28
(1) IIT - JEE 2014 (Advanced) CODE : 1

phamphuc
• Category

## Documents

• view

221

1

### Transcript of IIT - JEE 2014 (Advanced) -...

(1)

CODE : 1

(2) Vidyalankar : IIT JEE 2014 − Advanced : Question Paper & Solution

(2)

IIT JEE 2014 Advanced : Question Paper & Solution (Paper – I) (3)

(3)

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

πω

, 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

πω

is 1 × 10−3 C.

(B) The current in the left part of the circuit just before t = 7

6

πω

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

= 2 × 10−3 C

Charge on upper plate at t = 7

6

πω

= 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) Vidyalankar : IIT JEE 2014 − Advanced : Question Paper & Solution

(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

12 2 1n m

2

λλ =

n2600 = 1

400m

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 cos6 3

π π (B) y(t) = x 100 t

A sin cos3 3

π π

(C) y(t) = 5 x 250 t

A sin cos6 3

π π (D) y(t) = 5 x

A sin cos t2

π 250 π

3. (A), (C), (D)

ν = k

ω

1A 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

IIT JEE 2014 Advanced : Question Paper & Solution (Paper – I) (5)

(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 = 204 rσπ (B) r0 =

2

λπσ

(C) E1(r0/2) = 2E2(r0/2) (D) E2(r0/2) = 4E3(r0/2)

5. (C)

202 0 00 0 00

20 02

0 00

1 q2 r q 2 r 2 r4 2 rr

1 qq 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 7M 20,

20 10

⎛ ⎞ = = ⎜ ⎟

⎝ ⎠ (B) Nitrogen 10 3

M 28,28 5

⎛ ⎞ = = ⎜ ⎟

⎝ ⎠

(C) Oxygen 10 9M 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) Vidyalankar : IIT JEE 2014 − Advanced : Question Paper & Solution

(6)

for neon : 7

64010

× = 448 ms−1

for Nitrogen : 3

5905

× = 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 = 2ε

tR

M = 0.5 S ΔT = 40K.

ΔQ = 0.5 × S × 40 = 2ε

× 4 minR

R = ρL

πd4

2 R1 =

2

ρL R=

4(Rd)π

4

Req. series R R R

+ =4 4 2

∴ ΔQ = 2 2ε × 4 ε

= × tRR2

t = 2 min

Req Parallel =

R R× R4 4 =

R R 8+4 4

ΔQ = 2ε

× 4 R

= 2ε

× tR8

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

IIT JEE 2014 Advanced : Question Paper & Solution (Paper – I) (7)

(7)

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 θ2

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 +R2

−−′

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

(8) Vidyalankar : IIT JEE 2014 − Advanced : Question Paper & Solution

(8)

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

IIT JEE 2014 Advanced : Question Paper & Solution (Paper – I) (9)

(9)

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 × 10−2m 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 × 10−5m. The maximum percentage error in the Young’s modulus of the wire is

12. [8] M.s.d = 0.05 × 10−2m = 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 ×100Δx% =

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) Vidyalankar : IIT JEE 2014 − Advanced : Question Paper & Solution

(10)

13. [2]

τnet = 3τ1 = 3 F r sin 30 = 3 (0.5) (0.5) 1

2=

3

8N-m

I = 21.5(0.5) 3

2 16 =

α = I

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

X3,

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

μπ

B2 = 0

1 2

I 1 1

2 x x

⎛ ⎞μ + ⎜ ⎟π ⎝ ⎠ = 0

0

9 I

4 x

μπ

⇒ 1 2

2 1

R B3

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 3a3

2

ML T 1ML

secL

−− ⎛ ⎞ ⎛ ⎞

⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

1 a b 3a 3b cL M L T+ − − −=

−3a = 1 ⇒ a = − 1

3

0.5

30°

X0

X1

IIT JEE 2014 Advanced : Question Paper & Solution (Paper – I) (11)

(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 2n249

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 2492n .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 4990 Ig

Rg

R 1.5 − .006

.006

(12) Vidyalankar : IIT JEE 2014 − Advanced : Question Paper & Solution

(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 0.04 164.01 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

IIT JEE 2014 Advanced : Question Paper & Solution (Paper – I) (13)

(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 3002.

Q 150 = =

PART II : CHEMISTRY

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.

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) Vidyalankar : IIT JEE 2014 − Advanced : Question Paper & Solution

(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 OCH 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)

IIT JEE 2014 Advanced : Question Paper & Solution (Paper – I) (15)

(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) Vidyalankar : IIT JEE 2014 − Advanced : Question Paper & Solution

(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

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).

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

IIT JEE 2014 Advanced : Question Paper & Solution (Paper – I) (17)

(17)

NaBH4CH3 - 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) Vidyalankar : IIT JEE 2014 − Advanced : Question Paper & Solution

(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.

IIT JEE 2014 Advanced : Question Paper & Solution (Paper – I) (19)

(19)

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

20 01

1 2

MX M 2X+ −

− α α α

⎯⎯→ +

i = 1 + 2α = 1 + 2 × 0.5 = 2

PART III – MATHEMATICS

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.

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 N2M = N (NM) = N (MN) = (NM) N = (MN) N = MN2 …… (1) (M − N2) (M + N2) = M2 + MN2 − N2M − N4 = M2 − N4 (by (1)) = 0

As 2 2M N 0 M N 0− ≠ ⇒ + = …… (2)

Now, ( )2 2 2M MN M M N+ = +

= 2M M N+

= 0 (by (2))

Since 2M N 0+ = so B option is correct.

(20) Vidyalankar : IIT JEE 2014 − Advanced : Question Paper & Solution

(20)

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 )xt1

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) + f1

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 tt

1/x

dte

t

⎛ ⎞− +⎜ ⎟⎝ ⎠∫

f(x) =

1t

x t

1/x

edt

t

⎛ ⎞− +⎜ ⎟⎝ ⎠

1 1x x

x x

2

e e 1f '(x) 1

x 1/ x x

⎛ ⎞ ⎛ ⎞− + − +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎛ ⎞= ⋅ − −⎜ ⎟

⎝ ⎠

1

xx 1 1

f '(x) ex x

⎛ ⎞− +⎜ ⎟⎝ ⎠ ⎛ ⎞= +⎜ ⎟

⎝ ⎠

1x

x2ef '(x)

x

⎛ ⎞− +⎜ ⎟⎝ ⎠

=

f(x) is monotonically increasing on [1, ∞)

f(x) = 1x tt

1/x

e dt⎛ ⎞− +⎜ ⎟⎝ ⎠∫ …………. (i)

IIT JEE 2014 Advanced : Question Paper & Solution (Paper – I) (21)

(21)

11/x tt

x

1f e dt

x

⎛ ⎞− +⎜ ⎟⎝ ⎠⎛ ⎞ =⎜ ⎟

⎝ ⎠∫

1x tt

1/x

1f e dt

x

⎛ ⎞− +⎜ ⎟⎝ ⎠⎛ ⎞ = −⎜ ⎟

⎝ ⎠∫ ………….. (ii)

equation (i) + (ii)

f(x) + 1

fx

⎛ ⎞⎜ ⎟⎝ ⎠

= 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 aLt 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

(22) Vidyalankar : IIT JEE 2014 − Advanced : Question Paper & Solution

(22)

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)

( ) ( ){ }3f x log sec x tan x= +

( ) ( ){ }3f x log sec x tan x− = −

=3

1log

sec x tan x

⎧ ⎫⎛ ⎞⎨ ⎬⎜ ⎟+⎝ ⎠⎩ ⎭

= ( ){ }3log sec x tan x− +

= − f (x) ⇒ odd function

( ) ( ){ } { }2 21f ' x 3 log sec x tan x . sec x tan x sec x

sec x tan x= + +

+

= ( ){ }23 log sec x tan x sec x 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.

IIT JEE 2014 Advanced : Question Paper & Solution (Paper – I) (23)

(23)

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 12

⋅ = ° = ⋅ ⋅ =� � � �

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

(24) Vidyalankar : IIT JEE 2014 − Advanced : Question Paper & Solution

(24)

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

Mb c

⎡ ⎤= ⎢ ⎥⎣ ⎦

T a bM

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.

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).

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 2a 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 23

+ + = +

a + ar2 = 2ar + 6 ⇒ a (r2 − 2r +1) = 6 a (r − 1)2 = 6

( )2 6r 1

a− =

6

r 1 r 2a

− = ± ⇒ = (as ‘r’ is integer) ⇒ a = 6, b = 12, c = 24

So, 2 2a a 14 6 6 14 28

4a 1 6 1 7

+ − + −= = =+ +

IIT JEE 2014 Advanced : Question Paper & Solution (Paper – I) (25)

(25)

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 52

− = ⇒ =

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 22

0

d4x (1 x ) dx

dx

⎧ ⎫ − ⎨ ⎬

⎩ ⎭∫ is

55. [2]

( )( )51 2 2

32

0

d 1 x dx4x

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) Vidyalankar : IIT JEE 2014 − Advanced : Question Paper & Solution

(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 2y x x 1 x− = +

( )2 5y x 1 x x= + +

( ) ( )2 4dy 11 x x 2x 5x

dx 2 x= + + +

at x = 1, dy

8dx

=

57. The largest value of the non-negative integer a for which

1 x

1 x

x 1

ax sin (x 1) alim

x sin(x 1) 1

−−

⎧ ⎫− + − + ⎨ ⎬ + − −⎩ ⎭ =

1

4 is

57. [2]

1 x

1 x

x 1

ax sin(x 1) a 1lim

x sin(x 1) 1 4

−−

⎧ ⎫− + − + =⎨ ⎬+ − −⎩ ⎭

Let, f(x) = sin(x 1) a(x 1)

(x 1) sin(x 1)

− − −− + −

x 1

1 alim f (x)

2→

−= and g(x) = 1 x

1 x

−−

x 1lim g(x) 2

→=

∴ 2

1 a 1 1 a 1a 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′

IIT JEE 2014 Advanced : Question Paper & Solution (Paper – I) (27)

(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

42 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

42

⇒ 2 k 2 2 ≤ ≤

∴ Required area = ( ) ( )2 22 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 p a .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

p2 2

+ +

2 a b c 2p q r⎡ ⎤⇒ = + +⎣ ⎦ …… (2)

Take dot product with c�

c a b p a.c q b.c r c.c⎡ ⎤ = + +⎣ ⎦

� � � � �

= 1 q

p. r2 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 p4

q p

+ + + += =

(28) Vidyalankar : IIT JEE 2014 − Advanced : Question Paper & Solution

(28)