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IIT JEE ADVANCED – 2013 / Paper-1
JEE ADVANCED
-2013
[1]
Paper - I
PHYSICS
Part I: Physics
Section – I (Single Correct Answer Type)
This section contains 10 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of
which ONLY ONE is correct.
1.
The diameter of a cylinder is measured using a Vernier callipers with no zero error. It is found that the zero
of the Vernier scale lies between 5.10 cm and 5.15 cm of the main scale. The Vernier scale has 50
divisions equivalent to 2.45 cm. The 24th division of the Vernier scale exactly coincides with one of the
main scale divisions. The diameter of the cylinder is
(A) 5.112 cm
(B) 5.124 cm
(C) 5.136 cm
(D) 5.148 cm
Ans. [B]
Value of one
Vernier division
Value of main
scale division
V
2.45
cm = 0.049 cm
50
, S  0.05 cm
Least count, V C  S  V  0.00 1 cm

b
g
.  24 0.001  5124
.
cm
Hence, diameter  510
2.
A ray of light travelling in the direction
along the direction
(A) 30o
d
d
i
1 
i  3 j is incident on a plane mirror. After reflection, it travels
2
i
1 
i  3 j . The angle of incidence is
2
(B) 45o
(C) 60o
Ans. [A] Angle  between the incident and reflected ray will be
FG 1 IJ FG 1 IJ  FG 3 IJ FG  3 IJ
H 2K H 2K H 2 K H 2 K   1
cos 
2
b1gb1g

  60 0
Hence, angle made by incident ray with normal,   30 0
(D) 75o
[2]
IIT JEE ADVANCED – 2013 / Paper-1
PHYSICS
In the Young's double slit experiment using a monochromatic light of wavelength  , the path difference (in
terms of an integer n) corresponding to any point having half the peak intensity is
3.
g 2
b
b
(A) 2n  1
Ans. [B] Intensity at a point, I  I m cos2
So, for I 
b
b
FG IJ
H K
FG  IJ
H 2K
Im

1
or
, cos2

2
2
2
g 8
(C) 2n  1
b
b
g 16
(D) 2n  1
g 2  2 L
  2n  1
g 4
L  2n  1
or
4.
g 4
(B) 2n  1
Two non-reactive monoatomic ideal gases have their atomic masses in the ratio 2 : 3. The ratio of their
partial pressures, when enclosed in a vessel kept at a constant temperature, is 4 : 3. The ratio of their
densities is
(A) 1 : 4
(B) 1 : 2
(C) 6 : 9
(D) 8 : 9
Ans. [D]
From pM 0  RT , p 

(as T = constant)
M0
p1  1 M 02 8



p2  2 M 01 9
5.
Two rectangular blocks, having identical dimensions, can be arranged either in configuration I or in
configuration II as shown in the figure. One of the blocks has thermal conductivity k and the other 2k .
The temperature difference between the ends along the x-axis is the same in both the configurations. It
takes 9 s to transport a certain amount of heat from the hot end to the cold end in the configuration I. The
time to transport the same amount of heat in the configuraiton II is
(A) 2.0 s
Ans. [A] RI 
(B) 3.0
L
L
3L


kA 2 kA 2 kA
1
kA 2 kA 3kA
L



 RII 
RII
L
L
L
3kA
Q  Ht 
so
T
tt R
R
t1 RI
9

  t2  2s
t2 RII 2
(C) 4.5 s
(D) 6.0 s
[3]
IIT JEE ADVANCED – 2013 / Paper-1
PHYSICS
6.
A pulse of light of duration 100 ns is absorbed completely by a small object initially at rest. Power of the
pulse is 30 mW and the speed of light is 3 × 108 m s–1. The final momentum of the object is
(A) 0.3  1017 kg ms1
(B) 10
.  10 17 kg ms1 (C) 3.0  1017 kg ms1
(D) 9.0  1017 kg ms1
Ans. [B] The total energy absorbed,
E  Pt  30  10 3  100  10 9  3  10 9 J
Hence momentum delivered i.e. the final momentum of object,
p
7.
E 3  10 9

 10
.  10 17 kg m s 1
c
3  108
A particle of mass m is projected from the ground with an initial speed u0 at an angle  with the horizontal.
At the highest point of its trajectory, it makes a completelyinelastic collision with another identical particle,
which was thrown vertically upward from the ground with the same initial speed u0 . The angle that the
composite system makes with the horizontal immediately after the collision is




(A)
(B)  
(C)  
(D)
4
4
4
2
Ans. [A] At the highest point,

v2 

v1  u cosi
L
eu  2 ghj j  MMu
N
2
2
Fu
 2 gG
H
2
sin 2 
2g
I OP
JK PQ
1
2
j  u cos j
Applying conservation of linear momentum,
 
 v v
v
v  1 2 so tan   2  1    450
2
v1
8.
OP
y h
c x  y h PQ
stant of appropriate dimensions), when the particle is taken from the point ba, 0g to the point b0, a g along
The work done on a particle of mass m by a force, K
LM
MN c x
x
2
2
i 
3/ 2
y
2
2 3/ 2
j
(K being a con-
a circular path of radius a about the origin in the x  y plane is
(A)
2K
a
(B)

Ans. [D] Here F  K
LM xi  yj
MM e x  y j
N
2
2 3/ 2
K
a
OP Kr
PP  r
Q
3
(C)
K
2a
(D) 0

i.e., at all points F is acting radially outwards from origin while particle is
 
moving tengentially. Thus F  v and work done is zero.
[4]
IIT JEE ADVANCED – 2013 / Paper-1
PHYSICS
9.
One end of a horizontal thick copper wire of length 2L and radius 2R is welded to an end of another
horizontal thin copper wire of length L and radius R. When the arrangement is stretched by appyling forces
at two ends, the ratio of the elongation in the thin wire to the that in the thick wire is
(A) 0.25
(B) 0.50
(C) 2.00
(D) 4.00
Ans. [C] Both rod are joined end to end, hence tension is same in both elongation,  
F
, so
AY
1 1 A2 1


  0.50
 2  2 A1 2

10.
 2
2
1
The image of an object, formed by a plano-convex lens at a distance of 8 m behind the lens, is real and is
one third the size of the object. The wavelength of light inside the lens is
2
times the wavelength in free
3
space. The radius of the curved surface of the lens is
(A) 1 m
(B) 2 m
(C) 3 m
Ans. [C] Here, v  8, n 
1

3
 and m   . So u  24 cm, applying lens equation
3
0 2
b g FGH R1  R1 IJK
1
1
1

  n 1
8 24
f
b g
1 F 1I F 1 I
G J G J  R 3m
6 H 2K H RK
1
2
(D) 6 m
[5]
IIT JEE ADVANCED – 2013 / Paper-1
PHYSICS
Section–II Multiple Correct Answer Type
This section contains 5 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of
which ONE or MORE THAN ONE is / are correct.
11.
A horizontal stretched string, fixed at two ends, is vibrating in its fifth harmonic according to the equation,
b g b
g c
c
h
h
. , the correct statement(s) is
y x , t  0.01 m sin 62.8 m 1 x cos 628 s1 t . Assuming   314
(are)
(A) The number of nodes is 5.
(B) The length of the string is 0.25 m.
(C) The maximum displacement of the midpoint of the string, from its equilibrium position is 0.01 m.
(D) The fundamental frequency is 100 Hz.
Ans. [B, C] For fifth harmonic number of nodes are 6.
f 
5v
2
1

 L   0.25 m
2 L 628
4
midpoint of the string is antinode. So maximum displacement is 0.01 m
f0 
f
 20 Hz.
5
x=0
x=L
x=L/2
A solid sphere of radius R and density is attached to one end of a mass-less spring of force constant k.
The other end of the spring is connected to another solid sphere of radius R and density 3. The complete
arrangement is placed in a liquid of density2 and is allowed to reach equilibrium. The correct statement(s)
is (are)
12.
(A) the net elongation of the spring is
4R 3 g
.
3k
(C) the light sphere is partially submerged.
(B) the net elongation of the spring is
8R 3 g
3k
(D) the light sphere is completely submerged.
Ans. [A, D] The average density of complete system is equal to that of liquid so both the spheres will be completely
submerged. For equilibrium of system shown in figure,
b g
b g
Fsp  2  Vg  3 Vg
i.e.
Fsp  Vg and hence extension x 
Vg 4R 3 g

k
3k
[6]
IIT JEE ADVANCED – 2013 / Paper-1
PHYSICS

A particle of mass M and positive charge Q, moving with a constant velocity u1  4i ms1 , enters a region
of uniform static magnetic field normal to the x-y plane. The region of the magnetic field extends from
x = 0 to x = L for all values of y. After passing through this region, the particle emerges on the other side

1
after 10 milliseconds with a velocity u2  2 3i  j ms . The correct statement(s) is (are)
13.
d
i
(A) The direction of the magnetic field is -z direction.
(B) The direction of the magnetic field is +z direction.
(C)
The magnitude of the magnetic field
50M
units.
3Q
(D) The magnitude of the magnetic field is
100M
units.
3Q
Ans. [A, C] From the given situation it is apparent that the magnetic field is in - z direction.
v2
T
Particle deflects by 30° so duration is
.
12
FG 2M IJ  1  10  10
H qB K 12

B
or
o
30
3
y
50M
units.
3Q
x
v1
Two non-conducting solid spheres of radii R and 2R, having uniform volume charge densities 1 and 2
respectively, touch each other. The net electric field at a distance 2R from the centre of the smaller sphere,
14.
1
along the line joining the centres of the spheres, is zero. The ratio  can be
2
(B) 
(A) – 4
32
25
(C)
32
25
(D) 4
Ans. [B, D] The situation described can correspond to two situations as shown in figure. For
first,
for
E p1  0, field of both must be equal and opposite i.e. both must have same polarity of
P1
charge
i.e.
LM
N
1
4
 R 3
3
b g
4  0 2 R
OP
Q R
2
3 0
2
2R

 1 4
2
Similarly in second case, for fields to cancel, polarity of charge must be opposite on
them.
LM 
N
1
4
 R 3
3
b g
4  0 2 R

1
32

2
25
OP LM 
QN
2
2
b g OPQ
b5Rg
4
  2R
3
4  0
2
3
P2
2R
5R
R
[7]
IIT JEE ADVANCED – 2013 / Paper-1
PHYSICS
15.
In the circuit shown in the figure, there are two parallel plate capacitors each of capacitance C. The switch
S1 is pressed first to fully charge the capacitor C1 and then released. The switch S2 is then pressed to
charge the capacitor C2. After some time, S2 is released and then S3 is pressed. After some time,
S1
S2
C1
S3
C2
2V0
V0
(A) the charge on the upper plate of C1 is 2CV0.
(B)
the charge on the upper plate of C1 is CV0.
(C)
the charge on the upper plate of C2 is 0.
(D) the charge on the upper plate of C2 is – CV0.
Ans. [B, D] On pressing S1, C1 gets fully charged with potential 2V0. Now on releasing S1 and pressing S2, the charge of C1
gets redistributed to get same potential on C2 and C1 given by
V
b g V
C1 2V0
C1  C2
0
Thus charge on upper plates of C1 and C2 = + CV0 and on lower is – CV0. Now on releasing S2 and pressing S3, C1
becomes isolated and charge on its plates gets unchanged while due to battery C2 now gets charged by potential
V0 in a manner that its upper plate now has – CV0 charge
[8]
IIT JEE ADVANCED – 2013 / Paper-1
PHYSICS
Section – III: Integer Answer Type
This section contains 5 questions. The answer to each of the questions is a single-digit integer, ranging from
0 to 9. (both enclusive)
16.
The work functions of Silver and Sodium are 4.6 and 2.3 eV, respectively. The ratio of the slope of the
stopping potential versus frequency plot for Silver to that of Sodium is ______
Ans. [1]
hv    eVo or Vo 
h

v
e
e
slope of graph for both metals is same

FG h IJ
H eK
Ratio of slopes is 1.
A freshly prepared sample of a radioisotope of half-life 1386 s has activity 103 disintegrations per second.
Given that ln 2 = 0.693, the fraction of the initial number of nuclei (expressed in nearest integer percentage)
that will decay in the first 80 s after preparation of the sample is______
17.
Ans. [4] T1/ 2  1386

t  80 s  T1/ 2 , so the activity remains nearly constant for the given duration. Percentage of nuclei decayed
FG
H
IJ
K
N
At
ln 2

 100 
 t 100  4
N0
A/
T
18.
A particle of mass 0.2 kg is moving in one dimension under a force that delivers a constant power 0.5 W
to the particle. If the initial speed (in m s–1) of the particle is zero, the speed (in m s–1) after 5 s is_______
1
2
Ans. [5] Pt  K  mv 2 or
or
19.
0.5  5 
b g
1
0.2 v 2
2
v = 5 m / s.
A uniform circular disc of mass 50 kg and radius 0.4 m is rotating with an angular velocity of 10 rad s–1
[9]
IIT JEE ADVANCED – 2013 / Paper-1
PHYSICS
about its own axis, which is vertical. Two uniform circular rings, each of mass 6.25 kg and radius 0.2 m,
are gently placed symmetrically on the disc in such a manner that they are touching each other along the
axis of the disc and are horizontal. Assume that the friction is large enough such that the rings are at rest
relative to the disc and the system rotates about the original axis. The new angular velocity (in rad s–1) of
the system is_________
Ans. [8]
Applying angular momentum conservation
o
M
m
F MR I  F MR
GH 2 JK GH 2
2
0
2
e
I
jJK
 2 2mr 2  '
or   = 8 rad / s.
20.
A bob of mass m, suspended by a string of length l1 is given minimum velocity required to complete a full
circle in the vertical plane. At the highest point, it collides elastically with another bob of mass m suspended
by a string of length l2 , which is initially at rest. Both the strings are mass-less and inextensible. If the
second bob, after collision acquires the minimum speed required to complete a full circle in the vertical
l1
plane, the ratio l is________.
2
Ans. [5] At the highest point, speed of bob 1 is v1  g1 . For 1  D, elastic collision between objects of equal masses,
velocities get interchanged.
v2  gl1  5gl2 (for complete circle),
or
l1
 5.
l2
[10]
IIT JEE – 2013 / Paper-I
CHEMISTRY
Part II
Section–I : Straight Correct Answer Type
This section contains 10 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out
of which ONLY ONE is correct.
21.
The compound that does NOT liberate CO 2 , on treatment with aqueous sodium bicarbonate solution, is
(A) Benzoic acid
(C) Salicylic acid
Ans: (D)
22.
(B) Benzenesulphonic acid
(D) Carbolic acid (Phenol)
Concentrated nitric acid upon long standing, turns yellow-brown due to the formation of
(A) NO
(B) NO2
(C) N 2O
(D) N 2 O4
Ans: (D)
23.
Methylene blue, form its aqueous solution, is adsorbed on activated charocal at 25C . For this process,
the correct statement is
(A) The adsorption requires activation at 25C
(B) The adsorption is accompanied by a decrease in enthalpy.
(C) The adsorption increases with increase of temperature.
(D) The adsorption is irreversible.
Sol: (B)
24.
Sulfide ores are common for the metals
(A) Ag , Cu and Pb
(B) Ag, Cu and Sn
Sol: (A)
25.
(C) Ag, Mg and Pb
(D) Al, Cu and Pb
The arrangement of X  ions around A  ion in solid AX is given in the figure (not drawn to scale). If the
radius of X  is 250 pm, the radius of A  is
(A) 104 pm
Sol: (A) r 
e
(B) 125 pm
j
2  1 r  104 pm
(C) 183 pm
(D) 57 pm
[11]
IIT JEE – 2013 / Paper-I
CHEMISTRY
26.
Upon treatment with ammoniacal H 2S , the metal ion that precipitates as a sulfide is
b g
b g
(A) Fe III
bg
(B) Al III
(C) Mg II
bg
(D) Zn II
Sol: (D)
27.
The standard enthalpies of formation of CO 2 (g) H 2O ( ) and glucose (s) at 25C are
400 kJ / mol,300 kJ / mol and 1300 kJ / mol , respectively. The standard enthalpy of combustion
per gram of glucose at 25C is
(A)  2900 kJ
Sol: (C)
(B)  2900 kJ
(C)  1611
. kJ
(D)  16.11 kJ
C6 H12 O 6 (s)  6O 2 (g) 
 6CO 2 (g)  6H 2 O ( )
b
g b
g b
g
H oC  6  400  6 300  1300
 2900 kJ / mol

2900
kJ / g
180
 1611
. kJ / g
28.
Consider the following complex ions, P, Q and R.
P  FeF6
3
b g
, Q  V H 2O
6
2
b g
and R  Fe H 2O
2
6
The correct order of the complex ions, according to their spin-only magnetic moment values (in B.M.) is
(A) R  Q  P
Sol: (B)
(B) Q  R  P
(C) R  P  Q
(D) Q  P  R
[12]
IIT JEE – 2013 / Paper-I
CHEMISTRY
29.
In the reaction
P Q 
 R  S
the time taken for 75% reaction of P is twice the time taken for 50% reaction of P. The concentration of Q
varies with reaction time as shown in the figure. The overall order of the reaction is
(A) 2
Sol: (D)
30.
(B) 3
(C) 0
(D) 1
is independent of intial concentration of P. First order with respect to P and zero order with
respect to Q. (From given figure)
t 1/ 2
KI in acetone, undergoes SN 2 reaction with each of P, Q R and S. The rates of the reaction vary as
H 3C  Cl
P
Cl
Q
Cl
R
(A) P  Q  R  S
(B) S  P  R  Q
(C) P  R  Q  S
(D) R  P  S  Q
Sol: (B)
S
[13]
IIT JEE – 2013 / Paper-I
CHEMISTRY
Section–II
Multiple Correct Answer(s) Type
This section contains 5 multiple choice questions. Each question four 4 choices (A), (B), (C) and (D), out of
which ONE or MORE are correct.
31.
The pairs (s) of coordination complexes/ions exhibiting the same kind of isomerism is (are)
b g
Cob NH g Cl
b g
Ptb NH g b H OgCl
(A) Cr NH 3 5 Cl Cl 2 and Cr NH 3 4 Cl 2 Cl
(B)
3 4
(C) CoBr2Cl 2
2

2
and
3 2
and PtBr2Cl 2

2
2
b g bNO g Cl and PtbNH g Cl Br
(D) Pt NH 3
3
3
3 3
Sol: (B, D)
32.
Among P, Q, R and S, the aromatic compound(s) is/are
AlCl 3
 

P
NaH
 

Q
b NH g CO
(A) P
Sol: (A, B, C, D)
4 2
3
100


115C
R
HCl
 

S
(B) Q
(C) R
(D) S
[14]
IIT JEE – 2013 / Paper-I
CHEMISTRY
33.
The hyperconjugative stabilities of tert-butyl cation and 2-butene, respectively, are due to
(A)   p (empty) and    * electron delocalisations.
(B)    * and    electron delocalisations
(C)   p (filled) and    electron delocalisations.
(D) p (filled)   * and    * electron delocalisations
Sol: (A)
34.
Benzene and naphthalene form an ideal solution at room temperature. For this process, the true statement(s)
is (are)
(A) G is positive
(B) S system is positive
(C) Ssurroundings  0
(D) H  0
Sol: (B, C, D)
35.
b g
b
g b
g
The initial rate of hydrolysis of methyl acetate 1M by a weak acid HA , 1M is HA , 1M of that of
b
g
a strong acid HX, 1M , at 250C. The Ka of HA is
(B) 1  10 5
(A) 1  104
Sol: (A)
H
H
HA


(C) 1  106
1
100
HX

H

Ka 
HA
 10
.  10 2 M
H A
HA
e1.0  10 je1.0  10 j  1  10
2
~

2
1
4
(D) 1  10 3
[15]
IIT JEE – 2013 / Paper-I
CHEMISTRY
Section–III
Integer Answer Type
This section contains 5 questions. The answer to each is a single-digit integer, ranging from 0 to 9 (both
inclusive).
36.
The total number of carboxylic acid groups in the product P is

. H 3O , 
21

.O
3
3. H 2O2
Sol: (2)
37.
A tetrapeptide has COOH groups on alanine. This produces glycine (Gly), valine (Val), phenyl alanine
(Phe) and alanine (Ala), On complete hydrolysis. For this tetrapeptide, the number of possible sequences
(primary structures) wih NH 2 group attached to a chiral center is
Sol: (4)
38.
Phe – Gly – Val – Ala
Phe – Val – Gly– Ala
Val – Phe – Gly– Ala
Val – Gly – Phe – Ala
EDTA 4 is ethylenediaminetetraacetate ion. The total number of N  Co  O bond angles in
b
Co EDTA
g
1
complex ion is
Sol: (8)
39.
The total number of lone pairs of electrons in melamine is
Sol: (6)
40.
The atomic masses of He and Ne are 4 and 20 amu, respectively. The value of the Broglie wavelenth of He
gas at 73 C is "M" times that of the de Broglie wavelength of Ne at 7270C M is
Sol: (5)

M


MV
 He

 Ne

3 KT
M
M


3 K MT
M Ne TNe


M He THe
20 1000

5
4 200
[16]
IIT JEE ADVANCED – 2013 / Paper-I
MATHEMATICS
PART III MATHEMATICS
Section – 1 (Only One option correct Type)
This section contains 10 multiple choice questions Each question has four choices (A), (B), (C) and (D), out
of which ONLY ONE is correct.
41.
Let complex number  and
b
1
lie on circles x  x 0

g  by  y g
2
b
2
 r 2 and x  x 0
0
g  by  y g
2
0
2
 4r 2 ,
2
respectively. If z 0  x 0  iy 0 satisfies the equation 2 z 0  r 2  2 , then  
(A)
Sol.
1
2
1
2
1
7
(C)
1
3
(D)
[C]
z  z0
2
 r2

  z0
2
z  z0
2
 2r 2

1
 z0

2
also

1   z 0  2 z 0  cos   4 r 2  . . . (ii)
(i) – (ii)   1  z 0
e
F1  r  2 I e   1j  r e1  4  j
GH 2 JK
2
2
1  z0

42.
(B)
2
je
   z 0  2 z 0  cos   r 2 . . . (i)
 4r 2

2
2
j e
 1  r2 1 4 
2
j
1
2
2
2
2
  1  1 4   1   2  8 
2
e
2
2
 r2
j
1

2
2
 z0  2 z 0
e
2
7
2
1
2
j
2
2
 
1
cos   4 r 2

e1   j  r e1  4  j
2
2
2
2
1
7
Four persons independenly solve a certain problem correctly with probabilities
1 3 1 1
, , , . Then the
2 4 4 8
probability that the problem is solved correctly by at least one of them is
(A)
Sol.
[A]
235
256
b
(B)
g
21
256
b gb gb gb g
P A  B  C  D  1  P A ' P B' P C' P D'
 1
1 1 3 7
21 235
    1

2 4 4 8
256 256
(C)
3
256
2
(D)
253
256
[17]
IIT JEE ADVANCED – 2013 / Paper-I
MATHEMATICS
43.
LM 1 , 1OP  R (the set of all real numbers) be a positive, non-constant and differentiable function
N2 Q
F 1I
such that f ' b xg  2 f b xg and f G J  1 . Then the value of z f b xgdx lies in the interval
H 2K
Let f :
1
1
2
b
(A) 2e  1, 2e
Sol.
[D]


also
g
b
bg bg
d
ee f bxgj  0
dx
f ' x  2f x
e
bg
f x e
2
(C)
FG e  1 , e  1IJ
H2
K

FG 1 IJ 1
H 2 K f FG IJ
2 x
bg
bg
1
e
 0 f x e
2 x 1
bg
 f x  e 2 x 1
z zbg z
1
bg
f x 0

1

0
z
1
2
44.
Le
f b xg dx  M
MN 2
1
2
zbg
1
 0  f x dx 
1
2
1
2
b
1
2
e 1
2
g
The number of points in  ,  , for which x 2  x sin x  cos x  0, is
(A) 6
Sol.
OP
PQ
2 x 1 1
1
0 dx  f x dx  e 2 x 1dx
1
2
1
2 x
2 x
2 x
 e f x 
H 2K
FG
H
(D) 0,
bg bg
 e f ' b x g  2e f b x g  0
e f b xg is strictly decreasing
 f ' x  2f x  0
2 x
2 x
g
(B) e  1, 2e  1
[C]
(B) 4
(C) 2
bg
f ' b xg  2x  b x cos x  sin xg  sin x  xb2  cos xg
Let f x  x 2  x sin x  cos x
f’
f
bg
f 0  1
bg
also f is continuous and lim I x  
x

bg
f x  0 for exactly two values of x.
+
–
0
(D) 0
IJ
K
e 1
2
[18]
IIT JEE ADVANCED – 2013 / Paper-I
MATHEMATICS
45.
d
(A) 4
Sol.
LM  OP is
N 2Q
2 d 2  1i
The area enclosed by the curves y  sin x  cos x and y  cos x  sin x over the interval 0,
[B]
i
2 1
(B) 2 2

2
A
z cb
d
g
i
d
2 1
(C) 2
i
2 1
(D) 2
h
cos x  sin x  cos x  sin x dx
0

4

z cb
g

4
h
cos x  sin x  cos x  sin x dx  2
0
g b
gh
cos x  sin x  cos x  sin x dx
0

4
z
b
 4 sin x dx  4  cos x
g

4
0
FG
H
 4 1
0
46.
z cb
IJ  2 2 e
2K
1
j
2 1
FG  IJ . Let the slope of the curve at each point (x,
H 6K
A curve passes through the point 1,
FG IJ
H K
1
F yI
(A) sinG J  log x 
H xK
2
F 2y I
(C) secG J  log x  2
H xK
y
y
 sec
, x  0 . Then the equation of the curve is
x
x
Sol.
[A]
dy y
y
  sec
dx x
x

x
dt
 sec t
dx
 cos t dt 

FG y IJ  log x  2
H xK
1
F 2y I
(D) cosG J  log x 
H xK
2
(B) cosec
by  tx g
dx
x
sin t  ln x  k
x  1, y 

6
FG y IJ  ln y  k
H xK
 sin
 k
1
2
FG y IJ  ln x  1
H xK
2
 sin
y) be
[19]
IIT JEE ADVANCED – 2013 / Paper-I
MATHEMATICS
F  cot FG1   2kIJ I is
GH
H
K JK
23
47.
The value of cot
n
1
n 1
(A)
23
25
(B)
[B]
25
23
(C)
(D)
24
23
n
1
cot
n 1
23
k 1
23
1
n 1
e
1
1
n 1
j
 cot tan 1 24  tan 1 1  cot tan 1
48.
23
24
F
F
II
GH  cot GH1   2kJK JK
F
I F
I
 cot G  cot c1  nb n  1ghJ  cot G  e tan b n  1g  tan njJ
H
K H
K
23
Sol.
k 1
23 25

25 23
For a  b  c  0, the distance between (1, 1) and the point of intersection of the lines ax  by  c  0
and bx  ay  c  0 is less than 2 2. Then
(A) a  b  c  0
Sol.
[A]

49.
Intersection of given lines
FG  c , c IJ
H a  b a  bK
c
1
ab
(C) a  b  c  0
(B) a  b  c  0

FG
H
2 1
(D) a  b  c  0
IJ
K
c
2 2
ab
 abc
Perpendiculars are drawn from points on the line
 abc  0
x  2 y 1 z

 to the plane x  y  z  3. the feet of
2
1
3
perpendiculars lie on the line
(A)
Sol.
[D]
x y 1 z  2


5
8
13
(B)
x y 1 z  2


2
3
5
(C)
x y 1 z  2


4
3
7
i.e.
FG1,  5 , 9 IJ
H 2 2K
Intersection of line and plane
x  2 y 1 z
2  1  3 3

 

2
1
3
2 1 3 2
Foot of perpendicular of point (–2, –1, 0)
x  2 y 1 2
2  1  3

 
1
1
1
11 1
Equation of projection line is given by
x0
y 1
z2


1 0  5  1 9  2
2
2

x y 1 z  2


2
7
5
i.e. (0, 1, 2)
(D)
x y 1 z  2


2
7
5
[20]
IIT JEE ADVANCED – 2013 / Paper-I
MATHEMATICS
50.

Let PQ  3i  j  2 k and SQ  i  3j  4 k determine diagonals of a parallelogram PQRS and


PT  i  2 j  3k be another vector. Then the volume of the parallepiped determined by the vectors



PT, PQ and PS is
(A) 5
Sol.
(B) 20



(C) 10



[C]
PR  PQ  PS  3i  j  2 k , SQ  PQ  PS  i  3j  4 k

3i  j  2 k j  e i  3j  4 k j
e
PQ 
 2 i  j  3k ,

2
volume of parallelopiped
1
L
O
 M PT PQ PSP  2
N
Q 1



2
(D) 30

PS  i  2 j  k
3
1 3  10
2
1
SECTION – 2 : (One or more options correct Type)
The section contains 5 multiple choice questions. Each questions has four choices (A), (B), (C) and (D) out
of which ONE or MORE are correct
4n
51.
b g
Let S n   1
k 1
b g
k k 1
2
k 2 . Then S n can take value(s)
(A) 1056
(B) 1088
4n
Sol.
[A, D] S n 
 b1g
b g
k k 1
2
(C) 1120
k 2  12  2 2  32  4 2  52  62  ...
k 1
n 1

r 0
n 1

n 1
n 1
n 1
 b4r  1g   b4r  2g   b4r  3g   b4r  4g
2
r 0
 b32r  20g  32 
r 0
S8  1056
2
r 0
2
r 0
bn  1gn  20bng  4nb4n  1g
2
S 9  1332
2
(D) 1332
[21]
IIT JEE ADVANCED – 2013 / Paper-I
MATHEMATICS
52.
For 3 × 3 matrices M and N, which of the following statement(s) is(are) NOT correct?
N T M N is symmetric or skew symmetric, according as M is symmetric or skew symmetric
(A)
(B) M N – N M is skew symmetric for all symmetric matrices M and N
(C) M N is symmetric for all symmetric matrices M and N
(D) (adj M) (adj N) = adj (M N) for all invertible matrices M and N
Sol.
e
[C, D](A) N T M N
b g
(C) M N
T
j
T
b
g  N M M
(D) badjMgbadjN g  adj b N Mg
 N T MT N
(B) M N  N M
 NT MT  N M
bg
T
T
T
T
NT  N M  M N
bg
53.
Let f x  x sin x, x  0 . Then for all natural numbers n, f ' x vanishes at
Sol.
FG 1 IJ
H 2K
(C) a unique point in the interval b n, n  1g
[B, C] f ' bA g  sin x  x cos x  0
(A) a unique point in the interval n, n 
FG
H
IJ
K
1
(B) a unique point in the interval n  , n  1
2
b
g
(D) two points in the interval n, n  1
 tan x  x  0
FG
H
x  n, n 
FG
H
IJ
K
1
2 , tan x  x  0
IJ
K
1
x  n  , n  1 , tan x  x  0 (once)
2
54.
A rectangular sheet of fixed perimeter with sides having their lengths in the ratio 8 : 15 is converted into an
open rectangular box by folding after removing squares of equal area from all four corners. If the total area
of removed squares is 100, the resulting box has maximum volume. Then the lengths of the sides of the
rectangular sheet are
(A) 24
(B) 32
(C) 45
(D) 60
Sol.
[A, C]
55.
A line l passing through the origin is perpendicular to the lines
b g b g b g
: b3  2sgi  b3  2sgj  b2  sgk ,    s  
1 : 3  t i  1  2 t j  4  2 t k ,    t  
2
Then, the coordinate(s) of the point(s) on l 2 at a distance of 17 from the point of intersection of l and l1
is(are)
(A)
FG 7 , 7 , 5IJ
H 3 3 3K
b
g
(B) 1,  1, 0
(C) (1, 1, 1)
(D)
FG 7 , 7 , 8 IJ
H 9 9 9K
[22]
IIT JEE ADVANCED – 2013 / Paper-I
MATHEMATICS
Sol.
[B, D] d. r of line 
i
j k
1 2 2  2 i  3j  2 k
2 2 1
Equation of line 
x
y
z


2 3 2
The point of intersection of  and 1 is (2, –3, 2).
b
g
Any point of  2 is 3  2s, 3  2s, 2  s
b3  2s  2g  b3  2s  3g  b2  3  2g  17
 b9s  10gbs  2g  0
9s  28s  20  0
2

2
2
 s
2
10
, 2
9
SECTION – 3 : (Integer value correct Tyep)
Integer Answer Type
This section contains 5 questions. The answer to each question is a single digit integer, ranging from 0 to 9
(both inclusive)
b g
56.
The coefficients of three consecutive terms of 1  x
Sol.
[6]

n 5
Cr :
br  1g
bn  r  5g
n 5
n 5


n 5
C r 1
C r 2
b r  2g
C r 1 :


n 5
C r  2  5 : 10 : 14
1
 2r  2  n  r  5
2
5

are in the ratio 5 : 10 : 14. Then n =
bn  5g!  br  1g!bn  4  r g!  1
r !b n  5  r g!
2
bn  5g!
 3r  n  3
bn  5g!  br  2g!bn  3  r g!  5
7
b gb
g
bn  5g!
10
14
b n  r  4g  7
5n  4b n  3g  6  0
n 5
r 1 ! n  r  4 !
 7r  14  5n  5r  20
 5n  12 r  6  0
 5n  4 n  12  6  0
 n6
[23]
IIT JEE ADVANCED – 2013 / Paper-I
MATHEMATICS
57.
A pack contains n cards numbered from 1 to n. Two consecutive numbered cards are removed from the
pack and the sum of the numbers on the remaining cards is 1224. If the smaller of the numbers on the
removed cards is k, then k  20 
Sol.
[5]
b g  1224  k  bk  1g
n n 1
2
By observation n  50

58.
50  51
 1224  k  k  1
2

k  25

k  20  5
Of the three independent events E1 , E 2 and E 3 , the probability that only E 1 occurs is , only occurs in
 and only E 3 occurs is  . Let the probability p that none of events E 1 , E 2 or E 3 occurs satisfy the
b
g
b
g
equations   2 p   and   3 p  2 . All the given probabilities are assumed to lie in the
interval (0, 1).
Pr obability of occurrence of E1

Pr obability of occurrence of E 3
Then
Sol.
[6]
Let probability of E1 , E 2 and E 3 is X1 , X 2 and X 3 respectively then
  X b1  X gb1  X g
b gb g
  X b1  X gb1  X g and
P  b1  X gb1  X gb1  X g
Now b  2gP  
cX b1  X gb1  X g  2X b1  X gb1  X ghb1  X gb1  X gb1  X g

 X b1  X gb1  X gX b1  X gb1  X g
  X1 1  X 2 1  X 3
3
1
1

l.ly

2
3
1
3
1
3
2
2
1
3
1
1
2
2
3
3
X1  2 X 2 . . . (i)
b  3 gP  2r
cX b1  X gb1  X g  3bX gb1  X gb1  X ghb1  X gb1  X gb1  X g
 2 X b1  X gb1  X gX b1  X gb1  X g
2
1
2

2
2
1
2
3
1
3
3
X 2  3X 3 . . . (ii)
using (i) and (ii)
X1  6 X 3 
X1
6
X3
3
1
1
2
2
1
2
3
3
[24]
IIT JEE ADVANCED – 2013 / Paper-I
MATHEMATICS
59.
A vertical line passing through the point (h, 0) intersects the ellipse
x2 y2

 1 at the points P and Q. Let
4
3
bg
bg
h
the tangents to the ellipse at P and Q meet at the point R. If  h  area of the triangle PQR, 1  max
1
2
bg
8
 1  8 2 
5
and  2  1min  h , then
2
Sol.
 h 1
 h 1
[9]
(h, k)
(4/h, 0)
O
2
½ 1
(h, –k)
bg
Max
IJ
K
1
4
 2k
h 
2
h
1  
FG 1 IJ , 
H 2K
8
5

60.
b gFGH
 h 
2
 1  8 2 
8 3
5

3 4  h2
2
F 4  h I  3 e4  h j
GH h JK 2h
3
2 2
2
bg
 min    1
8
5

F 1I
3G 4  J
H 4K
3
2
8
bg
3
3
2
3
2
15 15
3
8
 3 3  45  36  9
8
2
o
l qt
Consider the set of eight vectors V  ai  bj  ck : a , b, c  1, 1 . Three noncoplanar vectors can
be chosen from V in 2p ways. Then p is
Sol.
Consider a cube whose centre is at origin and of side length two unit and faces are parallel to coordinate axes then all of the vertices of cube represent the given eight vectors. Now three vectors of them will
be coplaner if we select two vertices along body diagonal and third vertex from remaining.
[5]
Number of ways to select 3 noncoplanar vectors  8 C3  4. 6 C1  32  25
[1]
JEE Advanced – 2013/ Paper - 2
PHYSICS
Part I
Section–I : One or More Options Correct Type
The Section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of
which ONE or MORE are correct.
1.
A steady current I flows along an infinitely long hollow cylindrical conductor of radius R. This cylinder is
placed coaxially inside an infinite solenoid of radius 2R. The solenoid has n turns per unit length and carries
a steady current I. Consider a point P at a distance r from the common axis. The correct statement(s) is are
(A) In the region 0 < r < R, the magnetic field is non-zero.
(B) In the region R < r < 2R, the magnetic field is along the common axis.
(C) In the region R < r < 2R, the magnetic field is tangential to the circle of radius r, centered on the axis.
(D) In the region r > 2R, the magnetic field is non-zero.
Ans. [A, C, D]
For cylinder, BC  0, r  R,

0I
 to axis r  R
2r
For solenoid, BS   0 ni along the axis, 0  r  2 R;
 0, r  2 R
Accordingly, options A, C and D are correct.
2.
Two vehicles, each moving with speed u on the same hoizontal straight road, are approaching each other.
Wind blows along the road with velocity w. One of these vehicles blows a whistle of frequency f1. An
observer in the other vehicle hears the frequency of the whistle to be f2. The speed of sound in still air is V.
The correct statement(s) is are
(A) If the wind blows from the observer to the source, f2 > f1.
(B) If the wind blows from the source to the observer, f2 > f1.
(C) If the wind blows from observer to the source, f2 > f1.
(D) If the wind blows from the source to the observer, f2 > f1.
Ans. [A, B] When wind blows from observer to source,
f2 
FG V  v
HV  v
IJ
K
u
f1  f1
w u
w
When wind blows from source to observer,
f2 
FG V  v
HV  v
w
w
IJ
K
u
f1  f1 .
u
[2]
JEE Advanced – 2013/ Paper - 2
PHYSICS
3.
Two bodies, each of mass M, are kept fixed with a separation 2L. A particle of mass m is projected from
the midpoint of the line joining their centres, perpendicular to the line. The gravitational constant is G. The
correct statement(s) is (are)
(A) The minimum initial velocity of the mass m to escape the gravitational field of the two bodies is
4
GM
.
L
GM
.
L
(C) The minimum initial velocity of the mass m to escape the gravitational field of the two bodies is
(B) The miminum velocity of the mass m to escape the gravitational field of the two bodies is 2
2GM
.
L
(D) The energy of the mass m remains constant.
Ans. [B, D] For m to escape, its total energy must become zero or positive
v
M
×
o m
M
2L
LM
N
OP
Q
i.e.
1 2
2GM
mv  m 
0
2
L
or
v2
GM
L
Also during the motion, its total energy remains constant.
4.
A particle of mass m is attached to one end of a mass-less spring of force constant k, lying on a frictionless
horizontal plane. The other end of the spring is fixed. The particle starts moving horizontally from its
equilibrium position at time t = 0 with an initial velocity u0. When the speed of the particle is 0.5 u0, it
collides elasticallywith a rigid wall. After this collision
(A) the speed of the particle when it returns to its equilibrium position is u0.
(B) the time at which the particle passes through the equilibrium position for the first time is t  
(C) the time at which the maximum compression of the spring occurs is t 
4
3
m
.
k
m
.
k
(D) the time at which the particle passes through the equilibrium position for the second time is t 
5
3
m
.
k
[3]
JEE Advanced – 2013/ Paper - 2
PHYSICS
Ans. [A, D] Since the particle collides elastically so its speed first after collision is 0.5 u0.
So option (A) is correct.
bg
u0
T
t  .
2
6
So the particle passes through the equilibrium position first time at
v t  u0 cos t 
t2
FG T IJ  T  2
H 6K 3 3
m
k
so option (B) is wrong.
Maximum compression of the spring occurs at
T T 7T 7 m
 

4 3 12
6 k
so option (C) is wrong.
Particle passes through the equilibrium position for the second time at
t
t
T T 5T 5
 

2 3
6
3
m
k
Hence, option (D) is correct.
5.
The figure below shows the variation of specific heat capacity (C) of a solid as a function of temprature
(T). The temperature is increased continuously from 0 to 500 K at a constant rate. Ignoring any volume
change, the following statement(s) is (are) correct to a reasonable approximation.
(A) the rate at which heat is absorbed in the range 0-100 K varies linearly with temperature T.
(B) heat absorbed in increasing the temperature from 0-100 K is less than the heat required for increasing
the temperature from 400-500 K.
(C) there is no change in the rate of heat absorption in the range 400-500 K.
(D) the rate of heat absorption increases in the range 200-300 K.
Ans. [A, B, C, D] From observing the figure, options B, C and D are correct. Also C is varying approximately linearly with
temperature in range 0-100 K.

dQ
dT
dT
 mC
and since
= constant
dt
dt
dt
dQ
 C and is varying linearly with temperature T.
dt
[4]
JEE Advanced – 2013/ Paper - 2
PHYSICS
6.
The radius of the orbit of an electron in a Hydrogen-like atom is 4.5 a0, where a0 is the Bohr radius. Its
3h
. It is given that h is Planck constant and R is Rydberg constant. The
2
possible wavelength(s), when the atom de-excites, is(are)
orbital angular momentum is
(A)
9
32R
Ans. [A, C] L 
(B)
9
16R
(C)
9
5R
(D)
4
3R
nh
suggests electron is in third orbit.
2
n2
a0  z = 2.
2
Hence possible transitions are
Also using rn 
(i)
(ii)
(iii)
LM OP
N Q
1
L1 1O
9
 4RM  P   
3  2,

5R
N4 9Q
1
L 1O
3
 4 R M1  P   
2  1,

R
N 4Q
3  1,
1
1
9
 4R 1


9 
32R
Hence [A, C] are correct.
7.
Using the expression 2d sin  = , one calculates the values of d by measuring the corresponding angles 
in the range 0 to 90°. The wavelength is exactly known and the error in is constant for all values of .
As increases from 0°,
(A) the absolute error in d remains constant.
(B) the absolute error in d increases.
(C) the fractional error in d remains constant.
(D) the fractional error in d decreases.

bg
d d

 cosec  cot 
Ans. [D] Given d  2 sin  so
d
2
Absolute error in d,
Fractional error in d,
b g 2 cosec cot d
d bd g
 cot d
d d 
d
Both decrease with increase in  .
[5]
JEE Advanced – 2013/ Paper - 2
PHYSICS
8.
Two non-conducting spheres of radii R1 and R2 and carrying uniform volume charge densities   and
  , respectively, are placed such that they partially overlap, as shown in the figure. At all points in the
overlapping region,
(A) the electrostatic field is zero.
(B) the electrostatic potential is constant.
(C) the electrostatic field is constant in magnitude. (D) the electrostatic field has same direction.
Ans. [C, D] The field at any point say P is given by
r1
P
r2
d

 r1
  r2
E

3 0
3 0
b g

 
 r1  r2
d


3 0
3 0
i.e. field is same at all points.
Hence, its magnitude and direction both remains same. As electric field is non-zero so potential is not constant.
[6]
JEE Advanced – 2013/ Paper - 2
PHYSICS
Section–II : Paragraph Type
This section contains 4 paragraphs each describing theory, excperiment, data etc. Eight questions relate to
four paragraphs with two questions on each paragraph. Each question of a paragraph has only one correct
answer among the four choices. (A), (B), (C) and (D).
Paragraph for Questions 9 and 10
A thermal power plant produces electric power of 600 kW at 4000 V, which is to be transported to a place
20 km away from the power plant for consumers usage. It can be transported either directly with a cable of large
current carrying capacity or by using a combination of step-up and step-down transformers at the two ends. The
drawback of the direct transmission is the large energydissipation. In the method usingtransformers, the dissipation
is much smaller. In this method, a step-up transformer is used at the plant side so that the current is reduced to a
smaller value. At the consumers end, a step-down transformer is used to supply power to the consumers at the
specified lower voltage. It is reasonable to assume that the power cable is purely resistive and the transformers
are ideal with a power factor unity. All the currents and voltages mentioned are rms values.
9.
If the direct transmission method with a cable of resistance 0.4  km 1 is used, the power dissipation
(in %) during transmission is
(A) 20
(B) 30
(C) 40
(D) 50
Ans. [B]
10.
In the method using the transformers, assume that the ratio of the number of turns in the primary to that in
the secondary in the step-up transformer is 1 : 10. If the power to the consumers has to be supplied at
200 V , the ratio of the number of turns in the primaryto that in the secondary in the step-down transformer
is
(A) 200 : 1
(B) 150 : 1
(C) 100 : 1
(D) 50 : 1
Ans. [A] For step up transformer at plant,
VP N P

VS
NS
VS  40,000V

At consumer side
40,000 N P

200
NS 
N P 200

NS
1
Also current flowing during transmission
iT 
P
VT

(assuming first transformer is 100% efficient)
600  10 3
 15A
40,000
b g b
2
g
Hence PLoss  iT 2 R  15  0.4  20  1800 W

% loss 
1800
600  103
 100%  30% .
[7]
JEE Advanced – 2013/ Paper - 2
PHYSICS
Paragraph for Questions 11 and 12
A small block of mass 1 kg is released from rest at the top of a rough track. The track is a circular arc of radius
40 m. The block slides along the track without toppling and a frictional force acts on it in the direction opposite
to the instantaneous velocity. The work done in overcoming the friction up to the point Q, as shown in the figure
below, is 150 J. (Take the acceleration due to gravity, g = 10 ms–2).
11.
The speed of the block when it reaches the point Q is
(A) 5 ms1
(B) 10 ms1
(C) 10 3 ms1
(D) 20 ms1
Ans. [B]
12.
The magnitude of the normal reaction that acts on the block at the point Q is
(A) 7.5 N
(B) 8.6 N
(C) 11.5 N
(D) 22.5 N
Ans. [A] For the motion till point Q, applying work energy theorem
b
g
1
mVQ 2  Wmg  W f  mg R sin 30  150 J
2
VQ  10 m / s.

For the circular motion,

N  mg sin 30 
mv 2
R
N  mg sin 30 
mv 2
 7.5N .
R
[8]
JEE Advanced – 2013/ Paper - 2
PHYSICS
Paragraph for Questions 13 and 14
The mass of a nucleus AZ X is less than the sum of the masses of (A – Z) number of neutrons and Z number of
protons in the nucleus. The energy equivalent to the corresponding mass difference is known as the binding
energy of nucleus. A heavy nucleus of mass M can break into two light nuclei of masses m1 and m2 only if
bm  m g  M . Also two light nuclei of masses m and m can undergo complete fusion and form a heavy
nucleus of mass M  only if bm  m g  M  . The masses of some neutral atoms are given in the table below.
1
2
3
3
2
1
6
3
152
64
H
Li
4
2
1
7
3
206
82
1007825
.
u
6.015123 u
H
Li
Gd 151919803
.
u
4
2.014102 u
7.016004 u
Pb 205.974455 u
3
1
70
30
209
83
H
Zn
3.016050 u
69.925325 u
Bi 208.980388 u
4
2
82
34
210
84
He
Se
4.002603 u
81916709
.
u
.
Po 209.982876 u
c1 u  932 MeV / c h
2
13.
The correct statement is
(A) The nucleus 63 Li can emit an alpha particle
210
(B) The nucleus 84
Po can emit a proton
(C) Deuteron and alpha particle can undergo complete fusion.
(D) The nuclei
70
30
can undergo complete fusion
Zn and 82
34 Se
Ans. [C]
14.
210
The kinetic energy (in keV ) of the alpha particle, when the nucleus 84
Po at rest undergoes alpha decay,
is
(A) 5319
(B)5422
(C) 5707
(D) 5818
Ans. [A] For  decay of
e
Q m
214
84 Po
214
84 Po
j  me
Now energy taken
away by a particle
j  me Hej c = 5422 KeV.
F A  4 IJ Q  5319 KeV
G
H AK
206
82 Pb
K
, energy released
4
2
2
For second part, only deutron and alpha particle undergo fission is feasible.
[9]
JEE Advanced – 2013/ Paper - 2
PHYSICS
Paragraph for Questions 15 and 16
A point charge Q is moving in a circular orbit of radius R in the x-y plane with an angular velocity  . This can be
Q
. A uniform magnetic field along the positive
2
z-axis is now switched on, which increases at a constant rate from 0 to B in one second. Assume that the radius
of the orbit remains constant. The application of the magnetic field induces an emf in the orbit. The induced emf
is defined as the work done by an induced electric field in moving a unit positive charge around a closed loop.
It is known that, for an orbiting charge, the magnetic dipole moment is proportional to the angular momentum
with a proportionality constant  .
considered as equivalent to a loop carrying a steady current.
15.
The magnitude of the induced electric field in the oribit at any instant of time during the time interval of the
magnetic field change is
(A)
BR
4
(B)
BR
2
(C) BR
(D) 2BR
Ans. [B]
16.
The change in the magnetic dipole moment associated with the orbit, at the end of the time interval of the
magnetic field change, is
(A)  BQR
BQR 2
(B) 
2
2
BQR 2
(C) 
2
(D)  BQR 2
Ans. [B] The time varying increasing magnetic field in + z direction produces induced electric field as shown. The magnitude
of induced electric field is E 
R dB RB

.
2 dt
2
Z
E
The angular impulse imported by induced electric field is equal to change in the angular momentum of moving charge
particle.
z
z
L  QER dt  Q
FG R dB IJ Rdt  QR B
H 2 dt K
2
As   L so   L  
2
QR 2 B
2
[10]
JEE Advanced – 2013/ Paper - 2
PHYSICS
Section – III
Matching List Type
This section contains 4 questions. Each question has four statements (A, B, C and D) given in Column I and
five statements (p, q, r, s and t) in Column II. Any given statement in Column I can have correct mathcing
with ONE or MORE statement(s) given in Column II. For example, if for a given question, statement B
matches with the statements given in q and r, then for that particular question, against statement B, darken the
bubbles corresponding to q and r in the ORS.
17.
Match List I of the nuclear processes with List II containing parent nucleus and one of the end products of
each process and then select the correct answer using the codes given below the lists:
List I
List II
O  15
7 N . . .
(P)
Alpha decay
(1)
15
8
(Q)
  decay
(q)
238
92
U
(R) Fission
(r)
185
83
Bi  184
82 Pb . . .
(S)
(s)
239
94
Pu  140
57 La . . .
Proton emission
P
Q
R
S
(A) 4
2
1
3
(B) 1
3
2
4
(C) 2
1
4
3
(D) 4
3
2
1
Ans. [C]
234
90
Th . . .
[11]
JEE Advanced – 2013/ Paper - 2
PHYSICS
18.
One mole of a monatomic ideal gas is taken along two cyclic processes E  F  G  E and
E  F  H  E as shown in the pV diagram. The processes involved are purely isochoric, isobaric,
isothermal or adiabatic
Match the paths in List I with the magnitudes of the work done in list II and select the correct answer using
the codes given below the lists:
Column I
Column II
GE
(1)
160 p0V0 ln2
(Q) G  H
(q)
36 p0V0
(R)
FH
(r)
24 p0V0
(S)
FG
(s)
31 p0V0
(P)
Codes
P
Q
R
S
(A) 4
3
2
1
(B) 4
3
1
2
(C) 3
1
2
4
(D) 1
3
2
4
Ans. [A]
[12]
JEE Advanced – 2013/ Paper - 2
PHYSICS
19.
Math List I with List II and select the correct answer using the codes given below the lists :
List I
List II
P.
Boltzmann constant
1.
ML2 T 1
Q.
Coefficient of viscosity
2.
ML1T 1
R.
Planck constant
3.
MLT 3 K 1
S.
Thermal conductivity
4.
ML2 T 2 K 1
Codes : P
(A)
3
(B)
3
(C)
4
(D)
4
Q
1
2
2
1
R
2
1
1
2
S
4
4
3
3
Ans. [C]
20.
A right angled prism of refractive index 1 is placed in a rectangular block of refractive indux 2, which is
surrounded bya medium of refractive index 3, as shown in the figure. A rayof light e enters the rectangular
block at normal incidence. Depending upon the relationships between 1, 2 and 3, it takes one of the four
possible paths ef , eg , eh or ei .
Match the paths in List I with conditions of refractive indices in List II and select the correct answer using
the codes given below the lists :
List I
List II
P.
1.
e f
1  2  2
Q.
e g
2.
 2   1 and  2   3
R.
eh
3.
1   2
S.
ei
Codes : P
(A)
2
(B)
1
(C)
4
4.
 2   1  2  2 and  2   3
Q
3
2
1
R
1
4
2
S
4
3
3
(D)
3
4
1
Ans. [D]
2
[13]
JEE Advanced – 2013/ Paper -2
CHEMISTRY
Part II
SECTION – I : One or more options correct Type
This section contains 8 multiple choice questions. Each question has fcour choices (A), (B), (C) and (D), out
of which ONE or MORE are correct.
21.
The major product(s) of the following (are)
OH
Br2 b 3.0 equivalents g
aqueous

 ?
SO3H
(A) P
Ans: (B)
22.
(B) Q
(C) R
(D) S
After completion of the reaction (I and II), the organic compound(s) in the reaction mixtures is (are)
Reaction I:
Reaction II :
Br2 b1.0 mol g
aqueous


/ NaOH
2 b1.0 mol g
Br

CH 3COOH
CHBr3
P
Q
R
S
(A) Reaction I : P and Reaction II : P
(B) Reaction I : U, acetone and Reaction II : Q acetone
(C) Reaction I : T, U, acetone and Reaction II : P
(D) Reaction I : R, acetone and Reaction II : P
Ans: (A, C)
T
U
[14]
JEE Advanced – 2013/ Paper -2
CHEMISTRY
23.
The Ksp of Ag 2 CrO 4 is 11
.  1012 at 298 K. The solubility(in mol/L) of Ag 2 CrO 4 in a 0.1 M AgNO 3 is
solution is
(A) 11
.  1011
Ans: (B)
CrO 24  
(B) 11
.  1010
K eq
Ag
24.
 2

11
.  10 12
. g
b010
2
 11
.  10 10
(C) 11
.  1012
(D) 11
.  109
M
In the following reaction, the product(s) formed is(are)
CHCl

3  ?
OH 
P
(A) P (major)
Ans: (B, D)
25.
Q
(B) Q (major)
R
(C) R (minor)
S
(D) S (major)
The carbon-based reduction method is NOT used for the extraction of
(A) tin from SnO 2
(B) iron from Fe 2 O 3
(C) aluminium from Al 2 O 3
(D) magnesium from MgCO 3  CaCO 3
Ans: (C, D)
26.
The thermal dissociation equilibrium of CaCO 3 ( s) is studied under different conditions.
CaCO 3 (s)
CaO ( s)  CO 2 (g)
For this equilibrium, the correct statement(s) is (are)
(A) H is dependent on T
(B) K is independent of the initial amount of CaCO 3
(C) K is dependent on the pressure of CO 2 at a given T
(D) H is independent of the catalyst, if any
Ans: (A, B, D)
[15]
JEE Advanced – 2013/ Paper -2
CHEMISTRY
27.
The correct statement(s) about O3 is (are)
(A) O  O bond lengths are equal.
(B) Thermal decomposition of O3 is endothermic
(C) O3 is diamagnetic in nature.
(D) O3 is diamagnetic in nature
Ans: (A, C, D)
28.
In the nuclear transmutation
9
4
Be  X 
 84 Be  Y
bX, Yg is (are)
(A) b  , ng
b g
(B) p, D
b g
(C) n, D
b g
(D)  , p
Ans: (A, B)
Section–II : Paragrah Type
This section contains 4 paragraphs each describing theory, experiment, data etc. Eight questions relate to
four paragraphs with two questions on each paragraph. Each question of a paragraph has only one correct
answer among the four choices (A), (B), (C) and (D).
Paragraph for Questions 29 and 30
P and Q are isomers of dicarboxylic acid C4 H 4 O 4 . Both decolorize Br2 / H 2 O . On heating, P forms the cyclic
anhydride.
Upon treatment with dilute alkaline KMnO 4 , P as wall as Q could produce one or more than one from S, T and
U.
S
29.
T
U
Compounds formed from P and Q are, respectively
(A) Optically active S and optically active pair (T, U)
(B) Optically inactive S and optically inactive pair (T, U)
(C) Optically active pair (T, U) and optically active S
(D) Optically inactive pair (T, U) and optically inactive S
Ans: (B)
[16]
JEE Advanced – 2013/ Paper -2
CHEMISTRY
30.
In the following reaction sequences V and W are, respectively
2 / Ni
Q H
 V

3 b anhydrous g
+ V AlCl


. Zn  Hg / HCl
1
 W
2 . H 3PO4
(A)
and
(B)
(C)
and
(D)
and
and
Ans: (A)
Paragraph for Questions 31 and 32
An aqueous solution of a mixture of two inorganic salts, when treated with dilute HCl, gave a precipitate (P) and
a filtrate (Q). The precipitate P was found to dissolve in hot water. The filtrate (Q) remained unchanged, when
treated with H 2S in a dilute mineral acid medium. However, it gave a precipitate (R) with H 2S in an ammoniacal
medium. The precipitate R gave a coloured solution (S), when treated with H 2 O 2 in an aqueous NaOH medium.
31.
The precipitate P contains
(A) Pb2
(B) Hg2
2
(C) Ag 
(D) Hg2
(C) ZnSO4
(D) Na 2 CrO 4
Ans: (A)
32.
The coloured solution S contains
b g
(A) Fe SO4
3
(B) CuSO4
Ans: (D)
Paragraph for Questions 33 and 34
The reactions of Cl 2 gas with cold-dilute and hot concentrated NaOH in water give sodium salts of two (different) oxoacids of chlorine, P and Q, respectively. The Cl 2 gas reacts with SO 2 gas, in presence of charcoal, to
give a product R. R reacts with white phosphorus to give a compounds S. On hydrolysis, S gives an oxoacid of
phosphorus, T.
[17]
JEE Advanced – 2013/ Paper -2
CHEMISTRY
33.
P and Q, respectively, are the sodium salts of
(A) hypochlorus and chloric acids
(C) chloric and perchloric acids
Ans: (A)
(B) hypochlorus and chlorus acids
(D) chloric and hypochlorus acids
SO 2  Cl 2 
 SO 2 Cl 2
10 SO 2 Cl 2 ()  P4 
 4 P Cl5  10 SO 2
PCl5  4 H 2 O 
 H 3PO 4  HCl
34.
R, S and T, respectively, are
(A) SO 2 Cl 2 , PCl5 and H 3PO4
(B) SO 2 Cl 2 , PCl 3 and H 3PO3
(C) SOCl 2 , PCl 3 and H 3PO2
(D) SOCl 2 , PCl5 and H 3PO4
Ans: (A)
Paragraph for Questions 35 and 36
A fixed mass 'm' of a gas is subjected to transformation of states from K to L to M to N and back to K as shown
in the figure
L
N
M
Pressure
K
Volume
35.
The succeeding operations that enable this transformation of states are
(A) Heating, cooling, heating, cooling
(B) Cooling, heating, cooling, heating
(C) Heating, cooling, cooling, heating
(D) Cooling, heating, heating, cooling
Ans: (C)
36.
The pair of isochoric processes among the transformation of states is
(A) K to L and L to M
(B) L to M and N to K
(C) L to M and M to N
(D) M to N and N to K
Ans: (B)
[18]
JEE Advanced – 2013/ Paper -2
CHEMISTRY
Section–III : Matching List Type
This section contains 4 multiple choice questions. Each question has matching lists. The codes for the lists
have choices (A), (B), (C) and (D), out of which ONLY ONE is correct.
37.
The standard reduction potential data at 250C is given below.
c
h
E c Fe , Feh  0.44 V
E cCu , Cuh  0.34 V;
E cCu , Cuh  0.52 V
E O bgg  4 H  4e  2 H O  1.23 V;
E O bgg  2 H O  4e  4OH  0.40 V
E cCr , Cr h  0.74 V;
E cCr , Cr h  0.91 V
E 0 Fe 3 , Fe 2   0.77 V;
0
2
0
2
0


0

2
2

0
2

2
0
3
0
3
Match E0 of the redox pair in list I with the values given in List II and select the correct answer using the
code given below the lists.
List I
List II
c
(P) E 0 Fe 3 , Fe
h

0
 4H  4OH 
(Q) E 4H 2 O 
c
E cCr
(R) E 0 Cu 2   Cu  2Cu 
(S)
0
3
, Cr 2 

h
h
Codes:
(A)
(B)
(C)
(D)
Ans: (D)
P
4
2
1
3
Q
1
3
2
4
R
2
4
3
1
S
3
1
4
2
1.
018
. V
2.
0.4 V
3.
0.04 V
4.
0.83 V
[19]
JEE Advanced – 2013/ Paper -2
CHEMISTRY
38.
An aqueous solution of X is added slowlyto an aqeuous solution of Y as shown in the List I. The variationin
conductivity of these reactions is given in List II. Match List I with List II and select the correct answer
using the code given below the lists.
List I
List II
b
g
(P) C2 H 5 3 N  CH 3COOH
X
Y
(Q) CH 3COOH  KOH
X
Y
(R) CH 3COOH  KOH
X
Y
(S) NaOH  HI
X
Y
Codes:
P
Q
R
(A)
3
4
2
(B)
4
3
2
(C)
2
3
4
(D)
1
4
3
1. Conductivity decreases and then increases
2. Conductivity increases and then does not change much
3. Conductivity increases and then does not change much
4. Conductivity does not change much and then incresaes
S
1
1
1
2
Ans: (A)
39.
Match the chemical conversions in List I with the appropriate reagents in List II and select the correct
answer using the code given below the lists.
List I
List II
bg b gb g
(P)
1. i Hg OAc , ii NaBH 4
(Q)
2. NaOEt
(R)
3. Et  Br
(S)
4. i BH 3 , ii H 2 O 2 / NaOH
bg
Codes:
(A)
(B)
(C)
(D)
Ans: (A)
P
2
3
2
3
Q
3
2
3
2
R
1
1
4
4
S
4
4
1
1
bg
[20]
JEE Advanced – 2013/ Paper -2
CHEMISTRY
40.
The unbalanced chemical reactions given in List I show missing reagent or condition (?) which are provided in List II. Match List I with List II and select the correct answer using the code given below the lists.
List I
List II
(P)
PbO2  H 2SO4 
? PbSO4  O2  other product
1. NO
(Q)
Na 2S2O3  H 2 O 
? NaHSO4  other product
2. I2
(R)
N2H4 
? N 2  other product
3. Warm
(S)
XeF2 
? Xe  other product
4. Cl 2
Codes:
(A)
(B)
(C)
(D)
Ans: (D)
P
4
3
1
3
Q
2
2
4
4
R
3
1
2
2
S
1
4
3
1
[21]
JEE ADVANCED – 2013 / Paper -2
MATHEMATICS
Part III
Section – 1 : (One or more options correct Type)
This section contains 8 multiple choice questions. Each question has fcour choices (A), (B), (C) and (D), out
of which ONE or MORE are correct.
41.
Two lines L1 : x  5,
y
z
y
z

and L 2 : x   , 
are coplanar. Then  can take value(s)
3   2
1 2
(A) 1
Sol.
[A,D]
(B) 2
(C) 3
(D) 4
x5
y
z


0
3   2
x
y
z


0
1 2  
Given lines are coplanar.


5
0
0
0
3 
2
0
1
2
0
b  5g b3  gb2  g  2  0
b  5ge  5  4j  0
2


b  5g 6    5  2  0
b  5gb  4gb  1g  0
2
  1, 4, 5
42.
Sol.
1
In a triangle PQR, P is the largest angle and cos P  . Further the incircle of the triangle touches the sides
3
PQ, QR and RP at N, L and M respectively, such that the lengths of PN, QL and RM are consecutive
even integers. Then possible length(s) of the side(s) of the triangle is(are)
(A) 16
(B) 18
(C) 24
(D) 22
[B, D ] cosP 
1
3
NP  s  p  2n  0
QL  s  q  2 n  2  0

RM  s  r  2 n  4  0
 s  6n  6
3s  2s  6n  6
p  4n  6, q  4n  4, r  4n  2
cos P 
1 q 2  r 2  p2

3
2qr
[22]
JEE ADVANCED – 2013 / Paper -2
MATHEMATICS
b
g b g b
b gb g
bn  1g  2
1

3 4b n  1gb2 n  1g
2
g  b2n  2g  b2n  1g  b2n  3g
4b n  1gb2 n  1g
2
4 n  4  4n  2  4n  6
1

3
2 4n  4 4n  2
2
2
2
2
b g b gb g
4b n  1gb2n  1g
2

4 n  1  4 n  4 2
2 n  1  3n  9  2 n  8  n  4
43.
For a R (the set of all real numbers), a  1,
c1  2  ... n h
1
lim

bn  1g bna  1g  bna  2g  ...bna  ng 60
a
a
a
a 1
n 
then a 
(A) 5
Sol.
[B, D]
(B) 7
lim
n 

 ra
n
a 1
IJ z
K
1
1
FG a  1
H 2
x a dx 
0
2
1
2a 2  3a  119  0

a 2a  17  7 2a  17  0
g b
(D)
17
2
1
60
1

b

15
2
1
60

a  7, a 
Sol.
FG1  1 IJ LMa  1 FG1  1 IJ OP
H nK N 2 H nKQ
b2a  1g  ba  1g  60
120  b2a  1gba  1g

44.
1
a 1
(C)
g
17
2
Circle(s) touching x–axis at a distance 3 from the origin and having an intercept of length 2 7 on y–axis
is(are)
(A) x 2  y 2  6x  8y  9  0
2
2
(B) x  y  6x  7 y  9  0
(C) x 2  y 2  6x  8y  9  0
(D) x 2  y 2  6x  7 y  9  0
[A, C] r  9  7  4
b
C  3,  4
g
[23]
JEE ADVANCED – 2013 / Paper -2
MATHEMATICS
equation of circle
bx  3g  by  4g
2

45.
2
 42
x 2  y 2  6x  8y  9  0
Let  be a complex cube root of unity with   1 and P  p ij be a n × n matrix with p ij   i  j . Then
P 2  0, when n =
(A) 57
(B) 55
(C) 58
(D) 56
Ans. [B,C,D]
LM 
MM 

PM
MM .
MM .
N
2
3
4
5
3
.....  n 1
.
4.
n 1
.
.
.
.
 2n
.
OP
PP
PP
PP
PQ
P2  0
(2)2 + (3)2 + (4)2 + (5)2 + (6)2 + . . . + (n+1)2  0
 + 1 + 2 +  + 1 + 2 ... 0
n must not be multiple of 3.
46.
bg
The function f x  2 x  x  2  x  2  2 x has a local minimum or a local maximum at x =
(A) –2
(B)
2
3
(C) 2
R| 2x  x  2  cbx  2g  2xh, x  2
||2x  x  2  bx  2  2xg,  2  x   23
2
f b xg  S
2 x  x  2  b x  2  2 x g,
x0
||
3
|| 2x 2xxx21bx bx222x2g,xg, 0xx2 2
T
R| 2x  4, x  2
||2x  4,  2  x   23
 S 4 x,  2  x  0
||
3
4x 0  x  2
|T 2x  4, x  2
(D)
Ans. [A, B]
Points of local maxima or minima are 2, 
2
,0
3
2
3
Y
–2
– 2/3
O
X
2
[24]
JEE ADVANCED – 2013 / Paper -2
MATHEMATICS
47.
3 i
2
w
Let
and
RS
T
H 2  z  C : Re z 
o
t
P  w n : n  1, 2 , 3,... .
Further
UV
W
RS
T
H 1  z  C : Re z 
1
2

2
(B)

6
(C)
2
3
(D)
5
6

Ans. [C, D]  
i
3i
e 6
2
Y
H2
H1
o
X
Total 12 points in P
Required angles = 00, 1200, 1500, 1800
48.
If 3x  4 x 1 , then x =
2 log 3 2
(A) 2 log 2  1
3
Ans. [A, B, C]
3x  4 x 1
b g
x log 2 3  x  1 2
x log 2 3  2x  2
2  x 2  log 2 3
x

2
2  log2 3
2 log 2 3
2 log 2 3  1
and
1
, where C is the set of all complex numbers. If z1  P  H 1 , z 2  P  H 2 and
2
O represents the origin, then z1Oz 2 
(A)
UV
W
2
(B) 2  log 3
2
1
(C) 1  log 3
4
2 log 2 3
(D) 2 log 3  1
2
[25]
JEE ADVANCED – 2013 / Paper -2
MATHEMATICS
Section – 2 : (Paragrah Type)
This section contains 4 paragraphs each describing theory, experiment, data etc. Eight questions relate to
four paragraphs with two questioons on each paragraph. Each question of a paragraph has only one correct
answer among the four cohices (A), (B), (C) and (D), out of which ONE ONE is correct.
Paragraph for Questions 49 and 50
Let PQ be a focal chord of the parabola y 2  4ax. The tangents to the parabola at P and Q meet a point lying
on the line y  2 x  a , a  0 .
49.
50.
Length of chord PQ is
(A) 7a
(C) 2a
(D) 3a
2
If chord PQ subtends an angle  at the vertex of y  4ax, then tan  
(A)
Sol.
(B) 5a
2
7
3
(B)
2
7
3
(C)
2
5
3
(D)
e
j
Qeat , 2at j
[B, D] P at 12 , 2at 1
2
2
2
For focal chord t 1t 2  1
b
c
gh
c
b
Intersectionof tangents at P and Q is at 1t 2 , a t 1  t 2 i.e.  a , a t 1  t 2
y  a

t 1  t 2  1
bt
g  bt  t g  4t t  b1g  4b1g  5
PQ  a e t  t j  c2b t  t gh  a 5b t  t g
1
 t2
2
2
1
2
1
2
2
1 2
2 2
2
2
1
2
2 2

2 t 2  t1
t1 t 2
2 5
tan  


2 2
t 1t 2  4
3
1 
t1 t 2
b
b  is obtuseg
2
1
g
2
 5a
gh
2
5
3
[26]
JEE ADVANCED – 2013 / Paper -2
MATHEMATICS
Paragraph for Questions 51 and 52
Let f : 0, 1  R (the set of all real numbers) be a function. Suppose the function f is twice differentiable,
bg
b g bg
bg bg
f 0  f 1  0 and satisifes f " x  2 f ' x  f x  e x , x  0, 1 .
51.
Which of the following is true for 0  x  1 ?
bg
(B) 
(A) 0  f x  
(C) 
bg
bg
1
1
f x 
2
2
bg
1
 f x 1
4
(D)   f x  0
bg bg bg
cf bxg  f bxghe  e cf bxg  f bxgh  1
d
e c f b xg  f b xgh  1
dx
Ans. [D] f  x  2f  x  f x  e x

or

x
x
x
d2
dx 2
ee . f bxgj  1
x
bg
bg
Let g x  e  x x then

but



52.
bg
gb0g  gb1g  0
gb xg  0  x  b0, 1g
e f b xg  0  x  b0, 1g
f b xg  0   x b0, 1g
d2
dx 2
cgbxgh  1,
g x is concave up
x
bg
If the function e  x f x assumes its minimum in the interval [0, 1] at x 
(A) f ' x  f x ,
bg bg
1
3
x
4
4
bg bg
1
4
(C) f ' x  f x , 0  x 
bg
FG 1 IJ
H K
Ans. [C] g  x  0, x  0, 4
1
, which of the following is true?
4
bg bg
1
4
bg bg
3
 x 1
4
(B) f ' x  f x , 0  x 
(D) f ' x  f x ,
[27]


JEE ADVANCED – 2013 / Paper -2
MATHEMATICS
F 1I
e f b xg  e f b xg  0, x  G 0, J
H 4K
F 1I
f b xg  f b xg , x  G 0, J
H 4K
x
x
Paragraph for Questions 53 and 54
A box B1 contains 1 white ball, 3 red balls and 2 black balls. Another box B2 contains 2 white balls, 3 red balls
and 4 black balls. A third box B3 contains 3 white balls, 4 red balls and 5 black balls.
53.
If 1 ball is drawn from each of the boxes B1 , B2 and B3 , the probability that all 3 drawn balls are of the
same colour is
(A)
82
648
90
648
(B)
(C)
558
648
(D)
566
648
Ans. [A] E: Three balls drawn are of same colour
bg
PE 


54.
1
Favourable case
Sample space
C1 .2 C1 .3 C1  3C1 .. 3C1. 4 C1  2 C1.4 C1 .5 C1
6
C1 . 9 C1
12
C1
82
648
If 2 balls are drawn (without replacement) from a randomly selected box and one of the balls is white and
the other ball is red, the probability that these 2 balls are drawn from box B2 is
(A)
Ans. [D]
116
181
(B)
126
181
(C)
E1: balls are drawn from B1
E 2 : balls are drawn from B2
E 3 : balls are drawn from B3
E : One white and one red ball is drawn
FG E IJ PdE i
HE K
FE I
PG J 
H E K PbE g. PF E I  PbE g . PF E I  PbE g . PF E I
GH E JK
GH E JK
GH E JK
P
2
2
2
1
2
3
1

1
3
F
GH
1
.
3
1
C13C1
6
C2
I  1F
JK 3 GH
F
GH
2
2
2
C1 . 3 C 1
9
C2
3
3
I
JK
I F
JK GH
C1 . C 1
1

9
3
C2
3
C1  C1
12
C2
4
I
JK

55
181
65
181
(D)
55
181
[28]
JEE ADVANCED – 2013 / Paper -2
MATHEMATICS
Paragraph for Questions 55 and 56
Let
S  S1  S2  S3 , where
m
r
S1  z  C : z  4
R|
S|
T
LM z  1  3i OP  0U|V
MN 1  3i PQ |W
l
q
S 2  z  C : Im
and S 3  z  C : Re z  0
55.
Area of S =
(A)
56.
10
3
(B)
20
3
(C)
16
3
(D)
32 
3
(B)
2 3
2
(C)
3 3
2
(D)
3 3
2
min 1  3i  z 
z S
(A)
2 3
2
F Z  1e1  i 3 j I
Ans. [B, C] ImG
GH 1  i 3 JJK  0
F Z IJ  1  0
ImG

H 1 i 3 K
If
Z  x  iy
then
Im


S
bx  iyg  0
e1  i 3j
F F x  iy IJ ei  i 3jIJ  0
ImG G
HH 4 K
K
o
P (1, –3)
yx 3  0
Required area S 
60
e j FGH 6 IJK
1 2
4
2
FG 5 IJ  20
H 6K 3
8
b g
Now Min of |1  3i  Z|  Min of | Z  1  3i |
= minimum distance between region S and point P (1, –3)
4, 0
y  x 3
[29]
JEE ADVANCED – 2013 / Paper -2
MATHEMATICS
= Perpendicular distance from (1, –3) to y  x 3  0

3  3 3  3

2
2
Section – 3 : (Matching list Type)
This section contains 4 multiple choice questions. Each question has matching lists. The codes for the lists
have choices (A), (B), (C) and (D), out of which ONLY ONE is correct.
57.
x4 y3 z3
x 1 y z  3
L2 :




,
and the planes
1
1
2
2
1
1
P1 : 7 x  y  2 z  3 , P2 : 3x  5y  6z  4. Let ax  by  cz  d be the equation of the plane passing
Consider the lines L1 :
through the point of intersection of lines L1 and L 2 , and perpendicular to planes P1 and P2 .
Match List – I with List – II and select the correct answer using the code given below the lists:
List I
List II
P.
a=
1.
13
Q.
b=
2.
–3
R.
c=
3.
1
S.
d=
4.
–2
Codes :
Sol.
P
Q
R
S
(A) 3
2
4
1
(B) 1
3
4
2
(C) 3
2
1
4
(D) 2
4
1
3
[A]
Point of intersection of L1 and L 2 is
b2  1,   ,   3g  b  4,   3, 2  3g

2  1    4 ,      3 and   3  2  3

2    3 ,      3
Solving 3  6
2

    3  1
[30]

JEE ADVANCED – 2013 / Paper -2
b
MATHEMATICS
g
Point of intersection of lines L1 and L 2  5,  2,  1
normal to req. plane
i
j
 7 1
k
2  16i  48j  32 k
3 5 6
d.r. of normal to plane
 a, b, c > is  1,  3,  2 
Hence equation of required plane is
x  3y  2 z  d
b
g
which passes through P 5,  2,  1
58.
b g b g

5  3 2  2 1  d  d  13

a  1, b  3, c  2 d  13
Match List – I with List – II and select the correct answer using the code given below the lists:
List I
P.
 

Volume of parallelepiped determined by vectors a , b and c is 2.
Then the volume of the parallelepiped determined by vectors
   
 
2 a  b ,3 b  c and c  a is
d id i
Q.
1.
100
2.
30
3.
24
4.
60
b g
 

Volume of parallelepiped determined by vectors a , b and c is 5.
Then the volume of the parallelepiped determined by vectors
   
 
3 a  b , b  c and 2 c  a is
d id i
R.
List II
b g


Area of a triangle with adjacent sides determined by vectors a and b is 20.
 
Then the volume of the parallelepiped determined by vectors 2a  3b and
 
a  b is
d
i
d i
S.


Area of parallelogram with adjacent sides determined by vectors a and b is 30.
 

Then the area of the parallelogram determined by vectors a  b and a is
d i
[31]
JEE ADVANCED – 2013 / Paper -2
MATHEMATICS
Codes :
Sol.
P
Q
R
S
(A) 4
2
3
1
(B) 2
3
1
4
(C) 3
4
1
2
(D) 1
4
3
2
[C]

 
 
2 ab 3 bc
Q
  
a b c 5

 
3 ab
 
e j e j bc  ag  6 a  b b  c c  a
e

 
6a bc
2
 6  2 2  24
    
  
b  c c  a  6  2 a b c  12  5  60
j e b  c j 2bc  a g  6 a  b
S
1 
 
a  b  20  a  b  40
2
1     1
  1
2a  b  3b  a  5 a  b  5  40  100
Req. area 
2
2
2
 
a  b  30

 

 
Area of required parallelogram  a  b  a  a  b  3 0
R
59.
 
a bc 2
P

e
j
b g
A line L : y  mx  3 meets y–axis at E 0, 3 and the arc of the parabola y 2  16x, 0  y  6 at the
b
g
b
g
b g
point F x 0 , y 0 . The tangent to the parabola at F x 0 , y 0 intersects the y–axis at G 0, y1 . The slope
m of the line L is chosen such that the area of the triangle EFG has a local maximum.
Match List – I with List – II and select the correct answer using the code given below the lists:
List I
List II
P.
m=
1.
1
2
Q.
Maximum area of EFG is
2.
4
R.
y0 
3.
2
S.
y1 
4.
1
[32]
JEE ADVANCED – 2013 / Paper -2
MATHEMATICS
Codes :
P
Q
R
S
(A) 4
1
2
3
(B) 3
4
1
2
(C) 1
3
2
4
(D) 1
3
4
2
yt  x  4 t 2
Ans. [A] tangent at F is
Y
G : x  0  y  4t
b g
hence G 0, 4 t
0
Area of triangle EFG 
1
0
2 2
4t
3
1
4t 1
8t 1
e b gj
 2 t b3  4 t g
dA
 12 tb1  2 t g
dt

1 2
4t 3  4t
2
(0, 3)
E
F (4t2, 8t)
G (0, 4t)
X
O
2

area is maximum when t = 1/ 2
so
y1  4 t  2
bx , y g  e4t , 8t j  b1, 4g
2
0
0
y0  4
Area of triangle EFG  2
FG 3  1 IJ  2FG 3  2 IJ  1
H 4 2K H 4 K 2
and y  mx  3 passes through (1, 4).

m=1
dA
dt
+
–
½
[33]
JEE ADVANCED – 2013 / Paper -2
MATHEMATICS
60.
Match List – I with List – II and select the correct answer using the code given below the lists:
List I
List II
P.
F 1 F cosctan yh  y sinctan yhI
GG y G cot sin y  tan sin y J
H H c h c hK
Q.
If cos x  cos y  cos z  0  sin x  sin y  sin z then possible
1
1
2
value fo cos
R.
1
2
I
y J
JK
4
1
1/ 2
takes value
1.
1 5
2 3
2.
2
xy
is
2
FG   xIJ cos 2x  sin x sin 2x sec x  cos x sin 2x sec x  cosFG   xIJ cos 2x
H4 K
H4 K
If cos
3.
1
2
4.
1
then possible value fo sec x is
S.
e
e d ij
j
1
2
1
If cot sin 1  x  sin tan x 6 , x  0 ,
then possible value fo x is
Codes :
P
Q
R
S
(A) 4
3
1
2
(B) 4
3
2
1
(C) 3
4
2
1
(D) 3
4
1
2
Ans. [B]
LM e j e j OP
MN cot esin yj  tanesin yj PQ
1
1
1 cos tan y  y sin tan y
[P]
y2
e
1
j
1
e
j
e
j
for
cos tan 1 y  y sin tan 1 y
let
y  tan then
e
j
cos tan 1 y  y sin tan 1 y =
2
 y4
[34]
JEE ADVANCED – 2013 / Paper -2
MATHEMATICS
bg
1
 sec   1  y 2
cos 
cos   tan  sin  
and
e
j
e
1
1
for cot sin y  tan sin y
j
let sin 1 y  
than
cot   tan  
1
1

sin  cos  y 1  y 2
F
G
1 G 1 y
1
y G
GH y 1  y
2
so
[Q]

2
2
I
JJ
JJ
K
2
e1  y j  y
 y4 
4
4
1
cos x  cos y   cos z
. . .(1)
sin x  sin y   sin z
. . .(2)
(1) 2  ( 2) 2 gives
b
g
2  2 cos x cos y  sin x sin y  1




[R]



b g
2 c1  cosb x  y gh  1
F x  y IJ  1
4 cos G
H 2 K
F x  y IJ   1
cosG
H 2 K 2
F I
F I
cosG  xJ cos 2x  cosG  xJ cos 2x  sin 2xb1  tan xg
H4 K
H4 K
bcos x  sin xg
cos 2 xe 2 sin xj  sin 2 x
2  2 cos x  y  1
2
cos x
e 2 sin x cos xj  sin 2xbcos x  sin xg
bsin 2 xgFGH cos22x  bcos x  sin xgIJK  0
cos 2 x

bsin 2xgbcos x  sin xgLMFGH cos x 2sin x IJK  1OP  0
N
Q

x  0,

4
Posible value of sec x is 1, 2
[35]
JEE ADVANCED – 2013 / Paper -2
MATHEMATICS
[S]


F F 1  x I I  sinetan ex 6 jj
KK
H H
F F x6
F
x I
cot G cot
 sinG sin G
J
H
1 x K
H H 1  6x
cot sin 1
1
2
1
1
2
x
1 x
2

x 6
1  6x 2

e1  6x j  6e1  x j b x  0g

12 x 2  5

x
2
2
5
1 5

12
2 3
2
II
JK JK