JEE Main Online Exam 2019 - Career Pointcareerpoint.ac.in/studentparentzone/2019/jee-main/JEE...PHYSICS Q.1 A shell is fired from a fixed artillery gun with an initial speed u such

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JEE Main Online Paper

JEE Main Online Exam 2019

Questions & Solutions 12th April 2019 | Shift - I

PHYSICS

Q.1 A shell is fired from a fixed artillery gun with an initial speed u such that it hits the target on the ground at a distance R from it. If t1 and t2 are the values of the time taken by it to hit the target in two possible ways, the product t1t2 is -

(1) 2R/g (2) R/2g (3) R/g (4) R/4g Ans. [1] Sol.

t2 t1 R

90–

For & 90 – angle of projection, range will be same

Time of flight for : t1 = gsinu2

Time of flight for 90 – : t2 = g

)90sin(u2 = gcosu2

t1t2 = 2

2

gcossinu4 =

g2sin

gu2 2

=

g

2sinug2 2

= gR2

Q.2 A circular disc of radius b has a hole of radius a at its centre (see figure). If the mass per unit area of the disc

varies as

r0 , then the radius of gyration of the disc about its axis passing through the centre is:

a

b

(1) 3

ba (2) 3

abba 22 (3) 2

abba 22 (4) 2

ba

Ans. [2]

CAREER POINT



Sol.

a

b

r dm =

r0 (2rdr) = 20dr

I = mk2 : k = radius of gyration

b

a

b

a

22 dmkr)dm(

b

a

b

a0

220 dr2kr)dr2(

)ab(2k3

ab2 02

33

0

)ab(k3

)abab)(ab( 222

k = 3

abba 22

Q.3 Shown in the figure is a shell made of a conductor. It has inner radius a and outer radius b, and carries charge

Q. At its centre is a dipole P

as shown. In this case;

p

(1) surface charge density on the inner surface is uniform and equal to 2a4)2/Q(

(2) surface charge density on the outer surface depends on p

(3) surface charge density on the inner surface of the shell is zero everywhere (4) electric field outside the shell is the same as that of a point charge at the centre of the shell Ans. [4] Sol. The charge distribution at equilibrium on the conductor will be like :

+ ++

+

+ +

+

++

+ + + + +

––– ––

Net charge on the outer surface = Q Total charge on the inner surface = 0 So for any observer outside the shell, the resultant electric field is due to Q uniformly distributed on the outer

surface only.

CAREER POINT



Q.4 A thin ring of 10 cm radius carries a uniformly distributed charge. The ring rotates at a constant angular

speed of 40 rad s–1 about its axis, perpendicular to its plane. If the magnetic field at its centre is 3.8 × 10–9

T, then the charge carried by the ring is close to (0 = 4× 10–7 N/A2). (1) 7 × 10–6 C (2) 4 × 10–5 C (3) 2 × 10–6 C (4) 3 × 10–5 C Ans. [4] Sol.

+ ++

+

++

+

+ +

+R

O

Q

R = 10 cm = 10–1 m = 40 rad/sB0 = 3.8 × 10–9T

I =

2Q

tQ

B0 = R2

2Q

R2I

0

0

B0 = R4

Q0

Q =

0

0 R4B

= )40)(104(

)10)(108.3(7–

1–9–

= 7–

1–9–2–

1014.340101010380

= 6.125

380 × 10–12+7 = 3 × 10–5 C

Q.5 A uniform rod of length is being rotated in a horizontal plane with a constant angular speed about an axis

passing through one of its ends. If the tension generated in the rod due to rotation is T(x) at a distance x from the axis, then which of the following graphs depicts it most closely?

(1)

T(x)

x

(2)

T(x)

x

(3)

T(x)

x

(4)

T(x)

x

Ans. [4]

CAREER POINT



Sol.

T(x)

x m, dL

x dm

L

T(x)

T(x) =

x

2L)dm(

=

x

2LdLm

=

x

22

2Lm

T(x) = 2

m 2 ( 22 x )

T(x) = A – Bx2

(Parabola, mouth down)

T(x)

x

Q.6 The stopping potential V0 (in volt) as a function of frequency () for a sodium emitter, is shown in the figure.

The work function of sodium, from the data plotted in the figure, will be: (Given: Planck’s constant (h) = 6.63 × 10–34 Js, electron charges e = 1.6 × 10–19 C)

2 4 6 8 10

1.0 2.0 3.0

V0

(1014 Hz)

(1) 1.95 eV (2) 2.12 eV (3) 1.82 eV (4) 1.66 eV Ans. [4] Sol.

V0 (stopping potential)

(V0) Threshold frequency

(1014 hz) frequency

Work function = 0 = h0

= 19

1434

106.1)104)(1063.6(

eV = 1.66 eV

CAREER POINT



Q.7 Two identical parallel plate capacitors, of capacitance C each, have plates of area A, separated by a distance d. The space between the plates of the two capacitors, is filled with three dielectrics, of equal thickness and dielectric constants K1, K2 and K3. The first capacitor is filled as shown in fig.I, and the second one is filled as shown in fig II.

If these two modified capacitors are charged by the same potential V, the ratio of the energy stored in the two, would be (E1 refers to capacitor (I) and E2 to capacitor (II)):

K1 K2 K3

K1 K2 K3

(I) (II)

(1) 321

211332321

2

1KKK

)KKKKKK)(KKK(EE

(2) )KKKKKK)(KKK(

KKKEE

211332321

321

2

1

(3) 321

211332321

2

1KKK9

)KKKKKK)(KKK(EE

(4) )KKKKKK)(KKK(

KKK9EE

211332321

321

2

1

Ans. [4] Sol.

K1 K2 K3

K1 K2 K3

(I) (II)

K1

K2

K3

K1 K2 K3

(I) (II)

3/dAk

1

3/dAk

1

3/dAk

1C1

03

02

01

1

32101 k1

k1

k1

A3d

C1

CAREER POINT



C1 = )kkkkkk(

kkkd

A3133221

3210

C2 = d

3/Akd

3/Akd

3/Ak 03

02

01

= d3A0 [k1 + k2 + k3]

)kkkkkk)(kkk(

kkk9

VC21

VC21

EE

133221321

321

22

21

2

1

Q.8 In a double slit experiment, when a thin film of thickness t having refractive index . is introduced in front of

one of the slits, the maximum at the centre of the fringe pattern shifts by one fringe width. The value of t is ( is the wavelength of the light used) :

(1) )1(2

(2) )12(

(3) )1(

2 (4)

)1(

Ans. [4] Sol. Normal YDSE without slab

First bright x = Central bright (C.B.) x=0

P central bright (x = 0) t

x =

S1

S2

For central bright at the position of first bright S2P – S1P = 0

( + ) – (– t + t) = 0 optical path length

+ – + t – t = 0 = t (– 1)

1

t

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Q.9 The value of numerical aperature of the objective lens of a microscope is 1.25. If light of wavelength 5000 Å is used, the minimum separation between two points, to be seen as distinct, will be :

(1) 0.12 m (2) 0.38 m (3) 0.24 m (4) 0.48 m Ans. [3] Sol. Numerical aperture of the microscope is given as

NA = d61.0

d = minimum separation between two points to be seen as distinct

d = NA61.0 =

25.1)105000)(61.0( 10 = 2.4 ×10–7 m = 0.24 m

Q.10 The resistive network shown below is connected to a D.C. source of 16 V. The power consumed by the

network is 4 Watt. The value of R is: 4R

4R 12R

6RRR

= 16 V (1) 16 (2) 1 (3) 8 (4) 6 Ans. [3] Sol.

4R

4R 12R

6RRR

= 16 V

RR

= 16 V

2R 4R

= 16 V

8R

P = RV2

4 = R81616

8R

Q.11 The trajectory of a projectile near the surface of the earth is given as y = 2x – 9x2. If it were launched at an

angle 0 with speed v0 then (g = 10 ms–2) :

(1) 0 = cos–1

51 and v0 = 3

5 ms–1 (2) 0 = cos–1

52 and v0 = 5

3 ms–1

(3) 0 = sin–1

52 and v0 = 5

3 ms–1 (4) 0 = sin–1

51 and v0 = 3

5 ms–1

Ans. [1]

CAREER POINT



Sol. y = x tan

Rx1 …..(i)

Given eqn of trajectory : y = 2x – 9x2 = 2x

2x91 = 2x

92x1 …. (2)

Comparing equation (1) & (2)

tan = 2 & R = 92

5

cos = 5

1 = cos–1

51

92

g2sinu2

10tan1tan2u 2

2

= 92

u2

54 =

92 ×10

u2 = 49

5102

u = 6

10 = 35 m/s

Q.12 A person of mass M is, sitting on a swing of length L and swinging with an angular amplitude 0. If the

person stands up when the swing passes through its lowest point, the work done by him, assuming that his center of mass moves by a distance (<<L), is close to;

(1) mg (1 + 02) (2) mg (3) mg

21

20 (4) mg(1 – 2

0 )

Ans. [1] Sol.

L

vmax

The force acting on the man at the lowest point

F = mg + L

mv2max

CAREER POINT



= 2max )V(

Lmmg

= 2]A[Lmmg

[T = 2 = 2

gL =

Lg ]

= 20 ]

Lg)L[(

Lmmg

= mg + mg 20

= mg (1 + 20 )

Work done = (F) (displacement)

= [mg(1 + 20 )][]

= mg(1 + 20 )

Q.13 A point dipole xpp 0

is kept at the origin. The potential and electric field due to this dipole on the

y-axis at a distance d are, respectively: (Take V= 0 at infinity)

(1) 30

20 d4

p,d4

p

(2) 30 d4p,0

(3) 30

20 d4

p,d4

p

(4) 30 d4p,0

Ans. [2] Sol.

O

A

y E

d

p =–p0 x x

A is an equatorial point w.r.t the dipole :

Electric field at A = 3rpK

=

3

0

0 d)xp(

41

= xd4

p3

0

0

Electric potential at A = 0

CAREER POINT



Q.14 A galvanometer of resistance 100 has 50 divisions on its scale and has sensitivitv of 20 A/division. It is to be converted to a voltmeter with three ranges of 0-2V, 0-10 V and 0-20 V. The appropriate circuit to do so is

(1) R1 R2 R3

G

2V 10V 20V

R1 = 1900R2 = 8000R3 = 10000

(2) R1 R2 R3

G

20V 10V 2V

R1 = 19900R2 = 9900R3 = 1900

(3) R1 R2 R3

G

2V 10V 20V

R1 = 1900R2 = 9900R3 = 19900

(4) R1 R2 R3

G

2V 10V 20V

R1 = 2000R2 = 8000R3 = 10000

Ans. [1] Sol. sensitivity = 20 A/div Total division = 50 maximum current through galvanometer can be = Imax = (50) (20 A) = 10–3 A

R1G

2V

RG = 100

0V

Imax = 1R100

2

= 10–3

3102

= 100 + R1

R1 = 2000 – 100 R1 = 1900

R1G

10V

RG = 100

0V

R2

Imax = 21G RRR

10

10–3 = 2R1900100

10

R2 + 2000 = 31010

R2 = 10000 – 2000 = 8000

R1G

20V

RG = 100

0V

R2 R3

Imax = 3R80001900100

20

10000 + R3 = 20000 R3 = 10000

CAREER POINT



Q.15 The truth table for the circuit given in the fig. is:

YAB

(1)

111101110100YBA

(2)

011001110100YBA

(3)

111101010000YBA

(4)

011001010100YBA

Ans. [2] Sol.

YA

B

D

C(NAND)(OR)

Effectively D = A & C is output of 'OR' gate Y is output of 'NAND' gate

01111011011011010000YDCBA

Q.16 The figure shows a square loop L of side 5 cm which is connected to a network of resistances. The whole

setup is moving towards right with a constant speed of 1 cm s–1. At some instant, a part of L is in a uniform magnetic field of 1 T, perpendicular to the plane of the loop. If the resistance of L is 1.7 , the current in the loop at that instant will be close to :

23

21

1B

D

A C

5 cm

L

B

v=1 cm/sec

(1) 115 A (2) 150 A (3) 170 A (4) 60 A

Ans. [3] Sol.

2

3

21

1B

D

A C

v

Loop(L)

B a

CAREER POINT



V = 1 cm/s = 10–2 m/s Rloop = 1.7 A = 5 cm = 5 × 10–2 m equivalent circuit : Rtotal = Rloop + Rwheatstone

Req = 24)2)(4(

=

68 =

34 = 1.3

RTotal = 1.7 + 1.3 RTotal = 3 Induced emf = VB

current = I = TotalR

)VB(

= 3

)105)(1)(10( 22–

= 35 ×10–4

= 1.67 ×10–4 = 167 ×10–6 170 A Q.17 A sample of an ideal gas is taken through the cyclic process abca as shown in the figure. The change in the

internal energy of the gas along the path ca is –180 J. The gas absorbs 250 J of heat along the path ab and 60 J along the path bc. The work done by the gas along the path abc is:

c

V

b a

P

(1) 140 J (2) 130 J (3) 100 J (4) 120 J Ans. [2] Sol.

c

V

b a

P

Uca = – 180J Uac = + 180J (state function) Q = U + W Qac = Uac + Wac Qab + Qbc = 180 + Wac 250 + 60 = 180 + Wac Wac = 310 – 180 = 130 J

CAREER POINT



Q.18 An electromagnetic wave is represented by the electric field )]z8y6(tsin[nEE 0

. Taking unit

vectors in x, y and z directions to be k,j,i , the direction of propagation s , is :

(1) 5

j3k4s (2)

5k3j4s

(3)

5k4j3s (4)

5j4i3s

Ans. [3]

Sol. )]z8y6(t[nEE 0

)]y6z8(–t[nEE 0

]10.j106k

108t[nEE 0

]kst[nEE 0

s = direction of propagation

s =

10

j6k8

=

5

j3k4

= 5

k4j3

Q.19 At 40º C, a brass wire of 1 mm radius is hung from the ceiling. A small mass, M is hung from the free end of

the wire. When the wire is cooled down from 40ºC to 20ºC it regains its original length of 0.2 m. The value of M is close to :

(Coefficient of linear expansion and Young’s modulus of brass are 10–5/°C and 1011 N/m2, respectively; g= 10 ms–2)

(1) 1.5 kg (2) 0.5 kg (3) 9 kg (4) 0.9 kg Ans. [3] Sol. r = 1 mm = 10–3 m L0 = 0.2 m

M

L

L=L0×T Y =

)L/L()A/F(0

AYF

LL0

L = AYFL0

L0 T = AYFL0

CAREER POINT



T = AYMg

M = gTAY

= )10(

)10)(10)(20)(10( 1165

= 2 kg = 2 × 3.14 kg = 6.28 kg (closest to 9)

Q.20 To verify Ohm's law, a student connects the voltmeter across the battery as, shown in the figure. The

measured voltage is plotted as a function of the current, and the following graph is obtained; V

Internal Resistance Ammeter

R V

1000 mA

1.5 V

V0 I

If V0 is almost zero, identify the correct statement : (1) The value of the resistance R is 1.5 (2) The emf of the battery is l.5 V and its internal resistance is 1.5 (3) The emf of the battery is l.5 V and the value of R is 1.5 (4) The potential difference across the battery is 1.5 V when it sends a current of 1000 mA Ans. [2] Sol.

V

1000 mA = 1A

1.5 V

O I

(voltmeter reading)

V

Ammeter

R

iE

r

When voltmeter reading is zero E – ir = 0

E – rrR

E

= 0

CAREER POINT



1 – rR

r

= 0

R + r – r = 0 R = 0 : when voltmeter reading is zero

Req = r (for circuit)

i = rE

1000 mA = r5.1 (E = 1.5V from graph)

1 = r5.1

r = 1.5 Q.21 A concave mirror has radius of curvature of 40 cm. It is at the bottom of a glass that has water filled up to 5

cm (see figure). If a small particle is floating on the surface of water, its image as seen, from directly above the glass, is at a distance d from the surface of water. The value of d is dose to: (Refractive index of water = 1.33)

5 cm

Particle

(1) 11.7 cm (2) 6.7 cm (3) 13.4 cm (4) 8.8 cm Ans. [4] Sol.

5 cm

Particle Object is just inside the water

= 4/3 H

R = 40 cm f = – 20 cm

I (virtual image)

Observer

f

f1

u1

V1

201

)5(1

V1

201

51

V1

203

2014

V1

3

20V cm

CAREER POINT



H = 5 + 320 =

335 cm

Happerent = H =

34335

= 43

335

= 435 = 8.8 cm

Q.22 Two moles of helium gas is mixed with three moles of hydrogen molecules (taken to be rigid). What is the

molar specific heat of mixture at constant volume ? (R = 8.3 J/mol K) (1) 21.6 J/mol K (2) 19.7 J/mol K (3) 15.7 J/mol K (4) 17.4 J/mol K Ans. [4]

Sol. (CV)mix = 21

V2V1

nnCnCn

21

= 32

2fR3

2fR)2(

diaatomicmono

= 5

2R53

2R32

= 10

R21

= 10

3.821 = 17.4 J/mol-K

Q.23 A magnetic compass needle oscillates 30 times per minute at a place where the dip is 45°, and 40 times per

minute where the dip is 30º. If B1 and B2 are respectively the total magnetic field due to the earth at the two places, then the ratio B1/ B2 is best given by :

(1) 1.8 (2) 2.2 (3) 0.7 (4) 3.6 Ans. [3] Sol.

= angle of dip

B

= Bm

sinmB mB (small angular displacement) )cosB(mI

IcosmB

2 = IcosmB

T =

cosmBI2

CAREER POINT



11

22

2

1cosBcosB

TT

11

22cosBcosB

40603060

11

222

cosBcosB

34

1

2

2

1BB

coscos

916

2

1BB =

1223

169

= 32

69

= 32

44.29

= 3222 = 0.7

Q.24 Which of the following combinations has the dimension of electrical resistance (0 is the permittivity of

vacuum and 0 is the permeability of vacuum)?

(1) 0

0 (2)

0

0 (3)

0

0 (4)

0

0

Ans. [4] Sol. [0] = (M–1 L–3 T4 A2) [0] = (M L T–2 A–2) [R] = (M L2 T–3 A–2)

R = 0

0

Q.25 The transfer characteristic curve of a transistor, having input and output resistance 100 and 100 k

respectively is shown in the figure. The voltage and power gain, are respectively: (1) 5 × 104, 5 × 105 (2) 5 × 104, 5 × 106 (3) 5 × 104, 2.5 × 106 (4) 2.5 × 104, 2.5 × 106 Ans. [3]

Sol.

I0)

IC

(100,5) (200,10)

(300,15) (mA)

CAREER POINT



B = B

CII = 9

3

10100105

= 50

Voltage gain = AV = input

output

VV

= ininput

0output

RIRI

=

in

0

b RR

II

C

= ()

in

0RR

= (50)

100

10100 3

= 5 × 104

Power gain = 2 =

i

0RR

= (50)2 × 103 = 25 × 102 × 103 = 2.5 × 106 Q.26 When M1 gram of ice at –10°C (specific heat = 0.5 cal g–1ºC–1) is added to M2, gram of water at 50°C, finally

no ice is left and the water is at 0°C. The value of latent heat of ice, in cal g–1 is :

(1) 5MM50

1

2 (2) 1

2MM50 (3) 5

MM5

1

2 (4) 50MM5

2

1

Ans. [1] Sol.

M1

50ºC water

–10ºC ice

M2 M1+M2

(water)

0ºC

At equilibrium Using energy conservation Erelesed by water = Eused by ice M2SW(T)water = M1Sice(T)ice + M1Lfusion

M2(1)(50) = M1

21 (10) +M1Lfusion

50M2 – 5M1 = M1Lfusion

Lf = 1

12M

M5M50

Lf = 5MM50

1

2

CAREER POINT



Q.27 A man (mass = 50 kg) and his son (mass =20 kg) are standing on a frictionless surface facing each other. The man pushes his son so that he starts moving at a speed of 0.70 ms–1 with respect to the man. The speed of the man with respect to the surface is :

(1) 0.28 ms–1 (2) 0.47 ms–1 (3) 0.20 ms–1 (4) 0.14 ms–1 Ans. [3] Sol.

VA

A

B VB

50 kg 20 kg

Surface (=0)

s/m7.0V A/B

7.0VV S/AS/B

VB – (–VA) = 0.7 VB + VA = 0.7 VA + VB = 0.7 Momentum conservation )0F( ext

PC = Pf

0 = 20(+VB) + 50(–VA) 2VB = 5VA

VB = 2V5 A

VA + 2V5 A = 0.7

2V7 A = 0.7

VA= 0.7

72

VA = 0.2 m/s Q.28 A progressive wave travelling along the positive x-direction is represented by y(x,t) = Asin(kx – t + ). Its

snapshot at t = 0 is given in the figure. y

x

A

For this wave, the phase is :

(1) 2 (2) (3) 0 (4) –

2

Ans. [2]

CAREER POINT



Sol. Y = A sin (kx – t + ) At t = 0 Y = A sin (kx + )

x

y

Graph of : y = A sin(kx)

x

y

Graph of : y = –A sin(kx)

– A sin(kx) = A sin (kx + ) A sin(kx + ) = A sin (kx + ) = Q.29 An excited He+ ion emits two photons in succession, with wavelengths 108.5 nm and 30.4 nm, in making a

transition to ground state. The quantum number n, corresponding to its initial excited state is (for photon of

wavelength , energy E = )nmin(

eV1240

.

(1) n = 4 (2) n = 6 (3) n = 5 (4) n = 7 Ans. [3] Sol.

n E1

E2 2

1

n = 1

E = E1 + E2

13.6 (Z2)

22 n1

11 =

5.1081240 +

4.301240

13.6 × 4

2n11 = 11.43 + 40.79

2n11 =

4.5422.52

2n1 = 1 –

4.5422.52

CAREER POINT



2n1 =

4.5418.2

n2 = 18.24.54

n2 = 25 n = 5 Q.30 A submarine (A) travelling at 18 km/hr is being chased along the line of its velocity by another submarine

(B) travelling at 27 km/hr. B sends a sonar signal of 500 Hz to detect A and receives a reflected sound of frequency . The value of is close to: (Speed of sound in water =1500 ms–1)

(1) 507 Hz (2) 504 Hz (3) 499 Hz (4) 502 Hz Ans. [4] Sol.

B

VB 7.5 m/s

f0 = 500Hz A

VA 5 m/s

v = speed of sound in water = 1500 m/s

frequency received by A = f' = 0B

A fVVVV

= 0f5.71500

51500

frequency received by B = f" = 'fVVVV

A

B

= f5.71500

51500515005.71500

f" =

5150051500

5.715005.71500 (500)

= 502 Hz

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JEE Main Online Exam 2019 - Career Pointcareerpoint.ac.in/studentparentzone/2019/jee-main/JEE...PHYSICS Q.1 A shell is fired from a fixed artillery gun with an initial speed u such