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Vidyamandir Classes
VMC/JEE Advanced 1 2014
JEE Advanced 2014 Paper 1 | Code 7
25 May, 2014 | 9:00 AM – 12 Noon
Answer key and solutions by Vidyamandir Classes
Vidyamandir Classes
VMC/JEE Advanced 2 2014
PART-A PHYSICS
1. A light source, which emits two wavelengths 1 400= nmλ and 2 600= nm,λ is used in a Young’s
double slit experiment. If recorded fringe widths for 1λ and 2λ are 1β and 2β and the number of
fringes for them within a distance y on one side of the central maximum are m1 and m2, respectively,
then :
(A) 2 1>β β
(B) 1 2>m m
(C) From the central maximum, 3rd
maximum of 2λ overlaps with 5th minimum of 1λ
(D) The angular separation of fringes for 1λ is greater than 2λ
Answer
(ABC)
D y, m ,
d d
λ λβ = = θ =
β
As 1 2λ < λ option
⇒ 1 2β < β ⇒ option A is correct
⇒ 1 2m m> ⇒ option B is correct
⇒ 1 2θ < θ
Moreover, 3rd
max. of 22
3D
d
λλ =
5th
min. of 11
9
2
D
d
λλ =
and 2 13 9
2
D D
d d
λ λ= ⇒ option C is correct
2. A parallel plate capacitor has a dielectric slab of dielectric constant K between its plates that covers
1/3 of the area of its plates, as shown in the figure. The total capacitance of the capacitor is C while that
of the portion with dielectric in between is C1. When the capacitor is charged, the plate area covered by
the dielectric gets charge Q1 and the rest of the area gets charge Q2. The electric field in the dielectric is
E1 and that in the other portion is E2. Choose the correct option/options, ignoring edge effects.
(A) 1
2
1=E
E (B)
1
2
1=
E
E K
(C) 1
2
3=
Q
Q K (D)
1
2 +=
C K
C K
Answer
(AD)
1 2 1 2 1 2V V E d E d E E= ⇒ = ⇒ =
1 2
0 0 03 2 3
Q QQE
K A K A / A /= ⇒ =
∈ ∈ ∈
⇒ 1
2 2
Q K
Q=
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01
3
K AC
d
∈=
0 01 2
2
3 3
K A . AC C C
d d
∈ ∈= + = +
= ( ) 023
AK
d
∈+
⇒ 1
2C K
C K
+=
3. At time t = 0, terminal A in the circuit shown in the figure is connected to B by a key and an alternating
current I(t) = I0 cos ( ( )tω , with I0 = 1A and 1500 −= rad sω starts flowing in it with the initial
direction shown in the figure. 7
6=At t ,
π
ω the key is switched from B to D. Now onwards only A and D
are connected. A total charge Q flows from the battery to charge the capacitor fully. If
20=C F ,µ 10= ΩR and the battery is ideal with emf of 50 V, identify the correct statement(s).
(A) Magnitude of the maximum charge on the capacitor before 7
6=t
π
ω is 31 10−× C.
(B) The current in the left part of the circuit just before 7
6=t
π
ω is clockwise.
(C) Immediately after A is connected to D, the current in R is 10A.
(D) 32 10−= ×Q C.
Answer
(CD)
( )0 02
c c ci I cos t V I X cos tπ = ω ⇒ = ω −
( )0C
IV sin t
C= ω
ω
( ) ( )0 1
500
Iq CV sin t sin t= = ω = ω
ω
= ( )32 10 sin t−× ω
Max. charge = 32 10−×
00
37 7
6 6 2
Ii t I cos
π π = = ω × = − ω ω
Hence it is anticlockwise.
Just after connecting to D.
3 37
2 10 106
q sin− −π = × = −
.
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( )50 10 0q
ic
− − =
10i A=
Final charge = 31 10 C−×
( )3 3 31 10 1 10 2 10Q
− − −= × − − × = ×
4. Let E1 (r), E2(r) and E3 (r) be the respective electric field at a distance r from a point charge Q, an
infinitely long wire with constant linear charge density λ , and an infinite plane with uniform surface
charge density .σ If ( ) ( ) ( )1 0 2 0 3 0E r E r E r= = at a given distance r0, then :
(A) 204=Q rσπ (B) 0
2=r
λ
πσ
(C) ( ) ( )1 0 2 02 2 2=E r / E r / (D) ( ) ( )2 0 3 02 4 2=E r / E r /
Answer
(C)
1 204
QE
r=
π∈
202
Er
λ=
π∈
302
Eσ
=∈
At r = r0
1 2 020 00 0
224
QE E Q r
rr
λ= ⇒ = ⇒ = λ
π∈π∈ …(i)
2 3 00 0 0
22 2
E E rr
λ σ= ⇒ = ⇒ λ = π σ
π∈ ∈ …(ii)
From (i) and (ii) ⇒ 202Q r= σπ
01 2 2
0 0 0 0
4
2 4
r Q QE
r r
= =
π∈ π∈
02
0 0 0 0
2
2 2
rE
r r
λ λ = = π∈ π∈
03
02 2
rE
σ = ∈
⇒ 0 01 22
2 2
r rE E
=
0 01 34
2 2
r rE E
=
0 02 32
2 2
r rE E
=
5. A student is performing an experiment using a resonance column and a tuning fork of frequency
1244 s− . He is told that the air in the tube has been replaced by another gas (assume that the column
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VMC/JEE Advanced 5 2014
remains filled with the gas). If the minimum height at which resonance occurs is ( )0 350 0 005. . m,± the
gas in the tube is :
(Useful information : 1 2 1 2 1 2 1 2167 640 140 590/ / / /RT J mole ; RT J mole .− −= = The molar masses
M in grams are given in the options. Take the value of 10
Mfor each gas given there.)
(A) None 10 7
2020 10
M ,
= =
(B) Nitrogen10 3
2828 5
M ,
= =
(C) Oxygen 10 9
3232 16
M ,
= =
(D) Argon 10 17
3636 32
M ,
= =
Answer
(D)
1244f s−=
0 35 0 005min . .= ±ℓ
0 35 0 0054
. .λ
= ±
Using C f= λ
1000
244 0 35 4RT
.M
γ ×= × ×
For Monoatomic, 5
1 673
.γ = =
5 1000 10 10
167 6403
RTRT
M M M
×= × =
⇒ 10 244 0 35 4
0 533640
..
M
× ×= =
For diatomic, 7
1 45
.γ = =
7 1000 10 10
140 5905
RTRT
M M M
×= × =
⇒ 10 24 0 35 4
0 578590
..
M
× ×= =
6. One end of a taut string of length 3m along the x axis is fixed at x = 0. The Speed of the waves in the
string is 1100ms− . The other end of the string is vibrating in the y direction so that stationary waves are
set up in the string. The possible waveform(s) of these stationary waves is(are) :
(A) ( ) 50
6 3
x ty t Asin cos
π π= (B) ( ) 100
3 3
x ty t Asin cos
π π=
(C) ( ) 5 250
6 3
x ty t Asin cos
π π= (D) ( ) 5
2502
xy t Asin cos t
ππ=
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Answer
(ACD)
Equation of stationary wave will be.
y Asin kx cos t= ω
Possible values of λ are given by
32 4
nλ λ
+ = (n = 0, 1, 2,……)
( )2 1 34
nλ
+ =
⇒ ( ) ( )2 2 1 2 112 2
2 1 12 6
n n, K
n
π + π +πλ = = = =
+ λ
3 5 7
6 6 6 6K , , ,
π π π π= , ……..
7. A transparent thin film of uniform thickness and refractive index
1 1 4n .= is coated on the convex spherical surface of radius R at
one end of a long solid glass cylinder of refractive index
2 1 5n . ,= as shown in the figure. Rays of light parallel to the axis
of the cylinder traversing through the film from air to glass get
focused at distance 1f from the film, while rays of light
traversing from glass to air get focused at distance 2f from the
film. Then :
(A) 1 3f R= (B) 1 2 8f . R= (C) 2 2f R= (D) 2 1 4f . R=
Answer
(AC)
Refraction from the thin film will not matter hence
For air to glass
1
1 5 1 1 5 1. .
f R
−− =
∞
⇒ 1 3f R=
For glass to air
2
1 1 5 1 1 5. .
f R
−− =
∞ −
⇒ 2 2f R=
8. Heater of an electric kettle is made of a wire of length L and diameter d. It takes 4 minutes to raise the
temperature of 0.5 kg water by 40 K. This heater is replaced by a new heater having two wires of the
same material, each of length L and diameter 2d. The way these wires are connected is given in the
options. How much time in minutes will it take to raise the temperature of the same amount of water by
40 K?
(A) 4 if wires are in parallel (B) 2 if wires are in series
(C) 1 if wires are in series (D) 0.5 if wires are in parallel
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Answer
(BD)
0 2R
d
ρ
π=ℓ
( )2
0
4V
H minR
= ×
For the new wire
( )0
2 42
RR'
d= =
ℓρ
π
For series 0
2
RR =
( )2
0 2s
VH t
R /=
⇒ 2st min=
For parallel
0
8
RR =
( )2
0 8p
VH t
R /=
⇒ 0 5pt . min=
9. Two ideal batteries of emf V1 and V2 and three resistances R1, R2 and R3
are connected as shown in the figure. The current in resistance R2 would
be zero if :
(A) 1 2V V= and 1 2 3R R R= =
(B) 1 2V V= and 1 2 32R R R= =
(C) 1 22V V= and 1 2 32 2R R R= =
(D) 1 22V V= and 1 2 32R R R= =
Answer
(ABD)
If 2
0Ri =
⇒ VA = VB
⇒ 1 1 0V iR− = &
⇒ 1 2
1 3
V V
R V=
A, B, D satisfy the above criteria.
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VMC/JEE Advanced 8 2014
10. In the figure, a ladder of mass m is shown leaning against a wall.
It is in static equilibrium making an angle θ with the horizontal
floor. The coefficient of friction between the wall and the ladder
is 1µ and that between the floor and the ladder is 2µ .
The normal reaction of the wall on the ladder is N1 and that of
the floor is N2. If the ladder is about to slip, then :
(A) 1 20 0µ µ= ≠ and 22
mgN tanθ = (B) 1 20 0µ µ≠ = and 1
2
mgN tanθ =
(C) 1 20 0µ µ≠ ≠ and 2
1 21
mgN
µ µ=
+ (D) 1 20 0µ µ= ≠ and 1
2
mgN tanθ =
Answer
(CD)
1 20 0,µ = µ ≠
( )1 0AN sin mg cos× θ = × θ τ =ℓ
12
mgN
tan=
θ
2N mg=
1 20 0,µ ≠ µ =
⇒ The rod can’t stay in equilibrium
1 20 0,µ ≠ µ ≠
[ ]2 2 2 02
BN cos N sin mg . cos× θ = µ θ + θ τ =ℓ
ℓ ℓ
[ ]2 22
N cos sin mg . cosθ − µ θ = θℓ
ℓ ℓ …(i)
2 1 1N N mg+ µ = …(ii)
1 2 2N N= µ …(iii)
From (ii) and (iii), 21 21
mgN =
+ µ µ
11. A rocket is moving in a gravity free space with a constant acceleration of 22ms− along + x direction
(see figure). The length of a chamber inside the rocket is 4m. A ball is thrown from the left end of the
chamber in + x direction with a speed of 10 3. ms− relative to the rocket. At the same time, another ball
is thrown in x− direction with a speed of 10 2. ms− from its right end relative to the rocket. The time in
seconds when the two balls hit each other is ___________.
A
B N1
N2
mg
θ
1µ
1 1 1f N=µ
2 2 2f N=µ2µ
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Answer
(2)
12. A galvanometer gives full scale deflection with 0.006 A current. By connecting it to a 4990Ω
resistance, it can be converted into a voltmeter of range 0 30V− . If connected to a 2
249
nΩ resistance,
it becomes an ammeter of range 0 1 5. A− . The value of n is __________.
Answer
(5)
Ig = 0.006A
For voltmeter conversion.
( )g g sV I R R= +
( )30 0 006 4990g. r= + …(i)
For ammeter conversion:
( )g g g sI R I I R= −
( )0 006 1 5 0 006g s. R . . R'= − …(ii)
From (i) and (ii)
10
5249
sR' n= ⇒ =
13. A horizontal circular platform of radius 0.5 m and mass 0.45 kg is free to rotate about its axis.
Two massless spring toy-guns, each carrying a steel ball of mass 0.05 kg are attached to the platform at
a distance 0.25 m from the centre on its either sides along its diameter (see figure). Each gun
simultaneously fires the balls horizontally and perpendicular to the diameter in opposite directions.
After leaving the platform, the balls have horizontal speed of 19 ms− with respect to the ground.
The rotational speed of the platform in 1rad s−
after the balls leave the platform is:
G I Ig
gI I−
R's
Ig Rs G
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Answer
(4)
From conservation of angular momentum.
( )( )( ) ( ) ( )20 5
0 05 9 0 25 2 0 452
.. . .× = ω
⇒ 4 rad / sω =
14. Consider an elliptically shaped rail PQ in the vertical plane with OP = 3
m and OQ = 4 m. A block of mass 1 kg is pulled along the rail from P to
Q with a force of 18 N, which is always parallel to line PQ (see the
figure given). Assuming no frictional losses, the kinetic energy of the
block when it reaches Q is (n × 10) Joules. The value of n is ______.
(Take acceleration due to gravity = 210 ms− ).
Answer
(5)
FdW F . dr=
= ( ) ( )ˆ ˆ ˆ ˆF cos i F sin j · dx i dy j− θ + θ +
3 4
5 5F
F FdW dx dy= − +∫ ∫ ∫ .
( ) ( )3 43 4
5 5F
F FW = − − + +
= 25 25 18
905 5
FJ
×= =
From energy conservation.
1 10 4 40FW U K K K= ∆ + ∆ = × × + ∆ = + ∆
⇒ 50 5K J n∆ = ⇒ =
15. A uniform circular disc of mass 1.5 kg and radius 0.5 m is initially at rest
on a horizontal frictionless surface. Three forces of equal magnitude
F = 0.5 N are applied simultaneously along the three sides of an
equilateral triangle XYZ with its vertices on the perimeter of the disc
(see figure). One second after applying the forces, the angular speed of
the disc in 1rad s− is _________.
Answer
(2)
( )3 3 30Fr F R sinτ = =
Iτ = α
( )
2 2
3 30 3 0 5 0 5 0 5 2
2 1 5 0 5
F R sin . . .
I mR / . .
τ × × × ×α = = =
×
= 2 rad/s2
2 1 2t rad / sω = α = × =
P
Q
4m
3m
90
O
O
F
F
F
X
Y
Z
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16. Two parallel wires in the plane of the paper are distance X0 apart. A point charge is moving with speed
u between the wires in the same plane at a distance X1 from one of the wires. When the wires carry
current of magnitude I in the same direction, the radius of curvature of the path of the point charge is
R1. In contrast, if the currents I in the two wires have directions opposite to each other, the radius of
curvature of the path is R2. If 0
1
3x
x= , the value of 1
2
R
R is _______.
Answer
(3)
muR
qB=
If the current is in same direction
( )
0 01
1 0 12 2
I IB
x x x
µ µ= −
π π −
If the current is in opposite direction.
( )
0 02
1 0 12 2
I IB
x x x
µ µ= −
π π −
Give x0 = 3x1
⇒ B2 = 3B1
1 2
2 1
13
R BR
B R B∝ ⇒ = =
17. To find the distance d over which a signal can be seen clearly in foggy conditions, a railways engineer
uses dimensional analysis and assumes that the distance depends on the mass density ρ of the fog,
intensity (power/area) S of the light from the signal and its frequently f. The engineer finds that d is
proportional to 1 / nS . The value of n is __________.
Answer
(3)
Let a b cd s f= ρ
( ) ( ) ( ) ( )3 3 1a b C
L ML MT T− − −=
( ) ( ) ( ) ( )3 3a b a b cL M L T
+ − − −=
Equating Power.
3 1 1 3a a /− = ⇒ = −
0 1 3a b b a /+ = ⇒ = − =
⇒ 1 3 3/d s n∝ ⇒ =
18. Airplanes A and B are flying with constant velocity in
the same vertical plane at angles 300 and 60
0 with
respect to the horizontal respectively as shown in figure.
The speed of A is ms .−1100 3 At time t = 0s, an
observer in A finds B at a distance of 500 m.
This observer sees B moving with a constant velocity
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perpendicular to the line of motion of A. If at t = t0 ,
A just escapes being hit by B, t0 in seconds is _______.
Answer
(5)
BA B AV V V= −
At t = 0, AB = 500 m
If at t = t0 A just escapes being hit by B & is toBA AV V⊥
30B AV co s V=
⇒ 200BV m / s=
⇒ 500
530B
t s.V sin
= =
19. A thermodynamic system is taken from an initial state i with internal energy 100iU J= to the final state f
along two different paths and ,iaf ibf as schematically shown in the figure. The work done by the system along
the paths , andaf ib bf are 200 50 100, andaf ib bfW J W J W J= = = respectively. The heat supplied to the
system along the path , and are , andiaf ib bfiaf ib bf Q Q Q respectively. If the internal energy of the system in
the state b is 200 500and ,b iafU J Q J= = the ratio /af ibQ Q is _____.
Answer
(2)
100 200i bU J U J= , =
100 50 100af ib bfW J W W J= , = , =
500iafQ J=
( )∆ 200 100 50 150ib ib ibQ U W J= + = − + =
∆bf bf bfQ U W= +
Now
∆ ∆ 500 20 300ibf iaf iaf iaf iaf afU U Q W Q W J= = − = − = − =
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∆ ∆ 300ib bfU U+ =
⇒ ( )200 100 ∆ 300 ∆ 200bf bfU U− + = ⇒ =
⇒ 200 100 300bfQ J= + =
⇒ 300 150 2bf ibQ / Q /= =
20. During Searle’s experiment, zero of the Vernier scale lies between 23 20 10−×. m and
23 25 10−×. m of the main
scale. The 20th
division of the Vernier scale exactly coincides with one of the main scale divisions. When an
additional load of 2 kg is applied to the wire, the zero of the Vernier scale still lies between 23 20 10−×. m and
23 25 10−×. m of the main scale but now the 45th
division of Vernier scale coincides with one of the main scale
divisions. The length of the thin metallic wire is 2 m and its cross-sectional area is 7 28 10 .m−× The least count of
the Vernier scale is 51.0 10 .m−× The maximum percentage error in the Young’s modules of the wire is _____.
Answer
(8)
yAx
ω=ℓ
(x = elongation)
y x
y x
∆ ∆=
2 1x x x= −
45 20x L .C L .C= × − ×
= 525 25 1 10L .C m−× = × ×
52 1 2 2 10| x | | x | | x | L .C m−∆ = ∆ + ∆ = × = ×
⇒ 100 100y x
y x
∆ ∆× = ×
= 5
5
2 10100 8
25 10%
−
−
×× =
×
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PART-B CHEMISTRY
21. An ideal gas in a thermally insulated vessel at internal pressure = P1, volume = V1 and absolute
temperature = T1 expands irreversibly against zero external pressure, as shown in the diagram.
The final internal pressure, volume and absolute temperature of the gas are 2 2 2P , V and T , respectively.
For this expansion,
(A) q = 0 (B) 2 1T T= (C) 2 2 1 1P V P V= (D) 2 2 1 1P V P Vγ γ=
Answer
(ABC)
q = 0 (∵ Vessel is thermally insulated)
extw P V 0= − ∆ =
U q ( w) 0∆ = − − = T 0⇒ ∆ = (Hence T2 = T1)
Now for ideal gas PV = nRT is always valid.
⇒ 1 1 2 2
1 1 2 2 2 11 2
P V P VP V P V ( T T )
T T= ⇒ = =∵
Note : 1 21 2P V P Vγ γ≠ (∵ the process has been carried out irreversibly)
22. Hydrogen bonding plays a central role in the following phenomena:
(A) Ice floats in water
(B) Higher Lewis basicity of primary amines than tertiary amines in aqueous solutions
(C) Formic acid is more acidic than acetic acid
(D) Dimerisation of acetic acid in benzene
Answer
(ABD)
23. Upon heating with 2Cu S , the reagent(s) that give copper metal is(are)
(A) CuFeS2 (B) CuO (C) Cu2O (D) 4CuSO
Answer
(BCD)
2 2 2Cu S Cu O Cu SO+ → +
2 2Cu S 2CuO 4Cu SO+ → +
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24. The correct combination of names for isomeric alcohols with molecular formula 4 10C H O is(are) :
(A) tert-butanol and 2-methylpropan-2-ol (B) tert-butanol and 1, 1-dimethylethan-1-ol
(C) n-butanol and butan-1-ol (D) Isobutyl alcohol and 2-methylpropan-1-ol
Answer
(A,C,D)
3
3
3
CH|
H C C OH|CH
− − 3 2 2 2H C CH CH CH OH− − − − 3 2
3
H C C CH OH|CH
− − −
Trivial name (tert butanol) n-butanol isobutyl alcohol
IUPAC name 2 methyl propan-2-ol Butan-1-ol 2-methyl propan-1-ol
∴Only correct combination is (A,C,D).
25. The reactivity of compound Z with different halogens under appropriate conditions is given below:
The observed pattern of electrophilic substitution can be explained by :
(A) The steric effect of the halogen
(B) The steric effect of the tert-butyl group
(C) The electronic effect of the phenolic group
(D) The electronic effect of the tert-butyl group
Answer
(ABC)
)
As the size of halogens decreases, poly-halogenation can take place.
Though t-Butyl also activates the ring via +I effect but it is the steric effect of t-Butyl group which is more
dominant and causes crowding at ortho positions. Hence, in the given molecule the directing property is
being controlled by the OH group which exerts +M effect (o-p directing effect)
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26. The pair(s) of reagents that yield paramagnetic species is(are) :
(A) Na and excess of 3NH (B) K and excess of 2O
(C) Cu and dilute 3HNO (D) 2O and 2-ethylanthraquinol
Answer
(ABC)
( ) ( ) ( ) ( ) ( ) ( ) ( )3 3 3x y
Paramagnetic (Blue solution)
Na s x y NH l Na NH am. e NH am.−+ + + → +
2 2Paramagnatic
K O KO+ →
( ) ( )3 3 2 ParamagneticCu HNO dil Cu NO NO+ → +
2 2 2(not Paramagnetic)
O 2 ethylanthraquinol H O+ − →
27. The correct statement(s) for orthoboric acid is(are) :
(A) It behaves as a weak acid in water due to self ionization
(B) Acidity of its aqueous solution increases upon addition of ethylene glycol
(C) It has a three dimensional structure due to hydrogen bonding
(D) It is a weak electrolyte in water
Answer
(BD)
28. In the reaction shown below, the major product(s) formed is(are) :
(A) (B)
(C) (D)
Answer
(A)
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Since lone pairs of amide is involved in conjugation
∴ There is very less possibility of the lone pairs taking part in the reaction as a nucleophile.
29. In a galvanic cell, the salt bridge :
(A) Does not participate chemically in the cell reaction.
(B) Stops the diffusion of ions from one electrode to another.
(C) is necessary for the occurrence of the cell reaction.
(D) Ensures mixing of the two electrolytic solutions.
Answer
(AB)
30. For the reaction:
3 2 4 2 4I ClO H SO Cl I HSO− − − −+ + → + +
The correct statement(s) in the balanced equation is(are) :
(A) Stoichiometric coefficient of 4HSO− is 6 (B) Iodide is oxidized
(C) Sulphur is reduced (D) 2H O is one of the products
Answer
(ABD)
3 2 4 4 2I ClO H SO Cl HSO I− − − −+ + → + +
22I I 2e 3− −→ + ×
3 26H ClO 6e Cl 3H O 1+ − − −+ + → + ×
3 2 2ClO 6I 6H 3I Cl 3H O− − + −+ + → + +
3 2 4 2 2 4ClO 6I 6H SO 3I Cl 3H O 6HSO− − − −+ + → + + +
31. The total number(s) of Stable conformers with non-zero dipole moment for the following compound is(are)
Answer
(3)
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VMC/JEE Advanced 18 2014
Now the stable conformers of the given molecule are
32. Among 2 2 3 2PbS, CuS, HgS, MnS, Ag S, NiS, CoS, Bi S and SnS , the total number of BLACK coloured.
sulphides is _________.
Answer
(7)
2 2 3PbS, CuS, HgS, Ag S, NiS, CoS and Bi S are the black colour sulphides.
33. A list of species having the formula 4XZ is given below :
( ) [ ]224 4 4 4 4 3 44
XeF ,SF ,SiF , BF , BrF , [Cu NH ] , FeCl ,−− − +
[ ] [ ]2 2
4 4CoCl and PtCl .− −
Defining shape on the basis of the location of X and Z atoms, the total number of species having a square planar
shape is ______.
Answer
(4)
2 24 4 3 4 4XeF , BrF [Cu(NH ) ] and [PtCl ]− + − are square planar species
34. Consider all possible isomeric ketones, including stereoisomers of MW = 100. All these isomers are independently
reacted with NaBH4 (NOTE: stereoisomers are also reacted separately). The total number of ketones that give a
racemic product (s) is(are) ______.
Answer
(5)
4NaBH3 2 2 2 3
O||
CH CH CH CH C CH− − − − − → Racemic Mixture
4* NaBH
3 2 3|
3
O||
CH CH C H C CH
CH
− − − − → Diastereomeric Mixture
(2 isomers)
4NaBH3 2 3
|
3
O||
CH C H CH C CH
CH
− − − − → Racemic Mixture
4
3| ||
NaBH3 3
|
3
CH O
CH C C CH
CH
− − − → Racemic Mixture
4NaBH3 2 2 2 3
O||
CH CH CH C CH CH− − − − − → Racemic Mixture
Vidyamandir Classes
VMC/JEE Advanced 19 2014
4NaBH3 2 3
|
3
O||
CH C H C CH CH
CH
− − − − → Racemic Mixture
35. Consider the following list of reagents:
Acidified K2Cr2O7, Alkaline 4, 4 2 2 2 3 3 3 2 2 3KMnO CuSO , H O , Cl ,O , FeCl , HNO and Na S O .
The total number of reagents that can oxidise aqueous iodide to iodine is _____.
Answer
(7)
2 2 7K Cr O2I I− → ……….1
4KMnO / OH OH2 3I I I IO− − −→ → +
22 2 2I Cu Cu I I− ++ → + ……....2
2 2 2 3 3 3H O | Cl | O | FeCl | HNO2
3 4 5 6 7I I− →
2 22 2 3 4 6I S O I S O− − −+ → +
36. A compound H2X with molar weight of 80 g is dissolved in a solvent having density of 0.4 1gml− .
Assuming no change in volume upon dissolution, the molality of a 3.2 molar solution is _______.
Answer
(8)
Molarity = 3.2 M
Volume of solvent = 1000 ml
∴ weight of solvent = 400 g
∴ molality = 3.2
1000 8400
× =
37. In an atom, the total number of electrons having quantum numbers n = 4, 1 sm 1 and m 1/ 2= = −
is _______.
Answer
(6)
n = 4
l = 0,1, 2, 3
4s, 4p, 4d, 4f
1m 1=
1m 1⇒ = ±
Subshell l m
4s 0 0
4p 1 –1, 0, +1
4d 2 –2, –1, 0, 1, 2
4f 3 –3, –2, –1, 0, 1, 2, 3
Vidyamandir Classes
VMC/JEE Advanced 20 2014
38. MX2 dissociates into 2M + and X− ions in an aqueous solution, with a degree of dissociation ( α ) of
0.5. The ratio of the observed depression of freezing point of the aqueous solution to the value of the
depression of freezing point in the absence of ionic dissociation is
Answer
(2)
22MX M 2X+ −+
1 0 0
1-α α 2α
i = 1+2α = 2
Therefore the ratio of observed depression of freezing point of the aqueous solution to the value of the
depression of freezing point in the absence of ionic dissociation is 2 : 1.
39. The total number of distinct naturally occurring amino acids obtained by complete acidic hydrolysis
of the peptide shown below is ________.
Answer
(1) Total number of distinct amino acids obtained is 4.
Out of the four distinct products formed only glycine exists naturally.
Vidyamandir Classes
VMC/JEE Advanced 21 2014
40. If the value of Avogadro number is 23 16.023 10 mol−× and the value of Boltzmann constant is
23 11.380 10 JK− −× , then the number of significant digits in the calculated value of the universal gas
constant is
Answer
(4)
The value of R calculated from the given values will be written upto 4 significant digits.
Vidyamandir Classes
VMC/JEE Advanced 22 2014
PART-C MATHEMATICS
41. Let : (0, )f ∞ → ℝ be given by
1
1( )tx t
x
dtf x e
t
− + = ∫
Then :
(A) ( )f x is monotonically increasing on [ )1, ∞
(B) ( )f x is monotonically decreasing on (0, 1)
(C) 1
( ) 0f x fx
+ =
, for all (0, )x∈ ∞
(D) (2 )xf is an odd function of x onℝ
Answer
42. Let M and N be two 3 3× matrices such that MN = NM. Further, if 2M N≠ and
2 4M N= , then
(A) Determinant of 2 2( )M MN+ is 0
(B) There is a 3 3× non-zero matrix U such that 2 2( )M MN+ U is the zero matrix
(C) Determinant of 2 2( ) 1M MN+ ≥
(D) For a 3 3× matrix U, if 2 2( )M MN U+ equals the zero matrix then U Is the zero matrix
Answer
Vidyamandir Classes
VMC/JEE Advanced 23 2014
43. Let a∈ℝ and let :f →ℝ ℝ be given by :
5( ) 5f x x x a= − +
Then :
(A) ( )f x has three real roots if a > 4
(B) ( )f x has only one real root if a > 4
(C) ( )f x has three real roots if a < -4
(D) ( )f x has three real roots if –4 < a < 4
Answer
Vidyamandir Classes
VMC/JEE Advanced 24 2014
44. From a point ( , , )P λ λ λ , perpendiculars PQ and PR are drawn respectively on the lines , 1y x z= = and
, 1y x z= − = − . If P is such that QPR∠ is a right angle, then the possible value(s) ofλ is(are) :
(A) 2 (B) 1 (C) 1− (D) 2−
Answer
45. Let M be a 2 2× symmetric matrix with integer entries. Then M is invertible if :
(A) The first column of M is the transpose of the second row of M
(B) The second row of M is the transpose of the first column of M
(C) M is a diagonal matrix with nonzero entries in the main diagonal
(D) The product of entries in the main diagonal of M is not the square of an integer
Vidyamandir Classes
VMC/JEE Advanced 25 2014
Answer
46. Let ,x y and z
be three vectors each of magnitude 2 and the angle between each pair of them is 3
π.
If a
is a nonzero vector perpendicular to x and y z and b×
is a nonzero vector perpendicular to
y and z x×
, then :
(A) ( . )( )b b z z x= −
(B) ( . )( )a a y y z= −
(C) . ( . )( . z)a b a y b= −
(D) ( . )( )a a y z y= −
Answer
47. For every pair of continuous functions [ ]0 1 →ℝf , g : , such that max
[ ] [ ] ( ) : 0,1 max ( ) : 0.1 ,∈ = ∈f x x g x x the correct statement(s) is (are):
(A) 2 2( ( )) 3 ( ) ( ( )) 3 ( ) for some [0,1]+ = + ∈f c f c g c g c c
(B) 2 2( ( )) ( ) ( ( )) 3 ( ) for some [0,1]+ = + ∈f c f c g c g c c
(C) 2 2( ( )) 3 ( ) ( ( )) ( ) for some [0,1]+ = + ∈f c f c g c g c c
(D) 2 2( ( )) ( ( )) for some [0,1]= ∈f c g c c
Vidyamandir Classes
VMC/JEE Advanced 26 2014
Answer
48. Let :[ , ] [1, )→ ∞f a b be a continuous function and let : →ℝ ℝg be defined as
0 if ,
( ) ( ) if ,
( ) if .
<
= ≤ ≤
>
∫
∫
x
a
b
a
x a
g x f t dt a x b
f t dt x b
Then :
(A) g(x) is continuous but not differentiable at a
(B) g(x) is differentiable on ℝ
(C) g(x) is continuous but not differentiable at b
(D) g(x) is continuous and differentiable at either a or b but not both
Answer
Vidyamandir Classes
VMC/JEE Advanced 27 2014
49. Let :2 2
− →
ℝf ,
π π be given by f (x) = ( )( )3
log +sec x tan x .
Then :
(A) f (x) is an odd function (B) f (x) is a one-one function
(C) f (x) is an onto function (D) f (x) is an even function
Answer
50. A circle S passes through the point (0, 1) and is orthogonal to the circles ( )2 21 16− + =x y and
2 2 1+ =x y . Then :
(A) Radius of S is 8 (B) Radius of S is 7
(C) Centre of S is ( )7 1− , (D) Centre of S is ( )8 1− ,
Answer
Vidyamandir Classes
VMC/JEE Advanced 28 2014
51. Let 2n ≥ be an integer. Take n distinct points on a circle and join each pair of points by a line
segment. Colour the line segment joining every pair of adjacent points by blue and the rest by red. If the
number of red and blue line segments are equal, then the value of n is _______.
Answer
52. The value of ( )2 51
3 2
204 1
dx x dx
dx
−
∫ is _______.
Answer
53. Let 1 2 3 4 5n n n n n< < < < be positive integer such that 1 2 3 4 5 20n n n n n+ + + + = . Then the number of
such distinct arrangements ( )1 2 3 4 5n ,n ,n ,n ,n is _______.
Vidyamandir Classes
VMC/JEE Advanced 29 2014
Answer
Alternate approach/solution
Consider
2 1
3 2 11
4 3 1
5 4 1
, , , , 0
n n a
n n b n a bhere n a b c d
n n c n a b c
n n d n a b c d
= + = + = + + >
= + = + + + = + = + + + +
Given Equation reduces to
15 4 3 2 20n a b c d+ + + + =
Number of Solution will be
= Coeff. of 20x in (Integral Solutions)
Vidyamandir Classes
VMC/JEE Advanced 30 2014
5 10 4 8 12 3 6 9 2 4 6 8 2 3 4 5 6( ) ( ) ( ) ( ) ( )x x x x x x x x x x x x x x x x x x+ × + + × + + × + + + × + + + + +
=Coeff. of 20x in
15 5 4 8 3 6 2 4 6 2 3 4 5 (1 )(1 ) (1 ) (1 ) (1 )x x x x x x x x x x x x x x× + + + × + + × + + + × + + + + +
=Coeff. of 5x in
5 4 8 3 6 2 4 6 2 3 4 5(1 ) (1 ) (1 )(1 ) (1 )x x x x x x x x x x x x x+ × + + × + + + + + × + + + + +
=7
54. Let anda, b c
be three non-coplanar unit vectors that the angle between every pair of them is 3
π.
If a b b c pa qb rc ,× + × = + +
where andp, q r are scalars, then the value of 2 2 2
2
2p q r
q
+ +
is _______.
Answer
55. The slope of the tangent to the curve ( ) ( )2 25 21y x x x− = + at the point (1, 3) is
Answer
Vidyamandir Classes
VMC/JEE Advanced 31 2014
56. Let [ ] [ ]0 4 0→f : , ,π π be defined by f (x) = ( )1−cos cos x . The number of points [ ]0 4∈x , π satisfying
the equation ( )10
10
−=
xf x is :
Answer
57. Let a, b, c be positive integers such that b
a is an integer. If a, b, c are in geometric progression and the
arithmetic mean of a, b, c is 2+b , then the value of
2 14
1
+ −
+
a a
a is _______.
Answer
58. The largest value of the non-negative integer a for which( )
( )
1
1
1
1 1
1 1 4
x
x
x
ax sin x alim
x sin x
−
−
→
− + − + =
+ − − is ______.
Vidyamandir Classes
VMC/JEE Advanced 32 2014
Answer
59. Let : and :f g→ →ℝ ℝ ℝ ℝ be respectively given by ( ) ( ) 21 and 1f x x g x x= + = + .
Define :h →ℝ ℝ by ( )( ) ( ) ( ) ( )
0
0
max f x , g x if x ,h x
min f x , g x if x .
≤=
>
The number of points at which ( )h x is not differentiable is _______.
Answer
Vidyamandir Classes
VMC/JEE Advanced 33 2014
60. For a point P in the plane, let ( ) ( )1 2andd P d P be the distances of the point P from the lines
0 and 0x y x y− = + = respectively. The area of the region R consisting of all points P lying in the first
quadrant of the plane and satisfying ( ) ( )1 22 4d P d P ,≤ + ≤ is _______.
Answer
Q.No. 1 A,B,C Q.No.21 A,B,C Q.No.41 A,C,D
Q.No. 2 A,D Q.No.22 A,B,D Q.No.42 A,B
Q.No. 3 C,D Q.No.23 B,C,D Q.No.43 B,D
Q.No. 4 C Q.No.24 A,C,D Q.No.44 C
Q.No. 5 D Q.No.25 A,B,C Q.No.45 C,D
Q.No. 6 A,C,D Q.No.26 A,B,C Q.No.46 A,B,C
Q.No. 7 A,C Q.No.27 B,D Q.No.47 A,D
Q.No. 8 B,D Q.No.28 A Q.No.48 A,C
Q.No. 9 A,B,D Q.No.29 A,B Q.No.49 A,B,C
Q.No. 10 C,D Q.No.30 A,B,D Q.No.50 B,C
Q.No. 11 2 Q.No.31 3 Q.No.51 5
Q.No. 12 5 Q.No.32 7 Q.No.52 2
Q.No. 13 4 Q.No.33 4 Q.No.53 7
Q.No. 14 5 Q.No.34 5 Q.No.54 4
Q.No. 15 2 Q.No.35 7 Q.No.55 8
Q.No. 16 3 Q.No.36 8 Q.No.56 3
Q.No. 17 3 Q.No.37 6 Q.No.57 4
Q.No. 18 5 Q.No.38 2 Q.No.58 2
Q.No. 19 2 Q.No.39 1 Q.No.59 3
Q.No. 20 8 Q.No.40 4 Q.No.60 6
JEE Advanced 2014
Paper 125 May 2014 | 9:00AM - 12:00 PM
Code - 7
Answer keyPHYSICS CHEMISTRY MATHS