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JEE Main Online Paper

JEE Main Online Exam 2019

Questions & Solutions 12th April 2019 | Shift - II

PHYSICS

Q.1 In the given circuit, the charge on 4 µF capacitor will be :

4 µF 5 µF

1 µF

10 V

3 µF

(1) 5.4 µC (2) 9.6 µC (3) 13.4 µC (4) 24 µC

Ans. [4] Sol.

4 µF

5 µF

1 µF

10 V

3 µF

4 µF 6 µF

10 V

QQ

10 = 4Q +

6Q

10 = 2Q

31

21 10 =

2Q

65 µC24Q

Q.2 A particle is moving with speed v= xb along positive x-axis. Calculate the speed of the particle at

time t = (assume that the particle is at origin t = 0)

(1) 2

b2 (2) b2 (3) 2

b2 (4) 4

b2

Ans. [3]

Sol. v = xb

dtdx = bx1/2

x

0

2/1– dxx =

0

dtb

CAREER POINT



x2 = b x = 2b

v = b2b =

2b2

2

bv2

Q.3 A transparent cube of side, made of a material of refractive index µ2, is immersed in a liquid of refractive

index µ1(µ1 < µ2). A ray is incident on the face AB at an angle (shown in the figure). Total internal reflection takes place at point E on the face BC.

C

DA

µ1

µ2

EB

Then must satisfy :

(1) > sin–1 1–µµ

21

22 (2) < sin–1 1–

µµ

21

22 (3) < sin–1

2

1

µµ (4) > sin–1

2

1

µµ

Ans. [2] Sol.

µ1

µ2

E

90 –

By snell’s law µ1 sin = µ2sin … (1) For TIR µ2sin(90 – ) > µ1 From eq. (1)

sin = 2

1

µµ sin

cos = 2

221

22

µsinµ–µ

cos = 22

221

22

22

µsinµ–

µµ

CAREER POINT



cos = 222

21 sin

µµ–1 … (2)

sin(90 – ) > 2

1

µµ

cos > 2

1

µµ

using eq. (2)

222

21 sin

µµ–1 > 2

2

21

µµ

1 – 222

21 sin

µµ

> 22

21

µµ

22

21

µµ

sin2 <

22

21

µµ–1

sin2 <

1–

µµ

21

22

1–

µµsin 2

1

221–

Q.4 Let a total charge 2 Q be distributed in a sphere of radius R, with the charge density given by (r) = kr, where

r is the distance from the centre. Two charges A and B, of –Q each, are placed on diametrically opposite points, at equal distance, a from the centre. If A and B do not experience any force, then :

(1) a = 8–1/4 R (2) a = 2–1/4 R (3) a = 4/12R3 (4) a = R/ 3

Ans. [1] Sol. Total charge = 2 Q Charging density = kr Radius = R

aa

A B

dr

B a r

d(vol.) = 4r2dr dq = (d(vol.))

Force on charge at B will dq = R

0

kq 4r2 dr

be due to charge at A and 2Q = R

0

3drr4k

due to force applied by the 2Q = 4R4k 4

charge in sphere k = 4RQ2

…(1)

CAREER POINT



Fsphere FBA FBA= Fsphere Force on charge B due to element

dF = 2aQ)dq(k = 2

3

adr)r4K(kq

F = 2a4kQK

a

0

3drr = 4

a4kQK 2

F = kQKa2 FBA = Fsphere

2

2

)a2(kQ = kQ4Ka2 By replace value of K from (1)

2

2

a4Q = 2

4

2a

RQ2

a4 =8

R4

a = R 8–1/4 Q.5 A Carnot engine has an efficiency of 1/6. When the temperature of the sink is reduced by 62ºC, its efficiency

is doubled. The temperatures of the source and the sink are, respectively, (1) 99ºC, 37ºC (2) 124ºC, 62ºC (3) 37ºC, 99ºC (4) 62ºC, 124ºC

Ans. [3]

Sol. = 61

Solving eq. (1)

61 = 1 –

H

L

TT … (1)

61 =

H

LH

TT–T

TH = 6TH – 6TL

31 = 1–

H

L

T)62–T( … (2) 6TL = 5TH

TH = 5T6 L

Solving eq. (2)

31 =

H

LH

T)62–T(–T TH = 3TH – 3TL + 186

2TH = 3TL – 186

2 × 5T6 L = 3TL – 186

12TL = 15TL – 930 3TL = 930 TL = 310 K TL = 310 – 273 = 37 ºC Source temp. is higher & sink temp. is lower

TH = 5T6 L =

53106 = 372 K = 99 ºC

CAREER POINT



Q.6 A smooth wire of length 2r is bent into a circle and kept in a vertical plane. A bead can slide smoothly on

the wire. When the circle is rotating with angular speed about the vertical diameter AB, as shown in figure,

the bead is at rest with respect to the circular ring at position P as shown. Then the value of 2 is equal to -

r/2

O

A

r

B

P

(1) r/)3g( (2) 2g/r (3) r2g3 (4) )3r/(g2

Ans. [4] Sol.

r/2

mgcosmgsin

mgsin+ m2

2r cos

mgcos

N

m2

2r

m2

2r sin

m2

2r sin = mgcos

2= tanr

g2

tan = 2/r

4/r–r 22 =

2r4r3 2

= 3 2 = r3

g2

Q.7 A plane electromagnetic wave having a frequency v = 23.9 GHz propagates along the positive z-direction in

free space. The peak value of the Electric Field is 60 V/m. Which among the following is the acceptable

magnetic field component in the electromagnetic wave ?

(1) B = 2 × 10–7 sin(1.5 × 102x + 0.5 × 1011t) j (2)

B = 60 sin(0.5 × 103x + 1.5 × 1011t) k

(3) B = 2 × 10–7 sin(0.5 × 103z + 1.5 × 1011t) i (4)

B = 2 × 10–7 sin(0.5 × 103z – 1.5 × 1011t) i

CAREER POINT



Ans. [4] Sol. v = 23.9 GHz E0 = 60 V/m

0

0

BE

= C B0 = CE0 = 8103

60

= 2 × 10–7

Since the wave is propagating in positive z-direction So acceptable magnetic field component will be

B = 2 × 10–7 sin(0.5 × 103z – 1.5 × 1011t) i

Q.8 Consider the LR circuit shown in the figure. If the switch S is closed at t = 0 then the amount of charge that

passes through the battery between t = 0 and t = RL is :

i

E S

RL

(1) 2REL3.7 (2) 2R

EL7.2 (3) 2R3.7EL (4) 2R7.2

EL

Ans. [4]

Sol. I = Imax.

LRt–

e–1 Imax. = RE

dtdq =

LtR–

e–1RE

R4

0

dq =

R4

0

R– Lt

e–1RE dt

Q = RE

R/L

0

LtR–

eRLL

= RE

RL–e

RL

RL 1–

Q = eR

EL2 Q = 2R7.2

EL

Q.9 One kg of water, at 20ºC, heated in an electric kettle whose heating element has a mean (temperature

averaged) resistance of 20 . The rms voltage in the mains is 200 V. Ignoring heat loss from the kettle, time taken for water to evaporate fully, is close to :

[Specific heat of water = 4200 J/(kg ºC), Latent heat of water = 2260 kJ/kg] (1) 10 minutes (2) 22 minutes (3) 3 minutes (4) 16 minutes

CAREER POINT



Ans. [2]

Sol. R = 20 P = R

V2 =

20200200 = 2000 watt

V = 200 V 1 kg water 20ºC 1 kg water 100ºC Heat required Q1 = msT = (1)(4200)(80) = 336000 1 kg water 100ºC 1 kg vapour Heat required Q2 = mL = (1) 2260 × 1000 = 2260000 (power) (time) = total heat required 2000 × time = 2260000 + 336000 time = 1298 sec. time = 21.63 mint. –~ 22 minute Q.10 Consider an electron in a hydrogen atom revolving in its second excited state (having radius 4.65 Å). The

de-Broglie wavelength of this electron is : (1) 6.6 Å (2) 3.5 Å (3) 9.7 Å (4) 12.9 Å

Ans. [3] Sol. n = 3 (second exicted state) 2rn = ndB,

dB = nr2 3

= 3

65.414.32 –~ 9.7 Å

Q.11 In an amplitude modulator circuit, the carrier wave is given by, C(t) = 4 sin(20000 t) while modulating signal

is given by, m(t) = 2 sin (2000 t). The values of modulation index and lower side band frequency are : (1) 0.3 and 9 kHz (2) 0.5 and 10 kHz (3) 0.4 and 10 kHz (4) 0.5 and 9 kHz

Ans. [4] Sol. C(t) = 4sin(20000 t), Ac = 4, c = 10,000

m(t) = 2sin(2000t), Am = 2, m = 1000

Modulation index µ = c

m

AA =

42 =

21 = 0.5

Lower side bond frequency = L – m = 10,000 – 1000 = 9000 = 9 kHz Q.12 The electron in a hydrogen atom first jumps from the third excited state to the second excited state and

subsequently to the first excited state. The ratio of the respective wavelengths, 1/2, of the photons emitted in this process is :

(1) 22/5 (2) 7/5 (3) 9/7 (4) 20/7

CAREER POINT



Ans. [4] Sol.

n = 1

2

1

2 3 4 5 6 7

1

hc

= 13.6

161–

91

1

hc

= 13.6

91–

41

1

2

=

495169

7

1

2

=

549

1697

= 1

2

=

207

7

20

2

1

Q.13 A system of three polarizers P1, P2, P3 is set up such that the pass axis of P3 is crossed with respect to that of P1.

The pass axis of P2 is inclined at 60º to the pass axis of P3. When a beam of unpolarized light of intensity I0 is incident on P1, the intensity of light transmitted by the three polarizers is I. The ratio (I0/I) equals (nearly):

(1) 10.67 (2) 5.33 (3) 16.00 (4) 1.80 Ans. [1] Sol. When unpolarized light of intensity I0 passes through P1, P2 and P3, let the emergent light from P1, P2 and P3 and

I1, I2 & I3. Then from Malus law I = I0cos2 I0 = Incident Intensity = Angle between pass axes and incident light

So I1 = 2I0 < cos2 > =

21

I2 = 2I0 cos230º =

8I3 0

I3 = 8I3 0 cos260º =

32I3 0

So I = 32I3 0

I

I0 = 3

32 = 10.67

Q.14 A diatomic gas with rigid molecules does 10 J of work when expanded at constant pressure. What would be

the heat energy absorbed by the gas, in this process ? (1) 35 J (2) 40 J (3) 25 J (4) 30 J

CAREER POINT



Ans. [1] Sol. W = 10 J at constant pressure W = P(V2 – V1) = PV2 – PV1 10 = nR(T2 – T1) = nRT Q = W + U

= 10 J + 2

nf RT

= 10 J + 25 (10 J)

J35Q

Q.15 A block of mass 5 kg is (i) pushed in case (A) and (ii) pulled in case (B), by a force F = 20 N, making an

angle of 30º with the horizontal, as shown in the figures. The coefficient of friction between the block and floor is µ = 0.2. The difference between the accelerations of the blocks, in case (B) and case (A) will be : (g = 10 ms–2)

(A) (B)F = 20 N

30º30º

F = 20 N

(1) 3.2 ms–2 (2) 0.8 ms–2 (3) 0 ms–2 (4) 0.4 ms–2

Ans. [2] Sol.

N1 = mg + 10 (fs)max = 0.2 × (60) = 12 N

N2 = mg – 10 = 50 – 10 = 40 N (fs)max. = 0.2 × 40 = 8 N

20 N30º

30º20

20sin30º = 10

µ = 0.2

N1

20cos30º = 10 3 10 3

10N2

mg

Case-A Case-B

mg

aA = 5

12–310 = 5

12–32.17 aB = 5

8–310 = 5

8–32.17

aA = 532.5 aB =

532.9

difference between acceleration aB – aA = 51 (9.32 – 5.32) =

54

2s/m8.0a

CAREER POINT



Q.16 A moving coil galvanometer, having a resistance G, produces full scale deflection when a current Ig flows

through it. This galvanometer can be converted into (i) an ammeter of range 0 to I0(I0 > Ig) by connecting a shunt resistance RA to it and (ii) into a voltmeter of range 0 to V (V = GI0) by connecting a series resistance RV to it. Then,

(1) RARV = G2 and V

A

RR =

)I–I(I

g0

g

(2) RARV = G2

g0

g

I–II

and V

A

RR =

2

g

g0

II–I

(3) RARV = G2

g

g0

II–I

and V

A

RR =

2

g0

g

)I–I(I

(4) RARV = G2 and V

A

RR =

2

g0

g

I–II

Ans. [4] Sol. Galvanometer is converted into ammeter of range 0 to I0.

I0 –Ig S = RA

I0 GIg

IgG = (I0 – Ig) RA

RA = )I–I(

GI

g0

g … (1)

Galvanometer is converted into voltmeter of range 0 to V

RV

GIg

V = Ig(G + RV) GI0 = Ig(G + RV)

RV = g

0

IGI – G

RV = g

g0

I)I–I(G

… (2)

So from (1) & (2) RA RV = G2

V

A

RR =

2

g0

g

I–II

Q.17 A solid sphere, of radius R acquires a terminal velocity v1 when falling (due to gravity) through a viscous fluid having a coefficient of viscosity . The sphere is broken into 27 identical solid spheres. If each of these spheres acquires a terminal velocity, v2, when falling through the same fluid, the ratio (v1/v2) equals :

(1) 1/9 (2) 1/27 (3) 9 (4) 27 Ans. [3]

CAREER POINT



Sol. 3R34 = 27 × 3r

34

r3 = 3

3

3R

3Rr

V1 = R6

gR34)–( 3

.liq0

V2 =

3R6

g3R

34)–(

3

.liq0 =

31R6

271gR

34)–( 3

.liq0

V2 = 9V1

Q.18 Two sources of sound S1 and S2 produce sound waves of same frequency 660 Hz. A listener is moving from

source S1 towards S2 with a constant speed u m/s and he hears 10 beats/s. The velocity of sound is 330 m/s.

Then, u equals :

(1) 10.0 m/s (2) 5.5 m/s (3) 15.0 m/s (4) 2.5 m/s

Ans. [4] Sol.

S1 O S2

u

As observer goes away from source S1 so apparent frequency

1 = v

)v–v( 0 v = speed of round, v0 = speed of observer

=

330u–330 × 660

1 = 2 × 330 – 2u … (1)

As observer goes towards source S2 so apparent frequency

2 = v

)vv( 0

=

330u–330 × 660

2 = 2 × 330 + 2u … (2)

According to question

2 – 1 = 10

4u = 10

u = 2.5 m/s

CAREER POINT



Q.19 An electron, moving along the x-axis with an initial energy of 100 eV, enters a region of magnetic field B = (1.5 × 10–3

T) k at S (See figure). The field extends between x = 0 and x = 2 cm. The electron is detected at the point Q on a screen placed 8 cm away from the point S. The distance d between P and Q (on the screen) is :

(electron’s charge = 1.6 × 10–19C, mass of electron = 9.1 × 10–31 kg)

P

d

S

8 cm 2 cm

Q

(1) 2.25 cm (2) 12.87 cm (3) 1.22 cm (4) 11.65 cm

Ans. [2]

Sol. R = qBmk2

= 3–19–

19–31–

105.1106.1106.1100101.92

R = 2.24 cm

6 cm

h

R(1 – cos)x

R

R

sin = Rx =

24.22 = 0.89

1

0.45

0.89

tan = 45.089.0 –~ 1.97 =

6h

h = 11.82 cm cos = 0.45 d = h + 2 (1 – 0.45) = 11.82 + 2(1 – 0.45) –~ 12.9 cm

CAREER POINT



Q.20 Half lives of two radioactive nuclei A and B are 10 minutes and 20 minutes, respectively, If initially a sample has equal number of nuclei, then after 60 minutes, the ratio of decayed numbers of nuclei A and B will be :

(1) 9 : 8 (2) 1 : 8 (3) 8 : 1 (4) 3 : 8 Ans. [1]

Sol. A1 = 10/600

2A

= 60

2A

= 64A0

A2 = 20/600

2A

= 30

2A

= 8

A0

No. of decayed nuclei = 1A = A0 –

64A0 =

6463

2A = A0 –

8A0 =

87

89

78

6463Ratio

Q.21 A tuning fork of frequency 480 Hz is used in an experiment for measuring speed of sound (v) in air

by resonance tube method. Resonance is observed to occur at two successive lengths of the air column, l1 = 30 cm and 2 = 70 cm. Then, v is equal to -

(1) 338 ms–1 (2) 384 ms–1 (3) 379 ms–1 (4) 332 ms–1 Ans. [2] Sol. l1 = 30 cm l2 = 70 cm, = 480 = 2(l2 – l1) = 2(70 – 30) × 10–2 × 480 = 2 × 40 × 10–2 × 480 = 384 m/sec

Q.22 The number density of molecules of a gas depends on their distance r from the origin as, n(r) = 4r–

0en . Then the total number of molecules is proportional to :

(1) na–3/4 (2) 2/1

0n (3) n01/4 (4) n0

–3

Ans. [Bonus] Sol. n =

4r–0en

rdr

taken an element hollow sphere of thickness dr Vol. of element dV = (4r2)dr no. of molecules in elementary volume = n.e–r4r2dr

total no. of molecules = n.

0

r–2 drer44

= ?

Note : This equation can't be solved so it should be bonus and full marks should given to students.

CAREER POINT



Q.23 A small speaker delivers 2 W of audio output. At what distance from the speaker will one detect 120 dB intensity sound ? [Given reference intensity of sound as 10–12 W/m2]

(1) 20 cm (2) 10 cm (3) 40 cm (4) 30 cm Ans. [3]

Sol. L = 10 log 0II

120 = 10 log 0II

12 = log10 12–10I

12–10I = 1012

I = 1 = 2r4P

r2 = 14

2

r = 42 m = 100 ×

21 = 100 × 0.399 –~ 39.9 –~ 40 cm

Q.24 A uniform cylindrical rod of length L and radius r, is made from a material whose Young’s modulus of Elasticity equals Y. When this rod is heated by temperature T and simultaneously subjected to a net longitudinal compressional force F, its length remains unchanged. The coefficient of volume expansion, of the material of the rod, is (nearly) equal to :

(1) 3F/(r2YT) (2) 6F/(r2YT) (3) F/(3r2YT) (4) 9F(r2YT) Ans. [1]

Sol. Y =

2rF

Y = 2rF

×

= Yr

F2

change in length due to temperature change

= T

T = Yr

.F2

YTrF2

YTrF3Y 2

CAREER POINT



Q.25 Find the magnetic field at point P due to a straight line segment AB of length 6 cm carrying a current of 5A. (See figure) (µ0 = 4× 10–7 N-A–2)

A B

P

6 cm (1) 1.5 × 10–5 T (2) 3.0 × 10–5 T (3) 2.0 × 10–5 T (4) 2.5 × 10–5 T

Ans. [1] Sol.

P

3 cm

4 cm

1 5 cm

B =

d4iµ0

(sin1 + sin2)

= 7–2– 10

53

53

1045

4 cm = 4 × 10–2 m

= 45 × 2 2–

7–

105103

T105.1B 5– Q.26 A spring whose unstretched length is l has a force constant k. The spring is cut into two pieces of unstretched

lengths l1 and l2 where, l1 = nl2 and n is an integer. The ratio k1/k2 of the corresponding force constant, k1 and k2 will be ;

(1) 2n1 (2)

n1 (3) n2 (4) n

Ans. [2]

Sol. k,l = 2211 k,k, ll

given l1 = nl2 k1 = k/1 1 ll k2 = k/1 2

ll

k1 = ll1

k k2 = 2

1ll

k

2k1k =

1

2

ll

n1

kk

2

1

CAREER POINT



Q.27 The ratio of the weights of a body on the Earth’s surface to that on the surface of a planets is 9 : 4. The mass

of the planet is 91 th of that of the Earth. If 'R' is the radius of the Earth, what is the radius of the planet ?

(Take the planets to have the same mass density)

(1) 9R (2)

2R (3)

3R (4)

4R

Ans. [2]

Sol. p

e

WW =

49 Mp =

91 Me

We = m 2

e

RGM

p

e

WW = 2

e

RmGM ×

p

2

mGMR

Wp = m 2e

RGM

p

e

WW = 2

2

RR9

49 = 2

2

RR9 R2 =

4R 2

2RR

Q.28 Figure shows a DC voltage regulator circuit, with a Zener diode of breakdown voltage = 6V. If the

unregulated input voltage varies between 10 V to 16 V, then what is maximum Zener current ?

RL = 4 k

IZ

IL

RS = 2 k

IS

(1) 3.5 mA (2) 1.5 mA (3) 2.5 mA (4) 7.5 mA

Ans. [1] Sol.

4 k

I1

RS = 2 k

IS

V

V1

I

V2

I2

CAREER POINT



Case – I : V = 16 volt V2 = 6 V then V1 = 10 V

I = k2

10 = 5 × 10–3 amp.

I2 = 100046

= 1.5 × 10–3 Amp.

I1 = (5 – 1.5) × 10–3 Amp. = 3.5 × 10–3 Amp. Case – II : V = 10 volt V2 = 6 V & V1 = 4 vol.

I = 100024

= 2 × 10–3 Amp.

I2 = 100044

= 10–3 Amp.

I1 = (2 – 1) × 10–5 = 10–5 Amp. Maximum Zener current is in Case-I that is 3. 5 × 10–3 Amp.

Q.29 Three particles of masses 50 g, 100 g and 150 g are placed at the vertices of an equilateral triangle of side

1 m (as shown in the figure). The (x, y) coordinates of the centre of mass will be :

m3 = 150 g

m2 = 100 gX

Y

50 g = m1 O 0.5 m 1.0 m

60º

(1)

m

127,m

83 (2)

m

125,m

43 (3)

m

83,m

127 (4)

m

43,m

127

Ans. [4] Sol. m1 = 50 g m2 = 100 g m3 = 150 g x1 = 0 x2 = 1m x3 = 0.5 m

y1 = 0 y2 = 0 y3 = 23

xCOM = 321

332211

mmmxmxmxm

= 300

)5.0)(150(10)100(0

xCOM = 300

75100 = 300175 =

127 m.

yCOM = 321

332211

mmmymymym

= 300

23)150()0)(100(0

yCOM = 300

375 = 12

33 = 43 m

43,

127)y,x( COMCOM

CAREER POINT



Q.30 Two particles are projected from the same point with the same speed u such that they have the same range R, but different maximum heights, h1 and h2. Which of the following is correct ?

(1) R2 = h1h2 (2) R2 = 16 h1h2 (3) R2 = 4 h1h2 (4) R2 = 2h1h2 Ans. [2]

Sol.

Angle of projections must be , (90 – )

h1 = g2

sinu 22 , h2 = g2

cosu 22

R = g

cossinu2 2

h1h2 = 2

224

g4cossinu

R2 = 2

224

gcossinu4

212 hh16R

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