Career Point JEE Advanced Solution Paper1

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PAPER-1 (CODE-3) JEE-Advance 2013 EXAMINATION CAREER POINT
Date : 02 / 06 / 2013
Part I - PHYSICS
SECTION – 1 (Only One option correct 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.
Q.1 A particle of mass m is projected form the ground with an initial speed µ0 at an angle α with the horizontal.
At the highest point of its trajectory, it makes a completely inelastic 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)
4
π
(B)
4
π
+ α (C)
2
π
– α (D)
2
π
Ans. [A]
Sol. At highest point of projectile, velocity of particle is (u0 cos α) iˆ
max. height attained by particle H =
g2
sinu 2
0 α
For another particle thrown vertically upward
vy
2
= u0
2
– 2gH
vy
2
= u0
2
– 2g ×
g2
sinu 22
0 α
vy = u0 cos α
at the instant of collision
u0 cos α
u0 cos α
⇒ 45º
v
θ
v
u0 cos α
u0 cos α
tan θ = 1
θ = 45º

2 / 33
Q.2 The image of an object, formed by a plano-convex lens at a distance of 8m behind the lens, is real and is one
third the size of the object. The wavelength of light inside the lens is 2/3 times the wavelength in free space.
The radius of the curved surface of the lens is -
(A) 1m (B) 2m (C) 3m (D) 6m
Ans. [C]
Sol. Given that
position of image v = 8,
magnification m = –
3
1
=
f
vf −
3
4
f = 8
f = 6
here λ′ =
3
2
λ0
refractive index µ ∝
v
1
µ ∝
λ
1
µ =
2
3
we have
f
1
= (µ – 1) 





∞
−
1
R
1
R = 3 m
Q.3 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]
Sol. Here
1 MSD = 0.05 cm
1 VSD =
50
45.2
= 0.049 cm
Least count = 1 MSD – 1VSD
= 0.001 cm
diameter = 5.10 + (0.001) × 24
diameter = 5.124 cm

3 / 33
Q.4 The work done on a particle of mass m by a force , K 





+
+
+
jˆ
)yx(
y
iˆ
)yx(
x
2/3222/322
(K being a constant of
appropriate dimensions), when the particle is taken from the point (a, 0) to the point (0, a) along a circular
path of radius a about the origin in the x-y plane is
(A)
a
K2 π
(B)
a
Kπ
(C)
a2
Kπ
(D) 0
Ans. [D]
Sol. Here
x
F
y
F yx
∂
∂
=
∂
∂
that's why force is conservative and particle is moving on circle i.e. x2
+ y2
= a2
hence W = ∫
0
a
x dxF + ∫
0
a
y dyF
= ∫
0
a
3
dxx
a
K
+ ∫
a
0
3
dyy
a
K
= 0
Q.5 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 applying forces
at two ends, the ratio of the elongation in the thin wire to that in the thick wire is
(A) 0.25 (B) 0.50 (C) 2.00 (D) 4.00
Ans. [B]
Sol.
F F
2L, 2R L, R
We have, change in length
∆l =
yA
Frest l
Since the two rods are in series hence
(F1)rest = (F2)rest
A
l
l ∝∆ or ∆l = 2
R
l
(Q A = πR2
)
∴ 2
1
2
2
2
1
2
1
R
R
×=
∆
∆
l
l
l
l

4 / 33
Q.6 A ray of light travelling in the direction ( )jˆ3iˆ
2
1
+ is incident on a plane mirror. After reflection, it travels
along the direction ( )jˆ3iˆ
2
1
− . The angle of incidence is
(A) 30º (B) 45º (C) 60º (D) 75º
Ans. [A]
Sol. Here
i i
δ
For angle between incident ray and reflected ray i.e. δ
cos δ =
2.2
)jˆ3iˆ).(jˆ3i( −+
cos δ = –
2
1
δ = 120º
i.e. 180 – 2i = 120
i = 30º
Q.7 Two rectangular blocks , having identical dimensions, can be arranged either in configuration I or in
configuration II as show 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 9s 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 configuration II is
K 2K
2K
K
Configuration-I
Configuration-II
(A) 2.0 s (B) 3.0 s (C) 4.5 s (D) 6.0 s
Ans. [A]
Sol. In configuration 1
total thermal resistance Req =
KA2
3
KA2KA
lll
=+
In configuration 2

5 / 33
R′eq =
KA
2
1
1
2
1
1
l












+
×
=
KA3
1 l
We have
Q =
R
)TT( 21 −
t
t ∝ R
1
2
1
2
R
R
t
t
= ⇒ t2 =
9
2
× 9 = 2 sec
Q.8 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
ms–1
. The final momentum of object is
(A) 0.3 × 10–17
kg ms–1
(B) 1.0 × 10–17
kg ms–1
(C) 3.0 × 10–17
kg ms–1
(D) 9.0 × 10–17
kg ms–1
Ans. [B]
Sol. Energy absorbed by the object, E = P × t
E = 3 × 10–9
J
Linear momentum =
C
E
Linear momentum = 8
9
103
103
×
× −
Linear momentum = 1 × 10–17
kg ms–1
Q.9 In the Young's double slit experiment using a monochromatic light of wavelength λ, the path difference (in
terms of an integer n) corresponding to any point having half the peak intensity is
(A) (2n + 1)
2
λ
(B) (2n + 1)
4
λ
(C) (2n + 1)
8
λ
(D) (2n + 1)
16
λ
Ans. [B]
Sol. We have
I = 4I0 cos2
2
φ
2I0 = 4I0 cos2
2
φ
cos
2
φ
=
2
1
φ =
2
π
Path difference ∆x =
π
λ
2
× ∆φ

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∆x =
4
λ
Option B verifies the result hence option B is correct.
Q.10 Two non-reactive monatomic 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]
Sol. We have
PV = nRT
or
PM = ρRT
2
1
ρ
ρ
= 





2
1
P
P






2
1
M
M
2
1
ρ
ρ
=
3
2
×
3
4
2
1
ρ
ρ
=
9
8
SECTION – 2 (One or more options correct Type)
This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which
ONE or More are correct.
Q.11 Two non-conduction solid spheres of radii R and 2R, having uniform volume charge densities ρ1 and ρ2
respectively, touch each other. The net electric field at a distance 2R from the centre of the smaller sphere,
along the line joining the centers of the spheres, is zero. The ratio
2
1
ρ
ρ
can be
(A) –4 (B) –
25
32
(C)
25
32
(D) 4
Ans. [B, D]
Sol.
ρ1
+
+
+ R
ρ2+
+2R+
+
P′ P
at point P and P′, E = 0
2todue1todue EE ρρ + = 0

7 / 33
For point P
3
2
1
R
3
4
)R2(
k
π×
ρ
=
0
2
3
R
∈
×ρ
0
2
3
1
3R4
R
∈×
ρ
=
0
2
3
R
∈
×ρ
2
1
ρ
ρ
= 4 (option D)
For point P′
ρ1
+
+
+ R
ρ2–
–
–
2R–
–
2R
P′
2
1
R4
kρ
×
3
4
πR3
= 2
2
)R5(
kρ
×
3
4
π(8R3
)
2
1
ρ
ρ
= –
25
32
(option B)
Q.12 A horizontal stretched string, fixed at two ends, is vibrating in fifth harmonic according to the equation,
y(x, t) = (0.01 m) sin[(62.8 m–1
)x] cos[(628 s–1
)t] Assuming π = 3.14, the correct statement(s) is (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]
Sol.
L
y = 0.01 sin 6.2 8x cos 628t
k = 62.8 =
λ
π2
62.8 =
λ
× 14.32
λ =
8.62
28.6
=
10
1
2
5λ
= L

8 / 33
L = 5×
20
1
⇒
4
1
meter ⇒ 0.25 m
Antinode is at mid point of string
Maximum displacement at this point = 0.01
Fundamental frequency =
l2
v
v =
k
ω
⇒
8.62
28.6
= 10
Fundamental frequency ⇒
25.02
10010
×
×
(V = fλ)
⇒
5.0
10






π
ωλ
=
2
v
⇒
5
100
⇒ 20Hz
Q.13 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 S3
V0
C2C1
2V0
(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]
Sol. S1 is closed C1 get charged
Charge on C1 = C1 × 2V0
Now S2 is pressed and S1 is open
C2
–2C1V0
2C1V0
Vcommon =
21
01
CC
0VC2
+
+
⇒
21
01
CC
VC2
+

9 / 33
Charge on C2 =
21
012
CC
VC2C
+
×
⇒
C2
VC2 0
2
⇒ CV0 with upper plate positive.
Now S3 is pressed and S2 is open
∴ charge on C2 = CV0 with upper plate negative.
Q.14 A particle of mass M and positive charge Q, moving with a constant velocity 1u
r
= 1
msiˆ4 −
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 after 10
milliseconds with a velocity 1
2 ms)jˆiˆ3(2u −
+=
r
. The correct statement(s) is (are)
(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
Q3
M50π
units
(D) The magnitude of the magnetic field
Q3
M100π
units
Ans. [A,C]
Sol.
x = 0 x = L
x
y
2
π/6
32
π/6
× × × ×
× × × ×
× × × ×
× × × ×
× × × ×
× × × ×
tan θ =
32
2
=
3
1
θ = π/6
t =
qB6
mπ
10 × 10–3
=
QB6
M×π
B = 100
Q6
M
×
π
B =
Q3
M50π

10 / 33
Q.15 A solid sphere of radius R and density ρ 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 3ρ. The complete
arrangement is placed in a liquid of density 2ρ and is allowed to reach equilibrium. The correct statement(s)
is (are)
(A) the net elongation of the spring is
k3
gR4 3
ρπ
. (B) the net elongation of the spring is
k3
gR8 3
ρπ
.
(C) the light sphere is partially submerged. (D) the light sphere is completely submerged.
Ans. [A,D]
Sol.
R
ρ
3ρ
m1g
2ρ
V2ρg
R
kx
V2ρg
kx m2g
Weight of system ⇒ gR
3
4
3R
3
4 33






π×ρ+π×ρ
⇒
3
4
πR3
× 4ρ × g ⇒
3
Rg16 3
πρ
Buoyancy Force = V.2ρg
V.2ρg =
3
Rg16 3
πρ
V ⇒
3
R8 3
π
Submerged volume =
3
R8 3
π
, it mean both sphere are submerged completely
as total volume of both sphere is 33
R
3
4
R
3
4
π+π = 3
R
3
8
π
for spring elongation
kx + g2R
3
4 3
ρ×π = gR
3
4
3 3
×π×ρ
kx = ρg × 3
R
3
4
π
x = ρg ×
k
R
3
4 3
π

11 / 33
SECTION – 3 (Integer value correct Type)
This section contains 5 questions. The answer to each question is single digit integer, ranging from 0 to 9 (both
inclusive).
Q.16 A uniform circular disc of mass 50 kg and radius 0.4 m is rotating with a angular velocity of 10 rad s–1
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 large enough such that the axis of 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]
Sol. Conservation of angular momentum
I1ω1 = I2ω2
2
)4.0(50 2
×
× 10 = 2
2
2
4])2.0(25.6[
2
)4.0(50
′ω





××+
×
ω2 = 8
Q.17 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 Sodium is
Ans. [1]
Sol. hf = KEmax + φ
hf = eVs + φ
f =
h
e
Vs + φ
slope is
h
e
, slope does not depend on work function and it is same for all metals. So answer is 1.

12 / 33
Q.18 A bob a mass m, suspended by a string of length l1, is given a 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 acquired the minimum speed required to complete a full circle in the vertical plane, the
ratio
2
1
l
l
is
Ans. [5]
Sol.
Elastic
collision
l1
l1
l2
U = Vmin = 1g5 l
At the highest point velocity is
⇒ 1
2
2g2u l×−
⇒ )2(g2g5 11 ll −
⇒ 1gl
After collision at highest point.
As collision is elastic and both bodies have same mass so velocities of both get exchange.
∴ body velocity = 1gl
1gl = Vmin required to complete circle
1gl = 2g5 l
2
1
l
l
= 5
Q.19 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 ms–1
) of the particle is zero, the speed (in ms–1
) after 5 s is
Ans. [5]
Sol. P = 0.5 watt.
in 5 sec., W = Pt

13 / 33
⇒ 0.5 × 5
work ⇒ 2.5 joule
W = ∆KE (work energy theorem)
2.5 =
2
1
× 0.2[v2
– 0]
v = 5 ms–1
Q.20 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
Ans. [4]
Sol. T1/2 =
λ
2nl
1386 =
λ
693.0
λ =
10001386
693.0
×
⇒
2000
1
103
= λN0
N0 =
λ
3
10
= 1000 × 2000 ⇒ 2 × 106
N = N0e–λt
N0 – N0e–λt
= No. of decayed nuclei
% of decayed nuclei ⇒
0
t
00
N
eNN λ−
−
× 100
⇒ (1 – e–λ×t
) × 100
⇒ (1 – e–λ×80
) × 100
⇒ )e1( 2000
80
−
− × 100
= 3.9% ⇒ 4%

14 / 33
Part II - CHEMISTRY
Q.21 The standard enthalpies of formation of CO2(g), H2O(l) and glucose (s) at 25ºC are –400 kJ/mol, –300
kJ/mol and –1300 kJ/mol, respectively. The standard enthalpy of combustion per gram of glucose at 25ºC is -
(A) + 2900 kJ (B) –2900 kJ (C) –16.11 kJ (D) +16.11 J
Ans. [C]
Sol. C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l)
∆H = 6 × (–400) + 6 (–300) – (–1300)
= – 2900 kJ/mol
= –16.11 kJ/g
Q.22 KI in acetone, undergoes SN2 reaction with each of P, Q, R and S. The rates of the reaction vary as
H3C–Cl
P Q
ClCl
R S
O
Cl
(A) P > Q > R > S (B) S > P > R > Q (C) P > R > Q > S (D) R > P > S > Q
Ans. [B]
Sol. Reactivity order
C
O
CH2–Cl > CH3–Cl > Cl > Cl
Q.23 The compound that does NOT liberate CO2, on treatment with aqueous sodium bicarbonate solution, is -
(A) Bezoic acid (B) Benzensulphonic acid
(C) Salicylic acid (D) Carbolic acid (Phenol)
Ans. [D]
Sol. Phenol is less acidic than carbonic acid (H2CO3) so it does not release CO2 with NaHCO3.
Q.24 Consider the following complex ions, P, Q and R.
P = [FeF6]3–
Q = [V(H2O)6]2+
and R = [Fe(H2O)6]2+
The correct order of the complex ions, according to their spin-only magnetic moment values (in B.M.) is -
(A) R < Q < P (B) Q < R < P (C) R < P < Q (D) Q < P < R
Ans. [B]
Sol. Q < R < P
P = [FeF6]–3
Fe+3
= [Ar] 4sº 3d5
5 unpaired e–
Q = [V(H2O)6]+2
V+2
= [Ar] 4sº 3d3
3 unpaired e–
R = [Fe(H2O)6]+2
Fe+2
= [Ar] 4sº 3d6
4 unpaired e–

15 / 33
Q.25 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 -
[Q]0
[Q]
Time
(A) 2 (B) 3 (C) 0 (D) 1
Ans. [D]
Sol. P + Q → R + S
For Ist
order reaction t75% = 2 × t1/2
∴ order w.r.t. P is 1.
For zero order reaction
Integrated rate law is
(a0 – x) = – kt + a0
∴ Q follows zero order
Hence overall order is 1.
Q.26 Concentrated nitric acid, upon long standing, turns yellow-brown due to the formation of -
(A) NO (B) NO2 (C) N2O (D) N2O4
Ans. [B]
Sol. NO2
4HNO3(l) → νh ↑↑
+ )g(2)g(2 ONO4 + 2H2O
Q.27 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 -
X–
A+
(A) 104 pm (B) 125 pm (C) 183 pm (D) 57 pm
Ans. [A]
Sol.
−
+
r
r
= 0.414 for octahedral void.
r+ = 0.414 × 250 = 103.5 =~ 104 pm

16 / 33
Q.28 Upon treatment with ammoniacal H2S, the metal ion that precipitates as a sulfide is -
(A) Fe(III) (B) Al(III) (C) Mg(II) (D) Zn(II)
Ans. [D]
Sol. Zn(II)
Group 4 cations give precipitate with ammonical H2S.
Q.29 Methylene blue, from its aqueous solution, is adsorbed on activated charcoal at 25ºC. For this process, the
correct statement is -
(A) The adsorption requires activation at 25ºC.
(B) The adsorption is accompanied by a decrease in enthalpy.
(C) The adsorption increases with increase of temperature.
(D) The adsorption is irreversible.
Ans. [B]
Sol. This is a process of physical adsorption it results in release of energy.
Q.30 Sulfide ores are common for the metals -
(A) Ag, Cu and Pb (B) Ag, Cu and Sn (C) Ag, Mg and Pb (D) Al, Cu and Pb
Ans. [A]
Sol. Ag, Cu and Pb
Ag → Ag2S
Cu → CuFeS2
Pb → PbS
Q.31 Benzene and nepthalene form an ideal solution at room temperature. For this process, the true statement(s)
is(are) -
(A) ∆G is positive (B) ∆Ssystem is positive (C) ∆Ssurroundings = 0 (D) ∆H = 0
Ans. [B, C, D]
Sol. Condition for ideal solution
∆G < 0, ∆S > 0 & ∆H = 0
Q.32 The pair(s) of coordination complexes/ions exhibiting the same kind of isomerism is(are) -
(A) [Cr(NH3)5Cl]Cl2 and [Cr(NH3)4Cl2]Cl (B) [Co(NH3)4Cl2]+
and [Pt(NH3)2(H2O)Cl]+
(C) [CoBr2Cl2]2–
and [PtBr2Cl2]2–
(D) [Pt(NH3)3(NO3)]Cl and [Pt(NH3)3Cl]Br
Ans. [B, D]
Sol. [B] [Co(NH3)4Cl2]2
Ma4b2
[Pt(NH3)2(H2O)Cl]+
Ma2bc
Both can show geometrical isomerism
[D] [Pt(NH3)2(NO3)]Cl
[Pt(NH3)3Cl]Br
Both can show ionization isomerism

17 / 33
Q.33 The initial rate of hydrolysis of methyl acetate (1M) by a weak acid (HA, 1M) is 1/100th
of that of a strong
acid (HX, 1M), at 25ºC. The Ka of HA is -
(A) 1 × 10–4
(B) 1 × 10–5
(C) 1 × 10–6
(D) 1 × 10–3
Ans. [A]
Sol. CH3COOCH3 + H2O →
+
H
CH3COOH + CH3OH
Hydrolysis of ester
rate r1 = k[CH3COOCH3][H+
]SA
rate r2 = k[CH3COOCH3][H+
]WA
2
1
r
r
=
WA
SA
]H[
]H[
+
+
1
100
=
WA]H[
1
+
[H+
]WA = 0.01
We know [H+1
] = α C = Cka
or Cka = 0.01
ka × 1 = 10–4
ka = 1 × 10–4
Q.34 The hyperconjugative stabilities of tert-butyl cation and 2-butene, respectively, are due to -
(A) σ → p (empty) and σ → π* electron delocalisations
(B) σ → σ* and σ → π electron delocalisations
(C) σ → p (filled) and σ → π electron delocalisations
(D) p (filled) → σ → π* electron delocalisations
Ans. [A]
Sol. σ – π empty and σ – π*
electron delocalization.
Q.35 Among P, Q, R and S, the aromatic compound(s) is/are -
(A) P (B) Q (C) R (D) S
Cl
 → 3AlCl
P
 →NaH
Q
O O Cº115100
CO)NH( 324
−
 → R
O
→HCl
S
Ans. [A, B, C, D]

18 / 33
Sol. A = P Aromatic
B = Q Aromatic
C = R Aromatic
D = S Aromatic
P =
⊕
Q =
Θ
R = HO CH3
Aromatic ions
Θ
S =
OH
⊕
Q.36 The total number of lone-pairs of electrons in melamine is -
Ans. [6]
Sol.
NH2
N
N
N
NH2
NH2
Melamine
6 lone pairs are present.
Q.37 The total number of carboxylic acid groups in the product P is -
O
O
O
O
O
22
3
3
OH.3
O.2
,OH.1
 →
∆+
P
Ans. [2]
Sol.
O
O
O
O
O
 →
+
OH3
O
O
O
O
C–OH
OH
O
O
O
O
HO–C–CH2
HO–C–CH2
ozonolysis
Oxidative
OH/O 223
 ←
O
O
∆
–2CO2

19 / 33
Q.38 The atomic masses of He and Ne are 4 and 20 a.m.u., respectively. The value of the de Broglie wavelength of
He gas at – 73ºC is "M" times that of the de Broglie wavelength of Ne at 727ºC. M is -
Ans. [5]
Sol.
NeNe
Ne
HeHe
He
Vm
h
Vm
h
=λ
=λ
=
HeHe
NeNe
Vm
Vm
M =
4
20
×
M/T
M/T
=
4
20
×
4/200
20/1000
= 5 ×
50
50
= 5
Q.39 EDETA4–
is ethylenediaminetetraacetate ion. The total number of N–Co–O bond angles in [Co(EDTA)]1–
complex ion is -
Ans. [8]
Sol.
O–C–H2C
O–C–H2C
N–CH2–CH2–N
CH2–C–O
CH2–C–O
O
O
O
O
Co+3
Q.40 A tetrapeptide has –COOH group on alanine. This produces glycine (Gly), valine (Val), phenyl (Phe) and
alanine (Ala), on complete hydrolysis. For this tetrapeptide, the number of possible sequences (primary
structures) with –NH2 group attached to a chiral center is -
Ans. [4]
Sol. Tetrapeptide has 4 amino acid and –COOH group at alamine that means it should be at one end.
→ So possible 1º structure are
I II III IV V VI
G V P V G P
V G V P P G
P P G G V V
A A A A A A
In glycine –NH2 group is not attach chiral centre in other –NH2 is attach at chiral centre.
So II, III, IV & VI structure have –NH2
Group attach at chiral centre.

20 / 33
Part III - Mathematics
CODE : 3 02 / 06 / 2013
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.
Q.41 The value of cot
















+∑ ∑= =
−
23
1n
n
1k
1
k21cot is -
(A)
25
23
(B)
23
25
(C)
24
23
(D)
23
24
Ans. [B]
Sol. cot
















+ ∑∑ ==
−
n
1k
23
1n
1
k21cot
= cot








++∑=
−
))1n(n1(cot
23
1n
1
= cot














−+
++
∑=
−
n)1n(
)1n(n1
cot
23
1n
1
= cot








−+ −
=
−
∑ )n(cot)1n((cot 1
23
1n
1
= cot (cot–1
2 – cot–1
1 + cot–1
3 – cot–1
2 + ….. + cot–1
24 – cot–1
23)
= cot (cot–1
24 – cot–1
1)
=
)1cot(cot)24cot(cot
)1cot(cot)24cot(cot1
11
11
−−
−−
−
+
=
124
1241
−
×+
=
23
25
Q.42 Let kˆ2jˆiˆ3PR −+= and kˆ4jˆ3iˆSQ −−= determine diagonals of a parallelogram PQRS and
kˆ3jˆ2iˆPT ++= be another vector. Then the volume of the parallelepiped determined by the vectors PQ,PT
and PS is -
(A) 5 (B) 20 (C) 10 (D) 30

21 / 33
Ans. [C]
Sol.
R
QP
S
kˆ2jîˆ3RQPQ −+=+ ….(i)
kˆ4jˆ3iˆRQPQ −−=+ ….(ii)
kˆ4jˆ3iˆQRQP −−=− ….(iii)
2 kˆ6jˆ2iˆ4QR −−=
kˆ3jîˆ2QR −−= ….(iv)
)kˆ3jîˆ2()kˆ2jîˆ3(PQ −−−−+=
kˆjˆ2iˆPQ ++=
kˆ3jˆ2iˆPT ++= (given)
and
kˆjˆ2iˆPQ ++=
kˆ3jîˆ2QR −−=
∴ Volume =
312
121
321
−−
= 1(– 6 +1) – 2(– 3 –2) + 3(– 1 – 4)
= – 5 +10 – 15 = – 10
= 10
Q.43 Let complex numbers α and
α
1
lie on circles (x – x0)2
+ (y – y0)2
= r2
and (x – x0)2
+ (y – y0)2
= 4r2
,
respectively. If z0 = x0 + iy0 satisfies the equation 2 |z0|2
= r2
+ 2, then |α| =
(A)
2
1
(B)
2
1
(C)
7
1
(D)
3
1
Ans. [C]
Sol. α lies on |z – z0| = r
So |α – z0| = r ⇒ |α – z0|2
= r2
…(i)
α
1
lies on |z – z0| = 2r

22 / 33
So 0z
1
−
α
= 2r
|1 – |z0α = 2r || α
|1 – |z0α = 2r || α
⇒ |1 – |z0α 2
= 4r2
|α|2
…(ii)
Subtract (ii) from (i)
|1 – |z0α 2
– |α – z0|2
= r2
(4 |α|2
– 1)
⇒ 1 + |α|2
|z0|2
– |α|2
– |z0|2
= r2
(4 |α|2
– 1)
⇒ (1 – |α|2
) (1 – |z0|2
) – r2
(4|α|2
– 1) = 0
⇒ (1 – |α|2
) (1 – |z0|2
) + 2(1 – |z0|2
) (4|α|2
– 1) = 0
⇒ (1 – |z0|2
) (1 – |α|2
+ 8|α|2
– 2) = 0
⇒ (1 – |z0|2
) (7 |α|2
– 1) = 0
⇒ |α|2
= 1/7
⇒ |α| =
7
1
Q.44 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 22 . Then -
(A) a + b – c > 0 (B) a – b + c < 0 (C) a – b + c > 0 (D) a + b – c < 0
Ans. [A]
Sol. ax + by + c = 0
bx + ay + c = 0
Intersection point






+
−
+
−
ba
c
,
ba
c
Distance
8
ba
c
1
ba
c
1
22
<





+
++





+
+
2(a + b + c)2
< 8(a + b)2
(a + b + c)2
< (2a + 2b)2
(2a + 2b)2
– (a + b + c)2
> 0
(a + b – c) (3a + 3b + c) > 0
so, (a + b – c) > 0

23 / 33
Q.45 Perpendiculars are drawn from points on the line
3
z
1
1y
2
2x
=
−
+
=
+
to the plane x + y + z = 3. The feet of
perpendiculars lie on the line -
(A)
13
2z
8
1y
5
x
−
−
=
−
= (B)
5
2z
3
1y
2
x
−
−
=
−
=
(C)
7
2z
3
1y
4
x
−
−
=
−
= (D)
5
2z
7
1y
2
x −
=
−
−
=
Ans. [D]
Sol. Let point lies on given line is
(– 2, –1, 0)
Line ⊥ to plane and passing through (–2, –1, 0) is
1
2x +
=
1
1y +
=
1
z
= λ
General point on above line is
A(λ – 2, λ – 1, λ)
Now this point lies on plane so put point A in equation of plane so we get λ = 2
Point A (0, 1, 2)
Let second point on line is (0, –2, 3)
Let line ⊥ to plane and passing through point (0, –2, 3) is
1
x
=
1
2y +
=
1
3z −
= λ
General point on above line is B(λ, λ – 2, λ + 3)
Now this point lies on plane so we get λ = 2/3
So point B (2/3, –4/3, 11/3)
Clearly drs of line join foot of ⊥ i.e. A and B is (2/3, –7/3, 5/3) or (2, –7, 5)
Q.46 Four persons independently solve a certain problem correctly with probabilities
8
1
,
4
1
,
4
3
,
2
1
. Then the
probability that the problem is solved correctly by at least one of them is -
(A)
256
235
(B)
256
21
(C)
256
3
(D)
256
253
Ans. [A]
Sol. P(A) =
2
1
, P(B) =
4
3
, P(C) =
4
1
, P(D) =
8
1
Required probability = 1 – P( A )P( B )P( C )P( D )
= 1 –
2
1
×
4
1
×
4
3
×
8
7
= 1 –
256
21
=
256
235

24 / 33
Q.47 The area enclosed by the curves y = sin x + cos x and y = | cos x – sin x | over the interval 




 π
2
,0 is -
(A) )12(4 − (B) )12(22 − (C) )12(2 + (D) )12(22 +
Ans. [B]
Sol. Area = ∫
π
−−+
2/
0
|)xsinxcos|)xcosx((sin dx
= ∫
π
+
2/
0
)xcosx(sin dx – ∫
π
−
4/
0
)xsinx(cos dx – ∫
π
π
−
2/
4/
)xcosx(sin dx
= [– cos x + sin x 2/
0]π
– [sin x + cos x 4/
0]π
– [– cos x – sin x 4/
2/]π
π
= (1 + 1) – )21()12( +−−−
= 2 – 2 + 1 + 1 – 2
= 4 – 22
= 22 )12( −
Q.48 A curve passes through the point 




 π
6
,1 . Let the slope of the curve at each point (x, y) be 





+
x
y
sec
x
y
, x > 0.
Then the equation of the curve is -
(A)
2
1
xlog
x
y
sin +=





(B) 2xlog
x
y
eccos +=





(C) 2xlog
x
y2
sec +=





(D)
2
1
xlog
x
y2
cos +=





Ans. [A]
Sol.
dx
dy
=
x
y
+ sec
x
y
y = vx
dx
dy
= v + x
dx
dv
v + x
dx
dv
= v + sec v
∫∫ =
x
dx
vdvcos
sin v = ln|x| + c
sin
x
y
= ln|x| + c
As curve passes through 




 π
6
,1

25 / 33
So sin
6
π
= 0 + c ⇒ c =
2
1
So sin
x
y
= log x +
2
1
Q.49 Let f : 





1,
2
1
→ R (the set of all real numbers) be a positive, non-constant and differentiable function such
that f '(x) < 2 f(x) and 





2
1
f = 1. Then the value of ∫
1
2/1
dx)x(f lies in the interval -
(A) (2e – 1, 2e) (B) (e – 1, 2e – 1)
(C) 





−
−
1e,
2
1e
(D) 




 −
2
1e
,0
Ans. [D]
Sol. f '(x) – 2f(x) < 0 …(i)
Multiply equation (i) by e–2x
f '(x) e–2x
– 2e–2x
f(x) < 0
dx
d
(f(x) e–2x
) < 0
So f(x) e–2x
decreases
So f(x) e–2x
< f(1/2) e–1
; for x ∈ [1/2, 1]
f(x) e–2x
<
e
1
; for x ∈ [1/2, 1]
f(x) < e+2x–1
; for x ∈ [1/2, 1]
since f(x) > 0 (given)
so 0 < ∫
1
2/1
)x(f dx < ∫
−
1
2/1
1x2
e dx
so 0 < ∫
1
2/1
)x(f dx <
1
2/1
1x2
2
e







 −
dx
so, 0 < ∫
1
2/1
)x(f dx <
2
1e −
Q.50 The number of points in (–∞, ∞), for which x2
– x sin x – cos x = 0, is -
(A) 6 (B) 4 (C) 2 (D) 0
Ans. [C]
Sol. Let f(x) = x2
– x sin x – cos x
f '(x) = 2x – x cos x

26 / 33
f '(x) = x(2 – cos x)
as for x < 0, f '(x) < 0 so f(x) is decreasing.
and for x > 0, f '(x) > 0 so f(x) is increasing.
So, f(x) will be zero at 2 points.
SECTION – 2 (One or more options correct Type)
This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which
ONE or MORE are correct.
Q.51 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
Ans. [A,C]
Sol.
xx
8a – 2x
(15a – 2x)
xx
xx
xx
V = (15a – 2x) (8a – 2x)x
V = 4x3
– 46ax2
+ 120a2
x
dx
dV
= 12x2
– 92ax + 120a2
= 4(3x2
– 23ax + 30a2
)
at x = 5,
dx
dV
= 0
30a2
– 115a + 75 = 0
⇒ 6a2
– 23a + 15 = 0
⇒ (a – 3) (6a – 5) = 0
⇒ So, a = 3 or a =
6
5
Now a92x24
dx
Vd
2
2
−=

27 / 33
For a = 3, 2
2
dx
Vd
< 0
So, V is maximum for a = 3.
Hence lengths are 24 and 45.
Q.52 Let Sn = ∑=
+
−
n4
1k
2
)1k(k
)1( k2
. Then Sn can take value(s)
(A) 1056 (B) 1088 (C) 1120 (D) 1332
Ans. [A,D]
Sol. ∴ Sn = –12
– 22
+ 32
+ 42
– 52
– 62
+ 72
+ 82
………….. + 4n2
= [32
– 12
+ 72
– 52
……….. 2n terms] + [42
– 22
+ 82
– 62
………….. 2n terms]
= 2[1 + 3 + 5 + 7……….. 2n terms] + 2[2 + 4 + 6 + 8……… 2n terms]
= 2[2n/2 [2 + (2n – 1)2] + 2[2n/2 (4 + (2n – 1)2]
= 2n[4n] + 2n[4n + 2]
= 8n2
+ 8n2
+ 4n
= 16n2
+ 4n
Sn = 4n (4n + 1)
which gives option A and D for n = 8, 9
Q.53 A line l passing through the origin is perpendicular to the lines
l1 : (3 + t) iˆ + (–1 + 2t) jˆ + (4 + 2t) kˆ , – ∞ < t < ∞
l2 : (3 + 2s) iˆ + (3 + 2s) jˆ + (2 + s) kˆ , – ∞ < s < ∞
Then the coordinates(s) of the point(s) on l2 at a distance of 17 from the point of intersection of l and l1
is(are)
(A) 





3
5
,
3
7
,
3
7
(B) (–1, –1, 0) (C) (1, 1, 1) (D) 





9
8
,
9
7
,
9
7
Ans. [B,D]
Sol. )kˆ2jˆ2iˆ(t)kˆ4jˆiˆ3(1 +++++=l
)kˆjˆ2iˆ2(s)kˆ2jˆ3iˆ3(2 +++++=l
Drs of line ⊥ to both lines (2, –3, 2)
So line l is
2
x
=
3
y
−
=
2
z
Intersection point of line l and l1 is A (2, –3, 2)
General point on l2 is B (2k + 3, 2k + 3, k + 2)
Distance between A and B = 17
222
k)6k2()1k2( ++++ = 17

28 / 33
k = –2 and k = –
9
10
So point if k = – 2, is (–1, –1, 0)
if k =
9
10
− is 





9
8
,
9
7
,
9
7
Q.54 Let f(x) = x sin πx, x > 0. Then for all natural numbers n, f '(x) vanishes at -
(A) a unique point in the interval 





+
2
1
n,n
(B) a unique point in the interval 





++ 1n,
2
1
n
(C) a unique point in the interval (n, n + 1)
(D) two points in the interval (n, n + 1)
Ans. [B, C]
Sol. f(x) = x sin πx
f ′(x) = sin πx + πx cos πx
f′ 





+
2
1
n = 1; n ∈ even
= –1; n ∈ odd
O
y
1
/2
1 2
3
/2
x →
So f′(x) = 0 between 





++ 1n,
2
1
n
So correct options are B & C
Q.55 For 3 × 3 matrices M and N, which of the following statement(s) is (are) NOT correct ?
(A) NT
MN is symmetric or skew symmetric, according as M is symmetric or skew symmetric
(B) MN – NM is skew symmetric for all symmetric matrices M and N
(C) MN is symmetric for all symmetric matrices M and N
(D) (adj M) (adj N) = adj (MN) for all invertible matrices M and N
Ans. [C,D]
Sol. For option (A)
(NT
MN)T
= NT
MT
(NT
)T
= NT
MT
N

29 / 33
if M is symmetric then MT
= M
so NT
MN is also symmetric
if M is skew symmetric then MT
= – M
So, NT
MN is also skew symmetric
So (A) is correct
For option (B)
MT
= M, NT
= N
(MN – NM)T
= (MN)T
– (NM)T
= NT
MT
– MT
NT
= NM – MN
= – (MN – NM)
So, option (B) is correct.
For option (C)
MT
= M, NT
= N
(MN)T
= NT
MT
= NM
So, option (C) is not correct.
For option (D)
(adjM) (adjN) = adj(NM)
so, option (D) is not correct.
SECTION – 3 (Integer value correct Type)
This section contains 5 questions. The answer to each question is a single digit integer, ranging from 0 to 9 (both
inclusive)
Q.56 A vertical line passing through the point (h, 0) intersects the ellipse
3
y
4
x 22
+ = 1 at the points P and Q. Let
the tangents to the ellipse at P and Q meet at the point R. If ∆(h) = area of the triangle PQR,
∆1 = )h(max
1h2/1
∆
≤≤
and ∆2 = )h(min
1h2/1
∆
≤≤
, then 21 8
5
8
∆−∆ =
Ans. [9]
Sol.
h
height
Base
(h1, 0)








−
4
h
13,h
2
Q
P








−−
4
h
13,h
2
x = h
y
x
(0,0)

30 / 33
Line PQ is chord of contact
⇒
4
xh1
+ 0 = 1 ….(1)
x = h …..(2)
Compare (1) & (2)
h1 =
h
4
So area = 





− h
h
4
× 3
4
h
1
2
−
=
2
3
h
)h4( 2/32
−
regular decreasing
2
3
)Area(
2/1h
max =
= 2/1
h
1
4
2/3






−
,
2
3
)Area(
1h
min =
=
(3)3/2
=
2
3
( 15 )3/2
So,
5
8
∆1 – 8 ∆2 =
5
8
×
8
3
× ( 15 )3/2
– 8 ×
2
3
(3)3/2
= 5 × 9 – 4 × 9
= 45 – 36
= 9
Q.57 The coefficients of three consecutive terms of (1 + x)n + 5
are in the ratio 5 : 10 : 14. Then n =
Ans. [6]
Sol.
10
5
C
C
1r
5n
r
5n
=
+
+
+
⇒ n – 3r = – 3 …(1)
14
10
C
C
2r
5n
1r
5n
=
+
+
+
+
⇒ 5n – 12r = – 6 …(2)
solve (1) and (2)
r = 3
n = 6
Q.58 Consider the set of eight vectors V = }}1,1{c,b,a:kˆcjˆbiâ{ −∈++ . Three non-coplanar vectors can be
chosen from V in 2p
ways. Then p is
Ans. [5]
Sol. Total no. of vectors = 8
C3 = 56
Let consider following pairs of vectors
(i) kˆjîˆ ++ and kˆjîˆ −−−

31 / 33
(ii) – kˆjîˆ ++ and kˆjîˆ −−
(iii) kˆjîˆ −+ and – kˆjîˆ +−
(iv) kˆjîˆ +− and – kˆjîˆ −+
If we select any one pair out of these pairs and one vector from remaining 6 vectors then these 3 vectors will
be coplanar.
So, total no. of coplanar vectors = 4
C1 × 6
C1 = 24
So, total no. of non coplanar vectors = 56 – 24
= 32 = 25
∴ p = 5
Q.59 Of the three independent events E1, E2 and E3, the probability that only E1 occurs is α, only E2 occurs is β
and only E3 occurs is γ. Let the probability p that none of events E1, E2 or E3 occurs satisfy the equations
(α – 2β) p = αβ and (β – 3γ) p = 2βγ. All the given probabilities are assumed to lie in the interval (0, 1).
Then
3
1
EofoccurrenceofyProbabilit
EofoccurrenceofyProbabilit
=
Ans. [6]
Sol. Let probabilities of E1, E2 and E3 are p1, p2 and p3 respectively.
Given, p1(1 – p2) (1 – p3) = α
and p2 (1 – p1) (1 – p3) = β
and (1 – p1) (1 – p2)p3 = γ
also (1 – p1) (1 – p2) (1 – p3) = p
so
1
1
p1
p
p −
=
α
,
2
2
p1
p
p −
=
β
,
3
3
p1
p
p −
=
γ
also given that
(α – 2β) p = αβ
⇒ 1
p2p
=





α
−
β
…(i)
also (β – 3γ) p = 2βγ
2
p3p
=
β
−
γ
…(ii)
from (i) and (ii)
2
p6
3
p
=
α
−−
γ

32 / 33
α
−
γ
p6p
= 5
5
p
)p1(6
p
p1
1
1
3
3
=
−
−
−
⇒ 56
p
6
1
p
1
13
=+−−
⇒ 6
p
p
3
1
=
Q.60 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 =
Ans. [5]
Sol. Sum of n cards = 1 + 2 …….. + n =
2
)1n(n +
∴
2
)1n(n +
– (k + k + 1) = 1224
2
)1n(n +
= 1224 + 2k + 1 …(1)
∴ as k ≥ 1
so,
2
)1n(n +
≥ 1224 + 1 + 2
⇒ n2
+ n ≥ 2448 + 6
⇒
2
2
1n





 +
≥ 2448 + 6 +
4
1
≥





+
2
2
1
n (49.5)2
n ≥ 49 …(2)
also k ≤ n
∴
2
)1n(n +
≤ 1224 + n + n + 1
n2
+ n ≤ 2448 + 4n + 2
n2
– 3n ≤ 2450
4
9
2450
2
3
n
2
+≤





− < 502

33 / 33
n –
2
3
< 50
n < 51.5 ….(3)
from (2) and (3)
n can be 49, 50, 51
put n = 49 in (1) we get
49 × 25 = 1224 + 2k + 1
⇒ k = 0 not possible
At n = 50 we get
25.5 = 1224 + 2k + 1
k = 25
At n = 51 we get
51.26 = 1224 + 2k + 1
102 = 2k + 1
∴ k ∉ I
∴ k = 25 ⇒ k – 20 = 5

Career Point JEE Advanced Solution Paper1

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