he downward direct ion is (m 1
+ m 2 ) g s i n  t h u s a n gl e a t wh i c h
sl ippi ng start s
(m1 + m2) g sin  =  m2 g cos 
0.3  2
= 0.2
3
given tan 11.5° = 0.2
Thus  = 11.5°
Thus for angles less than 11.5° there won’t
be any slipping hence friction is static and
equal to (M1 + M2) g sin  for greater than
11.5° the friction is dynamic and is equal to
 m2 g cos .
tan  =
1
1

f1
2r
1 3
 1 1
   1   
f2  2
  r 
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CHEMISTRY
21.
Sol.
For the identification of -naphthol using dye
test, it is necessary to use
(A) dichloromethane solution of -napthol
(B) acidic solution of -naphthol
(C) neutral solution of -naphthol
(D) alkaline solution of -naphthol
D
Sol.
The correct order of their boiling point is :
(A) I > II > III
(B) III > II > I
(C) II > III > I
(D) III > I > II
B
Boiling point  surface area 
25.
Sol.
Assuming 2s-2p mixing is NOT operative, the
paramagnetic species among the following is:
(A) Be2
(B) B2
(C) C2
(D) N2
(C)
if 2s-2p mixing is not allowed then MOD will
be like O2.
 2p y 


1s * 1s 2s *2s 2px  || 
 2pz 
22.
Sol.
23.
Sol.
24.
For the elementary reaction M  N, the rate
of disappearance of M increases by a factor
of 8 upon doubling the concentration of M.
The order of the reaction with respect to M
is :
(A) 4
(B) 3
(C) 2
(D) 1
(B)
r = K [M]x
8r = K [2M]x

x=3
For the process
H2O (l)  H2O (g)
at T=100ºC and 1 atomsphere pressure, the
correct choice is :
(A) Ssystem > 0 and Ssurroundings > 0
(B) Ssystem > 0 and Ssurroundings < 0
(C) Ssystem < 0 and Ssurroundings > 0
(D) Ssystem > 0 and Ssurroundings < 0
(B)
H2O (l) + Heat  H2O (g)

Ssystem > 0
&
Ssurroundings < 0
26.
Sol.
27.
1
branching
*

  2p y 
 || 

 *2px.
*


 2pz 
The product formed in the reaction of SOCl2
with white phosphorous is :
(A) PCl3
(B) SO2Cl2
(D) SCl2
(D) POCl3
(A)
P4 + SOCl2  PCl3 + SO2 + S2Cl2
white
The major product in the following reaction
is
(A)
(B)
(C)
Sol.
(D)
D
Isomers of hexane, based on their branching, can be divided into three distinct classes
as shown in the figure.
(I)
(II)
(III)
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28.
Ans.
29.
Hydrogen peroxide in its reaction with KlO4
and NH2OH respectively, is acting as a
(A) reducing agent, oxidising agent
(B) reducing agent, reducing agent
(C) oxidising agent, oxidising agent
(D) oxidising agent, reducing agent
(A)
The acidic hydrolysis of ether (X), shown
below is fastest when
Ans.
30.
(A) one phenyl group is replaced by a methyl
group.
(B) one phenyl group is replaced by a paramethoxyphenyl group
(C) two phenyl groups are replaced by two
para-methoxyphenyl groups
(D) no structural change is made to X
C
Rate of SN1 reaction  stability of carbocation
. If two phenyl group is replaced by two pmethoxy phenyl group then it gives most
stable carbocation so fastest reaction.
Under ambient conditions, the total number
of gases released as products in the final
step of the reaction scheme shown below is
Complete
Hydrolysis
XeF6
Paragraph For Question 31 and 32
X and Y are two volatile liquids with molar weights
of 10g mol–1 and 40g mol–1 respectively. Two cotton
plugs, one soaked in X and the other soaked in Y,
are simultaneously placed at the ends of a tube of
length L = 24cm, as shown in the figure. The tube is
filled with an inert gas at 1 atmosphere pressure
and a temperature of 300 k. Vapurs of X and Y react
to form a product which is first observed at a distance d cm from the plug soaked in X. Take X and Y
to have equal molecular diameters and assume ideal
behaviour for the inert gas and the two vapous.
31.
Sol.
The value of d in cm (shown in the figure),
as estimated from Graham's law, is
(A) 8
(B) 12
(C) 16
(D) 20
(C)
L = 24 cm
Cotton wool
soaked in X
d
Cotton wool
soaked in Y
MM y
ratex

rate y
P + other product
Initial formation
of
the product
MM x
MM y

dx / t

dy /t

d
40

24  d
10


d = 48 – 2d
d = 16
MM x
-
OH /H2O
Q
-
slow disproportionation in OH /H2O
Products
Ans.
(A) 0
(C) 2
(C)
32.
(B) 1
(D) 3
Complete
 XeO3 + HF
XeF6 + H2O 
Hydrolysis
XeO3 + OH– (basic medium)
HXeO4–
Slow
 XeO64– +
HXeO4– + OH– 
Disproportionation
Xe  + O2 
+ H2O
Sol.
The experimental value of d is found to be
smaller than the estimate obtained using
Graham's law. This is due to
(A) larger mean free path for X as compared
to that of Y.
(B) larger mean free path for Y as compared
to that of X.
(C) increased collision frequency of Y with
the inert gas as compared to that of X with
the inert gas.
(D) increased collision frequency of X with
the inert gas as compared to that of Y with
the inert gas.
(B, D)
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Paragraph For Questions 33 and 34
Schemes 1 and 2 describe sequential transformation of alkynes M and N. Consider only the major
products formed in each step for both the schemes.
33.
Sol.
C
The product X is
(A)
It gives + Ve iodoform test and it is functional isomer if X .
Paragraph For Questions 35 and 36
An aqueous solution of metal ion M1 reacts separately with reagents Q and R in excess to give tetrahedral and square planar complexes, respectively.
An aqueous solution of another metal ion M2 always
forms tetrahedral complexes with these reagents.
Aqueous solution of M2 on reaction with reagent S
gives white precipitate which dissolves in excess of
S. The reactions are summarized in the scheme given
below:
SCHEME:
(B)
(C)
Tetrahedral
(D)
Sol.
R
Q
excess
A
HO–CH2–CH2–CCH NaNH2
C–Na+
CH3–I
CH3–CH2I
Tetrahedral
excess
R
Q
Tetrahedral
M2
excess
HO–CH2–CH2–CH
Square planar
M1
HO–CH2–CH2CC–CH2–CH3
excess
S, stoichiometric amount
CH3–O–CH2–CH2–CC–CH2–CH3
S
White precipitate
34.
The correct statement with respect to product Y is
(A) It gives a positive Tollens test and is a
functional isomer of X.
(B) It gives a positive Tollens test and is a
geometrical isomer of X.
(C) It gives a positive iodoform test and is a
functional isomer of X.
(D) It gives a positive iodoform test and is a
geometrical isomer of X.
Page 10
White precipitate
excess
H2/Lindlar's
catalst
35.
Ans.
M1, Q and R, respectively are
(A) Zn 2+, kCN and HCl (B) Ni2+, HCl and KCN
(C) Cd2+, KCN and HCl (D) Co2+, HCl and KCN
(B)
M1
2+
Ni
KCN [Ni(CN) ]2–
4
2
dsp sq.
planar
Ni2+ HCl
[NiCl4]2–
3
sp (tetrahedral)
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M2
Zn2+ + KCN
[Zn(CN)4]
2–
3
(Tetrahedral sp )
Zn
2+
+ HCl
(S)
[Zn(Cl4]2– + H+
3
(Tetrahedral sp )
OH–
Zn2+ + KOH  Zn(OH)2 

(white ppt.)
[Zn(OH)4]2–
(color less solution)
36.
Ans.
37.
(B) Na2HPO4
(D) KOH
Match each coordination compound in List-I
with an appropriate pair of characteristics
from List-II and select the correct answer
using the code given below the lists.
{en = H2NCH2CH2NH2; atomic numbers : Ti =
22, Cr = 24, Co = 27; Pt = 78}
List-I
(P) [Cr(NH3)4Cl2]Cl
(Q)[Ti(H2O)5Cl (NO3)2]
(R) [Pt(en)(NH3)Cl]NO3]
(S) [Co(NH3)4(NO3)2]NO3
Ans.
38.
Reagent S is
(A) K4[Fe(CN)6]
(C) K2CrO4
(D)
List-II
(1) Paramagnetic and exhibits ionisation isomerism
(2) Diamagnetic and exhibits cis-trans isomerism
(3) Paramagnetic and exhibits cis-trans isomerism
(4) Diamagnetic and exhibits ionisation isomerism
P
Q
R
S
(A)
4
2
3
1
(B)
3
1
4
2
(C)
2
1
3
4
(D)
1
3
4
2
B
Match the orbital overlap figure shown in ListI with the description given in List-II and
select the correct answer using the code
given below the lists.
List-I
Ans.
39.
List-II
(1) p–d  antibonding
(2) d–d  bonding
(3) p–d  bonding
(4) d–d  antibonding
P
Q
R
S
(A)
2
1
3
4
(B)
4
3
1
2
(C)
2
3
1
4
(D)
4
1
3
2
C
Different possible thermal decomposition
pathways for perosysters are shown below.
match each opathway form List with an appropriate structure from List II and select
the correct answer using the code given below the lists.
List-I
(P) Pathway P
(Q) Pathway Q
(R) Pathway R
(S) Pathway S
List-II
(1)
(2)
(P)
(3)
(Q)
(4)
(R)
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List-II
1. Scheme I
(i) KMnO4, HO–, heat (ii) H+, H2O
(iii) SoCl2 (iv) NH3
Code :
Sol.
P
Q
R
S
(A)
1
3
4
2
(B)
2
4
3
1
(C)
4
1
2
3
(D)
3
2
1
4
\
3. Scheme III
(i) red hot iron, 873 K
(ii) fuming HNO3, H2SO4 heat
(iii) H2S. NH3 (iv) NaNO2, H2SO4
(v) hydrolysis
A
Compound (I) follow the pathway (P) and
form formaldehyde as carbonyl. It remove CO2
in I step
Compound (II) follow the pathway (S) and
form formaldehyde as carbonyl. It remove CO2
in II step
Compound (III) follow the pathway (Q) and
does not form carbonyl. It remove CO2 in I
step
Compound (IV) follow the pathway (R) and
does not form carbonyl. It remove CO2 in II
step
40.
Match the four starting materials (P, Q, R, S)
given in List I with the corresponding reaction schemes (I, II, III, IV) provided in List-II
and select the correct answer using the code
given below the lists.
2. Scheme II
(i) Sn/HCl (ii) CH3COCl (iii) Conc. H2SO4
(iv) HNO3 (v) dil. H2SO4, heat (vi) HO–
Sol.
4. Scheme IV
(i) conc. H2SO4, 60ºC
(ii) conc. HNO3, conc. H2SO4
(iii)dil. H2SO4, heat
Code :
P
Q
R
S
(A)
1
4
2
3
(B)
3
1
4
2
(C)
3
4
2
1
(D)
4
1
3
2
C
(P) Acetylene
Benzene
Dinitrobenzene
m-nitro aniline
List-I
(P)
m-nitro benzene diazonium ion
(Q)
(Q)
m-nitro phenol
(C6H5NO3)
(R)
(S)
Page 12
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(R)
(S)
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MATHEMATICS
m4 + m2 – 2 = 0
 (m2 – 1) (m2 + 2) = 0
 m = ±1
R(2, 4) & S(2, –4)
P(–1, 1) Q(–1, –1)
SECTION – A
Single Correct
41.
In a triangle the sum of two sides is x and
the product of the same two sides is y.
If x2 – c2 = y, where c is the third side of the
triangle, then the ratio of the in-radius to
the circum-radius of the triangle is
3y
(A)
2x(x  c)
(C)
Sol.
43.
3y
(B)
2c(x  c)
3y
4x(x  c)
(D)
3y
4c(x  c)
B
a + b = x, ab = y
(a + b)2 – c2 = ab
Sol.
1
2
 cos C =
 C = 120°
also x – C =
Now
= 3·
42.
Sol.
y
xc
44.
(s  c) 3 · 3
r
(s  c) tan 60

=
c
c
R
2 sin120
y
1
·
2(x  c) c
The common tangents to the circle x2 + y2 =
2 and the parabola y2 = 8x touch the circle
at the points P, Q and the parabola at the
points R, S. Then the area of the quadrilateral PQRS is
(A) 3
(B) 6
(C) 9
(D) 15
D
y = mx +
00
2
m
m2  1
2
m
Sol.
R
Three boys and two girls stand in a queue.
The probability, that the number of boys
ahead of every girl is at least one more than
the number of girls ahead of her, is
P
A
(A)
1
2
(B)
1
3
(C)
2
3
(D)
3
4
B
 2
4
2
2
m (m2  1)
Six cards and six envelopes are numbered 1,
2, 3, 4, 5, 6 and cards are to be placed in
envelopes so that each envelope contains
exactly one card and no card is placed in the
envelope bearing the same number and
moreover the card numbered 1 is always
placed in envelope numbered 2. Then the
number of ways it can be done is
(A) 264
(B) 265
(C) 53
(D) 67
C
Required ways
1
1
1
1
1

6 ! 




2
!
3
!
4
!
5
!
6
! 

=
5
360  120  30  6  1
265
=
=
= 53
5
5
45.
 a 2a 
 a  2a 
R  2 , m  , S 2 , m 
m

m

1
1
(PQ + RS) · AB = (2 + 8)· 3
2
2
The quadratic equation p(x) = 0 with real
coefficients has purely imaginary roots.
Then the equation
p(p(x)) = 0 has
(A) only purely imaginary roots
(B) all real roots
(C) two real and two purely imaginary roots
(D) neither real nor purely imaginary roots
A
p(p(x)) = 0

p(x) = purely imaginary roots

No real value of x will satisfy

For real x, LHS is purely real
Area =
Q
S
Sol.
A
BGBGB BGBBG
BBGBG BBBGG
5 (3!  2 !) 1

5!
2
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Probability =
Page 14
BBGGB
1
46.
Let f : [0, 2]  R be a function which is
continuous on [0, 2] and is differentiable on
x2
(0, 2) with f(0) = 1. Let F(x) =  f( t) dt for
2x
 1 x2 dx
I.F.  e 2
I.F. = e
1
ln (1 x2 )
2

0
Sol.
x  [0, 2]. If F'(x) = f'(x) for all x  (0, 2),
then F(2) equals
(A) e2 – 1
(B) e4 – 1
(C) e – 1
(D) e4
B
F(x) =

x2
f( t ) dt
0

1  x2
x 4  2x
1  x2
1  x2 dx
x5
y 1  x2 =
+ x2
5
y=
differentiating
f'(x) = f(x) . 2x

y 1  x2 =
=
y=
f' (x)
 2x
f(x)
x5  5x2
5 1  x2
x5  5x2
5 1  x2

3 /2
n f(x) = x2 + k
f(0) = 1  k = 0
3 /2
x5

y=
 3 /2

dx +
5 1  x2
 3 /2
5x2
5 1  x2
dx
2
f(x) = ex
3 /2
F(x) =

x2
=2
t
e dt
0
2
=2
2
The function y = f(x) is the solution of the
x  2x
xy
dy
+ 2
=
dx
x –1
1 – x2
3
2
in (–1, 1) satisfying f(0) = 0. Then
(A)

3
–
3
2
(B)

3
–
3
4
(C)

3
–
6
4
(D)

3
–
6
2

1  x2

= 2 


3 /2


= 2 

3 /2

dx
3 /2
(1  x 2 ) dx 
0


0
(1  x 2 ) dx 
0
1
1  x2

dx 




3

 f(x) dx
–
is
dx
x2  1  1
0
4
Sol.
5 1  x2
3 /2
(ex  1) = e4 – 1
F(2) = Lim
x 2
differential equation

0
F(x) = [ ex – 1]
47.
5x2
3
2
3 /2


1

x

1  x2  sin1 x 
 
= 2  
 2
2
3
0


 


  3 1      
=2 
3
 4 2 6 
 

2

3
–
+
3
3
4
B
=–
x 4  2x
 x 
dy

– 
y
=
2
dx
1  x 
1  x2

3
–
3
4
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48.
Sol.
For x  (0, ), the equation sinx + 2sin 2x –
sin 3x = 3 has
(A) infinitely many solutions
(B) three solutions
(C) one solution
(D) no solution
D
sin x + 4 sin x cos x – 3 sin x + 4 sin3 x – 3 = 0
4 sin x (cos x + 1 – cos2 x) – 2 sin x – 3 = 0
1

 cos x  
2


2
+
Sol.
3
(cosecx – 1) = 0
4
Paragraph
Hence No solution
49.
Paragraph for Question Nos. 51 to 52

2
The following integral  (2 cos ec x)17 dx is

4
equal to
log(1  2)
2(eu  e–u )16 du
log(1  2)
(eu  e–u )17 du
(A) 0
(B) 0
log(1  2)
(C) 0
log(1 2)
(D) 0
Sol.
51.
2(eu – e–u )16 du
Sol.
x
cot
= eu
2
x = 2 cot–1 eu
2eu
1  e2u
I=

ln ( 2  1)

(C)
57
105
(D)
1
2
B
1, 2, 3
1, 2, 3, 4, 5,
Box (1)
Box (2)
=
2eu
17
 1  e2u 


 eu 


 2eu du 
. 
2u 

 1e

2(eu  eu)16 du
Coefficient of x 11 in the expansion of
(1 + x2)4 (1 + x3)7(1 + x4)12 is
(A) 1051
(B) 1106
(C) 1113
(D) 1120
Page 16
53
105
+ EEO
2 3 4
2 2 3
1 3 3
1 2 4
. .
+ . . +
. . + . .
3 5 7
3 5 7
3 5 7
3 5 7
52.
0
50.
(B)
A card is drawn from each Box
OOO +
OEE + EOE
du
log ( 2  1)
=
29
105
Box (3)
1  e2u
0
(A)
1, 2, 3, 4, 5, 6, 7
1
x
1  2u
1  tan2
2
e
cosec x =
=
x
2
2 tan
2
eu
=
Box 1 con t ai n s t h ree cards beari n g
numbers 1, 2, 3; box 2 contains five cards
bearing number 1, 2, 3, 4, 5; and box 3 contains seven cards bearing numbers 1, 2, 3,
4, 5, 6, 7. A card is drawn from each of the
boxes. Let xi be the number on the card drawn
from the ith box, i = 1, 2, 3.
The probability that x1 + x2 + x3 are in an
arithmetic progression, is
(eu – e–u )17 du
A
dx =
C
Given expression is
(1 + 4x2 + 6x4 + 4x6 + x8).
(1 + 7C1 x3 + 7C2 x6 + 7C3x9 +.....).
(1 + 12C1 x4 + 12C2 x8 + .......)
Coefficient of
x11 = 1 . 7C1 . 12C2 + 4 . 7C3 . 1 + 6 . 7C1 .
12C + 1. 7C .1
1
1
= 462 + 140 + 504 + 7
= 1113
Sol.
53
105
The probability that x1, x2, x3 are in an arithmetic progression, is
(A)
9
105
(B)
10
105
(C)
11
105
(D)
7
105
C
The only cases are
111, 123, 135, 147, 222, 234, 246, 321, 333,
345, 357
Required Probability =
11
11
=
357
105
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Paragraph for Question Nos. 53 to 54
53.
Let a, r, s, t be nonzero real number. Let
P(at2, 2at), Q(ar2, 2ar) and S(as2, 2as) be
distinct points on the parabola y2 = 4ax. Suppose that PQ is the focal chord and lines QR
and PK are parallel, where K is the point (2a, 0).
The value of r is
(A) –
(C)
Sol.
54.
If st = 1, then the tangent at P and the
normal at S to the parabola meet at a point
whose ordinate is
(A)
(C)
(t2  1)2
2t3
a(t2  1)2
t
2
1
t
(B)
t 1
t
Sol.
(D)
(D)
3
a(t2  1)2
2t3
a(t2  2)2
t3
C
x
+ at .....(1)
t
Normal at S is y = – sx + 2as + as3
Tangent at P is y =
2
1
t
(B)
t –1
t
D
Now st = 1 so s =
1
t
we get
at)
at ,2
(
P
)
r
2 2a
r,
2
a
R(
a
x
2a
+
+ 3
t
t
t
Adding (1) & (2)
y=–
.....(2)
k(2a,0)
F(a,0)
Q a

 t 2 , 2a 
t 

y = at +
=
a
2a
+ 3
t
t
at 4  2at2  a
t3
2
S(as , 2as)
mPK =
0  2at
2a  at2

Paragraph for Question Nos. 55 to 56
Given that for each a  (0, 1),
1h
Lim
h  0
a
(1  t)a  1 dt
exists. Let this limit be g(a). In addition,
it is given that the function g(a) is
differentiable on (0, 1).
55.
1
2
r  t
t
t
t2  1
t
t
h
1
r
t2  2
t

2
2t
r=
t3
2t
t2  2
2
2t
 2
1
t 2
r
t
1
r=t–
t
t3
a(t2  1)2
=
2
mQR=  1
,
r
t
a(t 4  2t2  1)
=
1
The value of g   is
2
(A) 
(C)
Sol.
(B) 2

2
(D)

4
A
1h
g(a) = hLim
 0
t
a
(1  t)a1 dt
h
1h
g(1/2) = hLim
 0
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t
1 / 2
(1  t)1 / 2 dt
h
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1h
= hLim
 0
1
=
1

t(1  t)
h
Matrix Match Type
dt
Matching List Type (Only One Option Correct)
57.
List I
List II
(P) The number of polynomials 1. 8
f(x) with non-negative
integer coefficients of
degree  2, satisfying
dt

t(1  t)
0
t = sin2 
dt = 2 sin  cos  d
 /2
=
2 sin  cos  d

sin  cos 

0
= 2

2 d
=

2
(C) –
(Q)
= 1, is
The number of points in
which f(x) = sin(x2) + cos(x2)
attains its maximum value is
3x2
2
(R)
–2
(S)
 1

1+x 
  21 cos 2x log 
dx 

 –

 1–x 
 2

equal
 1

1

x


  2 cos 2x log 
dx 
 0

1 – x 



(1  ex )
dx equals

2
(D) 0
1 h
g(a) =
Lim
t
h  0
a
=
t
h
a
(1  t)a 1 dt
0
1
 (t
g'(a) =
a
.ln t (1 – t)a–1
Sol.
(P)
P
Q
(A)
3
2
(B)
2
3
(C)
3
2
(D)
2
3
D
2
f(x) = ax + bx
0
+
R
4
4
1
1
S
1
1
4
4
1
t–a
(1 –
t)a–1
n (1 – t)) dt
1
g'(1/2) =
4. 0
(1  t)a 1 dt
1
 (t
1 / 2
 ax3 bx2 


 1
2 
 3
0
.ln t (1 – t)–1/2
0
+ t–1/2 (1 – t)–1/2 n (1 – t)) dt
1

3. 4
(B) 
D
=
2. 2
0
1
The value of g'   is
2
(A)
Sol.
 /2
the interval [ – 13, 13 ] at
 /2
0
56.
1
f(0) = 0 and 0 f(x) dx
t 1 / 2 (1 – t)–1/2 (–ln t + ln (1 – t))dt
a b
 1
3 2
a
b
1
3
2
0
........(1)
1
g'(1/2)=
 (1  t)
1 / 2
0
Adding (1) and (2)
g'(1/2) = 0
Page 18
a 2b

3
2
t–1/2 (–ln (1 – t)+ln t ) dt
........(2)
b=2–
N=
2a
 2 b
3
2a
3
b  2a
3
Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.)
a0b2
a3b0
(Q)
2
[ 13, 13] : f(x) = sin x + cos x
 9
,
4
4
2
x =
I=

3x2
2
2 1
2I =
I= 2
Sol.
9
4

,±
4
x=±
(R)
2

2
2

2
0
e
x

2
3ex x2
2
ex  1

(P)
P
Q
R
S
(A)
4
3
2
1
(B)
2
4
3
1
(C)
4
3
1
2
(D)
2
4
1
3
A

3
–1
y = sin(3cos x) 
2
 1x




–1
1  x2 y' = 3 sin(3 cos x)
3x2 dx
2x
3x2
y" –
2
1 x
2 1  x2
y'  9
cos(3 cos1 x)
1  x2
2
 3x3 
I =  3  = 8

0
numerator is odd function
=0
(S)
58.
(P)
1
(Q)

d y(x)
2


 
ak · ak 1
List II
1. 1
A1
dy(x) 
an
2. 2
2
n
a1
O
A2
a2
a3
(Q)
A3
a5
A4
A5

n 1

x2
y2
+
= 1 is perpendicular
6
3
to the line x + y = 8, then
the value of h is
Number of positive
solutions satisfying the
equation
 1 
 1 
tan–1 
+ tan–1 


 2x  1 
 4x  1 

 (a
k
, then the
minimum value of n is
If the normal from the
point P(h, 1) on the ellipse
 2 
= tan–1  2  is
x 
1  x2
2
x
Then y(x) (x – 1)

dx 
dx2

equal
Let A1, A2,...., An (n > 2)
be the vertices of a regular
polygon of n sides with its
centre at the origin. Let

ak be the position vector
of the point Ak, k = 1, 2,...,n.
 
n–1
If k 1 ak  ak 1 =
n–1
k 1
9y

(x – 1) y" + y' = 9y
3
.
2
2

(S)
1  x2
List I
Let y(x) = cos(3cos–1 x),
x  [–1,1], x  ±
(R)
(1  x2 )y " xy '

 ak 1 ) 
k 1
3. 8
n 1

 (a
k

·ak 1 )
k 1


2
(n – 1) | ak || ak 1 | sin
n


2
= (n – 1) | ak || ak 1 | cos
n
4. 9
 tan
2
=1
n
2 

n
4
n8
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(R)
y+x=8
slope of Normal = + 1




 1 


1
1
2
–1 
  tan 

x=–
tan  4
8



3


1


1



 3

 3

3x
dy
=–
6y
dx
tan
3 
1  3 

  tan 

 1 
 5 
–1 
–5
59.
f4 : R  [0, ) be defined by
Normal (h, 1)

slope at (h, 1)
is
| x | if x <0
f1 (x) =  x
if x  0
 e
6
=1
3h
f2 (x) =x2;
h =2

(S)
1
1


 2x  1  4x  1 
–1 

tan 
1
1 
1




2x  1 4x  1 

4x  1  2x  1
2
8x  6x  1  x

2
x
if x <0
 f2 (f1(x))
f3 (x) = 
f
(f
(x))

1
if x  0
2 1
(P)
f4 is
1. onto but not one-one
(Q)
f3 is
2. neither continuous nor one-one
(R)
f2of1 is
3. differentiable but not one-one
(S)
f2 is
4. continuous and one-one
2
6x  2
2
 3
2x(4x  3) x
3x  1 2
2

 3x + x = 8x + 6
4x  3 x
Sol.
x = 3, x = 
(P)
–1  2

 
9
R.H.S. tan
R
S
(A)
3
1
4
2
(B)
1
3
4
2
(C)
3
1
2
4
(D)
1
3
2
4
D
x x  0 

f1(x) =  x
e x  0 
x=3
tan
Q
f4 : R  [0, )
2
3
 1 1 



1
 13  7 
1  1 
1
–1    tan 
 tan  7 13   tan1 


tan
1
1
7
13
 
 
 90 
 1  

7 13 

P
f1 : R  R, f2 : [0, )  R, f3 : R  R
2
3x – 7x – 6 = 0
2
3x – 9x + 2x – 6 = 0
3x(x – 3) + 2(x – 3) = 0
Page 20
Let f1 : R  R, f2 : [0, )  R, f3 : R  R and
 f2 (f1(x))
x  0
f4(x) = f (f (x))  1 x  0
2 1

2
x  0
 x


= 2x
e  1 x  0
onto but not one-one
–1  2

 
9
 
Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.)
y
(Q)
(P)
f3(x) =
2k
10
Zk = 10
x
th
roots of unity
 P is true
(Q)
f3  differentiable but not one-one.
 x2
f2 0 f1(x) =  2x
e
(R)
i
Zk = e
x  0 

x  0 
k  {1, 2, 3, ...... 9}
i
2k
2
i (k 1)
Zk
e 10
10


e
Z= Z
2
i
1
e 10
Z obuiously has a solution in the set of
complex numbers. False
(R)
neither continuous
nor one-one
(S)
f2(x) = x
2
(x – z1)(x–z2)(x–z3)......(x–z9) =
y
x0
10
(x –1) = (x–1)(x–z1)(x–z2) ...... (x – z9)
y=x
x10  1
x 1
2
2
3
9
= 1 + x + x + x + ....... + x
x
Now put x = 1
continuous and one-one
(1 – z1) (1 – z2) ....... (1 – z9)
= 1 + 1 + 1 ....... + 1
 |1 – z1| |1 – z2| ......... |1 – z9| = 10
 2k 
 2k 
Let zk = cos 
+ i sin 

 ; k = 1,2,...,9.
 10 
 10 
60.
(P)
List I
List II
For each zk there
1. True
exists a zj such that

| 1  z1 || 1  z2 | ....... | 1  z9 |
1
10
9
(S)
1–
2k
 cos 10
k 1
zk · zj = 1
(Q)
There exists a
2. False
k  {1,2,...,9} such

2
4
6
18 
1 –  cos 10  cos 10  cos 10  ...........  cos 10 


that z1 · z = zk has no
solution z in the set of
complex numbers.
(R)
| 1 – z1 || 1 – z2 | ...| 1 – z9 |
equal
10
3. 1
(S)
 2k 
9
1 – k 1 cos 
 equal
 10 
4. 2
Sol.
P
Q
R
S
(A)
1
2
4
3
(B)
2
1
3
4
(C)
1
2
3
4
(D)
2
1
4
3
9


 sin 10 cos  


1– 

 = 1 + 1 = 2
 sin
10


C
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