Vidyamandir Jee Advanced 2013 Paper 1

Vidyamandir Classes
VMC/Solutions 1 Paper 1 ‐ Code 0 – JEE Advanced ‐ 2013

JEE‐Advanced‐2013
Paper 1
Code 0

2 June, 2013 | 9:00 AM – 12 Noon

Detailed solutions to the paper by Vidyamandir Classes

Vidyamandir Classes

Part 1 : PHYSICS

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
1. The work done on a particle of mass m by a force,
( ) ( )
3 2 3 2
2 2 2 2
⎡ ⎤
⎢ ⎥
+⎢ ⎥
⎢ ⎥+ +
⎢ ⎥⎣ ⎦
x y
ˆ ˆK i j
x y x y
(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)
2K
a
π
(B)
K
a
π
(C)
2
K
a
π
(D) 0
Ans.(D) ( )= = +∫ ∫
uurr
x yW F .dr F dx F dy
( ) ( )
3 2 3 2
2 2 2 2
= +
+ +
∫ / /
xdx ydy
W K
x y x y
=
( )
0
0
2 2 2 2
0 0
1 1
0
⎛ ⎞ ⎛ ⎞
⎜ ⎟ ⎜ ⎟− = =
⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠
∫
,a
( , a )
a, a,
K d
x y x y
2. Two rectangular blocks, having identical dimensions, can be arranged either in configuration I or in configuration
II as shown in the figure. One of the blocks has thermal conductivity κ and the other 2κ . 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 :
(A) 2.0s (B) 3.0s (C) 4.5s (D) 6.0s
Ans.(A)
Δ Δ
= = ⇒ =
dQ T T
H Q t
dt R R
Since Q and ΔT is same.
1 2
1 2
=
t t
R R
…(i)
1
2
= +
l l
R
KA KA
(configuration 1)
⇒ 1
3
2
=
l
R
KA
2
1 2 3
= + =
l l l
KA KA KA
R
(configuration 2)

Vidyamandir Classes
⇒ 2
3
=
l
R
KA
Putting in (i)
2
2
9
2
3 2 3
= ⇒ =
l l
t
t sec
/ KA / KA
3. Two non-reactive monoatomic ideal gases have their atomic masses in the ratio 2 : 3. The ratio of their partial
pressures, when enclosed in a vessel kept at a constant temperature, is 4 : 3. The ratio of their densities is :
(A) 1 : 4 (B) 1 : 2 (C) 6 : 9 (D) 8 : 9
Ans.(D)
1
2
2
3
=
M
M
1
2
4
3
=
P
P
Using = ρPM RT
1 1 2 2 1 1 1
1 2 2 2 2
4 2 8
3 3 9
ρ
= ⇒ = = × =
ρ ρ ρ
P M P M P M
P M
4. A particle of mass m is projected from the ground with an initial speed u0 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) From energy conservation both will have equal speeds at the point of collision.
0 cos=V u α
For perfectly inelastic collision, to conserve linear momentum
Final velocity of combined mass will be along the Resultant ⇒ 45θ = o
5. 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 8 1
3 10 ms−
× . The final momentum of the object is :
(A) 17 1
0 3 10. kg ms− −
× (B) 17 1
1 0 10. kg ms− −
×
(C) 17 1
3 0 10. kg ms− −
× (D) 17 1
9 0 10. kg ms− −
×
Ans.(B)
3
9
8
30 10
100 10
3 10
−
−× Δ ×
Δ = = = × ×
×
E power t
P
C C
= 17
10−
kgm / s

Vidyamandir Classes
6. 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) ( )2 1
2
n
λ
+ (B) ( )2 1
4
n
λ
+ (C) ( )2 1
8
n
λ
+ (D) ( )2 1
16
n
λ
+
Ans.(B) 2
04
2
Δφ
=I I cos
⇒ 2
0 02 4
2
Δφ
=I I cos
⇒ 2 1
2 2
Δφ
=cos
⇒ ( )2 1
2 4
Δφ π
= +n
⇒ ( ) ( )
1 2
2 1 2 1
2 4 4
π π λ
= + ⇒ = +
λ
p.d. n p.d n
7. The image of an object, formed by a plano-convex lens at a distance of 8 m behind the lens, is real and is one-
third the size of the object. The wavelength of light inside the lens is
2
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)
1 8 1
24
3 3
= ⇒ = ⇒ = −
−
V
u
u u
1 1 1 1 1 1
6
8 24
− = ⇒ = − ⇒ =
−
f
v u f f
0 0
0
2
3 2
3
λ λ
λ = ⇒ λ = ⇒ μ =
μ μ
/
1 1 1 1
1
μ − ⎛ ⎞
= −⎜ ⎟
∞⎝ ⎠f R
1 3 2 1 1
3
6 1
− ⎛ ⎞
= ⇒ =⎜ ⎟
⎝ ⎠
/
R
R
8. 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.(C) Force in both the wires will be same.
2
Δ
F F
x
Ay r yπ
= =
l l
2 22
1 1 2 1 2
2
2 2 2 11
Δ 2 1
Δ 2 2
x r r L R
.
x r L Rr
⎛ ⎞ ⎛ ⎞⎛ ⎞
⇒ = × = = =⎜ ⎟ ⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠⎝ ⎠
l l
l l
2
1
Δ
2 (ratio of elongation of thin to thick wire)
Δ
x
x
=
O I

Vidyamandir Classes
9. A ray of light travelling in the direction ( )
1
3
2
ˆ ˆi j+ is incident on a plane mirror. After reflection, it travels along
the direction ( )
1
3
2
ˆ ˆi j− . The angle of incidence is :
(A) 30° (B) 45° (C) 60° (D) 75°
Ans.(A)
1
2
−
= =
a.b
Cos
a b
θ
120θ = °
180
30
2
i
θ−
⇒ = = °
10. The diameter of a cylinder is measured using a Vernier callipers with no zero errors. 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
2.45cm. 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 (D) 5.136 cm (D) 5.148 cm
Ans.(B)
2 45
1 1 0 05
50
.
L.C MSD VSD . cm cm= − = −
0 05 0 049 0 001. . . cm= − =
5 10 24 0 001D . cm . cm= + ×
5 10 0 024 5 124. cm . cm . cm= + =

Vidyamandir Classes
Section – 2: (One or more options correct Type)
ONE or MORE are correct
11. 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,
(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 0CV−
Ans.(BD)
12. A particle of mass M and positive charge Q, moving with a constant velocity 1
1 4û i ms ,−
=
r
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 2 3ˆ û i j ms .−
= +
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
50
3
M
Q
π
units.
(D) The magnitude of the magnetic field is
100
3
M
Q
π
units.
Ans.(AC) 1 4û i=
r
2 2( 3 )ˆ û i j= +
r
2
30
2 3
Tanθ θ= ⇒ = °
B should be along directionz .−
M
t
QB
θ
=
3 610 10
.M
QB
π
−
⇒ × =
50
100
6 3
M M
B
Q Q
π π
⇒ = × = units

Vidyamandir Classes
13. A horizontal stretched string, fixed at two ends, is vibrating in its fifth harmonic according to the equation,
1 1
( ) (0 01 ) [(62 8 ) ] [628 ) ]y x,t . m sin . m x cos s t .− −
= Assuming 3 14. ,π = the correct statements(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.(BC) ( ) ( ) ( )1 1
0.01 sin 62.8 cos 628y m m x s t− −
⎡ ⎤ ⎡ ⎤= ⎣ ⎦ ⎣ ⎦
For 5th
harmonic
5
2
2 2 2
62.8
62.8
l
k m
λ
π π π
λ
λ λ
=
= ⇒ = ⇒ =
5 2 5 2 3.14 1
0.25
2 62.8 2 62.8 4
l m m
π ×
⇒ = × = × = =
Number of nodes = 6
At midpoint of string there is an antinode
maxy at midpoint = 0.01 m
0
/ 628/ 62.8 10
20
2 2 2 0.25 0.5
v k
f Hz
l l
ω
= = = = =
×
14. 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
3
4
3
R g
k
π ρ
(B) the net elongation of the spring is
3
8
3
R g
k
π ρ
(C) the light sphere is partially submerged.
(D) the light sphere is completely submerged.
Ans.(AD) In equilibrium let elongation of spring is x
Then
3
4
3
vg R g
kx vg x
k k
ρ ρπ
ρ= ⇒ = =
In equilibrium net force of weight and spring force on light sphere in
downward direction is
2 Bkx vg vg Fρ ρ+ = =
So it is fully submerged.

Vidyamandir Classes
15. Two non-conducting solid spheres of radii R and 2R, having uniform volume charge densities 1 2andρ ρ
respectively, touch each other, The net electric field at a distance 2R from the centre of the smaller sphere, along
the line joining the centres of the spheres, is zero. The ratio
1
2
ρ
ρ
can be.
(A) 4− (B)
32
25
− (C)
32
25
(D) 4
Ans.(BD)
0=PFor E
( )
( )
( )
3
1
1 2 2
2 2
0 00 0
4
3
3 34 2 4 2
⋅
= ⇒ =
R
Q R R
R R
ρ πρ ρ
ε επε πε
1
2
4=
ρ
ρ
Also for QE = 0
( ) ( )
1 2 1 1
2 2
2 20 0
4 32
0
25 254 2 4 5
Q Q Q
QR R
ρ
ρπε πε
+ = ⇒ = − ⇒ = −

Vidyamandir Classes
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)
16. A bob of 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 acquires the minimum speed required to complete a full circle in the vertical plane, the ratio 1
2
l
l
is
Ans.(5) 1 15=u gl
At highest point 1 1v gl= (from energy conservation)
After collision (identical mass and elastic collision)
2 1 1u v gl= =
To complete vertical circle
1
1 2
2
5 5
l
gl gl
l
= ⇒ =
17. 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 1
(in ms )−
of the particle is zero, the speed 1
(in )ms−
after 5 s is
Ans.(5) 0.5=P W
0.5 5 2.5= = = × =∫W Pdt Pt J
2 2
2
1 1
0.2 2.5
2 2
25 5 /
Δ =
⇒ = ⇒ × × =
⇒ = ⇒ =
K W
mv W v
v v m s
18. The work functions of Silver and Sodium are 4.6 and 2.3 eV, respectively. The ratio of the slope of the stopping
potential versus frequency plot for Silver to that of Sodium is
Ans.(1) 0
= + C
h w eVυ
0
⇒ = −C
wh
V
e e
υ
Slope of VC versus υ is Constant
h
e
=
Hence ratio = 1
19. 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) 31/2
0.693 0.693 1 1
1386
1386 20002 10
= = ⇒ = = =
×
T λ
λ

Vidyamandir Classes
0
0 0
0 0
(1 )
1
−
−
−
=
− −
= = = −
t
t
t
N N e
N N N e
f e
N N
λ
λ
λ
0.04
0.04
1 1
80 0.04
2000 25
1 (0.04) 0.96
(1 ) 1 0.96 0.04
Using e 1 for 1
−
−
−
= × = =
⎡ ⎤= − =
= − = − = ⎢ ⎥
⎢ ⎥− <<⎣ ⎦
x
t
e
f e
x x
λ
Percentage fraction = 4
20. A uniform circular disc of mass 50 kg and radius 0.4 m is rotating with an angular velocity of 10 rad 1
s−
about its
own axis, which is vertical. Two uniform circular rings, each of mass 6.25 kg and radius 0.2 m, are gently placed
symmetrically on the disc in such a manner that they are touching each other along the axis of the disc and are
horizontal. Assume that the friction is large enough such that the rings are at rest relative to the disc and the
system rotates about the original axis. The new angular velocity (in rad 1
s−
) of the system is
Ans.(8) From conservation of angular momentum
1 1
2 22 2
1
( 2 )
2
2 2 2 2
= +
⎡ ⎤⎛ ⎞⎛ ⎞ ⎛ ⎞
= + +⎢ ⎥⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎝ ⎠⎣ ⎦
I I I
MR MR R R
m m
ω ω
ω ω
2 2
2
1
1
1
2 2
50 50
10 6.25
2 2
8 rad/s
⎛ ⎞
= +⎜ ⎟⎜ ⎟
⎝ ⎠
⎛ ⎞
⇒ × = +⎜ ⎟
⎝ ⎠
⇒ =
MR MR
mRω ω
ω
ω

Vidyamandir Classes

Part 2 : CHEMISTRY

ONLY ONE is correct
21. Consider the following complex ions, P, Q and R.
[ ] ( ) ( )
2 23
6 2 26 6
+ +−
⎡ ⎤ ⎡ ⎤= = =⎣ ⎦ ⎣ ⎦P FeF , Q V H O and R Fe H O .
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) 3
6P [FeF ] −
=
3 0 5
Fe 4s 3d n 5+
⇒ ⇒ =
2
2 6Q [V(H O) ] +
=
2 0 3
V 4s 3d n 3+
⇒ ⇒ =
2
2 6R [Fe(H O) ] +
=
0 6
Fe 4s 3d n 42+
⇒ ⇒ =
Magnetic moment will be in order Q < R < P
22. The arrangement of X−
ions around A+
ion in solid AX is given in the figure (not drawn to scale). If the radius of
X−
is 250 pm, the radius of A+
is.
(A) 104 pm (B) 125 pm (C) 183 pm (D) 57 pm
Ans. (A)Using radius ratio for octahedral void (FCC)
r
0.414
r
+
−
=
∴ r 0.414 250 104+
= × =
23. 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) Ag2S Silver glance
Cu2S, CuFeS2 Copper glance & Copper pyrite
PbS Galena

Vidyamandir Classes
24. The standard enthalpies of formation of CO2(g), H2O(l) and glucose (s) at 25 Co
are
400 300 1300− − −kJ / mol, kJ / mol and kJ / mol , respectively. The standard enthalpy of combustion per gram
of glucose at 25 Co
is.
(A) + 2900 kJ (B) 2900− kJ (C) 16.11− kJ (D) + 16.11 kJ
Ans. (C) 6 12 6 2 2 2C H O 6O 6CO 6H O+ ⎯⎯→ +
As we know
f (Product) f (Reactants)H H HΔ = ∑ − ∑
2 2 6 12 6f (CO ) f (H O) f (C H O )6H 6H H= + − 6 ( 400) 6( 300) ( 1300)= × − + − − −
2400 1800 1300 2900 kJ= − − + = −
To calculate HΔ for one gram.
HΔ per gram =
2900
16.11 kJ
180
−
= − .
25. 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) 2
2H S Zn ZnS 2H+ +
+ ⎯⎯→ +
26. Methylene blue, from its aqueous solution, is absorbed on activated charcoal at 25 Co
. For this process, the
correct statement is.
(A) The adsorption requires activation at 25 Co
(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)Methylene blue on activated charcoal is an example of physiosorption. Since entropy decreases adsorption
therefore enthalpy has to be negative to make G 0Δ < .
27. KI in acetone, undergoes SN2 reaction with each of P, Q, R and S. The rates of the reaction vary as.
(A) P > Q > R > S (B) S > P > R > Q
(C) P > R > Q > S (D) R > P > S > Q
Ans. (B)Rate of SN2 :

Vidyamandir Classes
28. In the reaction,
P Q R S+ ⎯⎯→ +
the time taken for 75% reaction of P is twice the time taken for 50% reaction of P. The concentration of Q varies
with reaction time as shown in the figure. The overall order of the reaction is.
(A) 2 (B) 3
(C) 0 (D) 1
Ans. (D)Since time taken for 75% reaction of P is twice the time taken for 50% reaction of P. Hence reaction is first order
w.r.t. P.
Analysis of graph of Q w.r.t. time shows, reaction of Q is of zero order.
Hence overall order of the reaction is one.
29. Concentrated nitric acid, upon long standing, turns yellow-brown due to the formation of
(A) NO (B) NO2 (C) N2O (D) N2O4
Ans. (B)HNO3 on standing decomposes as 3 2 2 2
1
2HNO 2NO O H O
2
⎯⎯→ + +
30. The compound that does NOT liberate CO2, on treatment with aqueous sodium bicarbonate solution, is.
(A) Benzoic acid (B) Benzenesulphonic acid
(C) Salicylic acid (D) Carbolic acid (Phenol)
Ans. (D)Phenol is too weak to liberate CO2 with NaHCO3.

Vidyamandir Classes
31. The initial rate ofr hydrolysis of methyl acetate (1M) by a weak acid (HA, 1M) is 1/100th
of that of a strong acid
(HX, 1M), at 25o
C. The Ka of HA is
(A)
4
1 10−
× (B)
5
1 10−
×
(C)
6
1 10−
× (D)
3
1 10−
×
Ans. (A) 2R K[RCOOR][H O]=
Clearly
w w
s s
R K 1
R K 100
= = (Given) [Since (RCOOR) is common]
But Ea / Rt
K Ae−
= [Q Ea will not change as catalyst is same (H+
)]
So
w w
s s
K A 1
K A 100
= = . . . .(i)
Also, A depends on no. of particles i.e. A [H ]+
∝
So for 2 3
C C C C
HA H O H O A+ −
− α − α α
+ +
In equation (i)
C 1
0.01 ( C 1M)
1 100
α
= ⇒ α = =Q and
2
4
a
C
K 10
1
−α
= =
− α
32. The hyperconjugative stabilities of tert-butyl cation and 2-butene, respectively, are due to
(A) ( ) * .p empty and electron deloaclisationsσ σ π→ →
(B) * .and electron deloaclisationsσ σ σ π→ →
(C) ( ) .p filled and electron deloaclisationsσ σ π→ →
(D) ( ) * * .p filled and electron deloaclisationsσ σ π→ →
Ans. (A)Hyperconjugation in tert-butyl carbocation refers to delocalisation of σ -electrons over the empty p-orbitals of C+
.
In 2-butene, it refers to delocalisation of σ -electrons over the *π orbitals of C = C.
33. The pair(s) of coordination complexes/ions exhibiting the same kind of isomerism is(are)
(A) [ ] [ ]3 5 2 3 4 2( ) ( )Cr NH Cl Cl and Cr NH Cl Cl
(B) [ ] [ ]3 4 2 3 2 2( ) ( ) ( )Co NH Cl and Pt NH H O Cl
+ +
(C) [ ] [ ]
2 2
2 2 2 2CoBr Cl and PtBr Cl
− −
(D) [ ] [ ]3 3 3 3 3( ) ( ) ( )Pt NH NO Cl and Pt NH Cl Br
Ans. (BD)(B) 3 4 2 3 2 2[Co(NH ) Cl ] and [Pt(NH ) (H O) Cl]+ +
Both can show geometrical isomerism.
(D) 3 3 3 3 3[Pt(NH ) NO ]Cl and [Pt(NH ) Cl] Br
Both can show ionisation isomerism.

Vidyamandir Classes
34. Among P, Q, R AND S, the aromatic compound(s) is/are
(A) P (B) Q
(C) R (D) S
Ans. (ABCD) (A)
(B)
(C)
110 115 C
4 2 3 3 2 2(NH ) CO 2NH CO H O
°− °
⎯⎯⎯⎯⎯⎯→ + +
(D)
35. Benzene and naphthalene form an ideal solution at room temperature. For this process, the true statement(s) is
(are)
(a) is positiveGΔ
(b) is positivesystem
SΔ
(c) 0surroundings
SΔ =
(d) 0HΔ =
Ans. (BCD) Benzene and Naphthalene form an ideal solution at room temperature.
⇒ H 0Δ =
For mixing of liquids, systemSΔ is positive. (Entropy increase for mixing).
( )sys
Surr mix
q
S 0 Since for ideal solution H 0
T
⎛ ⎞
Δ = − = Δ =⎜ ⎟⎜ ⎟
⎝ ⎠

Vidyamandir Classes
36. 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 -73 0
C is “M” times that of the de Broglie wavelength of Ne at 727 0
C. M is
Ans. (5) As we know
2 ( )
h
m KE
λ =
As we know KEavg =3/2 KbT
20 1000
5
4 200
He Ne Ne
Ne He He
m T
M
m T
λ
λ
×
= = = =
×
So M = 5
37.
4
EDTA
−
is ethylenediaminetetraacetate ion. The total number of N-Co-O bond angles in [ ]
1
( )Co EDTA
−
complex ion is
Ans. (8)
Total number of N Co O− − bond angles = (4 2)× = 8
38. The total number of carboxylic acid groups in the product P is
Ans. (2)
39. A tetrapeptide has COOH− group on alanine. This produces glycine (Gly), valine (Val), phenyl alanine (Phe)
and alanine (Ala), on complete hydrolysis. For this tetrapeptide, the number of possible sequences (primary
structures) with 2NH− group attached to a chiral centre is
Co

Vidyamandir Classes
Ans. (4) (gly/valine/Ph-al)
3
O
||
NH C C OH
|
CH
− − − −
Primary structure
If glycine is at left corner, then NH2 will be at if valine at corner
2
O
|| gly
NH CH C NH alanine
Phenyl alanine|
i pr
⎛ ⎞
− − − − ⎯⎯⎜ ⎟
⎝ ⎠
−
2 arrangements
If Phenyl alanine is at left corner
2
O
|| gly
NH CH C NH alanine
Phenyl alanine|
Ph
⎛ ⎞
− − − − ⎯⎯⎜ ⎟
⎝ ⎠
2 arrangements
⇒ 4 possible structures.
40. The total number of lone-pairs of electrons in melamine is
Ans. (6) The structure of Melamine is
Hence number of lone pairs = 6

Vidyamandir Classes

Part 3 : MATHS

ONLY ONE is correct
41. For 0a b c ,> > > the distance between (1, 1) and the point of intersection of the lines 0ax by c+ + = is less than
2 2. Then
(A) 0a b c+ − > (B) 0a b c− + <
(C) 0a b c− + > (D) 0a b c+ − <
Ans.
42. The area enclosed by the curves y sin x cos x= + and y | cos x sin x |= − over the interval 0
2
π
,
⎡ ⎤
⎢ ⎥
⎣ ⎦
is
(A) 4( 2 1)− (B) 2 2( 2 1)−
(C) 2( 2 1)+ (D) 2 2( 2 1)+
Ans.
43. The number of points in ( , ),−∞ ∞ for which 2
0x x sin x cos x ,− − = is
(A) 6 (B) 4
(C) 2 (D) 0

Vidyamandir Classes
Ans.
44. The value of
23
2 1
1 1 2
n
n k
cot k
= =
⎛ ⎞⎛ ⎞
⎜ ⎟⎜ ⎟− +
⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠
∑ ∑ is
(A)
23
25
(B)
25
23
(C)
23
24
(D)
24
23
Ans.
45. A curve passes through the point 1
6
π
, .
⎛ ⎞
⎜ ⎟
⎝ ⎠
Let the slope of the curve at each point x,y be 0
y y
sec ,x
x x
⎛ ⎞
+ >⎜ ⎟
⎝ ⎠
.
Then the equation of the curve is
(A)
1
2
y
sin log x
x
⎛ ⎞
= +⎜ ⎟
⎝ ⎠
(B) 2
y
cosec log x
x
⎛ ⎞
= +⎜ ⎟
⎝ ⎠
(C)
2
2
y
sec log x
x
⎛ ⎞
= +⎜ ⎟
⎝ ⎠
(D)
2 1
2
y
cos log x
x
⎛ ⎞
= +⎜ ⎟
⎝ ⎠
Ans.

Vidyamandir Classes
46. Let
1
1
2
f : ,
⎡ ⎤
⎯⎯→⎢ ⎥
⎣ ⎦
(the set of all real numbers) be a positive, non-constant and differentiable function such
that 2f '(x) f ( x )< and
1
1
2
f .
⎛ ⎞
=⎜ ⎟
⎝ ⎠
Then the value of
1
1 2/
f ( x )dx∫ lies in the interval
(A) (2 1 2 )e , e− (B) ( 1 2 1)e , e− −
(C)
1
1
2
e
,e
−⎛ ⎞
−⎜ ⎟
⎝ ⎠
(D)
1
0
2
e
,
−⎛ ⎞
⎜ ⎟
⎝ ⎠
Ans.
47. Let 3 2ˆˆ ˆPR i j k= + −
r
and 3 4ˆˆ ˆSQ i j k= − −
r
determine diagonals of a parallelogram PQRS and 2 3ˆˆ ˆPT i j k= + +
r
be
another vector. Then the volume of the parallelepiped determined by the vectors PT ,PQ
r r
and PS
r
is
(A) 5 (B) 20
(C) 10 (D) 30
Ans.
48. Perpendiculars are drawn from points on the line
2 1
2 1 3
x y z+ +
= =
−
to the plane 3x y z .+ + = The feet of
perpendicular lie on the line.
(A)
1 2
5 8 13
x y z− −
= =
−
(B)
1 2
2 3 5
x y z− −
= =
−
(C)
1 2
4 3 7
x y z− −
= =
−
(D)
1 2
2 7 5
x y z− −
= =
−
Ans.

Vidyamandir Classes
49. Four persons independently solve a certain problem correctly with probabilities
1 3 1 1
2 4 4 8
, , , . Then the probability
that the problem is solved correctly by at least one of them is
(A)
235
256
(B)
21
256
(C)
3
256
(D)
253
256
Ans.
50. Let complex numbers α and
1
α
lie on circles 0
2 2 2
0( ) ( )x x y y r− + − = and 0
2 2 2
0( ) ( ) 4x x y y r ,− + − =
respectively. If 0 0 0
z x iy .= + satisfies the equation 0
2 2
2 2| z | r ,= + then | |α =
(A)
1
2
(B)
1
2

Vidyamandir Classes
(C)
1
7
(D)
1
3
Ans.

Vidyamandir Classes
51. A line l passing through the origin is perpendicular to the lines
( ) ( ) ( )1
ˆˆ ˆ: 3 1 2 4 2 ,l t i t j t k t+ + − + + + − ∞ < < ∞
( ) ( ) ( )2
ˆˆ ˆ: 3 2 3 2 2 ,l s i s j s k s+ + + + + − ∞ < < ∞
Then, the coordinate(s) of the point(s) on l2 at a distance of 17 from the point of intersection of l and l1 is (are)
(A) 7 7
, ,
3 3 3
5⎛ ⎞
⎜ ⎟
⎝ ⎠
(B) ( )1, 1, 0− −
(C) ( )1, 1, 1 (D) 7 7 8
, ,
9 9 9
⎛ ⎞
⎜ ⎟
⎝ ⎠
Ans.
52. Let ( ) sin , 0f x x x xπ= > . Then for all natural numbers n, f’(x) vanishes at
(A) a unique point in the interval 1
,
2
n n
⎛ ⎞
+⎜ ⎟
⎝ ⎠
(B) a unique point in the interval 1
, 1
2
n n
⎛ ⎞
+ +⎜ ⎟
⎝ ⎠
(C) a unique point in the interval ( ), 1n n +
(D) two points in the interval ( ), 1n n +

Vidyamandir Classes
Ans.
53. Let ( )
( 1)
4 2
2
1
1 .
k k
n
n
k
S k
+
=
= −∑ Then Sn can take value(s)
(A) 1056 (B) 1088
(C) 1120 (D) 1332
Ans.
54. For 3x3 matrices M and N, which of the following statement(s) is(are) NOT correct?
(A) NT
M N is symmetric or skew symmetric, according as M is symmetric or skew symmetric
(B) M N – N M is skew symmetric for all symmetric matrices M and N
(C) M N is symmetric for all symmetric matrices M and N
(D) (adj M) (adj N) = adj (M N) for all invertible matrices M and N
Ans.
55. 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

Vidyamandir Classes
Ans.

Vidyamandir Classes
56. Consider the set of eight vectors { }{ }ˆˆ ˆ : , , 1,1 .V ai bj ck a b c= + + ∈ − Three non-coplanar vectors can be
chosen from V in 2
p
ways. Then p is
Ans.
57. Of the 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
( ) and ( 3 ) 2p pα β αβ β γ βγ− 2 = − = .
All the given probabilities are assumed to lie in the interval (0,1).
1
3
Probability of occurrence of E
Then =
Probability of occurrence of E
Ans.
58. The coefficients of three consecutive terms of
5
(1 )
n
x
+
+ are in the ratio 5 : 10 : 14. Then n =

Vidyamandir Classes
Ans.
59. A pack contain 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.
60. A vertical line passing through the point (h, 0) intersects the ellipse
2 2
1
4 3
x y
+ = at the points P and Q. Let the
tangents to the ellipse at P and Q meet at the point R. If ( )hΔ =area of the triangle PQR, 1 1/2 1
max ( )
h
h
≤ ≤
Δ = Δ and
2 1/2 1
min ( ),
h
h
≤ ≤
Δ = Δ then 1 2
8
8
5
Δ − Δ =
Ans.

Vidyamandir Classes
JEE Advanced 2013
Answer Key
Paper 1 ‐ Code 0
PHYSICS

CHEMISTRY

MATHS
1 D 21 B 41 A
2 A 22 A 42 B
3 D 23 A 43 C
4 A 24 C 44 B
5 B 25 D 45 A
6 B 26 B 46 D
7 C 27 B 47 C
8 C 28 D 48 D
9 A 29 B 49 A
10 B 30 D 50 C
11 B,D 31 A 51 B,D
12 A,C 32 A 52 B,C
13 B,C 33 B,D 53 A,D
14 A,D 34 A,B,C,D 54 C,D
15 B,D 35 B,C,D 55 A,C
16 5 36 5 56 5
17 5 37 8 57 6
18 1 38 2 58 6
19 4 39 4 59 5
20 8 40 6 60 9

Vidyamandir Jee Advanced 2013 Paper 1

Recommended

Recommended

More Related Content

What's hot

What's hot (17)

Similar to Vidyamandir Jee Advanced 2013 Paper 1

Similar to Vidyamandir Jee Advanced 2013 Paper 1 (12)

More from askiitians

More from askiitians (20)

Recently uploaded

Recently uploaded (20)

Vidyamandir Jee Advanced 2013 Paper 1