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JEE Advance Exam 2015 (Solution)
Part I - PHYSICS
SECTION – 1 (Maximum Marks : 32)
• This Section contains EIGHT questions • The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both inclusive. • For each question, darken the bubble corresponding to the correct integer in the ORS • Marking scheme : + 4 If the bubble corresponding to the answer is darkened 0 In all other cases
Q.1 The densities of two solid spheres A and B of the same radii R vary with radial distance r as ρA(r) = k
⎟⎠⎞
⎜⎝⎛
Rr and ρB(r) = k
5
Rr
⎟⎠⎞
⎜⎝⎛ , respectively, where k is a constant. The moments of inertia of the individual
spheres about axes passing through their centres are IA and IB, respectively. If A
B
II =
10n , the value of n
is
Ans. [6]
Sol.
dr
r
Consider a shell of radius r and thickness dr
mass of element = ρ × 4πr2dr
Moment of inertial of shell = dI = 32 ρ × 4πr2dr r2
dI = 32 × ρ × 4πr4dr
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dI = 32 ×
Rkr × 4πr4 dr
dI = 32
Rk × 4πr5 dr
I = 32
Rk × ∫π
R
0
5 ]drr[4
IA = 32
Rk ×
6R4 6π =
32 k × 4π
6R5
For I2 = 32 × 5
5
Rkr × 4πr4dr
I2 ⇒ 32 × 5R
k × 4π ∫R
0
9drr
IB = 32 × 5R
k × 10R4 10π =
32 × k ×
10R4 5π
A
B
II =
106
Q.2 Four harmonic waves of equal frequencies and equal intensities I0 have phase angles 0, π/3, 2π/3 and π.
When they are superposed, the intensity of the resulting wave is nI0. The value of n is
Ans. [3]
Sol.
60º
A A
AA
I0 = kA2
Resultant amplitude = 3 A
IR = k3A2
IR = 3I0
n = 3
Q.3 For a radioactive material, its activity A and rate of change of its activity R are defined as A = – dtdN
and R = – dtdA , where N(t) is the number of nuclei at time t. Two radioactive sources P (mean life τ) and
Q (mean life 2τ) have the same activity at t = 0. Their rates of change of activities at t = 2τ are RP and
RQ, respectively. If Q
P
RR =
en , then the value of n is
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Ans. [2]
Sol. R = – dtdA
A = – dtdN
At time t
A = A0e–λt
R = – dtdA = – A0e–λt (–λ)
R = A0λe–λt
λP = τ1 λQ =
τ21
RP = A0 λP τ
τ− 21
e ⇒ A0λPe–2
RQ = A0λQ τ×
τ− 2
21
e = A0λQe–1
Q
P
RR =
Q
P
λλ 1
2
ee
−
−
⇒ e2
∴ n = 2
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Q.4 A monochromatic beam of light is incident at 60º on one face of an equilateral prism of refractive index
n and emerges from the opposite face making an angle θ(n) with the normal (see the figure). For
n = 3 the value of θ is 60º and dndθ = m. The value of m is
60ºθ
Ans. [2]
Sol.
60ºθ r (60–r)
A
B C Applying snell law at face AB.
sin 60º = n × sin r
sin r = n23 ....(1)
Applying snell law at face AC
n × sin (60 – r) = 1 × sin θ ....(2)
if n = 3 , θ = 60º
3 sin (60 – r) = sin 60
sin (60 – r) = 21
60 – r = 30
r = 30º
from (1) sin 30º = n23
n = 3
differentiating (2) with respect to n
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1 × sin (60 – r) + n × cos (60 – r) ⎥⎦⎤
⎢⎣⎡−
dndr = cos θ
dndθ ....(3)
differentiating (1) with respective to n
cos r dndr =
23 ⎥⎦
⎤⎢⎣⎡−
2n1 ....(4)
when n = 3 r = 30
cos 30º dndr =
23 ⎥⎦
⎤⎢⎣⎡−
31 Q –
dndr =
31
Put r = 30º, θ = 60º and – dndr =
31 in equation (3)
sin (60 – 30) + 3 × cos (60 – 30) ⎥⎦⎤
⎢⎣⎡31 =
21
dndθ ∴
dndθ = 2
Q.5 In the following circuit, the current through the resistor R (= 2Ω) is I Amperes. The value of I is
R (= 2Ω) 1Ω
2Ω 8Ω
6Ω4Ω
2Ω
10Ω
12Ω 4Ω
6.5V
Ans. [1]
Sol.
2Ω 1Ω
2Ω
6Ω4Ω
2Ω
10Ω
12Ω 4Ω
6.5V
balanced wheat stone bridge
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2Ω 1Ω
2Ω
6Ω4Ω
2Ω
10Ω
12Ω 4Ω
6.5V
2Ω
2Ω 6Ω
10Ω
12Ω
6.5V
4Ω
balanced wheat stone bridge
2Ω
2Ω
6.5V
4Ω
6Ω
12Ω
i
Req. = 6.5Ω
.Amp15.65.6i ==
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Q.6 An electron in an excited state of Li2+ ion has angular momentum 3h/2π. The de Broglie wavelength of
the electron in this state is pπa0 (where a0 is the Bohr radius). The value of p is
Ans. [2]
Sol. L = π2
nh
π2h3 =
π2nh
∴ n = 3
if n = 3
λ (λ = de Broglie wavelength)
3λ = length of orbit
length of orbit = 2πrn
= Z
na2 20π =
33a2 2
0π
= 6πa0
3λ = 6πa0
λ = 2πa0
So value of p = 2
Q.7 A large spherical mass M is fixed at one position and two identical point masses m are kept on a line
passing through the centre of M (see figure). The point masses are connected by a rigid massless rod of
length l and this assembly is free to move along the line connecting them. All three masses interact only
through their mutual gravitational interaction. When the point mass nearer to M is at a distance r = 3l
from M, the tension in the rod is zero for m = k ⎟⎠⎞
⎜⎝⎛
288M . The value of k is
r l
m mM
Ans. [7]
Sol. F1 = 29GMm
l – 2
2Gml
(Force between M and middle m)
F2 = 216GMm
l + 2
2Gml
(Force between M and extreme m)
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F1 = F2
29GMm
l – 2
2Gml
= 216GMm
l + 2
2Gml
M ⎥⎦⎤
⎢⎣⎡ −
161
91 = 2m
M ⎥⎦⎤
⎢⎣⎡ −
144916 = 2m
m = 2887 M
m = 7
Q.8 The energy of a system as a function of time t is given as E(t) = A2 exp(–αt), where α = 0.2 s–1. The
measurement of A has an error of 1.25 %. If the error in the measurement of time is 1.50 %, the
percentage error in the value of E(t) at t = 5 s is
Ans. [4]
Sol. E = A2e–αt
E = A2e–0.2t
ln E = 2lnA – 0.2t
E
dE = 2 A
dA – 0.2 dt
E
dE = 2 A
dA – 0.2 dt
t
dt = 1.5 % ∴ dt = 1.5 × 5 % = 1.5 × 5 = 7.5
= 2 × 1.25 + 0.2 × 7.5
= 2.5 + 1.5
= 4 %
SECTION – 2 (Maximum Marks : 32)
• This section contains EIGHT questions • Each questions has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four
options(s) is(are) correct • For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS • Marking scheme : +4 If only the bubble(s) corresponding to all the correct option(s) is (are) darkened 0 If none of the bubbles is darkened –2 In all other cases
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Q.9 A parallel plate capacitor having plates of area S and plate separation d, has capacitance C1 in air. When
two dielectrics of different relative permittivities (ε1 = 2 and ε2 = 4) are introduced between the two
plates as shown in the figure, the capacitance becomes C2. The ratio 1
2
CC is
S/2
S/2ε1
ε2
d/2
d
+ –
(A) 6/5 (B) 5/3 (C) 7/5 (D) 7/3
Ans. [D]
Sol. When no dielectric is filled then C1 = dS0ε
when dielectric is filled
ε1 = 2 ε2 = 42d
2d
2S
2S
ε1 = 2
d
C2 = d2
S10εε +
⎥⎦
⎤⎢⎣
⎡ε
+ε
ε
21
0
2d
2d2
S
= d2
S20 ×ε +
⎥⎦⎤
⎢⎣⎡
×+
×
ε
42d
22d2
S0
⇒ dS0ε +
d328S0
××ε
⇒ dS0ε ⎥⎦
⎤⎢⎣⎡
614 ⇒
dS0ε ⎥⎦
⎤⎢⎣⎡
37
1
2
CC =
37
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Q.10 An ideal monoatomic gas is confined in a horizontal cylinder by a spring loaded piston (as shown in the
figure). Initially the gas is at temperature T1, pressure P1 and volume V1 and the spring is in its relaxed
state. The gas is then heated very slowly to temperature T2, pressure P2 and volume V2. During this
process the piston moves out by a distance x. Ignoring the friction between the piston and the cylinder.
the correct statements (s) is (are)
(A) If V2 = 2V1 and T2 = 3T1, then the energy stored in the spring is 41 P1V1
(B) If V2 = 2V1 and T2 = 3T1, then the change in internal energy is 3P1V1
(C) If V2 = 3V1 and T2 = 4T1, then the work done by the gas is37 P1V1
(D) If V2 = 3V1 and T2 = 4T1, then the heat supplied to the gas is 6
17 P1V1
Ans. [A,B,C]
Sol.
P1 = P0
P0
x0
PP0
x0 + x
kx
(A) P = P1 + A
Kx
P2 = 23 P1 ⇒ x =
AV1
2P3 1 = P1 +
AKx
Kx = 2AP1
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Energy of spring
21 Kx2 =
4AP1 x =
4VP 11 Ans. A
(B) ΔU = 2f (P2V2 – P1V1)
= 3P1V1 Ans. B
(C) Pf = 3P4 1 KX =
3P1 A , X =
AV2 1
Wgas = – (atmPW + Wspring)
= (P1Ax + 21 Kx . x)
= + ⎟⎠⎞
⎜⎝⎛ +
AV2.
3AP.
21
AV2.AP 111
1
= 2P1V1 + 3VP 11 =
3VP7 11
(D) ΔQ = W + ΔU
= 3VP7 11 +
23 (P2V2 – P1V1)
= 3VP7 11 +
23 ⎟
⎠⎞
⎜⎝⎛ − 1111 VPV3.P
34
= 3VP7 11 +
29 P1V1 =
6VP41 11
Q.11 A fission reaction is given by U23692 → Xe140
54 + Sr9438 + x + y, where x and y are two particles. Considering
U23692 to be at rest, the kinetic energies of the products are denoted by KXe KSr, Kx(2MeV) and
Ky(2MeV), respectively. Let the binding energies per nucleon of U23692 , Xe140
54 and Sr9438 be 7.5 MeV,
8.5 MeV and 8.5 MeV, respectively. Considering different conservation laws, the correct option (s) is (are)
(A) x = n , y = n, KSr = 129 MeV, KXe = 86 MeV
(B) x = p , y = e–, KSr = 129 MeV, KXe = 86MeV
(C) x = p , y = n, KSr = 129 MeV, KXe = 86 MeV
(D) x = n , y = n, KSr = 86 MeV, KXe = 129 MeV
Ans. [A]
Sol. Q value of reaction, Kx + Ky + KXe + KSr – UK23692
Q = 2 + 2 + KXe + KSr – 0
Q = 4 + KXe + KSr
Q = 8.5 × 140 + 8.5 × 94 – 7.5 × 236
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Q = 219 MeV
219 – 4 = KXe + KSr
KXe + KSr = 215 MeV
From momentum conservation
heavy particle must have small kinetic energy.
Q.12 Two spheres P and Q of equal radii have densities ρ1 and ρ2, respectively. The spheres are connected by
a massless string and placed in liquids L1 and L2 of densities σ1 and σ2 and viscosities η1 and η2,
respectively. They float in equilibrium with the sphere P in L1 and sphere Q in L2 and the string being
taut (see figure). If sphere P alone in L2 has terminal velocity PVr
and Q alone in L1 has terminal
velocity QVr
, then
Q
PL1
L2
(A) |V||V|
Q
Pr
r
= 2
1
ηη (B)
|V||V|
Q
Pr
r
= 1
2
ηη
(C) PVr
. QVr
> 0 (D) PVr
. QVr
< 0
Ans. [A,D]
Sol.
T
σ1Vg
ρ1Vg
T
ρ2Vg
σ2 Vg
(σ1 + σ2) Vg = (ρ1 + ρ2) Vg σ1 + σ2 = ρ1 + ρ2
ρ1 – σ2 = σ1 – ρ2
If ρ1 > σ2 then σ1 > ρ2
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In separate liquid one body will get terminal velocity upward then other one will get downward terminal
velocity.
PV→
. QV→
< 0 (angle between pV→
and QV→
is π)
& for terminal velocity V ∝ η1
|V|
|V|
Q
P→
→
= ⎟⎟⎠
⎞⎜⎜⎝
⎛ηη
2
1 [because P in liquid L2 and Q in liquid L1]
Q.13 In terms of potential difference V, electric current I, permittivity ε0, permeability µ0, and speed of light
c, the dimensionally correct equation (s) is (are)
(A) µ0I2 = ε0V2 (B) ε0I = µ0V (C) I = ε0cV (D) µ0cI = ε0V
Ans. [A,C]
Sol. W = qV
ML2T–2 = ATV
V = M1L2T–3A–1
I ⇒ A
F = 04
1πε
2
2
rq
ε0 = 2
2
Frq
ε0 = 22
22
LMLTTA
×−
ε0 = M–1L–3T4A2
c = LT–1 Similarly µ0I2 = ε0V2
I = ε0cV
A = M0L0T0A1
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Q.14 Consider a uniform spherical charge distribution of radius R1 centred at the origin O. In this distribution,
a spherical cavity of radius R2, centred at P with distance OP = a = R1 – R2 (see figure) is made. If the
electric field inside the cavity at position rr is E
r( rr ), then the correct statement (s) is (are)
R1
O
aPR2
(A) Er
is uniform, its magnitude is independent of R2 but its direction depends on rr
(B) Er
is uniform, its magnitude depends on R2 and its direction depends on rr
(C) Er
is uniform, its magnitude is independent of a but its direction depends on ar
(D) Er
is uniform and both its magnitude and direction depend on ar
Ans. [D]
Sol.
ar
r1
a
r2
P
Strength of electric field at point P is given by Er
= )rr(3 21
0
rr−
ερ .... (1)
because in solid sphere Er
= r3 0
r
ερ
Q ar = 21 rr
rr−
∴by eq(1)
Er
= a3 0
r
ερ
Q | ar | = R1 – R2 = constant
∴ magnitude of Er
is constant and direction of Er
is depends on direction of ar
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Q.15 In plotting stress versus strain curves for two materials P and Q, a student by mistake puts strain on the
y-axis and stress on the x-axis as shown in the figure. Then the correct statements (s) is (are)
PQ
Stra
in
Stress (A) P has more tensile strength than Q
(B) P is more ductile than Q
(C) P is more brittle than Q
(D) The Young's modulus of P is more than that of Q
Ans. [A,B]
Sol. Actual graph is
PQSt
ress
Strain
Q Length of linear part of P is large then Q so P has more tensile strength than Q.
Q Curve part of P is large the curve part of Q so P is more ductile than Q.
Q (Slope)Q > (Slope)P
∴ Young's modulus of Q > Young's modulus of P
Q.16 A spherical body of radius R consists of a fluid of constant density and is in equilibrium under its own
gravity. If P(r), is the pressure at r(r < R), then the correct option (s) is (are)
(A) P(r = 0) =0 (B) )3/R2r(P)4/R3r(P
== =
8063
(C) )5/R2r(P)5/R3r(P
== =
2116 (D)
)3/Rr(P)2/Rr(P
== =
2720
Ans. [B,C]
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Sol.
dm EG
r
Pds
(P + dp)ds
R
Gravitational field
GE = 34 πρGr
EG ∝ r s
EG = Cr ⎥⎥⎦
⎤
⎢⎢⎣
⎡
=
πρ=
.constCG
34C
(P + dp) ds + dm EG = Pds
dpds = – dm EG
dpds = – ρ dsdr . EG
dp = – ρEGdr
∫P
0
dp = ∫ ρ−r
R
drCr [pressure at surface is zero]
P = ρC r
R
2
2r
⎥⎦
⎤⎢⎣
⎡−
P = 2Cρ [R2 – r2]
(1) P(r = 0) ≠ 0
(ii) ⎟⎠⎞
⎜⎝⎛ =
⎟⎠⎞
⎜⎝⎛ =
3R2rP
4R3rP
=
⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢
⎣
⎡
⎟⎠⎞
⎜⎝⎛−
⎟⎠⎞
⎜⎝⎛−
22
22
3R2R
4R3R
= 8063
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(iii) ⎟⎠⎞
⎜⎝⎛ =
⎟⎠⎞
⎜⎝⎛ =
5R2rP
5R3rP
= 22
22
5R2R
5R3R
⎟⎠⎞
⎜⎝⎛−
⎟⎠⎞
⎜⎝⎛−
= 2116
(iv) ⎟⎠⎞
⎜⎝⎛ =
⎟⎠⎞
⎜⎝⎛ =
3RrP
2RrP
= 22
22
3RR
2RR
⎟⎠⎞
⎜⎝⎛−
⎟⎠⎞
⎜⎝⎛−
= 3227
SECTION – 3 (Maximum Marks : 16)
• This section contains TWO paragraphs • Based on each paragraph, there will be TWO questions • Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four
option(s) is(are) correct • For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS • Marking scheme : +4 If only the bubble(s) corresponding to all the correct option(s) is(are) darkened 0 If none of the bubbles is darkened –2 In all other cases
PARAGRAPH 1
In a thin rectangular metallic strip a constant current I flows along the positive x-direction, as shown in
the figure. The length, width and thickness of the strip are l, w and d, respectively. A uniform magnetic
field Br
is applied on the strip along the positive y-direction. Due to this, the charge carriers experience
a net deflection along the z-direction. This results in accumulation of charge carriers on the surface
PQRS and appearance of equal and opposite charges on the face opposite to PQRS. A potential
difference along the z-direction is thus developed. Charge accumulation continues until the magnetic
force is balanced by the electric force. The current is assumed to be uniformly distributed on the cross
section of the strip and carried by electrons.
l
I
Q
I K
M z
x
y
RS
P d
w
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Q.17 Consider two different metallic strips (1 and 2) of the same material. Their lengths are the same, widths
are w1 and w2 and thicknesses are d1 and d2, respectively. Two points K and M are symmetrically
located on the opposite faces parallel to the x-y plane (see figure). V1 and V2 are the potential
differences between K and M in strips 1 and 2, respectively. Then, for a given current I flowing through
them in a given magnetic field strength B, the correct statement(s) is(are)
(A) If w1 = w2 and d1 = 2d2, then V2 = 2V1
(B) If w1 = w2 and d1 = 2d2, then V2 = V1
(C) If w1 = 2w2 and d1 = d2, then V2 = 2V1
(D) If w1 = 2w2 and d1 = d2, then V2 = V1
Ans. [A,D]
Sol. i = neAVd
Vd = neA
i
e–
B
Magnetic force on e = eVdB
due to magnetic force e drift towards face M. due to which an electric field is produce between two
faces.
Vd = BE
E = VdB
Potential difference between k & M = V
V = EW
V = VdBW
V = neAiBW ⇒
neWdiBW =
nediB
V∝ d1
∴ if d1 = 2d2 then V1 = 2
V2 or V2 = 2V1 Ans. (A)
if d1 = d2 then V1 = V2 Ans. (D)
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Q.18 Consider two different metallic strips (1 and 2) of same dimensions (length l, width w and thickness d)
with carrier densities n1 and n2, respectively. Strip 1 is placed in magnetic field B1 and strip 2 is placed
in magnetic field B2, both along positive y-directions. Then V1 and V2 are the potential differences
developed between K and M in strips 1 and 2, respectively. Assuming that the current I is the same for
both the strips, the correct option(s) is(are)
(A) If B1 = B2 and n1 = 2n2, then V2 = 2V1 (B) If B1 = B2 and n1 = 2n2, then V2 = V1
(C) If B1 = 2B2 and n1 = n2, then V2 = 0.5V1 (D) If B1 = 2B2 and n1 = n2, then V2 = V1
Ans. [A,C]
Sol. V = nediB
d is same for both strip
V ∝ nB
if B1 = B2
V ∝ n1
n1 = 2n2
V1 = 2
V2
V2 = 2V1 Ans. (A)
if n1 = n2
V ∝ B
if B1 = 2B2 then V1 = 2V2
V2 = 0.5 V1 Ans. (C)
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PARAGRAPH 2
Light guidance in an optical fiber can be understood by considering a structure comprising of thin solid
glass cylinder of refractive index n1 surrounded by a medium of lower refractive index n2. The light
guidance in the structure takes place due to successive total internal reflections at the interface of the
media n1 and n2 as shown in the figure. All rays with the angle of incidence i less than a particular value
im are confined in the medium of refractive index n1. The numerical aperture (NA) of the structure is
defined as sin im.
Air Cladding
Coreθ n1
n2
i
n1 > n2
Q.19 For two structures namely S1 with n1 = 4/45 and n2 = 3/2, and S2 with n1 = 8/5 and n2 = 7/5 and
taking the refractive index of water to be 4/3 and that of air to be 1, the correct option(s) is(are)
(A) NA of S1 immersed in water is the same as that of S2 immersed in a liquid of refractive index 153
16
(B) NA of S1 immersed in liquid of refractive index 15
16 is the same as that of S2 immersed in water
(C) NA of S1 placed in air is the same as that of S2 immersed in liquid of refractive index 154
(D) NA of S1 placed in air is the same as that of S2 placed in water
Ans. [A,C]
Sol.
n2 B Cladding
r'
rA
i
Medium (μ)
n1
at B
sin C = 1
2
nn
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cos C = 1
22
21
nnn −
For TIR at B –
r' ≥ C & r + r' = 90º
or cosr' ≤ cos C r = 90 – r'
cosr' ≤ 1
22
21
nnn −
sinr = cosr'
Now at A – (using snell's Laws)
μ × sin i = n1sinr
= n1 cos r'
sini ≤ μ− 2
221 nn
i.e. sin im = μ− 2
221 nn
& i ≤ im
i.e. NA = μ− 2
221 nn
using this relation we get
Ans. A & C
Q.20 If two structures of same cross-sectional area, but different numerical apertures NA1 and NA2
(NA2 < NA1) are joined longitudinally, the numerical aperture of the combined structure is
(A) 21
21
NANANANA
+ (B) NA1 + NA2 (C) NA1 (D) NA2
Ans. [D]
Sol. On the basis of discussion in Q.19, we can conclude that
if NA2 < NA1 then 2mi <
1mi
for transmission (i ≤ im)
hence i ≤ 2mi
i.e. NA should be NA2 for combined structure.
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Part II - CHEMISTRY
SECTION – 1 (Maximum Marks: 32)
• This Section contains EIGHT questions
• This answer to each questions is a SINGLE DIGIT INTEGER ranging from 0 to 9. both inclusive
• For each questions, darken the bubble corresponding to the correct integer in the ORS
• Marking scheme :
+ 4 If the bubbler corresponding to the answer is darkened 0 In all other cases
Q.21 In the complex acetylbromidodicarbonylbis(triethylphosphine)iron(II), the number of Fe – C bond (s) is Ans. [3]
Sol. 1232
|
3
2])PEt()CO)(Br)(OCCH(Fe[ +
+
Fe
BrPEt3
OH – CH
H
O+ ≡ C–
C– ≡ O+
PEt3
C
Q.22 Among the complex ions, [Co(NH2 – CH2 – CH2 – NH2)2Cl2]+, [CrCl2(C2O4)2]3–, [Fe(H2O)4(OH)2]+,
[Fe(NH3)2(CN)4]–, [Co(NH2 – CH2 – CH2 – NH2)2(NH3)Cl]2+ and [Co(NH3)4(H2O)Cl]2+, the number of complex ion(s) that show(s) cis-trans isomerism is
Ans. [6] Sol. [Co(NH2 – CH2 – CH2 – NH2)2Cl2]+ shows one cis and one trans isomer [CrCl2(C2O4)2]3– shows one cis and one trans isomer [Fe(H2O)4(OH)2]+ shows one cis and one trans isomer [Fe(NH3)2(CN)4]– shows one cis and one trans isomer [Co(NH2 – CH2 – CH2 – NH2)2(NH3)Cl]2+ shows one cis and one trans isomer [Co(NH3)4(H2O)Cl]2+ shows one cis and one trans isomer Q.23 Three moles of B2H6 are completely reacted with methanol. The number of moles of boron containing
product formed is Ans. [6]
Sol. 3B2H6 + 18MeOH → 6B(OMe)3 + 18H2
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Q.24 The molar conductivity of a solution of a weak acid HX (0.01 M) is 10 times smaller than the molar
conductivity of a solution of a weak acid HY (0.10 M). If 0Y
0X −− λ≈λ , the difference in their pKa values,
pKa(HX) – pKa(HY), is (consider-degree of ionization of both acids to be << 1) Ans. [3]
Sol. Degree of ionization of weak acid HX(αHX) = ºHX
HX
ΛΛ
Degree of ionization of weak acid HY(αHY) = ºHY
HY
ΛΛ
∴ αHX = 101 αHY
⎭⎬⎫
⎩⎨⎧ Λ
=Λ10
HYHXQ
Ka = α2C {for α << 1)
HXaK = 2
HYα × 100
1 × C1 = 2HYα ×
1001 × 0.01 {Q C1 = 0.01 M}
HYaK = 2
HYα × C2 = 2HYα × 0.1 {Q C2 = 0.1 M}
HXaK =
10001
HYaK
HYapK = – log
HXaK = – log 1000K
HYa = 3 – log HYaK
HYapK = – log
HYaK = – logHYaK
HYkapK –
HYapK = 3
Q.25 A closed vessel with rigid walls contains 1 mole of U23892 and 1 mole of air at 298 K. Considering
complete decay of U23892 to Pb206
82 , the ratio of the final pressure to the initial pressure of the system at
298 K is Ans. [9] Sol. Initial mole of gas ni = 1 92U238 → 82Pb206 + 8 2He4 + 6 –1e0 Final mole of gas nf = 8 + 1 = 9 At const T, & V ; p ∝ n
19
nn
PP
i
f
i
f == ∴ Answer is 9
Q.26 In dilute aqueous H2SO4, the complex diaquodioxalatoferrate(II) is oxidized by −4MnO , For this
reaction, the ratio of the rate of change of [H+] to the rate of change of [ −4MnO ] is
Ans. [8]
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Sol. OH6CO4Fe1Mn1H8])OX()OH(Fe[1OMn1 2
4
23
3222
2
3
22
)2(
4
)7(+++→++
++
++++−
++−
+
5 × 1 = 5 (1 × 1) + (1 × 4) = 51 1
∴ 18
]MnO[ofchangeofrate]H[ofchangeofrate
4
=−
+
Q.27 The number of hydroxyl group(s) in Q is
H
HO H3C CH3
heat
H⎯→⎯+
P Cº0
)excess(KMnOdiluteaqueous 4 ⎯⎯⎯⎯⎯⎯⎯⎯ →⎯ Q
Ans. [4]
Sol. First reaction involve dehydration of alcohol and second reaction is formation of syndiol rearrangement
H
HO H3C CH3
heat
H⎯→⎯+
HO
H
H3C CH3
+
H
Δ
–H2O +
H3C CH3CH3 CH3
–H⊕
H3C
HO HO CH3
OH
OH
(Q)
Rearrangement 1-2 Methyl Shift ⊕
CH3 CH3
(P)
0ºC, aq. & dil. KMnO4 (excess) Baeyer reagent (syn add.)
Total number of hydroxyl groups in Q ⇒ 4
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Q.28 Among the following, the number of reaction(s) that produce(s) benzaldehyde is
I.
CuCl/AlClAnhydrous
HCl,CO
3
⎯⎯⎯⎯⎯ →⎯
II. CHCl2
Cº100
OH2⎯⎯ →⎯
III. COCl
4
2
BaSOPd
H
−⎯⎯ →⎯
IV CO2Me
OH
Cº78,–Toluene
HDIBAL
2
⎯⎯⎯⎯ →⎯ −
Ans. [4]
Sol. All four reaction produce benzaldehyde
I.
(Benzene) + CO + HCl
CuCl
AlClAnhydrous 3⎯⎯⎯⎯⎯ →⎯
C – H
O
Benzaldehyde
(Gatter mann's koch aldehyde syn.)
II.
CH – Cl
Cl
Cº100
OH2⎯⎯ →⎯
CH – OH
OH
⎯⎯ →⎯− OH2
C – H
O
Benzaldehyde
III.
C – Cl
O
Benzoyal chloride
4
2
BaSOPd
HcatalystLindlar
−⎯⎯⎯⎯⎯⎯ →⎯
C – H
O
Benzaldehyde [Rosenmund reduction]
IV.
—C — O — Me
O
Methyl Benzoate
reductionerestrictivOH
Cº78,–Toluene
HDIBAL
2
⎯⎯⎯⎯ →⎯ −
C – H
O
Benzaldehyde
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SECTION – 2 (Maximum Marks: 32)
• This Section contains EIGHT questions
• Each question has FOUR options (A), (B), (C) and (D) ONE OR MORE THAN ONE of these four option
(s) is (are) correct
• For each questions, darken the bubble (s) corresponding to all the correct option (s) in the ORS
• Marking scheme :
+4 If only the bubble (s) corresponding to all the correct option (s) is (are) darkened
0 If none of the bubbles is darkened
–2 In all other cases
Q.29 Under hydrolytic conditions, the compounds used for preparation of linear polymer and for chain
termination, respectively, are
(A) CH3SiCl3 and Si(CH3)4 (B) (CH3)2SiCl2 and (CH3)3SiCl
(C) (CH3)2SiCl2and CH3SiCl3 (D) SiCl4 and (CH3)3SiCl
Ans. [B]
Sol. (i) (CH3)2SiCl2 undergoes hydrolysis by two sides so it form linear polymer
(ii) (CH3)3 SiCl undergoes hydrolysis by one side so it is chain termination.
Q.30 When O2 is adsorbed on a metallic surface, electron transfer occurs from the metal to O2. The TRUE
statement(s) regarding this adsorption is(are)
(A) O2 is physisorbed
(B) heat is released
(C) occupancy of *p2π of O2 is increased
(D) bond length of O2 is increased
Ans. [B,C,D]
Sol. M O = O
Vacant π* ABMO
(i) Due to bonding, energy is released
(ii) Due to back donation bond length of O2 is increased
(iii) Due to presence of vacant π*2p orbital it accept some electron cloud from metal.
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Q.31 One mole of a monoatomic real gas satisfies the equation p(V – b) = RT where b is a constant. The
relationship of interatomic potential V(r) and interatomic distance r for the gas is given by-
(A)
V(r)
0 r (B)
V(r)
0 r
(C)
V(r)
0 r (D)
V(r)
0 r
Ans. [C]
Sol. when only repulsive force act.
V(r) = ∞ r < σ
& V(r) = 0 r > σ
V(r)
0 rσ
Q.32 In the following reaction, the product S is –
H3C i. O3
ii. Zn, H2O
NH3R S
(A)
H3C N
(B)
H3C N
(C) H3C
N
(D)
H3C
N
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Ans. [A]
Sol.
H3C i. O3
ii. Zn, H2O
NH3R S
H3C
i. O3 O
OO
H3C
Zn, H2O , Reductive cleavage
HHO
O
(R)
H3CNH3 O H NH
C H3C H
H3C OH
–H2OH3C
NN H
Q.33 The major product U in the following reactions is –
CH2=CH–CH3, H+
high pressure heat
radical initiator, O2T U
(A)
O O
CH3
H
(B)
O
H3C CH3O
H
(C)
O O
CH2
H
(D)
O
CH2
OH
||
Ans. [B]
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Sol. This reaction is a part of formation of phenol
CH2=CH–CH3, H+
high pressure heat
radical initiator, O2
Free radical intermediate
CH3 CH3
H
C
CH3–CH–CH3 Carbocation Intermadiate
⊕
H3C CH3
O
C
O H
Cumene
[Cumene hydroperoxide]
Q.34 In the following reactions, the major product W is
0° C V
NH2
NaNO2, HCl
OH
, NaOHW
(A)
N=N OH (B)
N=N
OH
(C)
N=N
OH
(D)
N=N
HO
Ans. [A]
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Sol.
N=N OH
NaNO2 + HCl HO–N=O + NaCl
N= O ⊕
(Attacking species)[Nitrosonium ion]
| NH2
+ N= O ⊕ O°C
–H2O
N≡N ⊕
|
ClΘ
Benzene Diazonium chloride (BDC)
(coupling reaction)
V OH
, NaOH
W In coupling reaction if para position is blocked then attack occur at ortho position
Q.35 The correct statement(s) regarding, (i) HClO, (ii) HClO2, (iii) HClO3, and (iv) HClO4, is (are)
(A) The number of Cl = O bonds in (ii) and (iii) together is two
(B) The number of lone pairs of electrons on Cl in (ii) and (iii) together is three
(C) The hybridization of Cl in (iv) is sp3
(D) Amongst (i) to (iv), the strongest acid is (i)
Ans. [B, C]
Sol. (i)
H–O Cl
(ii)
H–O Cl||O
(iii) H–O
Cl||O
||O
(iv)
H–O Cl = O|| O
|| O
sp3 , hybridizationstrongest acid
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Q.36 The pair(s) of ions where BOTH the ions are precipitated upon passing H2S gas in presence of dilute HCl, is(are) -
(A) Ba2+, Zn2+ (B) Bi3+, Fe3+ (C) Cu2+, Pb2+ (D) Hg2+, Bi3+ Ans. [C, D] Sol. Group number (I) and (II) radicals give precipitate
SECTION – 3 (Maximum Marks : 16)
• This section contains TWO Paragraphs • Based on each paragraph, there will be TWO questions • Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four
option(s) is (are) correct • Four each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS. • Marking scheme : + 4 If only the bubble(s) corresponding to all the correct option(s) is (are) darkened 0 If none of the bubbles is darkened –2 In all other cases
PARAGRAPH 1
In the following reactions
C8H6Pd-BaSO4
C8H8i. B2H6
ii. H2O2, NaOH, H2OX
H2O
HgSO4,H2SO4
C8H8O i. EtMgBr, H2Oii. H+, heat
H2
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Q.37 Compound X is -
(A)
O
CH3 (B)
OH
CH3
(C) OH
(D) CHO
Ans. [C]
Q.38 The major compound Y is -
(A) CH3 (B) CH3
(C)
CH2 CH3
(D)
CH3
CH3
Ans. [D]
Sol. The following reactions
88
2
468 HC
HBaSOPdHC −
DBE = (x + 1) –1/2 = (8 + 1) –6/2 = 9 – 3 = 6
Compound - C8H6 is
C≡C–H
Now,
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C ≡ C–H
+ H2 Pd
BaSO4
C = C H H
H
HgSO4.H2SO4 Hydration
HO–C=CH2 |
CH2–CH2 |
Reduction ( Lindlar-catalyst)
Syn Addition B2H6
B
3
H–OH
Tautomerism
O=C–CH3 |
EtMgBr, H2O
HO–C–CH3 |
Et |
HO–OH NaOH, H2O
CH2– CH2 –OH|
H+ Δ
C |
Et |Me OH2
⊕
H3C
H3C–HC–C–CH3 |
⊕ |H
CH= C |
CH3
(Saytzeff product)
[X] Antimarkownikov alcohol
–H2O
–H⊕
PARAGRAPH 2
When 100 mL of 1.0 M HCl was mixed with 100 mL of 1.0 M NaOH in an insulated beaker at constant
pressure, a temperature increase of 5.7ºC was measured for the beaker and its contents (Expt.1).
Because the enthalpy of neutralization of a strong acid with a strong base is a constant (–57.0 kJ mol–1),
this experiment could be used to measure the calorimeter constant. In a second experiment (Expt. 2),
100 mL of 2.0 M acetic acid (Ka = 2.0 × 10–5) was mixed with 100 mL of 1.0 M NaOH (under identical
conditions to Expt. 1) where a temperature rise of 5.6 ºC was measured.
(Consider heat capacity of all solutions as 4.2 J g–1 K–1 and density of all solutions as 1.0 g mL–1)
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Q.39 Enthalpy of dissociation (in kJ mol–1) of acetic acid obtained from the Expt. 2 is -
(A) 1.0 (B) 10.0 (C) 24.5 (D) 51.4
Ans. [A]
Sol. After experiment 1
mass of solution = 200× 1.0 = 200 g {because density of all solution = 1g/ml}
let C is calorimetric constant
Q = m.s.ΔT + C.ΔT.
7.5C1000
7.52.42007.5 ×+××
= {for 0.1 mol}
⇒ C = 0.16
After experiment 2
6.516.01000
6.52.420010x
×+××
=
x = 56
ΔHN(WASB) = ΔHN(SASB) + ΔHdiss. of WA
57 = 56 + ΔHdiss. of WA {for 1 mol}
ΔHdiss. of WA = 1 KJ /mol.
Q.40 The pH of the solution after Expt. 2 is -
(A) 2.8 (B) 4.7 (C) 5.0 (D) 7.0
Ans. [B]
Sol. CH3COOH + NaOH CH3COONa + H2O
Initial 0.2 0.1 0
Final 0.1 0 0.1
pH = pKa + log ]acid[
]baseconjugate[
pH = pKa = 4.7
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Part III - MATHEMATICS
SECTION – 1 (Maximum Marks: 32)
• This section contains EIGHT questions
• The answer to each questions is a SINGLE DIGIT INTEGER ranging from 0 to 9, both inclusive
• For each questions, darken the bubble corresponding to the correct integer in the ORS
• Marking scheme :
+ 4 If the bubble corresponding to the answer is darkened
0 In all other cases
Q.41 The coefficient of x9 in the expansion of (1 + x) (1 + x2) (1 + x3) ……. (1 + x100) is
Ans. [8]
Sol. Required will be same as that of coefficient of x9 in (1 + x) (1 + x2) …….(1 + x9)
Possible cases are
(i) x0 x9
(ii) x1 x8
(iii) x2 x7
(iv) x3 x6
(v) x4 x5
(vi) x1 x2 x6
(vii) x2 x3 x4
(viii) x1 x3 x5
Each give (+1) to coefficient of x9 = 8
Q.42 Suppose that the foci of the ellipse 5
y9
x 22+ = 1 are (f1, 0 ) and (f2, 0) where f1 > 0 and f2 < 0. Let P1 and
P2 be two parabolas with a common vertex at (0, 0) and with foci at (f1, 0) and (2f2 , 0), respectively.
Let T1 be a tangent to P1 which passes through (2f2, 0) and T2 be a tangent to P2 which passes through
(f1, 0). If m1 is the slope of T1 and m2 is the slope of T2, then the value of ⎟⎟⎠
⎞⎜⎜⎝
⎛+ 2
221
mm1 is
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Ans. [4]
Sol. Foci of the ellipse are (2, 0) and (–2, 0) So f1 = 2, f2 = – 2 Equation of T1 :
y2 = –16x y2 = 8x
(–4, 0) (2, 0)
T1 : y = m1x + 1m
2
Q T1 passes through (– 4, 0)
so 21m =
21
T2 : y = 2
2 m4xm −
T2 passes through (2, 0) ∴ 2
2m = 2
∴ 21m
1 + 22m = 4
Q.43 Let m and n be two positive integers greater than 1.
If ⎟⎠⎞
⎜⎝⎛−=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
α
−α
→α 2eeelim m
)cos(
0
n
then the value of nm is
Ans. [2]
Sol. ⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
α
−α
→α m
cos
0
eelimn
= ⎥⎥⎦
⎤
⎢⎢⎣
⎡
−α
−−α
1cos1ee n
1cos n
. 2n
n
)(1cos
α
−α m
2n )(α
α
= e(lne) ⎟⎠⎞
⎜⎝⎛ −
21 a2n – m =
2e−
So 2n – m = 0 m = 2n
∴ nm = 2
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Q.44 If
α = ∫ ⎟⎟⎠
⎞⎜⎜⎝
⎛
+
+−+1
02
2xtan3x9 dx
x1x912)e(
1
where tan–1x takes only principal values, then the value of ⎟⎠⎞
⎜⎝⎛ π
−α+4
3|1|loge is
Ans. [9]
Sol. Let x = tan θ
dx = sec2 θ dθ
α = ( )∫π
θ+θ θθ+4/
0
2tan93 d)tan912.(e
α = ( )∫π
θ+θ θ+θ4/
0
2tan93 d)3sec9.(e
Let 3θ + 9tanθ = t
(3 + 9 sec2 θ) dθ = dt
α = ∫
π+ π
+
⎥⎥
⎦
⎤
⎢⎢
⎣
⎡−=
439
0
439t 1edte
α + 1 = 439
eπ
+
ln |1 + α | – 4
3π = 9
Q.45 Let f : R → R be a continuous odd function, which vanishes exactly at one point and 21)1(f = . Suppose
that F(x) = ∫−
x
1
dt)t(f for all x ∈ [–1, 2] and G(x) = dt|))t(f(f|tx
1∫−
for all x ∈ [–1, 2]. If )x(G)x(Flim
1x→=
141 ,
then the value of f ⎟⎠⎞
⎜⎝⎛
21 is
Ans. [7]
Sol. As f(x) is continuous odd function
So f(0) = 0
Given f(1) = 21
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Paper-2 [ CODE – 5 ]
F(x) = ∫−
x
1
dt)t(f ∀ x ∈ [– 1, 2]
G(x) = ∫−
x
1
dt|))t(f(f|t ∀ x ∈ [– 1, 2]
)x(G)x(Flim
1x→=
141
⇒
∫
∫
−
−→ x
1
x
11x
dt|))t(f(f|t
dt)t(f
lim = |))x(f(f|x
)x(flim1x→
{Using L' Hospital's Rule}
⇒ ⎟⎠⎞
⎜⎝⎛
21f
21
= 141
⇒ ⎟⎠⎞
⎜⎝⎛
21f = 7
Q.46 Suppose that randq,prrr are three non-coplanar vectors in R3. Let the components of a vector s
r along
randq,prrr be 4, 3 and 5, respectively. If the components of this vector s
r along )rqp(rrr
++− , )rqp(rrr
+−
and )rqp(rrr
+−− are x, y and z, respectively, then the value of 2x + y + z is
Ans. [9]
Sol. r5q3p4srrrr
++=
)rq–p(z)rq–p(y)rqp(xsrrrrrrrrrr
+−+++++−=
r)zyx(q)z–y–x(p)z–yx(srrrr
+++++−=
∴ – x + y – z = 4 ….(1)
x – y – z = 3 ….(2)
x + y + z = 5 ….(3)
Solving (1), (2) & (3)
x = 4 ; y – z = 8 and y + z = 1
∴ 2x + y + z = 8 + 1 = 9
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Paper-2 [ CODE – 5 ]
Q.47 For any integer k, let αk = cos ⎟⎠⎞
⎜⎝⎛ π
7k + i sin ⎟
⎠⎞
⎜⎝⎛ π
7k , where i = 1− . The value of the expression
∑
∑
=−
=+
α−α
α−α
3
1k2k41–k4
12
1kk1k
||
|| is
Ans. [4]
Sol. αk = cos ⎟⎠⎞
⎜⎝⎛ π
7k + i sin ⎟
⎠⎞
⎜⎝⎛ π
7k = 7
kie
π
Note :
|| n1n α−α + = 7ni
7)1n(i
eeππ
+−
= ⎟⎟
⎠
⎞
⎜⎜
⎝
⎛−
ππ
1ee 7i
7ni
= ⎟⎟⎠
⎞⎜⎜⎝
⎛−
π−
πππ14
i14
i14
i7
nieee.e
= ⎟⎠⎞
⎜⎝⎛ π
+π
147ni
e i14
sin2 π
= 2 sin 14π
∴
∑
∑
=−−
=+
α−α
α−α
3
1K2k41k4
12
1Kk1k
||
||=
14sin23
14sin212
π⋅
π⋅
= 4
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Q.48 Suppose that all the terms of an arithmetic progression (A.P.) are natural numbers. If the ratio of the
sum of the first seven terms to the sum of the first eleven terms is 6 : 11 and the seventh term lies in
between 130 and 140, then the common difference of this A.P. is
Ans. [9]
Sol. An A.P. of natural numbers
116
SS
11
7 = ⇒ 116
]d10a2[2
11
]d6a2[27
=+
+
⇒ 116
)d10a2(11)d6a2(7
=++
⇒ 14a + 42d = 12a + 60 d
⇒ 2a = 18 d
⇒ a = 9d
and 130 < a7 < 140
⇒ 130 < a + 6d < 140
⇒ 130 < 15 d < 140
⇒ 3
26 < d < 328
⇒ d = 9 (as 'd' is a natural number)
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Paper-2 [ CODE – 5 ]
SECTION – 2 (Maximum Marks : 32)
• This Section contains EIGHT questions • Each question has FOUR options (A), (B), (C) and (D) ONE OR MORE THAN ONE of these four option(s) is (are) correct • For each questions, darken the bubble (s) corresponding to all the correct option (s) in the ORS • Marking scheme : +4 If only the bubble (s) corresponding to all the correct option (s) is (are) darkened 0 If none of the bubbles is darkened –2 In all other cases
Q.49 Let f, g : [–1, 2] → R be continuous functions which are twice differentiable on the interval (–1, 2). Let
the values of f and g at the points – 1, 0 and 2 be as given in the following table :
x = – 1 x = 0 x = 2
f(x) 3 6 0
g(x) 0 1 –1
In each of the intervals (–1, 0) and (0, 2) the function (f – 3g)" never vanishes. Then the correct
statement(s) is (are)
(A) f'(x) – 3g'(x) = 0 has exactly three solutions in (– 1, 0) ∪ (0, 2)
(B) f'(x) – 3g'(x) = 0 has exactly one solution in (– 1, 0)
(C) f'(x) – 3g'(x) = 0 has exactly one solution in (0, 2)
(D) f'(x) – 3g'(x) = 0 has exactly two solutions in (– 1, 0) and exactly two solutions in (0, 2)
Ans. [B, C]
Sol. f(–1) = 3 f(0) = 6 f(2) = 0
g(–1) = 0 g(0) = 1 g(2) = – 1
h(x) = f(x) – 3g(x)
h (–1) = f(–1) – 3g (–1) = 3 – 3 (0) = 3
h (0) = f(0) – 3g (0) = 6 – 3 (1) = 3
h (2) = f(2) – 3g (2) = 0 – 3 (–1) = 3
By rolles theorem atleast one solution for h'(x) = 0 in (–1, 0)
& atleast one solution for h'(x) = 0 in (0, 2)
Now given at (–1, 0) & (0, 3)
(f – 3g)" = h"(x) ≠ 0
h'(x) is monotonic
so exactly one solution in (–1, 0)
& exactly one solution in (0, 2)
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Paper-2 [ CODE – 5 ]
Q.50 Let f(x) = 7 tan8 x + 7tan6 x – 3 tan4 x – 3tan2 x for all x ∈ ⎟⎠⎞
⎜⎝⎛ ππ−
2,
2. Then the correct expression(s) is
(are)
(A) ∫π
=4/
0121dx)x(xf (B) ∫
π
=4/
0
0dx)x(f
(C) ∫π
=4/
061dx)x(xf (D) ∫
π
=4/
0
1dx)x(f
Ans. [A, B]
Sol. ∫π 4/
0
dx)x(f
= ∫π 4/
0
26 xdxsecxtan7 – ∫π 4/
0
22 xdxsec.xtan3
Let tan x = t , sec2 xdx = dt
= ∫1
0
6 –dtt7 ∫1
0
2dtt3
= 10
7 ]t[ – 10
3]t[ = 1 – 1 = 0
Now
∫π 4/
0
dx)x(xf
= ∫π 4/
0 II
26I
dx)xsecxtan7(x – ∫π 4/
0 II
22I
dx)xsecxtan3(.x
= 4/0
7 ]xtan.x[ π – 4/0
34/
0
7 ]xtanx[–xdxtan ππ
∫ + ∫π 4/
0
3 xdxtan
=4
–dx)xtan–x(tan4
4/
0
37 π−
π∫
π
= ∫π
−−4/
0
43 dx)1x(tanxtan
∫π
−−4/
0
223 xdxsec)1x(tanxtan
Let tan x = t
∴ sec2xdx = dt
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= ∫ −−1
0
23 dt)1t(t
= – 1
0
46
4t
6t
⎥⎥⎦
⎤
⎢⎢⎣
⎡−
= – ⎥⎦⎤
⎢⎣⎡ −
41
61 =
121
So options (A, B) are correct
Q.51 Let f ′(x) = xsin2
x1924
3
π+for all x ∈ R with f ⎟
⎠⎞
⎜⎝⎛
21 = 0. If m ≤ ∫
1
2/1
dx)x(f ≤ M, then the possible values of m
and M are -
(A) m = 13, M = 24 (B) m = 41 , M =
21
(C) m = –11, M = 0 (D) m = 1, M = 12
Ans. [D]
Sol. f ′(x) = xsin2
x1924
3
π+ > 0 x ∈ ⎟
⎠⎞
⎜⎝⎛ 1,
21
f(x) monotonic increasing
min f ′(x) = 12
x192 3
+ =
3x192 3
(when sin4 πx = 1)
min f (x) = dxx64 3∫ = 4x64 4
= 16x4 at x =21
max f ′(x) = 02
x192 3
+ = 96x3 (when sin4 πx = 0)
max f(x) = 4
96 x4 = 24x4 at x = 1
16 ∫1
2/14 dx
21 ≤ ∫
1
2/1
dx)x(f ≤ ∫1
2/1
dx24
⎟⎠⎞
⎜⎝⎛ −
211 ≤ ∫
1
0
dx)x(f ≤ 24 ⎟⎠⎞
⎜⎝⎛ −
211
21 ≤ ∫
1
2/1
dx)x(f ≤ 12
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Q.52 Let S be the set of all non-zero real numbers α such that the quadratic equation αx2 – x + α = 0 has two
distinct real roots x1 and x2 satisfying the inequality |x1 – x2| < 1. Which of the following intervals is(are)
a subset(s) of S ?
(A) ⎟⎟⎠
⎞⎜⎜⎝
⎛5
1,–21– (B) ⎟⎟
⎠
⎞⎜⎜⎝
⎛0,
51–
(C) ⎟⎟⎠
⎞⎜⎜⎝
⎛5
1,0 (D) ⎟⎟⎠
⎞⎜⎜⎝
⎛21,
51
Ans. Ans. [A, D]
Sol. Roots are real so
D > 0
1 – 4α2 > 0
⇒ –21 < α <
21 ...(i)
also, x1 + x2 = α1 , x1x2 = 1
Given |x1 – x2| < 1
⇒ 212
21 xx4)xx( −+ < 1
⇒ 412 −
α < 1
⇒ 21
α – 4 < 1
⇒ 21
α – 5 < 0
⇒ (5α2 – 1) > 0
( 5 α – 1) ( 5 α + 1) > 0
⇒ α < 51− or α >
51 ...(ii)
from (i) & (ii)
α ∈ ⎟⎟⎠
⎞⎜⎜⎝
⎛ −−51,
21 ∪ ⎟⎟
⎠
⎞⎜⎜⎝
⎛21,
51
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Paper-2 [ CODE – 5 ]
Q.53 If α = 3sin–1 ⎟⎠⎞
⎜⎝⎛116 and β = 3cos–1 ⎟
⎠⎞
⎜⎝⎛
94 , where the inverse trigonometric functions take only the principal
values, then the correct option(s) is(are) -
(A) cos β > 0 (B) sin β < 0
(C) cos(α + β) > 0 (D) cos α < 0
Ans. [B,C,D]
Sol.
6 11
85
α/3
3α = tan–1
856
as 3
1 < 856 < 1
30º < 3α < 45º
⇒ 90º < α < 135º
So, cos α < 0 (Option (D) is correct)
Also
65 9
β/3
4
⇒ 3β = tan–1
465
as 465 > 3
So, 60º < 3β < 90º
∴ 180º < β < 270º
∴ cos β < 0, sin β < 0
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Paper-2 [ CODE – 5 ]
so option (A) is incorrect and option (B) is correct.
(α + β) will lie in 4th quadrant.
So, cos (α + β) > 0
Hence option (C) is correct.
Q.54 Let E1 and E2 be two ellipses whose centers are at the origin. The major axes of E1 and E2 lie along the
x-axis and the y-axis, respectively. Let S be the circle x2 + (y – 1)2 = 2. The straight line x + y = 3
touches the curves S, E1 and E2 at P, Q and R, respectively. Suppose that PQ = PR =3
22 . If e1 and e2
are the eccentricites of E1 and E2, respectively, then the correct expression(s) is (are) -
(A) 4043ee 2
221 =+ (B) e1e2 =
1027 (C)
85e–e 2
221 = (D) e1e2 = 4
3
Ans. [A, B]
Sol.
P
R
Q
E1
E2
x + y = 3
(0, 1)x + y = 3
P (1, 2)
Given PQ = PR = 3
22
∴ using parametric equation of line
∴ Q = ⎟⎟⎠
⎞⎜⎜⎝
⎛×−×+
21
3222,
21
3221
⇒ Q = ⎟⎠⎞
⎜⎝⎛
34,
35
Similarly; R = ⎟⎠⎞
⎜⎝⎛
38,
31
Now :
Let E1 : 21
2
21
2
by
ax
+ = 1
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Paper-2 [ CODE – 5 ]
Let point of contact is Q(x1, y1) then equation of tangent is 21
1
axx + 2
1
1
byy = 1
Comparing it with x + y = 3
Q(x1, y1) = ⎟⎟⎠
⎞⎜⎜⎝
⎛
3b,
3a 2
121 ≡ ⎟
⎠⎞
⎜⎝⎛
34,
35
21a = 5, 2
1b = 4
⇒ e1 = 5
1
Similarly let E2 : 22
2
ax + 2
2
2
by = 1
Then R ⎟⎟⎠
⎞⎜⎜⎝
⎛
3b,
3a 2
222 = ⎟
⎠⎞
⎜⎝⎛
38,
31
22a = 1, 2
2b = 8
⇒ e2 22
7
Now checking option
(A) 21e + 2
2e = 51 +
87 =
4043 (Correct)
(B) e1e2 = 1027 (Correct) & (D) Incorrect
(C) | 21e – 2
2e |= 87
51
− = 4027 (Incorrect)
Q.55 Consider the hyperbola H : x2 – y2 = 1 and a circle S with center N(x2, 0). Suppose that H and S touch
each other at a point P(x1, y1) with x1 > 1 and y1 > 0. The common tangent to H and S at P intersects the
x-axis at point M. If (l, m) is the centroid of the triangle ΔPMN, then the correct expression(s) is(are) -
(A) 1dx
dl = 1 – 21x3
1 for x1 > 1 (B) 1dx
dm = ⎟⎠⎞⎜
⎝⎛ 1–x3
x21
1 for x1 > 1
(C) 1dx
dl = 1 + 21x3
1 for x1 > 1 (D) 1dy
dm = 31 for y1 > 0
Ans. [A,B,D]
Sol. PN is normal to hyperbola & circle
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Paper-2 [ CODE – 5 ]
p(x1, y1)
x
y
OM
⎟⎟⎠
⎞⎜⎜⎝
⎛0,
x1
1
N(x2, 0)
Now normal at P(x1, y1) on hyperbola is
y – y1 = 1
1xy− (x – x1)
put y = 0
x = 2x1
∴ x2 = 2x1
N(2x1, 0)
Equation of tangent at P (x1, y1) on hyperbola is xx1 – yy1 = 1
Put y = 0
x = 1x
1
Point M ⎟⎟⎠
⎞⎜⎜⎝
⎛0,
x1
1
∴ l = 3
x1xx21
11 ++ & m =
30y0 1 ++
1dy
dm = 31 )1x( 1 >
l = x1 + 1x3
1
1dx
dl = 1 – 21x3
1 (x1 > 1)
Also m = 31 1x2
1 −
∴ 1dx
dm = ⎟⎠⎞⎜
⎝⎛ −1x3
x21
1 if x1 > 1
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Paper-2 [ CODE – 5 ]
Q.56 The option(s) with the values of a and L that satisfy the following equation is(are)
∫
∫π
π
+
+
0
46t
4
0
46t
dt)atcosat(sine
dt)atcosat(sine
= L ?
(A) a = 2, L =1–e1–e4
π
π
(B) a = 2, L = 1e1e4
++
π
π
(C) a = 4, L = 1–e1–e4
π
π
(D) a = 4, L =1e1e4
++
π
π
Ans. [A,C]
Sol.
1
0
46t
I
dt)atcosat(sine∫π
+ +
2
246t
I
dt)atcosat(sine∫π
π
+ +
3
3
2
46t
I
dt)atcosat(sine∫π
π
+ +
4
4
3
46t
I
dt)atcosat(sine∫π
π
+
I2 put t = π + α I3 put t = 2π + α I4 put t = 3π + α Numerator =
∫π
+0
46t dt)atcosat(sine + ∫π
π +0
46t dt)atcosat(sinee + ∫π
π +0
46t2 dt)atcosat(sinee + ∫π
π +0
46t3 dt)atcosat(sinee
= (1 + eπ+ e2π + e3π) ∫π
+0
46t dt)atcosat(sine
Now,
∫
∫π
ππππ
+
++++
0
46t
0
46t32
dt)atcosat(sine
dt)atcosat(sine)eee1(
L = 1. 1e1e4
−−
π
π
, a = 2, 4
SECTION – 3 (Maximum Marks: 16)
• This Section contains Two paragraphs • Based on each paragraph, there will be Two questions • Each question has Four options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four option(s)
is (are) correct • For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS • Marking scheme : +4 If only the bubble(s) corresponding to all the correct option(s) is (are) darkened 0 If none of the bubbles is darkened –2 In all other cases
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Paper-2 [ CODE – 5 ]
Paragraph # 1 (Questions 57 & 58)
Let n1 and n2 be the number of red and black balls, respectively, in box I. Let n3 and n4 be the number of
red and black balls, respectively, in box II.
Q.57 One of the two boxes, box I and box II, was selected at random and a ball was drawn randomly out of
this box. The ball was found to be red. If the probability that this red ball was drawn from box II is 31 ,
then the correct option(s) with the possible values of n1, n2, n3 and n4 is(are) -
(A) n1 = 3, n2 = 3, n3 = 5, n4 = 15 (B) n1 = 3, n2 = 6, n3 = 10, n4 = 50
(C) n1 = 8, n2 = 6, n3 = 5, n4 = 20 (D) n1 = 6, n2 = 12, n3 = 5, n4 = 20
Ans. [A, B]
Sol. Box I Box II
Red → n1 Red → n3
Black → n2 Black → n4
Event A = Red ball is drawn.
E1 : Box I selected
E2 : Box II selected
P ⎟⎠⎞
⎜⎝⎛
AE2 =
⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛
⎟⎟⎠
⎞⎜⎜⎝
⎛
22
11
22
EAP)E(P
EAP)E(P
EAP)E(P
31 =
43
3
21
1
43
3
nnn
21
nnn
21
nnn.
21
++
+
+
21
1
nnn+
+ 43
3
nnn+
= 43
3
nnn3+
(A) 43
205.3
41
21
205
63
==+=+ correct
(B) 60
)10(36010
93
=+ ⇒ 21
61
31
=+ ⇒ 21
21
= correct
(C) 25
5.3255
148
=+ ⇒ 53
51
74
≠+ incorrect
(D) 25
5.3255
186
=+ ⇒ 53
51
31
≠+ incorrect
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Paper-2 [ CODE – 5 ]
Q.58 A ball is drawn at random from box I and transferred to box II. If the probability of drawing a red ball
from box I, after this transfer, is 31 , then the correct option(s) with the possible values of n1 and n2
is (are) -
(A) n1 = 4 and n2 = 6 (B) n1 = 2 and n2 = 3 (C) n1 = 10 and n2 = 20 (D) n1 = 3 and n2 = 6
Ans. [C, D]
Sol. Case (I) : Red ball is transferred
Red → n1 – 1
Black → n2
Case (II) : Black ball is transferred
Red → n1
Black → n2 – 1
P = ⎟⎟⎠
⎞⎜⎜⎝
⎛+ 21
1
nnn
⎟⎟⎠
⎞⎜⎜⎝
⎛−+
−1nn
1n
21
1 + ⎟⎟⎠
⎞⎜⎜⎝
⎛+ 21
2
nnn ⎟⎟
⎠
⎞⎜⎜⎝
⎛−+ 1nn
n
21
1
↓ ↓ ↓ ↓
Probability of Red ball transferred
Probability of selecting Red ball
Probability of Black ball transferred
Probability of selecting Red ball
P = )1nn)(nn(
nnnn
2121
21121
−+++− =
)1nn)(nn()1nn(n
2121
211
−++−+ =
)nn(n
21
1
+=
31
(A) 104 ≠
31 Incorrect
(B) 52 ≠
31 Incorrect
(C) 3010 =
31 Correct
(D) 93 =
31 Correct
Paragraph For Questions 59 and 60
Let F : R → R be a thrice differentiable function. Suppose that F(1) = 0, F(3) = –4 and F′(x) < 0 for all x ∈ (1/2, 3). Let f(x) = xF(x) for all x ∈ R.
Q.59 The correct statement(s) is(are) -
(A) f ′(1) < 0 (B) f(2) < 0
(C) f ′(x) ≠ 0 for any x ∈ (1, 3) (D) f ′(x) = 0 for some x ∈ (1, 3)
CAREER POINT
Kota H.O. : Career Point Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744‐5151200 www.careerpoint.ac.in 52
Paper-2 [ CODE – 5 ]
Ans. [A,B,C]
Sol. F(1) = 0 F(3) = –4 F′(x) < 0, x ∈ ⎟⎠⎞
⎜⎝⎛ 3,
21
F(x) is decreasing in ⎟⎠⎞
⎜⎝⎛ 3,
21
f(x) = xF(x)
f ′(x) = F(x) + xF′(x) < 0, x ∈ (1, 3)
so f(x) is decreasing
(A) f ′(1) = F(1) + F′(1) = F′(1) < 0
So option (A) is correct
(B) f(1) = 1F(1) = 0
f(3) = 3F(3) = –12
f(2) < 0 correct
(C) f ′(x) < 0 so f ′(x) ≠ 0 any x ∈ (1, 3) correct
Q.60 If ∫ ′3
1
2 dx)x(Fx = –12 and ∫ ″3
1
3 dx)x(Fx = 40, then the correct expression(s) is(are) -
(A) 9f ′(3) + f ′(1) – 32 = 0 (B) ∫3
1
dx)x(f = 12
(C) 9f ′(3) – f ′(1) + 32 = 0 (D) ∫3
1
dx)x(f = –12
Ans. [C,D]
Sol. ∫ ″3
1
3 dx)x(Fx = 3
13 xFx )′( – ∫ ′
3
1
2 dx)x(Fx3
40 = 27F′(3) – F′(1) – 3(–12)
27F′(3) – F′(1) = 4 … (1)
f ′(x) = F(x) + xF′(x)
f ′(1) = F(1) + F′(1) ⇒ F′(1) = f ′(1)
f ′(3) = F(3) + 3F′(3)
f ′(3) = – 4 + 3F′(3) ⇒ 3F′(3) = f ′(3) + 4
Eqn (1) 9(f ′(3) + 4) – f ′(1) = 4
⇒ 9f ′(3) – f ′(1) + 32 = 0 so option (C) is correct
CAREER POINT
Kota H.O. : Career Point Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744‐5151200 www.careerpoint.ac.in 53
Paper-2 [ CODE – 5 ]
∫3
1
dx)x(f = ∫3
1
dx)x(xF = 3
1
2)x(F
2x
– ∫ ′3
1
2dx)x(F
2x
= 29 F(3) –
2)1(F –
21
∫ ′3
1
2 dx)x(Fx
=29 (–4) – 0 –
21 (–12)
= –18 + 6 = – 12 so option (D) is correct