IIT - JEE 2013 (Advanced)...IIT JEE 2013 Advanced : Question Paper & Solution (Paper – II) (3) 3...
Transcript of IIT - JEE 2013 (Advanced)...IIT JEE 2013 Advanced : Question Paper & Solution (Paper – II) (3) 3...
IIT - JEE 2013 (Advanced)
CODE : O
(2) Vidyalankar : IIT JEE 2013 − Advanced : Question Paper & Solution
2
IIT JEE 2013 Advanced : Question Paper & Solution (Paper – II) (3)
3
PART I : PHYSICS SECTION 1 : (One or more options correct Type)
This 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. The figure below shows the variation of specific heat capacity (C) of solid as a function of
temperature (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.
1. (A), (B), (C), (D)
dQ = mc dT
dQ
dT= m (aT + b) in the range (0 − 100 k)
∴ option (A) is correct. Considering (0 − 100 k), the graph to be linear using the
reasonable approximation given in the question.
During (400k − 500k) C is the maximum.
∴ option (B) is correct.
∴ Option (C) is correct as C is constant in the range (400k − 500k)
and option (D) is also correct.
2. The radius of the orbit of an electron in a Hydrogen-like atom is 4.5 a0, where a0 is the
Bohr radius. Its orbital angular momentum is 3h
2π. It is given that h is Planck constant and
R is Rydberg constant. The possible wavelength(s), when the atom de-excites, is (are)
(A) 9
32R (B)
9
16R (C)
9
5R (D)
4
3R
2. (A), (C)
rn = 2
0
na
z
4.5a0 = 2
0
na
z
∴ 2n
z = 4.5
∴ 9
z = 4.5 and z = 2
L = nh
2π
L = 3h
2π
∴ n = 3
100 200 300 400 500
C
T (K)
(4) Vidyalankar : IIT JEE 2013 − Advanced : Question Paper & Solution
4
Wavelength is given by
1
λ = 2
2 2
f i
1 1Rz
n n
⎛ ⎞−⎜ ⎟⎝ ⎠
1
λ =
2 2
f i
1 14R
n n
⎛ ⎞−⎜ ⎟⎝ ⎠
For transition, n = 3 → 1
λ = 1
14R 1
9
⎛ ⎞−⎜ ⎟⎝ ⎠
= 9
32R …(A)
For transition n = 3 → 2
λ = 1
1 14R
4 9
⎛ ⎞−⎜ ⎟⎝ ⎠
= 1 36
4R 5
××
= 9
5R … (C)
3. Using the expression 2d sin θ = λ, one calculates the values of d by measuring the
corresponding angles θ in the range 0 to 90º. The wavelength λ is exactly known and the
error in θ is constant for all values of θ. As θ increases from 0º.
(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.
3. (D)
2d sin θ = λ
⇒ δ (2d sin θ) = δλ
But δλ = 0
⇒ (2 sin θ) δd + 2d cos θ δθ = 0
⇒ d
d
δ + cot θ δθ = 0
⇒ d
d
δ = − cot θ (dθ)
δθ = constant
absolute error in d is |δd| = d cot θ (δθ)
⇒ |δd| = 2
cos| d |
2 sin
λ θ θθ
at θ increase from 0 to 90°, |δd| decreases.
Hence (B) is not true.
By the same fractional error decreases.
4. Two non-conducting spheres of radii R1 and R2 and carrying uniform volume charge
densities +ρ and −ρ, 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.
IIT JEE 2013 Advanced : Question Paper & Solution (Paper – II) (5)
5
4. (C), (D)
1 10
E O P3
ρ=∈
2 20
E PO3
ρ=∈
Net field at P
1 2E E E= +
1 20
[O P PO ]3
ρ= +∈
1 20
E O O3
ρ=∈
∴ E is constant (C)
And option (D) is also correct.
5. 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.
5. (A), (D)
n = N
L
Bd∫ = μ0I
B1 = 0I
2 R
μπ
along axis of cylinder
B2 = μ0nI along axis of solenoid
(i) 0 < r < R, magnetic field
At this point, there will be field only due to solenoid.
(ii) R < r < 2R
B = 0I
2 r
μπ
by the rod. (Along the tangent to the circular part around the cylinder)
Bsolenoid = μ0nI along axis
6. Two vehicles, each moving with speed u on the same horizontal 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 statements(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.
O1
E2
O2
E1P
(6) Vidyalankar : IIT JEE 2013 − Advanced : Question Paper & Solution
6
6. (A), (B)
uSW = u − w
uOW = u + w
f2 = v u w
v (u w)
+ +− −
f1 ⇒ f2 = v u w
v u w
+ +⎛ ⎞⎜ ⎟− +⎝ ⎠
f1
Clearly f2 > f1 irrespective of sign of w.
7. 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 GM
4L
.
(B) The minimum initial velocity of the mass m to escape the gravitational field of the two
bodies is GM
2L
.
(C) The minimum initial velocity of the mass m to escape the gravitational field of the two
bodies is 2GM
L.
(D) The energy of the mass m remains constant.
7. (B), (D)
(D) is clearly correct, since gravitational field is conservative.
2
e
2GMm 1mv
L 2− + = 0
⇒ ve = GM
2L
8. 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 elastically with 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 = m
kπ .
(C) the time at which the maximum compression of the spring occurs is t = 4 m
3 k
π.
(D) the time at which the particle passes through the equilibrium position for the second
time is t = 5 m
3 k
π.
8. (A), (D)
The impulse of the wall merely reverses the direction of motion is thout changing particle
speed. Hence (A)
S Ou u
W
IIT JEE 2013 Advanced : Question Paper & Solution (Paper – II) (7)
7
Displacement from mean position be x
x = 0u
w
⎛ ⎞⎜ ⎟⎝ ⎠
sin (ωt) ⇒ x = u0 cos (ωt)
when particle speed is 0u
2
⎛ ⎞⎜ ⎟⎝ ⎠
cos (ωt) = 1
2 ⇒ ωt =
3
π [for first time]
⇒ t = 3
πω
Time after which it passes the mean position = 2t = 3
3u
π=
2 m
3 K
π
Hence (B) is incorrect.
Time after which particle crosses mean position for second time = 2
3
π π+ω ω
= 5
3
πω
Hence (D).
SECTION 2 : (Paragraph 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 9 and 10
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
).
9. 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
9. (B)
Kinetic energy of block at Q = mgR sin 30° − 150 J
⇒ 21mv
2 = 10 × 40 ×
1
2 − 150 = 50 J
⇒ r = 2 50
1
× = 10 ms
−1
(8) Vidyalankar : IIT JEE 2013 − Advanced : Question Paper & Solution
8
10. 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
10. (A)
N − mg cos 60° = 2mv
r
⇒ N =mg cos 60° + 2mv
r
⇒ N = 10 1 100
2 40
×+ = 7.5 N
Paragraph for Questions 11 and 12
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 energy dissipation. In the method using transformers, 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.
11. If the direct transmission method with a cable of resistance 0.4Ω km−1
is used, the power
dissipation (in%) during transmission is
(A) 20 (B) 30 (C) 40 (D) 50
11. (B)
P = 600 kW
V = 4000 V
I = 3600 10
4000
× = 150A
Power Loss = I2R
= (150)2 × 0.4 × 20 = 180 kW
% Power Loss = 180
100600
× = 30%
12. 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 primary to
that in the secondary in the step-down transformer is
(A) 200 : 1 (B) 150 : 1 (C) 100 : 1 (D) 50 : 1
12. (A)
Step up transformer
1 : 10
4000 V → 40,000 V
Step down transform
40,000 V → 200
∴ turns ratio = 40000
200 = 200 : 1
30°
60°
mg
IIT JEE 2013 Advanced : Question Paper & Solution (Paper – II) (9)
9
Paragraph for Questions 13 and 14
A point charge Q is moving in a circular orbit of radius R in the x-y plane with an angular
velocity ω. This can be considered as equivalent to a loop carrying a steady current Q
2
ωπ
. A
uniform magnetic field along the positive 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 γ.
13. The magnitude of the induced electric field in the orbit 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
13. (B)
Magnitude of induced emf is given by = A dB
2 R dtπ =
2RB
2 R
ππ
= BR
2
14. 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) −γBQR2 (B)
2BQR
2−γ (C)
2BQR
2γ (D) γBQR
2
14. (B)
ωf − ωi = α t = ( )2
q ER1
MR− = ( )qE
1MR
−
E = BR
2
Δω = q BR
MR 2
⎛ ⎞− ⎜ ⎟⎝ ⎠
= qB
2M−
μ = γL = γIω
Δμ = ( )2MRγ Δω = 2 qBMR
2M
−⎛ ⎞γ ⎜ ⎟⎝ ⎠
= 2qBR
2
γ−
Paragraph for Questions 15 and 16
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 the nucleus. A heavy nucleus of mass M can
break into two light nuclei of masses m1 and m2 only if (m1 + m2) < M. Also two light nuclei
of masses m3 and m4 can undergo complete fusion and form a heavy nucleus of mass M′ only
if (m3 + m4) > M′. The masses of some neutral atoms are given in the table below :
11H 1.007825 u 2
1 Η 2.014102 u 31 H 3.016050 u 4
2 He 4.002603 u
63 Li 6.015123 u 7
3 Li 7.016004 u 7030 Zn 69.925325 u 82
34 Se 81.916709 u
15264 Gd 151.919803 u 206
82 Pb 205.974455 u 20983 Bi 208.980388 u 210
84 Po 209.982876 u
(1 u = 932 MeV/c2)
(10) Vidyalankar : IIT JEE 2013 − Advanced : Question Paper & Solution
10
15. The correct statement is
(A)The nucleus 63 Li can emit an alpha particle.
(B) The nucleus 21084 Po can emit a proton.
(C) Deuteron and alpha particle can undergo complete fusion.
(D)The nuclei 7030 Zn and 82
34Se can undergo complete fusion.
15. (C)
3 2 46 1 2Li H He⎯⎯→ +
M 6.015123=
m1 = 2.014102
m2 = 4.002603
m1 + m2 = 6.016705
m1 + m2 > M (not possible)
(B) 210 209 184 83 1Po Bi P⎯⎯→ +
M = 209.982876
m1 = 208.980388
m2 = 001.007825
m1 + m2 = 209.988213
M < m1 + m2 (not possible)
(C) 2 4 61 2 3H H Li+ ⎯⎯→
(m1 + m2) > M′ Possible
(D) 70 82 15230 34 64Zn Se Gd+ ⎯⎯→
m1 = 69.925325
m2 = 81.916709
m1 + m2 = 141.832034
M′ = 151.919803
m1 + m2 < M′ (not possible)
16. The kinetic energy (in keV) of the alpha particle, when the nucleus 21084 Po at rest
undergoes alpha decay, is
(A) 5319 (B) 5422 (C) 5707 (D) 5818
16. (A)
210 206 484 82 2Po Pb He⎯⎯→ +
M 209.982876=
1m 205.974455=
2m 4.002603=
m1 + m2 = 209.977058
1 2M (m m ) 209.982876− + =
− 209.977058
0.005818
2E mc (0.005818)932= Δ = 5422=
1 2K K E+ =
IIT JEE 2013 Advanced : Question Paper & Solution (Paper – II) (11)
11
2 2
1 2
P PE
2m 2m+ =
2P
E2
=μ
2
1 2
1 2
Em mP
2 m m=
+
2
21
1 1 2
Emp E(206) 103k E
2m m m 210 105= = = =
+
1
103k (5422) 5319
105= =
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
17. 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.
List I List II
P. G→E 1. 160 P0V0 1n2
Q. G→H 2. 36 P0V0
R. F→H 3. 24 P0 V0
S. F→G 4. 31 P0V0
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
17. (A)
F F G GP V P V=
0 032P V = 0 GP V
(12) Vidyalankar : IIT JEE 2013 − Advanced : Question Paper & Solution
12
G 0V 32V=
F F H HP V P Vγ γ=
0 0 0 H32P V P Vγ γ=
VH =
1
032 Vγ
= ( )3 5
0 032 V 8V/ =
(P) ( ) ( )GE 0 E G 0 0 0W P V V P V 32V= − = −
= 0 031P V− → 4.
(Q) ( )GH 0 0 0 0 0W P 8V 32V 24P V= − = − → (3)
(R) ( )0 0 0 0H H F F
FH
P 8V 32P VP V P VW
511
3
−−= =− γ −
= 0 00 0
24P V36P V 2
2
3
( )−
= →−
(A) G 0FG F F 0 0
F 0
V 32VW P V n 32P V n
V V= =
= 0 032P V n 32
= ( )5
0 032 P V n 2 = 0 0160P V n 2 1( )→
⇒ (P) → 4; (Q) → (3); (R) → (2); (S) → (1)
18. 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
P. Alpha decay 1. 15 158 7O N ..... → +
Q. β+ decay 2. 238 234
92 90U Th ..... → +
R. Fission 3. 185 18483 82Bi Pb ..... → +
S. Proton emission 4. 239 14094 57Pu La ..... → +
Codes :
P Q R S
(A) 4 2 1 3
(B) 1 3 2 4
(C) 2 1 4 3
(D) 4 3 2 1
18. (C)
P Q R S
2 1 4 3
IIT JEE 2013 Advanced : Question Paper & Solution (Paper – II) (13)
13
19. A right angled prism of refractive index μ1 is placed in a rectangular block of refractive
index μ2, which is surrounded by a medium of refractive index μ3, as shown in the figure.
A ray of light 'e' enters the rectangular block at normal incidence. Depending upon the
relationships between μ1, μ2 and μ3, 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. e → f 1. μ1 > 2 μ2
Q. e → g 2. μ2 > μ1 and μ2 > μ3
R. e → h 3. μ1 = μ2
S. e → i 4. μ2 < μ1 < 2 μ2 and μ2 > μ3
Codes :
P Q R S
(A) 2 3 1 4
(B) 1 2 4 3
(C) 4 1 2 3
(D) 2 3 4 1
19. (D)
(P) e → f → for 1 → 2 towards normal
⇒ μ2 > μ1
For 2 → 3 always from normal
⇒ μ3 < μ2
i.e., (2)
(Q) e → g → no deviation any where
μ1 = μ2 = μ3, i.e., (3)
(R) e → h → 1 → 2 away from normal.
μ2 < μ1
2 → 3 away from normal.
⇒ μ3 < μ2
And also at 1 − 2 no I.R.
⇒ sin 45 < 2
1
μμ
⇒ 1 22μ < μ
⇒ 2 1 22μ < μ < μ → i.e., (4)
(S) e → i → TIR at 1 → 2
(14) Vidyalankar : IIT JEE 2013 − Advanced : Question Paper & Solution
14
⇒ sin 45 > 2
1
μμ
⇒ 1 22μ > μ
i.e., 1
(P) → 2; (Q) → 3; (R) → (4); (S) → (1).
20. Match 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. [ML2T
−1]
Q. Coefficient of viscosity 2. [ML−1
T−1
]
R. Planck constant 3. [MLT−3
T−1
]
S. Thermal conductivity 4. [ML2T
−2K
−1]
Codes :
P Q R S
(A) 3 1 2 4
(B) 3 2 1 4
(C) 4 2 1 3
(D) 4 1 2 3
20. (C)
(P) u = B
3K T
2
[u] = BK T[ ]
2 2
B
ML TK
K
[ ][ ]
[ ]
−=
[KB] = 2 2 1ML T K 4[ ]− − →
(Q) F = 6π rη V
[η] = 2
1 1
1
F MLTML T 2
rv L LT[ ]
( )
−− −
−
⎡ ⎤⎡ ⎤ = = →⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦
(R) E = hν
[h] = 2 2
2 1
1
E ML TML T 1
T[ ] ( )
−−
−
⎡ ⎤⎡ ⎤ = = →⎢ ⎥⎢ ⎥ν⎣ ⎦ ⎣ ⎦
(S) Q
t = 2 1KA T T( )−
[K] = 2 3
2
ML TL
L K
[ ][ ]
[ ][ ]
−
= 3 1MLT K 3[ ] ( )− − →
(P) → 4; (Q) → (2); (R) → (1); (S) → (3)
IIT JEE 2013 Advanced : Question Paper & Solution (Paper – II) (15)
15
PART II : CHEMISTRY
SECTION 1 : (One or more options correct Type) This section contains 8 multiple choice questions. Each question has four choices (A), (B),
(C) and (D) out of which ONE or MORE are correct.
21. The carbon-based reduction method is NOT used for the extraction of
(A) tin from SnO2
(B) iron from Fe2O3
(C) aluminium from Al2O3
(D) magnesium from MgCO3 . CaCO3
21. (C), (D)
The oxides of the less electropositive methods are reduced by strongly heating them with
coke.
22. The thermal dissociation equilibrium of CaCO3(s) is studied under different conditions.
3 2CaCO (s) CaO(s) CO (g)+
For this equilibrium, the correct statement(s) is (are)
(A) ΔH is dependent on T
(B) K is independent of the initial amount of CaCO3
(C) K is dependent on the pressure of CO2 at a given T
(C) ΔH is independent of the catalyst, if any
22. (A), (B), (D)
K Equilibrium constant is independent of the partial pressure of CO2.
It is independent of the catalyst.
23. 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 has a bent structure
23. (A), (C), (D)
3 22O 3O 68 k calΔ ⎯⎯→ + i.e. Exothermic
24. In the nuclear transmutation
9 84 4Be X Be Y+ ⎯⎯→ +
(X, Y) is (are)
(A) (γ, n) (B) (p, D) (C) (n, D) (D) (γ, p)
24. (A), (B)
(A) 9 8 14 4 0Be Be n + γ ⎯⎯→ +
(B) 9 1 8 24 1 4 1Be H Be D + ⎯⎯→ +
(C) 9 1 8 24 0 4 1Be n Be D + ⎯⎯→ + (False)
(D) 9 8 14 4 1Be Be H + γ ⎯⎯→ + (False)
(16) Vidyalankar : IIT JEE 2013 − Advanced : Question Paper & Solution
16
2| |
B r (1m o l )3 3 N a O H(1 m o l )
O
C H C C H (T ) ( U )− − ⎯ ⎯ ⎯ ⎯→ +
25. The major product(s) of the following reaction is(are)
(A) P (B) Q (C) R (D) S
25. (B)
26. After completion of the reactions (I and II), the organic compound(s) in the reaction
mixtures is(are)
(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 : S, acetone
26. (C)
NaOH
SO3H
OH
2aq. Br ⎯⎯⎯⎯→
Br
OHBrBr
(Q)
2
3
Br (1mol)
CH COOH(p)⎯⎯⎯⎯⎯→
IIT JEE 2013 Advanced : Question Paper & Solution (Paper – II) (17)
17
27. The Ksp of Ag2CrO4 is 1.1 × 10−12
at 298 J. The solubility (in mol / L) of Ag2CrO4 in a
0.1 M AgNO3 solution is
(A) 1.1 × 10−11
(B) 1.1 × 10−10
(C) 1.1 × 10−12
(D) 1.1 × 10−9
27. (B)
22 4(s) 4Ag CrO 2Ag (aq) CrO (aq)+ −+
2s + 0.1 s
= 0.1
2 12spK (0.1) s 1.1 10−= = ×
10s 1.1 10 M−= ×
28. In the following reaction, the product(s) formed is (are)
(A) P (major) (B) Q (minor) (C) R (minor) (D) S (major)
28. (B), (D)
⎯⎯→CCl
2
..
CH3
O O
CH3
H
CCl2 ⎯⎯→
O
CH3
CHCl2
OH⎯⎯→
CH3
OH
CHO
major
2H O←⎯⎯⎯
CH3
OCHO
⎯⎯→
O O
HC3 CHCl
2
(minor)
(18) Vidyalankar : IIT JEE 2013 − Advanced : Question Paper & Solution
18
SECTION 2 : (Paragraph 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)..
Using the following Paragraph , Solve Q. No. 29 and 30
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 H2S in a dilute mineral acid medium.
However, it gave a precipitate (R) with H2S in an ammoniacal medium. The precipitate R
gave a coloured solution (S), when treated with H2O2 in an aqueous NaOH medium.
29. The precipitate P contains
(A) Pb2+
(B) 22Hg + (C) Ag
+ (D) Hg
2+
29. (A)
22
(p)
Pb HCl PbCl+ + ⎯⎯→ ↓ ⎯⎯→ soluble in hot water
30. The coloured solution S contains
(A) Fe2(SO4)3 (B) CuSO4 (C) ZnSO4 (D) Na2CrO4
22
(p)
Pb HCl PbCl+ + ⎯⎯→ ↓ ⎯⎯→ soluble in hot water
30. (D)
3 2 4 2(S)(R)
2Cr(OH) 4NaOH 3[O] 2Na CrO 5H O+ + ⎯⎯→ +
Using the following Paragraph , Solve Q. No. 31 and 32
P and Q are isomers of dicarboxylic acid C4H4O4. Both decolorize Br2 / H2O. On heating , P
forms the cyclic anhydride.
Upon treatment with dilute alkaline KMnO4 , P as well as Q could produce one or more than
one from S, T and U.
31. 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.
IIT JEE 2013 Advanced : Question Paper & Solution (Paper – II) (19)
19
31. (B)
C
C
H
H
COOH
COOH
2H O
Δ−⎯⎯⎯→ C
C
C
C
O
H
H
O
O
(P)
(cis)cyclic anhydride
C
C
H
H
COOH
HOOC
(Q)
H OH
H OH
COOH
COOH
(S)
(meso)
4dil. KMnOP
⎯⎯⎯⎯⎯→
32. In the following reaction sequences V and W are , respectively
(A) (B)
(C) (D)
4dil. KMnOQ
⎯⎯⎯⎯⎯→H OH
OH H
COOH
COOH
(T)
OH H
H OH
COOH
COOH
+
(U)
(20) Vidyalankar : IIT JEE 2013 − Advanced : Question Paper & Solution
20
32. (A)
2H /Ni(Q) Δ ⎯⎯⎯→
3anhy. AlCl ⎯⎯⎯⎯⎯→
CH2
CH2
C
O
C
O
O
2 2C CH CH COOH− − −
O
O
3 4H PO←⎯⎯⎯⎯2 2 2CH CH CH COOH− − −
+ (V)
(V)
(W)
Zn-Hg/conc.HCl
Using the following Paragraph , Solve Q. No. 33 and 34
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
33. The succeeding operations that enable this transformation of states are
(A) Heating, cooling, heating , cooling (C) Cooling , heating cooling, heating
(C) Heating, cooling, cooling, heating (D) Cooling, heating, heating, cooling
33. (C)
Factual
34. 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
34. (B)
Factual
Using the following Paragraph , Solve Q. No. 35 and 36
The reactions of Cl2 gas with cold − dilute and hot − concentrated NaOH in water give
sodium salts of two (different) oxoacids of chlorine, P and Q respectively. The Cl2 gas reacts
with SO2 gas , in presence of charcoal, to give a product R. R reacts with white phosphorus to
give a compound S. On hydrolysis, S gives an oxoacid of phosphorus , T.
IIT JEE 2013 Advanced : Question Paper & Solution (Paper – II) (21)
21
35. P and Q respectively, are the sodium salts of
(A) hypochlorus and chloric acids (B) hypoclorus and chlorus acids
(C) chloric and perchloric acids (D) chloric and hypochlorus acids
35. (A)
Cold
2 22NaOH Cl NaCl NaOCl(P) H O+ ⎯⎯⎯→ + +
hot
2 3 26NaOH 3Cl 5NaCl NaClO (Q) 3H O+ ⎯⎯→ + +
36. R, S and T , respectively are
(A) SO2Cl2 , PCl5 and H3PO4 (B) SO2Cl2 , PCl3 and H3PO3
(C) SOCl2 , PCl3 and H3PO2 (D) SOCl2 , PCl5 and H3PO4
36. (A)
2 2 2 2SO Cl SO Cl (R)+ ⎯⎯→
4 2 2 5 2P 10SO Cl 4PCl 10 SO+ ⎯⎯→ +
5 2 3 4PCl 4H O H PO 5HCl+ ⎯⎯→ +
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
37. An aqueous solution of X is added slowly to an aqueous solution of Y as shown in List I.
The variation in 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
P. 2 5 3 3(C H ) CH COOH
X Y
Ν + 1. Conductivity decreases and then increases
Q. 3KI (0.1M) AgNO (0.01M)
X Y
+ 2. Conductivity decreases and then does not
change much
R. 3CH C KOHYX
ΟΟΗ + 3. Conductivity increases and then does not
change much
S. NaOH HIYX
+ 4. Conductivity does not change much and then
increases
Codes :
P Q R S
(A) 3 4 2 1
(B) 4 3 2 1
(C) 2 3 4 1
(D) 1 4 3 2
37. (A)
Factual
38. The standard reduction potential data at 25° C is given below .
E° (Fe3+
, Fe2+
) = + 0.77 V;
E° (Fe2+
, Fe) = − 0.44 V
E° (Cu2+
, Cu) = + 0.34 V
E° (Cu+ , Cu) = + 0.52 V
E° [O2(g) + 4H+ + 4e
− → 2H2O] = + 1.23 V;
(22) Vidyalankar : IIT JEE 2013 − Advanced : Question Paper & Solution
22
E° [O2(g) + 2H2O + 4e−
→ 4OH−] = + 0.40 V
E° (Cr3+
, Cr) = −0.74 V ;
E° (Cr2+
, Cr) = −0.91 V
Match E° 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
P. E° (Fe3+
, Fe) 1. −0.18 V
Q. E° 2(4H O 4H 4OH )+ − +
2. −0.4 V
R. E° (Cu2+
+ Cu → 2Cu+) 3. −0.04 V
S. E° (Cr3+
, Cr2+
) 4. −0.83 V
Codes :
P Q R S
(A) 4 1 2 3
(B) 2 3 4 1
(C) 1 2 3 4
(D) 3 4 1 2
38. (D)
Use formula 0 0 03 3 1 1 2 2n F E n FE n FE= +
⇒ 0 0
0 1 1 23
3
n E n EE
n
+=
39. 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. ?2 2 4 4 2PbO H SO PbSO O other product + ⎯⎯→ + + 1. NO
Q. ?2 2 3 2 4Na S O H O NaHSO other product + ⎯⎯→ + 2. I2
R. ?2 4 2N H N other product ⎯⎯→ + 3. Warm
S. ?2XeF Xe other product ⎯⎯→ + 4. Cl2
Codes :
P Q R S
(A) 4 2 3 1
(B) 3 2 1 4
(C) 1 4 2 3
(D) 3 4 2 1
39. (D)
warm
2 2 4 4 2 2P PbO H SO PbSO O H O→ + ⎯⎯⎯→ + +
2Cl2 2 3 2 4Q Na S O H O NaHSo NaCl HCl→ + ⎯⎯⎯→ + +
2I2 4 2R N H N 4HI→ ⎯⎯→ +
NO
2S XeF Xe 2NOF→ ⎯⎯⎯→ +
IIT JEE 2013 Advanced : Question Paper & Solution (Paper – II) (23)
23
40. 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
P. 1. (i) Hg(OAc)2 ,; (ii) NaBH4
Q. 2. NaOEt
R. 3. Et − Br
S. 4. (i) BH3 ; (ii) H2O2 / NaOH
Codes :
P Q R S
(A) 2 3 1 4
(B) 3 2 1 4
(C) 2 3 4 1
(D) 3 2 4 1
40. (A)
P = 2, Q = 3, R = 1, S = 4
PART III : MATHEMATICS
SECTION 1 : (One or more options correct Type)
This section contains 8 multiple choice questions. Each question has four choices (A), (B),
(C) and (D) out of which ONE or MORE are correct.
41. Let ω be a complex cube root of unity with ω ≠ 1 and P = [pij] be a n × n matrix with
pij = ωi + j. Then P
2 ≠ 0, when n =
(A) 57 (B) 55 (C) 58 (D) 56
41. (B), (C), (D)
P2 = O only when n is multiple of 3.
Ex :
2 2
2 2
2 2
1 1
1 1
1 1
⎡ ⎤ ⎡ ⎤ω ω ω ω⎢ ⎥ ⎢ ⎥
ω ω ⋅ ω ω⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ω ω ω ω⎣ ⎦ ⎢ ⎥⎣ ⎦
=
0 0 0
0 0 0
0 0 0
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
∴ P2 ≠ 0, when n = 55, 56, 58.
42. 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 (D)
2
3
(24) Vidyalankar : IIT JEE 2013 − Advanced : Question Paper & Solution
24
42. (A), (B)
Given, f(x) = 2|x| + |x + 2| − ||x + 2| − 2|x||
Sign scheme of |x + 2| − 2|x| is
For 2
x 23
− ≤ ≤
f (x) = 4|x|
For x < 2
3− ∪ x > 2
f(x) = 2 |x + 2|
Now, f(x) = 4 |x|, 2
3− ≤ x ≤ 2
= 2|x + 2|, x < 2
3− ∪ x > 2
The point of maxima or minima are −2, 2
3− , 0.
43. Let 3 i
w2
+ = and P = {wn : n = 1, 2, 3, … }. Further H1 =
1z : Re z
2
⎧ ⎫ ∈ > ⎨ ⎬⎩ ⎭
and
2
1H z : Re z ,
2
−⎧ ⎫ = ∈ < ⎨ ⎬⎩ ⎭
where is the set of all complex numbers. If z1 ∈ P ∩ H1,
z2 ∈ P ∩ H2 and O represents the origin, then ∠ z1 O z2 =
(A) 2
π (B)
6
π (C)
2
3
π (D)
5
6
π
43. (C), (D)
ω = 3 i
2 2 + = cos i sin
6 6
π π⎛ ⎞ ⎛ ⎞+ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
ωn =
n ncos i sin
6 6
π π⎛ ⎞ ⎛ ⎞+ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ (by De Moivre’s Theorem)
If z1 ∈ P ∩ H1, then n
cos6
π⎛ ⎞ ⎜ ⎟⎝ ⎠
> 1
2
which is true for n = 0, 1, 11.
−2/3 2
8/38
y
x
−2/3 2
−ve −ve +ve
x
−2 2
8/38
y
−2/3 x
−2 2
8/38
y
−2/3 x
IIT JEE 2013 Advanced : Question Paper & Solution (Paper – II) (25)
25
If z2 ∈ P ∩ H2, then n
cos6
π⎛ ⎞⎜ ⎟⎝ ⎠
< 1
2
−, which is true for n = 5, 6, 7.
∴ ∠ z1 oz2 can be 2
3
π or
5.
6
π
44. If 3x = 4
x−1, then x =
(A) 3
3
2log 2
2log 2 1− (B)
2
2
2 log 3− (C)
4
1
1 log 3− (D) 2
2
2log 3
2log 3 1−
44. (A), (B), (C)
3x = 4
x−1
2x log 3 = (x−1) 2log 4
x 2log 3= 2x − 2
x (2 − 2log 3 ) = 2
∴ x = 2
2
2 log 3− (B)
3x = 4x−1
x 4log 3 = x − 1
⇒ x = 4
1
1 log 3− (C)
3x = 4
x−1
x 3log 3 = (x − 1) 3log 4
x = (x − 1) log3 4 = (x − 9) 2. log3 2
∴ x = 3
3
2.log 2
2.log 2 1− (A)
45. Two lines 1 2
y z y zL : x 5, and L : x ,
3 2 1 2= = = α =
− α − − − α are coplanar. Then α
can take values (s)
(A)1 (B) 2 (C) 3 (D) 4
45. (A), (D)
L1 : x 5
0
− =
y
3 − α =
Z
2− Line L1 passes (x1, y1, z1) ≡ (5, 0, 0)
L2 : x
0
− α =
y
1− =
z
2 − α Line L2 passes (x2, y2, z3) ≡ (α, 0, 0)
As Line co-planar ⇒ Shortest distance = 0
3 1,
2 2
⎛ ⎞− ⎜ ⎟⎝ ⎠
3 1,
2 2
⎛ ⎞ ⎜ ⎟
⎝ ⎠
3 1,
2 2
⎛ ⎞− − ⎜ ⎟⎝ ⎠
3 1,
2 2
⎛ ⎞− ⎜ ⎟
⎝ ⎠
(−1, 0) (1, 0)
(26) Vidyalankar : IIT JEE 2013 − Advanced : Question Paper & Solution
26
⇒
2 1 2 1 2 1
1 1 1
2 2 2
x x y y z z
m n
m n
− − − = 0
⇒
5 0 0
0 3 2
0 1 2
α − − α −− − α
= 0 ⇒ (α − 5) (α − 4) (α − 1) = 0
α = 1, 4, 5
46. In a triangle PQR, P is the largest angle and 1
cos P3
= . Further the incircle of the triangle
touches the sides 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
46. (B), (D)
cos p = 2 2 2(4x 4) (4x 2) (4x 6)
2(4x 4) (4x 2)
+ + + − + + +
1
3 =
216x 16
16(x 1) (2x 1)
− + +
1
3 =
2
2
x 1
2x 3x 1
− + +
⇒ 3x2 − 3 = 2x
2 + 3x + 1
⇒ 3x2 − 3x − 4 = 0
x = 4 or −1. Hence x = 4
So, possible lengths of sides are 18, 22.
47. For a ∈ R (the set of all real numbers), a ≠ −1,
( )
( ) ( ) ( ) ( )
a a a
a 1x
1 2 .... n 1lim
60n 1 na 1 na 2 ... na n−→∞
+ + +=
+ + + + + +⎡ ⎤⎣ ⎦
Then a =
(A) 5 (B) 7 (C) 15
2
− (D)
17
2
−
47. (B), (D)
Let, L = ( ) ( )( )
a a a
a 1n
1 2 ...... nlim
n 1 na 1 (na 2) ... (na n)−→∞
+ + + + + + + + + +
=
( )
a a aa
n a 1 2
1 2 nn ...
n n nlim
n(n 1)n 1 n a
2
→∞ −
⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞⋅ + + +⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠
+⎛ ⎞+ +⎜ ⎟⎝ ⎠
=
( )
ana
r 1
n a 1
1 rn n
n nlim
n 1n 1 n na
2
=
→∞ −
⎛ ⎞⎛ ⎞⋅ ⋅ ⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠+⎛ ⎞+ +⎜ ⎟
⎝ ⎠
∑ =
( )
1a
a0
a 1n
x dxn
limn 1n 1 na
2
−→∞ ++ +
∫
P
Q R
2x 2x
2x + 4
2x + 42x + 2
2x + 2
IIT JEE 2013 Advanced : Question Paper & Solution (Paper – II) (27)
27
Using limit of a sum we get an
r 1
1 r
n n =
⎛ ⎞ ⎜ ⎟⎝ ⎠
∑ =
1a
0
x dx∫
∴ L = ( )
a
a 1n
1
n a 1limn 1n 1 na
2
−→∞+ ++ +
= a 1n
11 a 1lim
1 11 a1 2 2nn
−→∞+ ⋅
⎛ ⎞ + ++⎜ ⎟⎝ ⎠
= ( )2
(a 1) 2a 1+ + Given L =
1
60
∴ (a + 1) (2a + 1) = 120
⇒ 2a2 + 3a − 119 = 0
⇒ a = 7 or a = 17
2−
∴ Answer is (B), (D)
48. 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) x2 + y
2 − 6x + 8y + 9 = 0 (B) x
2 + y
2 − 6x + 7y + 9 = 0
(C) x2 + y
2 − 6x − 8y + 9 = 0 (D) x
2 + y
2 − 6x − 7y + 9 = 0
48. (A), (C)
Let centre, C = (3, α)
and BC = 2 2 cα − = 2 7
⇒ α2 − c = 7 …(1)
Agai α2 = α2
+ 9 − c ⇒ c = 9 …(2)
From (1), α2 − 9 = 7
⇒ α = ± 4
Eq is x2 + y
2 − 6x ± 8y + 9 = 0
SECTION 2 : (Paragraph 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 Question 49 and 50
Let f : [0, 1] → R (the set of all real numbers) be a function. Suppose the function f is twice
differentiable, f(0) = f(1) = 0 and satisfies ( ) ( ) ( ) [ ]xf '' x 2f ' x f x e , x 0,1− + ≥ ∈ .
49. Which of the following is true for 0 < x < 1?
(A) ( )0 f x< < ∞ (B) ( )1 1f x
2 2− < < (C) ( )1
f x 14
− < < (D) ( )f x 0−∞ < <
49. (D)
C
B
e
α
A 0
(28) Vidyalankar : IIT JEE 2013 − Advanced : Question Paper & Solution
28
50. If the function e−x
f(x) assumes its minimum in the interval [0, 1] at 1
x4
= , which of the
following is true?
(A) ( ) ( ) 1 3f ' x f x , x
4 4< < < (B) ( ) ( ) 1
f ' x f x , 0 x4
> < <
(C) ( ) ( ) 1f ' x f x , 0 x
4< < < (D) ( ) ( ) 3
f ' x f x , x 14
< < <
50. (C)
Let g(x) = e−x
f(x) ⇒ g′(x) = e−x
(f′(x) − f(x))
If x = 1
4 is point of minima, then
e−x
(f′(x) − f(x)) < 0 ∀ x ∈ 1
0,4
⎛ ⎞ ⎜ ⎟⎝ ⎠
⇒ f′(x) < f(x) ∀ x ∈ 1
0,4
⎛ ⎞ ⎜ ⎟⎝ ⎠
Paragraph for Question 51 and 52
Let PQ be a focal chord of the parabola y2 = 4ax. The tangents to the parabola at P and Q
meet at a point lying on the line y = 2x + a, a > 0.
51. Length of chord PQ is
(A) 7a (B) 5a (C) 2a (D) 3a
51. (B)
P + T
Given point T lies on y = 2x + a
⇒ a(t1 + t2) = 2at1t2 + a
t1 + t2 = −2 + 1 (as focal chord t1t2 = −1)
t1 + t2 = −1
∴ ( )1 2t t− = ( )2
1 2 1 2t t 4t t+ − = 5
So, PQ = ( ) ( )2 22 2 2
1 2 1 2a t t 2at 2at− + −
= ( ) ( )2 22 2
1 2 1 2a t t 4 t t− + − = ( ) ( )2 2a ( 1) 5 4 5− ⋅ + = a 5 20 + = 5a
52. If chord PQ subtends an angle θ at the vertex of y2 = 4ax, then tan θ =
(A) 2
73
(B) 2
73
− (C)
25
3 (D)
25
3
−
52. (D)
Slope of OP = 121
2 at
at
=
1
2
t = m1
Slope of OQ = 2
2
t = m2
tanθ = 1 2
1 2
m m
1 m m
−+
= 1 2
1 2
2 2
t t
2 21
t t
−
+ ⋅ = 2 1
1 2
t t2
t t 4
−+
= 5
21 4− +
= 2 5
3−
as ‘θ’ is 2nd
quadrant.
( )( )1 2 1 2T at t , a t t +
( )21 1P at , 2at
( )22 2Q at , 2at
IIT JEE 2013 Advanced : Question Paper & Solution (Paper – II) (29)
29
Paragraph for Question 53 and 54
Let 1 2 3S S S S= ∩ ∩ , where
{ }1S z C : z 4= ∈ < , 2
z 1 3 iS z C : Im 0
1 3 i
⎧ ⎫⎡ ⎤− +⎪ ⎪= ∈ >⎨ ⎬⎢ ⎥−⎪ ⎪⎣ ⎦⎩ ⎭
and
{ }3S z C : Re z 0= ∈ >
53. Area of S =
(A) 10
3
π (B)
20
3
π (C)
16
3
π (D)
32
3
π
53. (B)
S1 = {z ∈ |z| < 4}
S2 = z 1 3 i
z c : Im 01 3 i
⎧ ⎫⎡ ⎤ − + ⎪ ⎪ ∈ > ⎨ ⎬⎢ ⎥ − ⎪ ⎪⎣ ⎦⎩ ⎭
Let z = x + iy
z 1 3 i
1 3 i
− + −
= (x iy) 1 3 i
1 3 i
+ − + −
= ( )(x 1) i y 3 (1 3 i)
1 3 i (1 3 i)
⎡ ⎤ − + + + ⎣ ⎦ × − +
= { } { }(x 1) 3 (y 3) i 3(x 1) (y 3
4
− − + + − + +
z 1 3 iIm
1 3 i
⎛ ⎞ − + ⎜ ⎟⎜ ⎟ − ⎝ ⎠
= 3 x y
4
+
z 1 3 iIm 0
1 3 i
⎛ ⎞ − + > ⎜ ⎟⎜ ⎟ − ⎝ ⎠
3 x y
04
+ >
3 x + y > 0
y
y′
xx′ A
B
C
D (0, −4)
(4, 0)
(0, 4)
(−4, 0)
x O
(30) Vidyalankar : IIT JEE 2013 − Advanced : Question Paper & Solution
30
S3 = {z ∈ c, Re(z) > 0}
x > 0
S1 ∩ S2 ∩ S3
Area of sector is = π × (4)2 ×
5 / 6
2
ππ
=16 5
6 2
π × × ×
= 20
3
π
54. min
1 3i zz S − − =∈
(A) 2 3
2
− (B)
2 3
2
+ (C)
3 3
2
− (D)
3 3
2
+
54. (C)
min |1 − 3i − z|
z ∈ s
|1 − 3i − z| = |z − (3i − 1)|
= |z − (−1 + 3i)|
PM = 3 3
2
− (Perpendicular distance is minimum distance)
y
xO
(4, 0)
(0, 4)x2 + y2 = 42
y + 3 x = 0
(2, −2 3 )
π/3
y
xO
(4, 0)
(0, 4)
(2, −2 3 )
y′
x′ M
P(1, −3)
IIT JEE 2013 Advanced : Question Paper & Solution (Paper – II) (31)
31
Paragraph for Question 55 and 56
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.
55. 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 (B)
90
648 (C)
558
648 (D)
566
648
55. (A)
1W
3R
2B
2W
3R
4B
3W
4R
5B
E1 = one white from each box
E2 = one red from each box
E3 = one black from each box
B1 B2 B3
E1, E2, E3 are mutually exclusive events
P(E) = P(E1) + P(E2) + P(E3)
= 1 2 3 3 3 4 2 4 5
6 9 12 6 9 12 6 9 12
⎛ ⎞ ⎛ ⎞ ⎛ ⎞× × + × × + × ×⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
= 82
648
56. 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) 116
181 (B)
126
181 (C)
65
181 (D)
55
181
56. (D)
2BP
E
⎛ ⎞⎜ ⎟⎝ ⎠
=
1 1
3 61 1 1 1 1 2
3 5 3 6 3 11
×
⎛ ⎞ ⎛ ⎞ ⎛ ⎞× + × + ×⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
= 55
181
P(B1) = 1
3
P(B2) = 1
3
P(B3) = 1
3
2
E 1P
B 6
⎛ ⎞ =⎜ ⎟⎝ ⎠
1
E 1P
B 5
⎛ ⎞ =⎜ ⎟⎝ ⎠
3
E 2P
B 11
⎛ ⎞ =⎜ ⎟⎝ ⎠
(32) Vidyalankar : IIT JEE 2013 − Advanced : Question Paper & Solution
32
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. Match List − I with List − II and select the correct answer using the code given below the
lists :
List I List II
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
1. 100
Q
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
2. 30
R
Area of a triangle with adjacent sides determined by vectors
a and b is 20. Then the area of the triangle with adjacent sides
determined by vectors ( ) ( )2a 3b and a b+ − is
3. 24
S
Area of a parallelogram with adjacent sides determined by vectors
a and b is 30. Then the area of the parallelogram with adjacent
sides determined by vectors ( )a b and a+ is
4. 60
Codes :
P Q R S
(A) 4 2 3 1
(B) 2 3 1 4
(C) 3 4 1 2
(D) 1 4 3 2 57. (C)
(P) Given volume of parallelepiped formed by a,b,c is equal to 2
⇒ [ a b c ] = 2 (1)
∴ volume of parallelepiped formed by 2 ( )a b× , 3 ( )b c× and c a× is given by
v = ( ) ( )2 a b , 3 b c , c a⎡ ⎤× × ×⎣ ⎦
or v = 6 [ a b c ]2
⇒ v = 24 ∴ p ⇒ 3
(Q) given [ a b c ] = 5
∴ volume of parallelepiped formed by
3 ( )a b+ , b c+ , and 2 ( )c a+ is given by
v = ( ) ( )3 a b , b c, 2 c a⎡ ⎤+ + +⎣ ⎦
v = 6 a b, b c, c a⎡ ⎤+ + +⎣ ⎦
v = 12 [ a b c ] ⇒ v = 60
Q ⇒ 4
IIT JEE 2013 Advanced : Question Paper & Solution (Paper – II) (33)
33
(R) given ( )1a b
2× = 20
⇒ a b× = 40 (1)
Area of triangle with adjacent sides
( )2a 3b+ and ( )a b− is given as
Δ = ( ) ( )12a 3b a b
2+ × −
Δ = ( )15 a b
2− × = 100
∴ R → 1
(S) given a b× = 30
∴ area of parallelogram with sides
a b+ and a is given as
area = ( )a b a+ ×
= b a× = 30
∴ S → 2
∴ P Q R S
3 4 1 2 Answer is (C)
58. Consider the lines 1 2
x 1 y z 3 x 4 y 3 z 3L : , L :
2 1 1 1 1 2
− + − + += = = =−
and the planes
P1 : 7x + y + 2z = 3, P2 : 3x + 5y −6z = 4. Let ax + by + cz = d be the equation of
the plane passing through the point of intersection of lines L1 and L2, 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 :
P Q R S
(A) 3 2 4 1
(B) 1 3 4 2
(C) 3 2 1 4
(D) 2 4 1 3
58. (A)
Let L1 : x 1
2
− =
y
1− =
z 3
1
+ = t
and L2 : x 4
1
− =
y 3
1
+ =
z 3
2
+ = s
∴ any point on L1 is M1 (2t + 1, −t, t−3)
and any point on L2 is M2 (s + 4, s − 3, 2s − 3)
(34) Vidyalankar : IIT JEE 2013 − Advanced : Question Paper & Solution
34
at the point on integration
2t + 1 = s + 4 …(1)
−t = s − 3 …(2)
t − 3 = 2s − 3 …(3)
Solving (1), (2) and (3) we get
t = 2 and s = 1
∴ M1 ≡ M2 ≡ (5, −2, −1)
Also if the plane is perpendicular to P1and P2 then
7a + b + 2c = 0 …(4)
and 3a + 5b − 6c = 0 …(5)
∴ b = −3a; c = −2a
hence the plane becomes,
ax − 3ay − 2az = d …(6)
using M1 or M2 in (6) we get
5a + 6a + 2a = d ⇒ d = 13a
∴ the plane becomes
ax − 3ay − 2az = 13a
or x − 3y − 2z = 13 (a ≠ 0)
∴ (P) a = 1 P → 3
(Q) b = −3 Q → 2
(R) c = −2 R → 4
(S) d = 13 S → 1
∴ P Q R S
3 2 4 1
59. Match List I with List II and select the correct answer using the code given below
the lists :
List I List II
P ( ) ( )( ) ( )
1 22
1 1
4
2 1 1
cos tan y ysin tan y1y
y cot sin y tan sin y
− −
− −
⎛ ⎞⎛ ⎞+⎜ ⎟⎜ ⎟ +⎜ ⎟⎜ ⎟+⎜ ⎟⎝ ⎠⎝ ⎠
takes value 1. 1 5
2 3
Q
If cos x + cos y + cos z = 0 = sin x + sin y + sin z then possible
value of x y
cos2
− is
2. 2
R
If cos x cos 2x sin x sin 2x sec x cos x sin 2x sec x4
cos x cos 2x4
π⎛ ⎞− + =⎜ ⎟⎝ ⎠π⎛ ⎞+ +⎜ ⎟⎝ ⎠
then possible value of sec x is
3. 1
2
S If ( ) ( )( )1 2 1cot sin 1 x sin tan x 6 , x 0− −− = ≠ ,
then possible value of x is
4. 1
IIT JEE 2013 Advanced : Question Paper & Solution (Paper – II) (35)
35
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
59. (B)
(P) Now,
( ) ( )( ) ( )
1/21 1
4
2 1 1
cos tan y y sin tan y1y
y cot sin y tan sin y
− −
− −
⎛ ⎞⎛ ⎞+ ⎜ ⎟⎜ ⎟ + ⎜ ⎟⎜ ⎟ + ⎜ ⎟⎝ ⎠⎝ ⎠
=
1/22
1 1
2 24
2 211
2
1 ycos cos y sin sin
1 y 1 y1y
y y1 y tancot cot tany 1 y
− −
−−
⎛ ⎞⎛ ⎞⎛ ⎞ ⎛ ⎞⎜ ⎟+ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟+ + ⎝ ⎠ ⎝ ⎠⎜ ⎟+⎜ ⎟⎜ ⎟⎛ ⎞ ⎛ ⎞⎜ ⎟−⎜ ⎟⎜ ⎟+ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟−⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠⎝ ⎠
=
1/22
2
2 24
2 2
2
1 y
1 y 1 y1y
y 1 y y
y 1 y
⎛ ⎞⎛ ⎞⎜ ⎟+⎜ ⎟⎜ ⎟+ +⎜ ⎟ +⎜ ⎟⎜ ⎟⎜ ⎟−⎜ ⎟+⎜ ⎟⎜ ⎟⎜ ⎟−⎝ ⎠⎝ ⎠
= 1
∴ P → 4
(Q) From the given equations we have
cos z = − (cos x + cos y) …(1)
sin z = −(sin x + sin y) …(2)
Squaring and adding (1) and (2) we get
1 = cos2x + cos
2y + 2 cos x cos y + sin
2x + sin
2y + 2 sin x sin y
⇒ 2 + 2 cos (x − y) = 1
⇒ cos (x − y) = 1
2−
⇒ 2 x y2 cos 1
2
−⎛ ⎞ −⎜ ⎟⎝ ⎠
= 1
2−
⇒ x y
cos2
−⎛ ⎞⎜ ⎟⎝ ⎠
= 1
2
Q → 3
(R) The given equation can be simplified as
2sin sin x cos 2x4
π = sin 2x sec x (cos x sin x) −
⇒ 2 sin x cos 2x = sin 2x sec x (cos x sin x) −
⇒ 1
sin 2x cos 2x2
= ( )sin 2x cos x sin x −
(36) Vidyalankar : IIT JEE 2013 − Advanced : Question Paper & Solution
36
⇒ ( )cos 2xsin 2x cos x sin x
2
⎛ ⎞ − − ⎜ ⎟⎝ ⎠
= 0
⇒ ( )2 2cos x sin x
sin x cos x sin x2
⎛ ⎞ − 2 − − ⎜ ⎟⎝ ⎠
= 0
⇒ cos x sin x
sin 2x (cos x sin x) 12
+ ⎛ ⎞ − − ⎜ ⎟⎝ ⎠
= 0
⇒ ( )sin 2x cos x sin x sin x 14
⎛ π ⎞⎛ ⎞ − + −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ = 0
⇒ x = 4
π ∴ sec x = 2
R → 2
(S) Now,
( )1 2cot sin 1 x− − = 1
2
xcot cot
1 x
−⎛ ⎞ ⎜ ⎟−⎝ ⎠
∴ ( )1 2cot sin 1 x− − = 2
x
1 x − …(1)
and ( )( )1sin tan x 6− = 1
2
x 6sin sin
1 6x
−⎛ ⎞⎜ ⎟⎜ ⎟+⎝ ⎠
∴ ( )( )1sin tan x 6− = 2
x 6
1 6x+ …(2)
From (1) and (2) we get
2
x
1 x− =
2
x 6
1 6x + (∵ x ≠ 0)
Solving this we x2 =
5
12
or x = 1 5
2 3
S → 1
∴ P Q R S
4 3 2 1
60. A line L : y = mx + 3 meets y − axis at E (0, 3) and the arc of the parabola
y2 = 16x, 0 ≤ y ≤ 6 at the point F(x0, y0). The tangent to the parabola at F(x0, y0)
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 and List II and select the correct answer using the code given below
the lists :
IIT JEE 2013 Advanced : Question Paper & Solution (Paper – II) (37)
37
List I List II
P m = 1. 1
2
Q Maximum area of ΔEFG is 2. 4
R y0 = 3. 2
S y1 = 4. 1
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
60. (A)
Tangent at P(x0, y0) to y2 = 16x is yy0 = 2 × 4(x + x0)
S0, G = 0
0
8x0,
y
⎛ ⎞⎜ ⎟⎝ ⎠
As F (x0, y0) lies on y2 = 16x
⇒ 2
0y = 16x0 … (1)
Δ = 1
2 × EG × x0
= 2
0 0
0
8x y13
2 y 16
⎛ ⎞× − ×⎜ ⎟⎝ ⎠
= 00 0
y1[3y 8x ]
2 16− ×
= 2 3
0 0
1[6y y ]
64− … (2)
F(x0, y0) = (1, 4)(0, y1) = 0
0
8x0,
y
⎛ ⎞⎜ ⎟⎝ ⎠
E
G
y2 = 16x
(38) Vidyalankar : IIT JEE 2013 − Advanced : Question Paper & Solution
38
Δ′ = 2
0 0
1[12y 3y ]
64−
Δ′ = 0 ⇒ 3y0 (4 − y0) = 0
y0 = 0 or y0 = 4
0y 4=
′′Δ = 0
1[12 6y ]
64− < 0
⇒ local maximum when y0 = 4
So, put in (2) y0 = 4,
Δmax = 2 31[6 4 4 ]
64× − =
1[6 4]
4− =
1
2
From (1), 2
0y = 16x0
⇒ 16 = 16x0
⇒ x0 = 1
y1 = 0
0
8x
y =
8.1
4 = 2
Line GF, y 3
x
− =
4 3
1 0
−−
⇒ y = x + 3
So, slope m = 1
Hence, (A) option correct.
IIT JEE 2013 Advanced : Question Paper & Solution (Paper – II) (39)
39