Jee advanced-2014-question-paper1-with-solution

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IIT - JEE 2014 (Advanced)
CODE : 1

(2) Vidyalankar : IIT JEE 2014 − Advanced : Question Paper & Solution
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IIT JEE 2014 Advanced : Question Paper & Solution (Paper – I) (3)
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PART I : PHYSICS
SECTION − 1 : (One or More Than One Options Correct Type)
This section contains 10 multiple choice questions. Each question has four choices (A),
(B), (C) and (D) out of which ONE or MORE THAN ONE are correct.
1. 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 (ωt), with I0 = 1A and ω = 500 rad s−1
starts flowing in
it with the initial direction shown in the figure. At t =
7
6
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 C = 20μF, R = 10 Ω and the battery is ideal with emf of 50V,
identify the correct statement (s).
(A)Magnitude of the maximum charge on the capacitor before t =
7
6
π
ω
is 1 × 10−3
C.
(B) The current in the left part of the circuit just before t =
7
6
π
ω
is clockwise.
(C) Immediately after A is connected to D, the current in R is 10A.
(D)Q = 2 × 10−3
C.
1. (C), (D)
If q represents the charge on capacitor’s upper plate:
I (t) = I0 cos(ωt) =
dq
dt
⇒ q(t) = 0I
ω
sin(ωt)
Max charge = 1
1A
500 rad s−
= 2 × 10−3
C
Charge on upper plate at t =
7
6
π
ω
= 0I
ω
sin
7
6
π⎛ ⎞
⎜ ⎟
⎝ ⎠
= 0I
2
−
ω
When capacitor is fully charged; charge on upper plate = 50 V × 20μF
= 1 × 10−3
C
∴ Q = 1 × 10−2
C − 0I
2
⎛ ⎞
−⎜ ⎟ω⎝ ⎠
= 2 × 10−3
C
Voltage across capacitor when A and D are connected =
3
6
1 10 C
20 10 F
−
−
×
× μ
= 50V
Total voltage across resistor = 100V
⇒ current =
100v
10Ω
= 10A
Current is anti clockwise.

(4)
2. A light source, which emits two wavelengths λ1 = 400 nm and λ2 = 600 nm, is used in a
Young's double slit experiment. If recorded fringe widths for λ1 and λ2 are β1 and β2 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)β2 > β1
(B) m1 > m2
(C) From the central maximum, 3rd
maximum of λ2 overlaps with 5th
minimum of λ1
(D) The angular separation of fringes for λ1 is greater than λ2
2. (A), (B), (C)
β = λ
D
d
1
2 2 1n m
2
λ
λ =
n2600 = 1
400
m
2
3n2 = m1
For n2 = 3
m1 = 9
which is 5th
minima.
3. 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 100 ms−1
. 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) y(t) =
x 50 t
A sin cos
6 3
π π
(B) y(t) =
x 100 t
A sin cos
3 3
π π
(C) y(t) =
5 x 250 t
A sin cos
6 3
π π
(D)y(t) =
5 x
A sin cos t
2
π
250 π
3. (A), (C), (D)
ν =
k
ω
1
A B C D 100ms−
ν = ν = ν = ν =
x = 3 is an antinode. This eliminates (B)
4. 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
E
E
= 1 (B) 1
2
E
E
=
1
K
(C) 1
2
Q
Q
=
3
K
(D)
1
C
C
=
2
K
+ Κ

(5)
4. (A), (D)
If c = oA
3d
ε
then c1 = kc c2 = 2c
c1 + c2 = C ⇒ (k + 2) c = C ⇒ c =
c
k 2+
∴ c1 =
kc
k 2+
c2 =
2C
k z+
If charging voltage is V: charges will be in the ratio of capacities and as potential
difference is same. Electric field should be equal.
5. Let E1(r), E2(r) and E3(r) be the respective electric fields 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 E1(r0) = E2(r0) = E3(r0) at a given distance
r0, then
(A)Q = 2
04 rσπ (B) r0 =
2
λ
πσ
(C) E1(r0/2) = 2E2(r0/2) (D)E2(r0/2) = 4E3(r0/2)
5. (C)
202 0 0
0 0 00
2
0 02
0 00
1 q
2 r q 2 r 2 r4 2 rr
1 q
q 2 r r
4 2r
λ ⎫
= ⇒ λ = ⇒ π σ = λ⎪πε πε ⎪
λ⎬
σ ⇒ σ =⎪= ⇒ = π σ π
⎪πε ε ⎭
1 0E (r / 2)
4
= 0E(r / 2)
2
6. A student is performing an experiment using a resonance column and a tuning fork of
frequency 244 s−1
. He is told that the air in the tube has been replaced by another gas
(assume that the column 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 : 167RT = 640 j1/2
mole−1/2
; 140RT = 590 J1/2
mole−1/2
. THe
molar masses M in grams are given in the options. Take the values of
10
M
for each gas
as given there)
(A)Neon
10 7
M 20,
20 10
⎛ ⎞
= =⎜ ⎟
⎝ ⎠
(B) Nitrogen
10 3
M 28,
28 5
⎛ ⎞
= =⎜ ⎟
⎝ ⎠
(C) Oxygen
10 9
M 32,
32 16
⎛ ⎞
= =⎜ ⎟
⎝ ⎠
(D) Argon
10 17
M 36,
36 32
⎛ ⎞
= =⎜ ⎟
⎝ ⎠
6. (D)
f = 244 Hz
4
λ
= 0.356 ± 0.005 ⇒ λ = 1.400 ± 0.020 ⇒ ν = (341.6 ± 4.88)
m
s
ν = λf = ν =
rRT
M
=
100 rRT
100M
=
3
100RrT
100 M 10 kg / mol−
× ×

(6)
for neon :
7
640
10
× = 448 ms−1
for Nitrogen :
3
590
5
× = 384 ms−1
for Oxygen : 590 ×
9
16
= 351.875 ms−1
for Argon : 640 ×
17
32
= 340 ms−1
7. 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
7. (B) (D)
MSΔT =
2
ε
t
R
M = 0.5 S ΔT = 40K.
ΔQ = 0.5 × S × 40 =
2
ε
× 4 min
R
R =
ρL
πd
4
2
R1 = 2
ρL R
=
4(Rd)
π
4
Req. series
R R R
+ =
4 4 2
∴ ΔQ =
2 2
ε × 4 ε
= × t
RR
2
t = 2 min
Req Parallel =
R R
×
R4 4 =
R R 8+
4 4
ΔQ =
2
ε
× 4
R
=
2
ε
× t
R
8
t = 0.5 min
8. 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 μ1 and that
between the floor and the ladder is μ2. 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)μ1 = 0, μ2 ≠ 0 and N2 tan θ =
mg
2
(B) μ1 ≠ 0, μ2 = 0 and N1 tan θ =
mg
2
(C) μ1 ≠ 0, μ2 ≠ 0 and N2 =
1 2
mg
1+μ μ
(D)μ1 = 0, μ2 ≠ 0 and N1 tan θ =
mg
2

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8. (C) (D)
N2 + μ1 N1 = Mg
N1 = μ2 N2
N2 + μ1μ2N2 = Mg
N2 =
1 2
Mg
1+ µ µ
N1 l sin θ = Mg cos θ
2
l
N1 tan θ =
Mg
2
9. A transparent thin film of uniform thickness and
refractive index n1 = 1.4 is coated on the convex
spherical surface of radius R at one end of a long solid
glass cylinder of refractive index n2 = 1.5, 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 f1 from the film, while rays of
light traversing from glass to air get focused at
distance f2 from the film. Then
(A)| f1 | = 3R (B) | f1 | = 2.8 R (C) | f2 | = 2R (D)| f2 | = 1.4R
9. (A) (C)
1.4 1 1.4 1
=
υ R
−
−
∞
1.4 4
=
υ R
7R
υ =
2
1.5 1.5 1.5 1.4
=
7Rυ +R
2
−
−
′
1.5 4 1
=
υ R R
−
′
1.5 5
=
υ R′
υ = 3R′ ∴ f1 = 3R
1.4 1.5 1.4 1.5
=
υ R
−
−
∞ −
1.4 1
=
υ R
−
−
υ =14f
1 1.4 1 1.4
=
υ 14R R
−
−
′ −
1 1 +0.4
=
υ 10R +R
−
′
1
υ′
=
0.4 0.1 0.5
+ =
R R R
R
υ =
0.5
′ = 2R ∴ f2 = 2R
N2
N1 μ1N1
μ2N2
θ

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10. 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)V1 = V2 and R1 = R2 = R3 (B) V1 = V2 and R1 = 2R2 = R3
(C) V1 = 2V2 and 2R1 = 2R2 = R3 (D)2V1 = V2 and 2R1 = R2 = R3
10. (A), (B), (D)
iR1 + i1R2 = V1
i1R2 − (i − i1) R3 = −V2
i1(R2 + R3) − iR3 = V2
i1(R2 + R3) − iR3 = −V2 × R1
i1R2 + iR1 = V1 × R3
i1(R1R2 + R1R3+R2R3) = V1R3 − V2R1
i1 = 1 3 2 1
1 2 1 3 2 3
V R V R
R R R R R R
−
+ +
V1 R3 = V2 R1
(a) V1 = V2 ⇒ R1 = R3
(b) V1 = V2 ⇒ R1 = R3
(d) 2V1 = V2 R3 = 2R1
SECTION − 2 : (One Integer Value Correct Type)
This section contains 10 questions. Each question, when worked out will result in one
integer from 0 to 9 (both inclusive).
11. Airplanes A and B are flying with constant velocity in the same vertical plane at angles
30° and 60° with respect to the horizontal respectively as shown in figure. The speed of A
is 100 3 ms−1
. At time t = 0 s, an observer in A finds B at distance of 500 m. This
observer sees B moving with a constant velocity perpendicular to the line of motion of A.
If a t = t0, A just escapes being hit by B, t0 in seconds is
i
i1
V1
i1
R1
i
V2
R3 i − i1

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11. [5]
B A| V V |− = VA tan 30
= 100 m/s
rel|S | = 500
and it must be along the joining them which is direction of relative velocity as it just
miss to hit which means distance of closest approach be zero.
So, t0 = rel
rel
|S |
| V |
= 5 sec
12. During Searle’s experiment, zero of the Vernier scale lies between 3.20 × 10−2
m and
3.25 × 10-2
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 3.20 × 10−2
m and 3.25 × 10−2
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 8 ×
10−7
m2
. The least count of the Vernier scale is 1.0 × 10−5
m. The maximum percentage
error in the Young’s modulus of the wire is
12. [8]
M.s.d = 0.05 × 10−2
m = 0.05 cm
= 0.5 mm
L. C. = 1.0 × 10−5
m
Li = (3.2 × 10−2
+ 20 × 1.0 × 10−5
) m
= (3.2 + 0.02) × 10−2
= 3.22 × 10−2
m
Lf = (3.2 + 45 × 10−3
) × 10−2
m
= 3.245 × 10−2
m
x = Lf − Li = 0.025 × 10−2
m
F
Y =
Ax
l
Δx = ΔL1 + ΔL2 = 2 × (1 × 10−5
) m
ΔY% = x%
5
2
2×10 ×100
Δx% =
0.025×10
−
−
= 8 %
ΔY% = 4%
We are assuming , F, A to be known with proper accuracy.
13. 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 rad s−1
is
30°
AV
BV
VA
500

(10)
13. [2]
τnet = 3τ1 = 3 F r sin 30 = 3 (0.5) (0.5)
1
2
=
3
8
N-m
I =
2
1.5(0.5) 3
2 16
=
α =
I
τ
= 2 rad/s2
ω = αt = 2 rad/s
14. 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
X
3,
X
= the value of 1
2
R
R
is
14. [3]
X1 = 0X
3
X2 = 02X
3
r =
mu
qB
1 2
2 1
R B
R B
=
B1 = 0
1 2
I 1 1
2 x x
⎛ ⎞μ
−⎜ ⎟
π ⎝ ⎠
= 0
0 0
I 3 3
2 x 2x
⎛ ⎞μ
−⎜ ⎟
π ⎝ ⎠
= 0
0
3 I
4 x
μ
π
B2 = 0
1 2
I 1 1
2 x x
⎛ ⎞μ
+⎜ ⎟
π ⎝ ⎠
= 0
0
9 I
4 x
μ
π
⇒ 1 2
2 1
R B
3
R B
= =
15. 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
frequency f. The engineer finds that d is proportional to S1/n
. The value of n is
15. [3]
d = va
sb
fc
d =
a b c
2
kg w 1
m3 secm
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
× ×⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
d = ( )
b c2 3a
3
2
ML T 1
ML
secL
−
− ⎛ ⎞ ⎛ ⎞
⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠
1 a b 3a 3b c
L M L T+ − − −
=
−3a = 1 ⇒ a = −
1
3
0.5
30°
X0
X1

(11)
a + b = 0 ∴ b =
1
3
−3b − c = 0 ∴ c = 1
∴ d =
1 1
13 3v s f
−
∴ d ∝ s1/n
∴ n = 3
16. A galvanometer gives full scale deflection with 0.006 A current. By connecting it to a
4990 Ω resistance, it can be converted into a voltmeter of range 0 − 30 V. If connected to
a
2n
249
Ω resistance, it becomes an ammeter of range 0 − 1.5 A. The value of n is
16. [5]
0.006 × (4990 + Rg) = 30
6 × (4990 + Rg) = 30000
∴ Rg = 10Ω
.006 × 10 = (1.5 − 006) ×
2n
249
2490 1.500
1 249
2n .006
= − =
10 = 2n
n = 5
17. 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 = 10 ms−2
)
17. [5]
w = 18 × 5 = 90 joules, u = −1 × 10 × 4 = −40 joules
wF + wg = ΔK = 90 − 40 = 50
∴ n × 10 = 50
∴ n = 5
18. A rocket is moving in a gravity free space with a constant acceleration of 2 ms−2
along + x
direction (see figure). The length of a chamber inside the rocket is 4 m. A ball is thrown
from the left end of the chamber in + x direction with a speed of 0.3 ms−1
relative to the
rocket. At the same time, another ball is thrown in −x direction with a speed of 0.2 ms−1
from its right end relative to the rocket. The time in seconds when the two balls hit each
other is
G
Rg 4990Ig
Rg
R
1.5 − .006
.006

(12)
18. [2]
w.r.t. rocket,
tf fo 1st
=
( )2 0.3
0.3 sec.
2
=
So it will keep on colliding in interval of 0.3 sec.
For other,
−4 = −0.2t − t2
t2
+ 0.2 t − 4 = 0
t = ( )0.2 0.04 16
4.01 0.1 1.9sec.
2
− ± +
= − ≈
So they must collide between interval of 1.8 to 1.9 sec while 7th
trip of 1st
ball. So it
should be close to 2 sec.
19. 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 9 ms−1
with respect to the ground. The rotational speed of the
platform in rad s−1
after the balls leave the platform is
19. [4]
2 × .05 × 9 ×.25 =
2
.45 .5
2
⎛ ⎞×
⎜ ⎟⎜ ⎟
⎝ ⎠
ω
2 × .05 × 9 ×.25 =
.45 .25
2
×
×ω
ω = 4 rad/sec
20. A thermodynamic system is taken from an initial state i with internal energy Ui = 100 J to
the final state f along two different paths iaf and ibf, as schematically shown in the figure.
The work done by the system along the paths af, ib and bf are Waf = 200 J, Wib = 50 J and
Wbf = 100 J respectively. The heat supplied to the system along the path iaf, ib and bf are
Qiaf, Qib and Qbf respectively. If the internal energy of the system in the state b is Ub = 200 J
and Qiaf =500 J, the ratio bf
ib
Q
Q
is
a
b
f
P

(13)
20. [2]
Waf = 200 j
Wbf = 100 j
Wib = 50 j
Wia = 0 j
Qiaf, Qib, Qbf
Ub = 200 j Qiaf = 500 j
ΔQiaf = ΔUiaf + Wiaf
500 = ΔUiaf + 200 j
ΔUiaf = 300 j
∴ ΔUibf = 300 j
Wibf = 150 j
∴ ΔQibf = 450 j = Qib + Qbf
Ui = 100 j Ub = 200 j
ΔUjb = 100 j Wib = 50 j
∴ Qib = 150 j
∴ 450 = Qib + Qbf
450 = 150 j + Qbf
Qbf = 300 j
bf
ib
Q 300
2.
Q 150
= =
PART II : CHEMISTRY
21. The correct combination of names for isomeric alcohols with molecular formula C4H10O
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
21. (A), (C), (D)
Factual.
22. The reactivity of compound Z with different halogens under appropriate conditions is
given below :

(14)
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
22. (A), (B) , (C)
Factual.
23. In the reaction shown below, the major product(s) formed is/are
23. (A)
NH2
O
NH2
2
2 2
Ac O
CH Cl
⎯⎯⎯⎯→
NH2
O
NHCOCH3
(major)
24. 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
P2 , V2 and T2, respectively. For this expansion,
(A) (B)
(C) (D)

(15)
(A)q = 0 (B) T2 = T1 (C) P2V2 = P1V1 (D) 2 2P Vγ
= 1 1P Vγ
24. (A), (B), (C)
For free expansion
W = 0, Q = 0, T2 = T1 and P1V1 = P2V2
25. 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.
25. (A), (B), (D)
Formic acid is more acidic than acetic acid due to electronic effect.
26. 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.
26. (A), (B), (C)
Properties of salt bridge.
27. Upon heating with Cu2S, the reagent(s) that give copper metal is/are
(A)CuFeS2 (B) CuO (C) Cu2O (D)CuSO4
27. (C)
Cu2S + 2Cu2O ⎯⎯→ 6Cu + SO2
28. 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.
28. (B), (D)
Properties of orthoboric acid.

(16)
29. For the reaction :
I−
+ CIO3
−
+H2SO4 → CI−
+ HSO4
−
+ I2
The correct statement(s) in the balanced equation is/are:
(A)Stoichiometric coefficient of HSO4
−
is 6.
(B) lodide is oxidized.
(C) Sulphur is reduced.
(D)H2O is one of the products.
29. (A), (B), (D)
3 2 4 2 2 4
(R.A) (O.A)
6I ClO 6H SO Cl 3I 3H O SO− − − −
+ + ⎯⎯→ + + + 6Η
30. The pair(s) of reagents that yield paramagnetic species is/are
(A)Na and excess of NH3
(B) K and excess of O2
(C) Cu and dilute HNO3
(D)O2 and 2-ethylanthraquinol
30. (A), (B), (C)
2 2K excess O KO (Superoxide)+ ⎯⎯→ → Paramagnetic
3 3 2 23Cu 8HNO 3Cu(NO ) 2NO 4H O+ ⎯⎯→ + + → Paramagnetic
31. 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
31. [5]
CnH2nO = 100 14n = 100 − 16 = 84 ⇒ n = 6
I)
||
C C C C C C
O
− − − − − II)
||
C C C C C C
O
− − − − −
III)
|| |
C C C C C
CO
− − − − IV)
||
|
C
C C C C C
O
− − − −
Et
OH
OH
2O
⎯⎯→
Et
O
O
H2O2+ → Dimagnetic

(17)
NaBH4
CH3 - CH - C - C - C -C
OH dl+-
NaBH4
- CH - C - C - C
OH dl+-
C - C
NaBH4
- CH - C - C
OH dl+-
C - C
C
NaBH4
- C - C - C
OH dl+-
C - CH
C
NaBH4
- C - C
OH dl+-
C - CH
C
C
V)
|
|| |
C
C C C C C
CO
− − − −
I)
II)
III)
IV)
V)
32. A list of species having the formula XZ4 is given below.
XeF4, SF4, SiF4, BF4
−
, [Cu(NH3)4]2+
, [FeCI4]2−
, [CoCl4]2−
and [PtCI4]2−
.
Defining shape on the basis of the location of X and Z atoms, the total number of species
having a square planar shape is
32. [4]
XeF4 , 4BrF−
, [Cu(NH3)4]+2
, [PtCl4]2−
33. Among PbS, CuS, HgS, MnS, Ag2S, NiS, CoS, Bi2S3 and SnS2, the total number of
BLACK coloured sulphides is
33. [7]
PbS , CuS, HgS, Ag2S, NiS, CoS, Bi2S3
34. The total number(s) of stable conformers with non-zero dipole moment for the following
compound is (are)

(18)
34. [3]
CH3
ClBr
ClBr
CH3
CH3
BrCl
CH3Cl
Br
CH3
BrCl
CH3Br
Cl
35. Consider the following list of reagents:
Acidified K2Cr2O7, alkaline KMnO4, CuSO4, H2O2, CI2, O3, FeCI3, HNO3 and Na2S2O3.
The total number of reagents that can oxidise aqueous iodide to iodine is
35. [7]
H+
/ K2Cr2O7 , 4OH / KMnO , CuSO4 , H2O2 , Cl2 , O3, HNO3
36. The total number of distinct naturally occurring amino acids obtained by complete
acidic hydrolysis of the peptide shown below is
36. [1]
2 2 2 2NH CH C OH HO C CH NH HO C CH NH− − − + − − − + − − −
O O O
CH2
2HO C CH NH CH COOH− − − − −
O
CH2
(Natural A.A.)
+
37. In an atom, the total number of electrons having quantum numbers n = 4, |m | = 1 and
ms = −1/2 is
37. [6]
4p → 2e−
4d → 2e−
4f → 2e−
38. If the value of Avogadro number is 6.023 × 1023
mol−1
and the value of Boltzmann
constant is 1.380 × 10−23
J K−1
, then the number of significant digits in the calculated
value of the universal gas constant is
38. [4]
Number of significant digits is 4.

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39. A compound H2X with molar weight of 80 g is dissolved in a solvent having density of
0.4 g ml−1
. Assuming no change in volume upon dissolution, the molality of a 3.2 molar
solution is
39. [8]
Let us consider, volume of solution = 1 litre = 1000 mL
wt. of solvent = 0.4 × 1000 = 400 g = 0.4 kg
wt. of solute = 80 g ; no. of moles of solute = 3.2 moles
molality of solution =
3.2
0.4
= 8
40. MX2 dissociates into M2+
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
40. [2]
2
2
0 01
1 2
MX M 2X+ −
− α α α
⎯⎯→ +
i = 1 + 2α = 1 + 2 × 0.5 = 2
PART III – MATHEMATICS
41. Let M and N be two 3 × 3 matrices such that MN = NM. Further, if M ≠ N2
and M2
= N4,
then
(A)determinant of (M2
+ MN2
) is 0
(B) there is a 3 × 3 non−zero matrix U such that (M2
+ MN2
) U is the zero matrix
(C) determinant of (M2
+ MN2
) ≥ 1
(D)for a 3 × 3 matrix U, if (M2
+ MN2
) U equals the zero matrix the U is the zero matrix.
41. (A), (B)
MN = NM
N2
M = N (NM)
= N (MN)
= (NM) N
= (MN) N
= MN2
…… (1)
(M − N2
) (M + N2
) = M2
+ MN2
− N2
M − N4
= M2
− N4
(by (1))
= 0
As 2 2
M N 0 M N 0− ≠ ⇒ + = …… (2)
Now, ( )2 2 2
M MN M M N+ = +
= 2
M M N+
= 0 (by (2))
Since 2
M N 0+ = so B option is correct.

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42. For every pair of continuous functions f, g: [0, 1] → » such that
max {f(x) : x ∈ [0, 1]} = max {g(x) : x ∈ [0, 1]},
the correct statement(s) is (are):
(A)(f(c))2
+ 3f(c) = (g(c))2
+ 3g(c) for some c ∈ [0, 1]
(B) (f(c))2
+ f(c) = (g(c))2
+ 3g(c) for some c ∈ [0, 1]
(C) (f(c))2
+ 3f(c) = (g(c))2
+ g(c) for some c ∈ [0, 1]
(D)(f(c))2
= (g(c))2
for some c ∈ [0, 1]
42. (A), (D)
Since f(x) and g(x) are continuous function and their maximum values are equal, their
graphs will intersect at atleast one point in [0, 1].
∴ f(c) = g(c) for some c ∈ [0, 1]
∴ Options (A) and (D) are correct.
Options (B), (C) can be eliminated by taking f(x) = 1 and g(x) = 1
43. Let f: (0, ∞) → » be given by
f(x) =
1
(t )x
t
1
x
e
− +
∫
dt
t
Then
(A) f(x) is monotonically increasing on [1, ∞)
(B) f(x) is monotonically decreasing on (0, 1)
(C) f(x) + f
1
x
⎛ ⎞
⎜ ⎟
⎝ ⎠
= 0, for all x ∈ (0, ∞)
(D)f(2x
) is an odd fuction of x on »
43. (A), (C), (D)
f : (0, ∞) → R
f(x) =
1x t
t
1/x
dt
e
t
⎛ ⎞
− +⎜ ⎟
⎝ ⎠
∫
f(x) =
1
t
x t
1/x
e
dt
t
⎛ ⎞
− +⎜ ⎟
⎝ ⎠
∫
1 1
x x
x x
2
e e 1
f '(x) 1
x 1/ x x
⎛ ⎞ ⎛ ⎞
− + − +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
⎛ ⎞
= ⋅ − −⎜ ⎟
⎝ ⎠
1
x
x 1 1
f '(x) e
x x
⎛ ⎞
− +⎜ ⎟
⎝ ⎠ ⎛ ⎞
= +⎜ ⎟
⎝ ⎠
1
x
x
2e
f '(x)
x
⎛ ⎞
− +⎜ ⎟
⎝ ⎠
=
f(x) is monotonically increasing on [1, ∞)
f(x) =
1x t
t
1/x
e dt
⎛ ⎞
− +⎜ ⎟
⎝ ⎠
∫ …………. (i)

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11/x t
t
x
1
f e dt
x
⎛ ⎞
− +⎜ ⎟
⎝ ⎠⎛ ⎞
=⎜ ⎟
⎝ ⎠
∫
1x t
t
1/x
1
f e dt
x
⎛ ⎞
− +⎜ ⎟
⎝ ⎠⎛ ⎞
= −⎜ ⎟
⎝ ⎠
∫ ………….. (ii)
equation (i) + (ii)
f(x) +
1
f
x
⎛ ⎞
⎜ ⎟
⎝ ⎠
= 0
If we put x = 2x
f(2x
) + f(2−x
) = 0
f(2−x
) = −f(2x
)
It means f(2x
) is odd function.
44. Let a ∈ » and let f: » → » be given by
f(x) = x5
− 5x + 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
44. (B), (D)
f(x) = x5
− 5x + a
f′(x) = 5x4
− 5 = 0 ⇒ x = ± 1
f(−1) = a + 4 and f(1) = a − 4
∴ f(x) has three real roots if −4 < a < 4
and f(x) has one real root if a < −4 or a > 4.
45. Let f : [a, b] → [1, ∞) be a continuous function and let g : » → » be defined as
g(x) =
x
a
b
a
0 if x a,
f (t) dt if a x b,
f (t) dt if x b.
⎧
<⎪
⎪
≤ ≤⎨
⎪
⎪ >
⎩
∫
∫
Then
(A)g(x) is continuous but not differentiable at a
(B) g(x) is differentiable on »
(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
45. (A), (C)
x a
Lt g(x) 0,
−→
=
a
x a a
Lt g(x) g(a) f (t)dt 0
+→
= = =∫ .
Similarly,
b
x b x b a
Lt g(x) g(b) Lt g(x)dx f (t)dt
− +→ →
= = = ∫
∴ g(x) is continuous at both x = a and x = b
(a − 4)
(a + 4)
−1 1

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Also, g′(a−
) = 0 and g′(a+
) = f(a)
g′(b−
) = f(b) and g′(b+
) = 0
But since f(a) and f(b) must be greater than equal to 1, g(x) is not differentiable at x = a
and x = b
46. Let f : ,
2 2
π π⎛ ⎞
−⎜ ⎟
⎝ ⎠
→ » be given by
f(x) = (log (sec x + tan x))3
.
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
46. (A), (B), (C)
( ) ( ){ }3
f x log sec x tan x= +
( ) ( ){ }3
f x log secx tan x− = −
=
3
1
log
sec x tan x
⎧ ⎫⎛ ⎞
⎨ ⎬⎜ ⎟
+⎝ ⎠⎩ ⎭
= ( ){ }3
log secx tan x− +
= − f (x)
⇒ odd function
( ) ( ){ } { }2 21
f ' x 3 log secx tan x . sec x tan x sec x
secx tan x
= + +
+
= ( ){ }2
3 log secx tan x secx 0= + > as x ,
2 2
π π⎛ ⎞
∈ −⎜ ⎟
⎝ ⎠
⇒ one − one
Range is R
So onto function
47. From a point P(λ, λ, λ), perpendiculars PQ and PR are drawn respectively on the lines y =
x, z = 1 and y = −x, z = −1. If P is such that ∠QPR is a right angle, then the possible
value(s) of λ is (are)
(A) 2 (B) 1
(C) −1 (D)− 2
47. (C)
From (λ, λ, λ), foot of the perpendicular on the line
x 0
1
−
=
y 0
1
−
=
z 1
0
−
= r is (λ, λ, 1)
Similarly. Foot of the perpendicular on the line
x 0
1
−
=
y 0
1
−
−
=
z 1
0
+
= r1 is (0, 0, −1)
Since QP ⊥ PR,
(λ − λ) (λ − 0) + (λ − λ) ⋅ (λ − 0) + (λ − 1) (λ + 1) = 0
⇒ λ = ± 1. At λ = 1, the point P lies on the 1st
line.
∴ λ = −1.

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48. 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 2
2 2 2
q
p 2q r+ +
is a nonzero vector perpendicular to y and z × x , then
(A) b = (b ⋅ z ) ( z − x ) (B) a = (a ⋅ y ) ( y − z )
(C) a ⋅ b = − (a ⋅ y ) ( b ⋅ z ) (D) a = (a ⋅ y ) (z − y )
48. (A), (B), (C)
Given | x | | y | | z | 2= = =
1
x y | x | | y | cos 60 2 2 1
2
⋅ = ° = ⋅ ⋅ =
Similarly, y z 1 & z x 1⋅ = ⋅ =
Also, x (y z) (x z) y (x y) z× × = ⋅ − ⋅
a y zλ = − …………….. (i)
Again, y (z x) (y x)z (y z)x× × = ⋅ − ⋅
b z xμ = − ……………. (ii)
From (i)
a y (y z) yλ ⋅ = − ⋅
a y y y z yλ ⋅ = ⋅ − ⋅
a y 2 1λ ⋅ = −
1
a y⋅ =
λ
Similarly
1
b z⋅ =
μ
1 1
a y b z⋅ = ⋅ =
λ μ
−
1
(a y) (b z)⋅ ⋅ = −
λμ
…………….. (iii)
y z z x
a b
⎛ ⎞− −⎛ ⎞
⋅ = ⋅⎜ ⎟ ⎜ ⎟
λ μ⎝ ⎠ ⎝ ⎠
y z z z y x z x
a b
⋅ − ⋅ − ⋅ + ⋅
⋅ =
λμ
1
a b⋅ = −
λμ
………………. (iv)
Hence (A), (B), (C) are correct.
49. A circle S passes through the point (0, 1) and is orthogonal to the circles (x − 1)2
+ y2
= 16
and x2
+ y2
= 1. 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)
49. (B), (C)
Given circles
x2
+ y2
− 2x − 15 = 0 &
x2
+ y2
− 1 = 0

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Let circle x2
+ y2
+ 2gx + 2fy + c = 0 …… (1)
Orthogonal ⇒2 [g1 g2 + f1 f2] = c1 + c2
⇒2 [(−1) g + 0. f] = −15 + c
⇒−2g = −15 + c …… (2)
& 2 [0. g + 0.f] = −1 + c
⇒ c = 1 …… (3)
From (2) & (3) g = 7
Circle is x2
+ y2
+ 14x + 2fy + 1 = 0
Given it passes (0, 1)
⇒ 1 + 2f + 1 = 0
f = −1
So, circle is
x2
+ y2
+ 14x − 2y + 1 = 0
centre (−7, 1) & r 49 1 1 7= + − =
50. 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
50. (C), (D)
a b
M
b c
⎡ ⎤
= ⎢ ⎥
⎣ ⎦
T a b
M
b c
⎡ ⎤
= ⎢ ⎥
⎣ ⎦
Det. M = ac − b2
≠ 0 (For invertible)
⇒ Option (C) is correct (as non−diagonal element must be 0)
Obviously option (D) is correct.
51. 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 b + 2, then the value of
2
a a 14
a 1
+ −
+
is
51. [4]
a, b, c in G. P. ⇒ a, b = ar, c = ar2
(r is integer given)
a b c
b 2
3
+ +
= +
a + ar2
= 2ar + 6 ⇒ a (r2
− 2r +1) = 6
a (r − 1)2
= 6
( )2 6
r 1
a
− =
6
r 1 r 2
a
− = ± ⇒ = (as ‘r’ is integer) ⇒ a = 6, b = 12, c = 24
So,
2 2
a a 14 6 6 14 28
4
a 1 6 1 7
+ − + −
= = =
+ +

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52. Let n ≥ 2 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.
52. [5]
Number of red line segments =
n(n 3)
2
−
Number of blue line segments = n
∴
n(n 3)
n n 5
2
−
= ⇒ =
53. Let n1 < n2 < n3 < n4 < ns be positive integers such that n1 + n2 + n3 + n4 + n5 = 20. Then
the number of such distinct arrangements (n1, n2, n3, n4, n5) is
53. [7]
Under given restrictions, following arrangements are possible :
{1, 2, 3, 4, 10}
{1, 2, 3, 5, 9}
{1, 2, 3, 6, 8}
{1, 2, 4, 5, 8}
{1, 2, 4, 6, 7}
{1, 3, 4, 5, 7}
{2, 3, 4, 5, 6}
54. Let f : R → R and g : R → R be respectively given by f(x) = |x| + 1 and g(x) = x2
+ 1.
Define h : R → R by
h(x) =
max {f (x), g(x)} if x
min {f (x), g(x)} if x 0.
≤ 0,⎧
⎨
>⎩
The number of points at which h(x) is not differentiable is
54. [3]
h (x) is not differentiable at 3 points because sharp edge.
55. The value of
1 2
3 2
2
0
d
4x (1 x ) dx
dx
⎧ ⎫
−⎨ ⎬
⎩ ⎭
∫ is
55. [2]
( )( )51 2 2
3
2
0
d 1 x dx
4x
dx
−
∫ = ( ) ( )
1
33 2 2
0
4x ×10 1 x 9x 1 dx− −∫
put x2
= t and 2xdx = dt
= ( )
1
5 4 3 2
0
20 dt9t 28t 30t 12t t− + − + −∫ = 2
1
2
n
(−1, 2)
(0, 1)
(1, 2)
h (x)
g (x)
f (x)

(26)
56. The slope of the tangent to the curve (y − x5
)2
= x(1 + x2
)2
at the point (1, 3) is
56. [8]
(y−x5
)2
= x (1 + x2
)2
at x = 1
(y − 1)2
= 4
(y − 1) = ± 2
y − 1 = + 2 ⇒y = 3
it means,
( )5 2
y x x 1 x− = +
( )2 5
y x 1 x x= + +
( ) ( )2 4dy 1
1 x x 2x 5x
dx 2 x
= + + +
at x = 1,
dy
8
dx
=
57. The largest value of the non-negative integer a for which
1 x
1 x
x 1
ax sin (x 1) a
lim
x sin(x 1) 1
−
−
→
⎧ ⎫− + − +
⎨ ⎬
+ − −⎩ ⎭
=
1
4
is
57. [2]
1 x
1 x
x 1
ax sin(x 1) a 1
lim
x sin(x 1) 1 4
−
−
→
⎧ ⎫− + − +
=⎨ ⎬
+ − −⎩ ⎭
Let, f(x) =
sin(x 1) a(x 1)
(x 1) sin(x 1)
− − −
− + −
x 1
1 a
limf (x)
2→
−
= and g(x) =
1 x
1 x
−
−
x 1
limg(x) 2
→
=
∴
2
1 a 1 1 a 1
a 0,2
2 4 2 2
− −⎛ ⎞
= ⇒ = ± ⇒ =⎜ ⎟
⎝ ⎠
58. Let f: [0, 4π] → [0, π] be defined by f(x) = cos−1
(cos x). the number of points. x ∈ [0,
4π] satisfying the equation
f(x) =
10 x
10
−
is
58. [3]
f(x) = 1 −
x
10
There are three intersections.
0 π 2π 3π 4π
x
y
π
x′

(27)
59. For a point P in the plane, let d1(P) and d2 (P) be the distances of the point P from the lines
x − y = 0 and x + y = 0 respectively. The area of the region R consting of all points P
lying in the first quadrant of the plane and satisfying 2 ≤ d1(P) + d2 (P) ≤ 4, is
59. [6]
2 ≤
h k h k
4
2 2
− +
+ ≤
(i) In 1st
quadrant, if h ≥ k, 2 ≤
2h
2
≤ 4
⇒ 2 h 2≤ ≤ 2
(ii) Int 1st
quadrant, if h < k, 2 ≤
2k
4
2
≤
⇒ 2 k 2 2≤ ≤
∴ Required area = ( ) ( )2 2
2 2 2−
= 6 sq. units.
60. Let a, b and c be three non-coplanar unit vectors such that the angle between every pair
of them is .
3
π
If a × b + b × c = pa q b rc× + , where p, q and r are scalars, then the
value of 2
2 2 2
q
p 2q r+ +
is
60. [4]
a b b c pa qb rc× + × = + +
Take dot product with b
0 pa.b q b.b r c.b= + +
1 1
0 p. q r.
2 2
= + +
⇒ p + 2q + r = 0 …… (1)
Take dot product with a
[ ]0 abc p a.a q a.b r a.c+ = + +
=
q r
p
2 2
+ +
2 a b c 2p q r⎡ ⎤⇒ = + +⎣ ⎦ …… (2)
Take dot product with c
c a b pa.c q b.c r c.c⎡ ⎤ = + +⎣ ⎦
=
1 q
p. r
2 2
+ +
2 a b c p q 2r⎡ ⎤ = + +⎣ ⎦ …… (3)
Solving (1), (2) & (3) p = −q = r
So,
2 2 2 2 2 2
2 2
p 2q r p 2p p
4
q p
+ + + +
= =

(28)

Jee advanced-2014-question-paper1-with-solution

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