final jee–main examination – april,2019 · f = ´ ´w 1 2 1 1 253i 2 2 36 = w2 11 25 i 48 Þ e...

1

ALLEN

Final JEE -Main Exam April,2019/10-04-2019/Morning Session

1. Figure shows charge (q) versus voltage (V)graph for series and parallel combination of twogiven capacitors. The capacitances are :

A

B

q( C)m

500

8010V V(Volt)

(1) 50 mF and 30 mF (2) 20 mF and 30 mF

(3) 60 mF and 40 mF (4) 40 mF and 10 mF

Official Ans. by NTA (4)Sol. As q = CV

Hence slope of graph will give capacitance.Slope will be more in parallel combination.Hence capacitance in parallel should be 50mF& in series combination must be 8mF.

Only in option 40mF & 10 mF

Cparallel = 40 + 10 = 50mF

Cseries = ´+

40 1040 10

= 8 mF

2. A current of 5 A passes through a copperconductor (resistivity = 1.7 × 10–8 Wm) of radiusof cross-section 5 mm. Find the mobility of thecharges if their drift velocity is 1.1 × 10–3 m/s.

(1) 1.3 m2/Vs (2) 1.5 m2/Vs

(3) 1.8 m2/Vs (4) 1.0 m2/Vs

Official Ans. by NTA (4)

Sol. m = dV

EE = rJ

= -

--

´

´ ´p´ ´

3

86

1.1 105

1.7 1025 10

= - -

-

´ ´ p´ ´»

´ ´

3 62

8

1.1 10 25 101.01m / Vs

1.7 10 5

FINAL JEE–MAIN EXAMINATION – APRIL,2019(Held On Wednesday 10th APRIL, 2019) TIME : 9 : 30 AM To 12 : 30 PM

PHYSICS TEST PAPER WITH ANSWER & SOLUTION

3. In a meter bridge experiment, the circuitdiagram and the corresponding observationtable are shown in figure

Resistancebox

KE

l

GUnknownresistance

R X

SI. No. R(W) l(cm) 1. 2. 3. 4.

1000 100 10 1

60 13 1.5 1.0

Which of the readings is inconsistent?

(1) 4 (2) 1 (3) 2 (4) 3


Sol. as x = - l

l

R(100 )

for (1) x = ´

»1000 (100 – 60)

66740

for (2) x = ´ -

»100 (100 13)

66913

for (3) x = ´ -

»10 (100 1.5)

65698.5

for (4) x = ´ -

»1 (100 1)

951

So option (4) is completely different hencecorrect Ans. (4)

Page No: 1

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4. One plano-convex and one plano-concave lensof same radius of curvature 'R' but of differentmaterials are joined side by side as shown inthe figure. If the refractive index of the materialof 1 is m1 and that of 2 is m2, then the focallength of the combination is :

1

2m1

m2

(1) - m - m1 2

R2 ( ) (2) m - m1 2

2R

(3) m - m1 2

R2( ) (4) m - m1 2

R


Sol. For 1st lens m - m -æ öæ ö= - =ç ÷ç ÷ ¥ -è øè ø

1 1

1

1 11 1 1

f 1 R R

for 2nd lens m - m -æ öæ ö= - = -ç ÷ç ÷ -è øè ø

2 2

2

1 11 10

f 1 R R

= +eq 1 2

1 1 1f f f

= + Þ =m - - m - m - meq 1 2 eq 1 2

1 R R 1 Rf 1 ( 1) f

Hence feq = m - m1 2

R5. A ball is thrown upward with an initial velocity

V0 from the surface of the earth. The motionof the ball is affected by a drag force equal tomgu2 (where m is mass of the ball, u is itsinstantaneous velocity and g is a constant).Time taken by the ball to rise to its zenith is :

(1) - æ ögç ÷ç ÷g è ø

10

1sin V

gg

(2) - æ ögç ÷ç ÷g è ø

10

1tan V

gg

(3) - æ ögç ÷ç ÷g è ø

10

1 2tan V

g2 g

(4)æ ög

+ç ÷ç ÷g è ø0

1ln 1 V

gg


Sol. –(g + gv2) = dvdt

–gdt = 2

g dvg

v

æ öç ÷ç ÷

g ç ÷+ç ÷gè ø

Integrating 0 ® t & V0 ® 0 :-

–gt = –1 0Vg

tang

æ öç ÷ç ÷- ç ÷gç ÷ç ÷gè ø

t = 1

01

tan Vgg

- æ ögç ÷ç ÷g è ø

6. A cylinder with fixed capacity of 67.2 litcontains helium gas at STP. The amount of heatneeded to raise the temperature of the gas by20°C is : [Given that R = 8.31 J mol–1 K–1]

(1) 748 J (2) 374 J (3) 350 J (4) 700 J


Sol. DQ = nCV DT = n32

RDT

= æ öæ ö´ç ÷ç ÷è øè ø

67.2 38.31 (20)

22.4 2

» 748 J

7. A thin disc of mass M and radius R has massper unit area s(r) = kr2 where r is the distancefrom its centre. Its moment of inertia about anaxis going through its centre of mass andperpendicular to its plane is :

(1)2MR

6(2)

2MR3

(3)22MR

3(4)

2MR2


Page No: 2

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Sol. rdr

IDisc = Þ = s pò òR R

2 2Disc

0 0

(dm)r I ( 2 rdr)r

IDisc = pòR

2 2

0

(kr 2 rdr)r Mass of disc

IDisc = p òR

5

0

2 k r dr M = pòR

2

0

2 rdr kr

IDisc = æ ö

p ç ÷è ø

R6

0

r2 k

6 M = p òR

3

0

2 k r dr

IDisc = p6R

2 k6

M = pR4

0

r2 k

4

IDisc = æ öp p

= ç ÷è ø

6 4 2kR kR R 23 2 3 M = p

4R2 k

4

IDisc = 2M2R

3

IDisc = 22MR

3

8. Two coaxial discs, having moments of inertia

I1 and 1I

2, are rotating with respective angular

velocities w1 and w1

2, about their common axis.

They are brought in contact with each other andthereafter they rotate with a common angularvelocity. If Ef and Ei are the final and initial totalenergies, then (Ef – Ei) is :

(1)w2

1 1I12

(2) w21 1

3I

8

(3)w2

1 1I6

(4) w2

1 1I24


Sol. Ei = w

´ w + ´2

2 1 11 1

I1 1I

2 2 2 4

= w æ ö = wç ÷

è ø

221 1

1 1

I 9 9I

2 8 16

I1w1 + w

= w1 1 1I 3I

4 2

w = w11 1

3I5I

4 2

w = w156

Ef = ´ ´ w211

3I1 252 2 36

= w21 1

25I

48

Þ Ef – Ei = -æ öw - = wç ÷

è ø2 2

1 1 1 125 9 2

I I48 16 48

= - w2

1 1I24

9. A particle of mass m is moving along atrajectory given byx = x0 + a cosw1ty = y0 + b sinw2tThe torque, acting on the particle about theorigin, at t = 0 is :

(1) m (–x0b + y0a) w21 k̂

(2) +my0a w21 k̂

(3) – w - w2 20 2 0 1

ˆm(x b y a )k

(4) ZeroOfficial Ans. by NTA (2)

Sol. F = –m w w + w w2 21 1 2 2

ˆ ˆ(a cos ti b sin tj)

= + w + + wr

0 1 0 2ˆ ˆr (x acos t)i (y bsin t) j

= ´ = - + w w wr rr 2

0 1 2 2ˆT r F m(x a cos t)b sin t k

+ + w w w20 2 1 1

ˆm (y bsin t)a cos t k

= maw12y0 k̂

Page No: 3

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10. A proton, an electron, and a Helium nucleus,have the same energy. They are in circularorbits in a plane due to magnetic fieldperpendicualr to the plane. Let rp, re and rHe betheir respective radii, then,

(1) re > rp > rHe (2) re < rp < rHe

(3) re < rp = rHe (4) re > rp = rHe


Sol. r = =mv 2mKqB qB

rHe = rp > re

11. The electric field of a plane electromagneticwave is given by

= wr

0ˆE E i cos(kz)cos( t)

The corresponding magnetic field rB is then

given by :

(1) = wr

0E ˆB jsin(kz)cos( t)C

(2) = wr

0E ˆB jsin(kz)sin( t)C

(3) = wr

0E ˆB k sin(kz)cos( t)C

(4) = wr

0E ˆB jcos(kz)sin( t)C


Sol. ´r r r

QE B || v

Given that wave is propagating along positive

z-axis and rE along positive x-axis. Hence

rB

along y-axis.

From Maxwell equation

¶Ñ ´ = -

¶

r r BE

t

¶ ¶= -

¶E B

i.e.Z dt

and B0 = 0E

C

12. 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 :

d

xI1 I2

AC

B

(1)æ ö

= =ç ÷- +è ø1 2

1 2 1 2

I Ix d and x d

I I (I I )

(2) = ±-1

1 2

I dx

(I I )

(3)æ ö

= =ç ÷+ -è ø1 2

1 2 1 2

I Ix d and x d

I I (I I )

(4)æ ö æ ö

= =ç ÷ ç ÷+ -è ø è ø2 2

1 2 1 2

I Ix d and x d

I I I I

Official Ans. by NTA (2)Sol.

xI1 I2

d – xI

Net force on wire carrying current I per unit

length is

m m+ =

p p -0 1 0 2I I I I

02 x 2 (d x)

=-

1 2I I

x x d

Þ x = -1

1 2

I dI I

Page No: 4

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13. A message signal of frequency 100 MHz andpeak voltage 100 V is used to executeamplitude modulation on a carrier wave offrequency 300 GHz and peak voltage 400 V.The modulation index and difference betweenthe two side band frequencies are :(1) 4; 1 × 108 Hz (2) 0.25; 1 × 108 Hz(3) 4; 2 × 108 Hz (4) 0.25; 2 × 108 HzOfficial Ans. by NTA (4)

Sol. fm = 100 MHz = 108 Hz, (Vm)0 = 100Vfc = 300 GHz , (Vc)0 = 400V

Modulaton Index = = = =m 0

c 0

(V ) 100 10.25

(V ) 400 4Upper band frequency (UBF) = fc + fmLower band frequency (LBF) = fc – fm\ UBF – LBF = 2fm = 2 × 108 Hz

14. In an experiment, the resistance of a materialis plotted as a function of temperature (in somerange). As shown in the figure, it is a straightline. One may conclude that :

lnR(T)

1/T2

(1) R(T) = 02

R

T

(2) -=2 2

0T /T0R(T) R e

(3) =2 20–T /T

0R(T) R e

(4) =2 2

0T /T0R(T) R e


Sol. + =l

l

2

020

1n(T)T 1

1 nR(T )T

lnR(T )0

lnR(T)

1/T21

1 T2

T20

Þæ ö

= -ç ÷ç ÷è ø

l l

20

0 2

TnR(T) [ nR(T )] 1

T

Þ R(T) = æ öç ÷-ç ÷è ø

202

T

T0R e

15. A ray of light AO in vacuum is incident on aglass slab at angle 60° and refracted at angle 30°along OB as shown in the figure. The optical pathlength of light ray from A to B is :

30°

60°a

b

A

O Vacuumglass

B

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

2a3

(3) +2 3

2ba

(4) +2b

2a3

Official Ans. by NTA (1)Sol. From Snell's law

1.sin60° = m sin30°

30°

60°

A

O Vacuumglass

B

Þ m = 3Optical path = AO + m(OB)

= +° °

a b3

cos60 cos30 = 2a + 2b

16. A transformer consisting of 300 turns in theprimary and 150 turns in the secondary givesoutput power of 2.2 kW. If the current in thesecondary coil is 10A, then the input voltageand current in the primary coil are :(1) 220 V and 10A(2) 440 V and 5A(3) 440 V and 20 A(4) 220 V and 20 AOfficial Ans. by NTA (2)

Sol. Given NP = 300, Ns = 150, P0 = 2200WIs = 10AP0 = V0I0 Þ 2200 = V0 × 10 Þ V0 = 220V

= Þ = ´ =Qpi

i0 s

NVV 2 220 440V

V N

Also P0 = ViIi

Þ Ii = =2200

5A440

Page No: 5

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17. In a photoelectric effect experiment thethreshold wavelength of the light is 380 nm. Ifthe wavelentgh of incident light is 260 nm, themaximum kinetic energy of emitted electronswill be:

Given E (in eV) = l1237

(in nm)

(1) 1.5 eV (2) 4.5 eV(3) 15.1 eV (4) 3.0 eVOfficial Ans. by NTA (1)

Sol. Kmax = -l l0

hc hc

Þ Kmax = æ öl - lç ÷llè ø

0

0

hc

Þ Kmax = (1237)-æ ö

ç ÷´è ø

380 260380 260

= 1.5 eV18. The displacement of a damped harmonic

oscillator is given byx(t ) = e–01.1t cos (10pt + f). Here t is in seconds.The time taken for its amplitude of vibration todrop to half of its initial value is close to :(1) 13 s (2) 7 s (3) 27 s (4) 4 sOfficial Ans. by NTA (2)

Sol. A = A0e–0.1t = 0A

2

ln2 = 0.1tt = 10 ln2 = 6.93 » 7 sec

19. A moving coil galvanometer allows a full scalecurrent of 10–4A. A series resistance of 2 MWis required to convert the above galvanometerinto a voltmeter of range 0-5 V. Therefore thevalue of shunt resistance required to convert theabove galvanometer into an ammeter of range0-10 mA is :(1) 200 W (2) 100 W(3) 10 W (4) 500 WOfficial Ans. by NTA (Dropped)

Sol. 200 + 10–4 G = 5G = –veSo answer is Bonus

20. 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) 16, 14 (2) 12, 18

(3) 12, 16 (4) 8, 18


Sol. f = 480 = -

´1300 v500

300

= - 11440

300 v5

v1 = =60

12m / s5

530 = +

´2300 v500

300

1590 = 1500 + 5v2

5v2 = 90

v2 = 18 m/s

21. Two radioactive materials A and B have decay

constants 10l and l, respectively. It 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)l

1110

(2) l1

9

(3)l

110

(4) l

111

Official Ans. by NTA (2)Sol. N1 = N0e–10lt

N2 = N0e–lt

- l= = 9 t1

2

N1e

e N

Þ 9lt = 1

t = l1

9

Page No: 6

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22. An npn transistor operates as a common emitteramplifier, with a power gain of 60 dB. The inputcircuit resistance is 100W and the output loadresistance is 10 kW. The common emittercurrent gain b is :(1) 60 (2) 104

(3) 6 × 102 (4) 102

Official Ans. by NTA (4)Sol. Av × b = Pgain

60 = 10æ öç ÷è ø

100

Plog

P

P = 106 = b ´2 out

in

R

R

= b ´4

2 10100

b2 = 104

b = 10023. In the given circuit, an ideal voltmeter

connected across the 10W resistance reads 2V.The internal resistance r, of each cell is :

1.5V, 1.5VrW rW

10W

15W2W

(1) 1W (2) 1.5W(3) 0W (4) 0.5WOfficial Ans. by NTA (4)

Sol. Req = ´

+ +15 10

2 2r25

= 8 + 2r

i = +3

8 2r

2 = i Req = ´+3

68 2r

16 + 4r = 18Þ r = 0.5 W

24. A 25 × 10–3 m3 volume cylinder is filled with1 mol of O2 gas at room temperature (300K).The molecular diameter of O2, and its rootmean square speed, are found to be 0.3 nm, and200 m/s, respectively. What is the averagecollision rate (per second) for an O2 molecule ?(1) ~1011 (2) ~1013

(3) ~1010 (4) ~1012

Official Ans. by NTA (4)Allen Ans. (3)

Sol. n = lavV

l = ps2A

RT

2 N P

s = 2 × .3 × 10–9

P = RTV

Þ l = ps2A

V

2 N

Vav = ´p rms8

V3

\ n = -

´ p´ s´

p´

2A

3

200 2 N 8325 10

= 17.68 × 108/sec. = .1768 × 1010/sec. ~ 1010

This answer does not match with JEE-Answer key25. n moles of an ideal gas with constant volume

heat capcity CV undergo an isobaric expansionby certain volume. The ratio of the work donein the process, to the heat supplied is :

(1) -V

4nRC nR (2) -V

nRC nR

(3) +V

nRC nR (4) +V

4nRC nR

Official Ans. by NTA (3)Sol. w = nRDT

DH = (Cv + nR) DT

=D +v

w nRH C nR

Page No: 7

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26. A uniformly charged ring of radius 3a and totalcharge q is placed in xy-plane centred at origin.A point charge q is moving towards the ringalong the z-axis and has speed u at z = 4a. Theminimum value of u such that it crosses theorigin is :

(1)æ öç ÷peè ø

1/22

0

2 1 qm 15 4 a


1/22

0

2 2 qm 15 4 a


1/22

0

2 4 qm 15 4 a

(4) æ öç ÷peè ø

1/22

0

2 1 qm 5 4 a

Official Ans. by NTA (2)Sol. Ui + Ki = Uf + Kf

+ =+

2 22

2 2

kq 1 kqmv

2 3a16a 9a

æ ö= - =ç ÷è ø

2 221 kq 1 1 2kq

mv2 a 3 5 15a

v = 24kq

15ma

27. The value of acceleration due to gravity atEarth's surface is 9.8 ms–2. The altitude aboveits surface at which the acceleration due togravity decreases to 4.9 ms–2, is close to :(Radius of earth = 6.4 × 106 m)(1) 1.6 × 106 m (2) 6.4 × 106 m(3) 9.0 × 106 m (4) 2.6 × 106 mOfficial Ans. by NTA (4)

Sol. =+ 2 2

GM GM

(R h) 2R

R + h = 2R

h = -( 2 1)R

´;62.6 10 m

28. Given below in the the left column are different

modes of communication using the kinds of

waves given the right column.

A. Optical Fibre communication

P. Ultrasound

B. Radar Q. Infrared Light C. Sonar R. Microwaves D. Mobile Phones S. Radio Waves (1) A-S, B-Q, C-R, D-P

(2) A-R, B-P, C-S, D-Q

(3) A-Q, B-S, C-R, D-P

(4) A-Q, B-S, C-P, D-R


Sol. Conceptual

29. The radtio of surface tensions of mercury and

water is given to be 7.5 while the ratio of thier

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) 3/5

(3) 2/5 (4) 4/5


Sol. h = q

r1 1

1 1

2S cos

r g

h = q

r2 2

2 2

2S cosr g

Þ =1

2

r 2r 5

Page No: 8

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30. 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 thecollision, they move along the indicateddirections with speeds u1 and u2, respectively.The values of u1 and u2 are nearly :

30° 30°45°45°

5m/s

2M M

u2

u110m/s

M 2M

(1) 3.2 m/s and 6.3 m/s

(2) 3.2 m/s and 12.6 m/s

(3) 6.5 m/s and 6.3 m/s

(4) 6.5 m/s and 3.2 m/s


Sol. M × 10 cos 30° + 2M × 5 cos 45°

= 2M × v1 cos 30° + M v2 cos 45°

+ = + 21

v35 3 5 2 2v

2 2

10 × M sin 30° – 2M × 5 sin 45°

= M v2 sin 45° – 2M v1 sin 30°

= -21

v5 – 5 2 v

2

Solving v1 = ;17.5

6.5m / s2.7

v2 » 6.3 m/s

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final jee–main examination – april,2019 · f = ´ ´w 1 2 1 1 253i 2 2 36 = w2 11 25 i 48 Þ e...

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