JEE Main Online Exam 2019careerpoint.ac.in/studentparentzone/2019/jee-main/JEE...Q.16 A transformer...

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JEE Main Online Exam 2019

Questions & Solutions 10th April 2019 | Shift - I

Physics Q.1 A 25 × 10–3 m3 volume cylinder is filled with 1 mol of O2 gas at room temperature (300 K). The molecular

diameter of O2, and its root mean square speed, are found to be 0.3 nm and 200 m/s, respectively. What is the average collision rate (per second) for an O2 molecule ?

(1) ~1012 (2) ~1010 (3) ~1013 (4) ~1011 Ans. [1]

Sol. collision frequency = avv

vav = 38 vrms

vav = 38 × 200

= NPd2

RT2

P = V

RT

= 2d2V

= 20–

3–

1094.11025

put values frequency ~– 0. 2 × 1010/sec Answer by NTA is given as 1012 per sec. Q.2 A thin disc of mass M and radius R has mass per unit area (r) = kr2 where r is the distance from its centre.

Its moment of inertia about an axis going through its centre of mass and perpendicular to its plane is :

(1) 3

MR2 2 (2)

6MR 2

(3) 3

MR 2 (4)

2MR 2

Ans. [1] Sol.

r O

dr

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assume a ring of radius r and with dr dI = (dm)r2

I = R

0

2rdm = R

0

2rdr)r2(

= 2k drrR

0

5

= 2k

6R 6

= 3

kR 6

mass of disc M = drr2kR

0

3

M = k2

4R 4

k = 4R2M4

put the value

I = 3

4R2M4 R6

= 3

MR2 2

Q.3 A moving coil galvanometer allows a full scale current of 10–4 A. A series resistance of 2 M is required to

convert the above galvanometer into a voltmeter of range 0-5 V. Therefore the value of shunt resistance required to convert the above galvanometer into an ammeter of range 0-10 mA is :

(1) 10 (2) 500 (3) 100 (4) 200 Ans. [Drop by NTA] Sol. ig = 0.1 mA , V = 5V, R = 2 × 106 V = ig (G + R) 5 = 0.1 × 10–3 (G + R) G + R = 5 × 104 G = 5 × 104 – 2 × 106 = Negative Not possible Q.4 An npn transistor operates as a common emitter amplifier, with a power gain of 60 dB. The input circuit

resistance is 100 and the output load resistance is 10 k. The common emitter current gain is . (1) 102 (2) 104 (3) 60 (4) 6 × 102 Ans. [1] Sol. Ri = 100 , Ro = 104

Power gain 60 = 10 log

i

o

PP

i

o

PP = 106 = 2

i

o

RR

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2 = 106 × o

i

RR

= 106 × 410100

2 = 104 = 100 Q.5 A current of 5 A passes through a copper conductor (resistivity = 1.7 × 10–8 m) of radius of cross-section

5 mm. Find the mobility of the charges if their drift velocity is 1.1 × 10–3 m/s. (1) 1.0 m2/Vs (2) 1.8 m2/Vs (3) 1.5 m2/Vs (4) 1.3 m2/Vs Ans. [1]

Sol. = 2ne

m = ne

1

i = neA vd

n = deAv

i

= i

Avd = 8–

6–3–

107.151025101.1

= 7.11027.17 1– 1.0 m2/Vs

Q.6 In the given circuit, an ideal voltmeter connected across the 10 resistance reads 2V. The internal resistance

r, of each cell is : 15

10

2

1.5 V, 1.5 Vr r

(1) 1 (2) 0.5 (3) 1.5 (4) 0 Ans. [2] Sol.

6 2

2r

3V

i

current i =

62 =

31

31 (8 + 2r) = 3

2r = 1 r = 0.5

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Q.7 In a meter bridge experiment, the circuit diagram and the corresponding observation table are shown in figure.

R

G

Resistance box

Unknown resistance

X

K E

Sl. No. R () (cm) 1. 2. 3. 4.

1000 100 10 1

60 13 1.5 1.0

Which of the reading is inconsistent ? (1) 3 (2) 4 (3) 2 (4) 1 Ans. [2]

Sol. By using RX =

–100

X = R

–100

The value of X for reading 4 is totally different Q.8 Two wires A & B are carrying currents I1 & I2 as shown in the figure. The separation between them is d. A

third wire C carrying a current I is to be kept parallel to them at a distance x from A such that the net force acting on it is zero. The possible values of x are :

A C B

I2

d

I1x

(1) x = dI–I

I

21

1

and x = d

)II(I

21

2

(2) x = ± )I–I(

dI

21

1

(3) x = dII

I

21

1

and x = d)I–I(

I

21

2

(4) x = dII

I

21

2

and x = d)I–I(

I

21

2

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Ans. [2] Sol.

A C B

I2

d

I1 x d –x

Fnet on C = 0

x2II10

+

)x–d(2II20

= 0

xI1 =

d–xI2

x = 21

1

I–IdI

Q.9 One plano-convex and one plano-concave lens of same radius of curvature 'R' but of different materials are

joined side by side as shown in the figure. If the refractive index of the material of 1 is 1 and that of 2 is 2, then the focal length of the combination is :

2

1

1

2

(1) )–(–2

R

21 (2)

)–(2R

21 (3)

21 –R

(4) 21 –

R2

Ans. [3]

Sol. 1r1 = (1 – 1)

R–

1–1 = R

1–1

2r1 =

R–1–2

Therefore eqr1 =

21 –R

Q.10 A message signal of frequency 100 MHz and peak voltage 100 V is used to execute amplitude modulation on

a carrier wave of frequency 300 GHz and peak voltage 400 V. The modulation index and difference between the two side band frequencies are :

(1) 0.25 ; 2 × 108 Hz (2) 4 ; 1 × 108 Hz (3) 4 ; 2 × 108 Hz (4) 0.25 ; 1 × 108 Hz Ans. [1] Sol. s = 2fs, fs = 100 × 106, Es = 100 V c = 2fc, fc = 300 × 109 Hz, Ec = 400 V

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modulation index m = 400100

EE

c

s = 0.25

side band F = Fmax – Fmin

= (Fs + Fc) – (Fc + Fs)

= )1–3000(–)13000(108 = 108 (2) = 2 ×108 Q.11 A cylinder with fixed capacity of 67.2 lit contains helium gas at STP. The amount of heat needed to raise the

temperature of the gas by 20°C is : [Given that R = 8.31 J mol–1 K–1] (1) 748 J (2) 350 J (3) 374 J (4) 700 J Ans. [1] Sol. Q = nCvdT = 748 Joule Q.12 A stationary source emits sound waves of frequency 500 Hz. Two observers moving along a line passing

through the source detect sound to be of frequencies 480 Hz and 530 Hz. Their respective speeds are, in ms–1, (Given speed of sound = 300 m/s)

(1) 12, 16 (2) 12, 18 (3) 16,14 (4) 8, 18 Ans. [2]

Sol. nA = vn (v – vA) =

300500 [300 – VA] = 480

300 – VA = 5

3480 = 288

VA = 300 – 288 = 12 m/s

similarly nB = vn (v + vB)

VB = 18 m/s Q.13 Given below in the left column are different modes of communication using the kinds of waves given in the

right column.

A. Optical Fibre communication

P. Ultrasound

B. Radar Q. Infrared Light C. Sonar R. Microwaves D. Mobile Phones S. Radio Waves

From the options given below, find the most appropriate match between entries in the left and the right column.

(1) A-R, B-P, C-S, D-Q (2) A-S, B-Q, C-R, D-P (3) A-Q, B-S, C-R, D-P (4) A-Q, B-S, C-P, D-R Ans. [4] Sol. According to frequency AQ; BS; CP ; DR

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Q.14 The electric field of a plane electromagnetic wave is given by

E

= E0 i cos(kz) cos (t) The corresponding magnetic field B

is then given by :

(1) B

= kCE0 sin (kz) cos (t) (2) B

= j

CE0 sin (kz) sin (t)

(3) B

= jCE0 sin (kz) cos (t) (4) B

= j

CE0 cos (kz) sin (t)

Ans. [2] Sol. E

= E0 coskz cost i

vibration of B = perpendicular to E

and propagation of electro magnetic wave is in z direction. Hence

the direction of vibration of B

should be along j .

0

0

BE

= C

B

= CE0 sin t sin kz j

Q.15 In an experiment, the resistance of a material is plotted as a function of temperature (in some range). As

shown in the figure, it is a straight line.

1/T2

ln R(T)

One may conclude that :

(1) 20

2 T/T0eR)T(R (2) R(T) = 2

0

TR

(3) 20

2 T/T–0eR)T(R (4)

220 T/T–

0eR)T(R

Ans. [4] Sol. y = – mx + c

ln (R) = – 2Tm + c

If R = R02

20

TT–

e

ln R = – 2

20

TT + ln (R0)

satisfies the equation Q.16 A transformer consisting of 300 turns in the primary and 150 turns in the secondary gives output power of

2.2 kW. If the current in the secondary coil is 10 A, then the input voltage and current in the primary coil are ;

(1) 440 V and 20 A (2) 440 V and 5 A (3) 220 V and 20 A (4) 220 V and 10 A

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Ans. [2] Sol. np = 300, ns = 150

p

s

ii =

s

p

nn

ip = is p

s

nn = 5A

Ps = Vs is Vs = 10

102.2 3 = 220

VP = Vs s

P

nn = 220 ×

150300

= 440V Q.17 In a photoelectric effect experiment the threshold wavelength of light is 380 nm. If the wavelength of

incident light is 260 nm, the maximum kinetic energy of emitted electrons will be :

Given E (in eV) = )nmin(

1237

(1) 3.0 eV (2) 1.5 eV (3) 4.5 eV (4) 15.1 eV Ans. [2]

Sol. 0 = 380

1237 , E = 260

1237 of photon

KE of electron kE = E – 0 = 1.5 eV Q.18 The ratio of surface tensions of mercury and water is given to be 7.5 while the ratio of their densities is 13.6.

Their contact angles, with glass, are close to 135° and 0°, respectively. It is observed that mercury gets depressed by an amount h in a capillary tube of radius r1, while water rises by the same amount h in a capillary tube of radius r2. The ratio (r1/r2), is then close to :

(1) 2/3 (2) 4/5 (3) 2/5 (4) 3/5 Ans. [3] Sol. hHg = hwater

h = gR

cosT2

w

Hg

RR

= 52 = 0.4

Q.19 A proton, an electron, and a helium nucleus, have the same energy. They are in circular orbits in a plane due

to magnetic field perpendicular to the plane. Let rp, re and rHe be their respective radii, then, (1) re < rp < rHe (2) re > rp = rHe (3) re < rp = rHe (4) re > rp > rHe Ans. [3]

Sol. r = BqmE2 E = same

r qm

proton p

p

q

m =

qm

, He+2

rp = rHe

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For electron e

e

qm

< p

p

q

m proton

re < rp = rHe Q.20 A particle of mass m is moving along a trajectory given by x = x0 + a cos 1t y = y0 + b sin 2t The torque, acting on the particle about the origin , at t = 0 is :

(1) – m (x022b – y0

21a ) k (2) Zero

(3) kamy 210 (4) k)aybx(–m 2

100

Ans. [3] Sol. x = x0 + a cos 1t y = y0 + b sin 2t

r

= x i + y j

acceleration a

= 2

2

dtrd

= r

× F

= )ar(

m

= k)amy( 210

Q.21 The value of acceleration due to gravity at Earth's surface is 9.8 ms–2. The altitude above its surface at which

the acceleration due to gravity decreases to 4.9 ms–2, is close to : (Radius of earth = 6.4 × 106 m) (1) 6.4 × 106 m (2) 2.6 × 106 m (3) 1.6 × 106 m (4) 9.0 × 106 m Ans. [2]

Sol. gh = 2

Rh1

g

, gh = 2g

2

Rh1

= 2

1+ Rh = 2

h = R ( 2 –1) = (1.4 –1) × 64 × 105 m = 2.6 × 106 m

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Q.22 Two particles, of masses M and 2M, moving, as shown, with speeds of 10 m/s and 5 m/s, collide elastically at the origin. After the collision, they move along the indicated directions with speeds v1 and v2, respectively. The values of v1 and v2 are nearly :

10 m/s

5 m/s

v1

v2

2M

30°

45°

30°

45°

M

M

2M

(1) 3.2 m/s and 12.6 m/s (2) 3.2 m/s and 6.3 m/s (3) 6.5 m/s and 6.3 m/s (4) 6.5 m/s and 3.2 m/s Ans. [3] Sol. using momentum conservation in x direction m × 10 cos 30° + 2 m cos 45° = mv2 cos 45° + 2m v1 cos 30° in y direction 2m (5) sin 45° – m × 10 sin 30° = 2m v1 sin 30° – mv2 sin 45° solving equation v1 = 6.5 m/s , v2 = 6.3 m/s Q.23 A ray of light AO in vacuum is incident on a glass slab at angle 60° and refracted at angle 30° along OB as

shown in the figure. The optical path length of light ray from A to B is :

30°

60° Vacuum

Glass

B

O a b

A

(1) 2a + 2b (2) b2a

32 (3)

3b2a2 (4)

3b2a2

Ans. [1] Sol.

30°

60°

C

C

A a

b

AC = 60cos

a , BC = 30cos

b , =

30sin60sin

path = AC + BC = 2a + 2b

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Q.24 Figure shows charge (q) versus voltage (V) graph for series and parallel combination of two given capacitors. The capacitance are :

500

80

10 V V (Volt)

A

B

q(C)

(1) 50 F and 30 F (2) 40 F and 10 F (3) 20 F and 30 F (4) 60 F and 40 F Ans. [2]

Sol. In series 21

21

CCCC

= Vq =

1080 = 8 × 10–6

In parallel C1 + C2 = Vq =

10500 = 50 × 10–6

C1C2 = 400 × 10–6 C1 + C2 = 50 × 10–6

C1 = 10 F, C2 = 40F Q.25 A ball is thrown upward with an initial velocity V0 from the surface of the earth. The motion of the ball is

affected by a drag force equal to mv2 (where m is mass of the ball, v is its instantaneous velocity and is a constant). Time taken by the ball to rise to its zenith is :

(1) 1–tang

1

0V

g

(2)

0

1– Vg2tan

g21

(3)

0

1– Vg

sing

1

(4)

0V

g1ln

g1

Ans. [1]

Sol. a = – (g +v2) = dtdv

V

V 20 vg

dv = – t

0dt

By integrating

t = g

1 tan–1

0V

g

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Q.26 A uniformly charged ring of radius 3a and total charge q is placed in xy-plane centred at origin. A point charge q is moving towards the ring along the z-axis and has speed v at z = 4a. The minimum value of v such that it crosses the origin is :

(1) 2/1

0

2

a4q

51

m2

(2)

2/1

0

2

a4q

152

m2

(3)

2/1

0

2

a4q

151

m2

(4)

2/1

0

2

a4q

154

m2

Ans. [2] Sol.

+

+

+

+

+ + +

+ + +

+

v q 4a

q

3a

21 mv2 = PE

= a15

kq2 2

v = a

kqm15

4 2

Q.27 Two coaxial discs, having moments of inertia I1 and 2I1 , are rotating with respective angular velocities 1

and 2

1 , about their common axis. They are brought in contact with each other and thereafter they rotate

with a common angular velocity. If Ef and Ei are the final and initial total energies, then (Ef –Ei) is :

(1) 6

I 211 (2) 2

11I83

(3) –12

I 211 (4) –

24I 2

11

Ans. [4] Sol. conservation of angular momentum Li = LF

and kinetic energy = 2I21

put the value KE = –24

I 211

Q.28 n moles of an ideal gas with constant volume heat capacity CV undergo an isobaric expansion by certain

volume. The ratio of the work done in the process, to the heat supplied is :

(1) nRC

nR

V (2)

nR–CnR

V (3)

nRCnR4

V (4)

nR–CnR4

V

Ans. [1] Sol. In isobaric process w = nRdT w = Q – U

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Q = nCpdT

Qw =

dTnCnRdT

p =

pCR =

RCR

V

= nRnC

nR

V

= nRC

nR

V

Q.29 Two radioactive materials A and B have decay constants 10 and , respectively. If initially they have the

same number of nuclei, then the ratio of the number of nuclei of A to that of B will be 1/e after a time

(1) 10

1 (2) 10

11 (3) 91 (4)

111

Ans. [3] Sol. 1 = 10 , 2 =

2

1

NN = t–

t10–

ee

= e–9t = e–1

t = 91

Q.30 The displacement of a damped harmonic oscillator is given by x(t) = e–0.1t cos (10t + ). Here t is in seconds. The time taken for its amplitude of vibration to drop to half of its initial value is close to : (1) 4 s (2) 13 s (3) 7 s (4) 27 s Ans. [3]

Sol. 2

A0 = A0e–0.1t

t = 10 ln (2) = 7 sec

JEE Main Online Exam 2019careerpoint.ac.in/studentparentzone/2019/jee-main/JEE...Q.16 A transformer...

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