FINAL JEE–MAIN EXAMINATION – APRIL,2019...2019/12/04 · 1 m 2 90–C C sin c = 1 2 m m m 1 sin q = m 2 sin (90º – C) sin q = 2 1 2 2 1 1 m m-m m q = sin–1 22 21 2 1 m -m

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Final JEE -Main Exam April,2019/12-04-2019/Evening Session

1. The electron in a hydrogen atom first jumpsfrom the third excited state to the secondexcited state and subsequently to the firstexcited state. The ratio of the respectivewavelengths, l1/l2, of the photons emitted inthis process is :(1) 9/7 (2) 7/5(3) 27/5 (4) 20/7

Official Ans. by NTA (4)

Sol. 2 2f i

1 1 1R

n n

æ ö= -ç ÷l è ø

2 21

1 1 1R

3 4æ ö= -ç ÷l è ø

1

1 7R

9 16æ ö= ç ÷l ´è ø

2 22

1 1 1R

2 3æ ö= -ç ÷l è ø

= R5

4 9æ öç ÷´è ø

1

2

5367

9 16

l=

l´

= 207

2. One kg of water, at 20ºC, is heated in an electrickettle whose heating element has a mean(temperature averaged) resistance of 20 W. Therms voltage in the mains is 200 V. Ignoring heatloss from the kettle, time taken for water toevaporate fully, is close to :[Specific heat of water = 4200 J/kg ºC),Latent heat of water = 2260 kJ/kg](1) 3 minutes (2) 22 minutes(3) 10 minutes (4) 16 minutes

FINAL JEE–MAIN EXAMINATION – APRIL,2019(Held On Friday 12th APRIL, 2019) TIME : 2 : 30 PM To 5 : 30 PM

PHYSICS TEST PAPER WITH ANSWER & SOLUTION


Sol. Q = P × t

Q = mcDT + mL

P = 2rmsVR

4200 × 80 + 2260 × 103 = 2(200)

t20

´

t = 1298 sec

t ; 22 min3. A tuning fork of frequency 480 Hz is used in

an experiment for measuring speed of sound(n) in air by resonance tube method. Resonanceis observed to occur at two successive lengthsof the air column, l1 = 30 cm and l2 = 70 cm.Then n is equal to :(1) 332 ms–1 (2) 379 ms–1

(3) 384 ms–1 (4) 338 ms–1


Sol. n = 2f (l2 – l1)

n = 2 × 480 × (70 – 30) × 10–2

n = 960 × 40 × 10–2

n = 38400 × 10–2 m/s

n = 384 m/s4. Two sources of sound S1 and S2 produce sound

waves of same frequency 660 Hz. A listener ismoving from source S1 towards S2 with aconstant speed u m/s and he hears 10 beats/s.The velocity of sound is 330 m/s. Then, uequals:(1) 2.5 m/s (2) 15.0 m/s

(3) 5.5 m/s (4) 10.0 m/s


Sol. f = 660 Hz, n = 330 m/s

n nu S2S1

f1 = fun -æ ö

ç ÷nè ø

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f2 = fun +æ ö

ç ÷nè ø

f2 – f1 = fn

[n + u – (n – u)]

10 = f2 – f1 = fn

[2u]

u = 2.5 m/s

5. A particle is moving with speed b xn = along

positive x-axis. Calculate the speed of theparticle at time t = t(assume that the particle isat origin at t = 0).

(1) 2b4

t(2)

2b2

t

(3) b2 t (4) 2b

2

t


Sol. n = b x

d b dxdt dt2 x

n=

a =b

2 x

n

a = ( )b b x

2 x

2d ba

dt 2n

= =

n = 2b

2t

6. Consider an electron in a hydrogen atom,revolving in its second excited state (havingradius 4.65Å). The de-Broglie wavelength ofthis electron is :

(1) 12.9 Å (2) 3.5 Å

(3) 9.7 Å (4) 6.6 Å


Sol. 2prn = nln

l3 = 102 (4.65 10 )

3

-p ´

l3 = 9.7 Å7. A moving coil galvanometer, having a

resistance G, produces full scale deflectionwhen a current Ig flows through it. Thisgalvanometer can be converted into (i) anammeter of range 0 to I0 (I0 > Ig) by connectinga shunt resistance RA to it and (ii) into avoltmeter of range 0 to V(V = GI0) byconnecting a series resistance RV to it. Then,

(1) g2A V

0 g

IR R G

I I

æ ö= ç ÷ç ÷-è ø

and

2

0 gA

V g

I IRR I

æ ö-= ç ÷ç ÷

è ø

(2) RARV = G2 and

2

gA

V 0 g

IRR I I


(3) RARV = G2 and gA

V 0 g

IRR (I I )

=-

(4) 0 g2

A Vg

I IR R G

I

æ ö-= ç ÷ç ÷

è ø and

2

gA

V 0 g

IRR I I



Sol. When galvanometer is used as an ammeter

shunt is used in parallel with galvanometer.

\

G

I –Ig0

RA

Ig

I0

\ IgG = (I0 – Ig)RA

\ RA = g

0 g

IG

I I

æ öç ÷ç ÷-è ø

When galvanometer is used as a voltmeter,resistance is used in series with galvanometer.

GIg RV

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

\ RV = 0 g

g

(I I )G

I

-

\ RARV = G2 &

2

gA

V 0 g

IRR I I


8. The number density of molecules of a gasdepends on their distance r from the origin as,

4r0n(r) n e-a= . Then the total number of

molecules is proportional to :

(1) n0a1/4 (2) n0a–3

(3) n0a–3/4 (4) 1/ 20n a

Official Ans. by NTA (3)Sol. Given number density of molecules of gas as

a function of r is

4r0n(r) n e-a=

\ Total number of molecule = 0

n(r)dV¥

ò

4r 20

0

n e 4 r dr¥

-a= pò

\ Number of molecules is proportional to

n0a–3/4

9. A smooth wire of length 2pr is bent into a circle

and kept in a vertical plane. A bead can slidesmoothly on the wire. When the circle is

rotating with angular speed w 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 w2 is

equal to :

(1) ( )g 3 / r

(2) 3 g2r

w

B

r/2

O

rA

P(3) 2g/r

(4) ( )2g / r 3


Sol.

N

N cos q

N sin q

q

N sin q = mr2

w2 ......(1)

N cos q = mg .......(2)

tan q = 2r

2gw

2r r2g3 r

22

w=

w2 = 2g

3 r

10. Find the magnetic field at point P due to a

straight line segment AB of length 6 cm

carrying a current of 5 A. (See figure)

(m0 = 4p × 10–7 N-A–2)

A B6 cm

5 cm

P

5 cm

(1) 3.0 × 10–5 T (2) 2.5 × 10–5 T

(3) 2.0 × 10–5 T (4) 1.5 × 10–5 T


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Sol.q q 5

3 3

B = 0I4 dmp

2 sin q

d = 4 cm

sin q = 35

11. A transparent cube of side d, made of a material

of refractive index m2, is immersed in a liquid

of refractive index m1(m1 < m2). A ray is incident

on the face AB at an angle q(shown in the

figure). Total internal reflection takes place at

point E on the face BC.

B

A

C

D

q m2

m1

E

The q must satisfy :

(1) 1 1

2

sin- mq <

m (2) 2

1 221

sin 1- mq < -

m

(3) 1 1

2

sin- mq >

m (4) 2

1 221

sin 1- mq > -

m


Sol. q

m1m2

90–C

C

sin c = 1

2

mm

m1 sin q = m2 sin (90º – C)

sin q =

21

2 22

1

1m

m -m

m

q = sin–1 2 22 1

21

m - mm

For TIR

q < sin–1 2221

1m

-m

12. Let a total charge 2Q be distributed in a sphere

of radius R, with the charge density given by

r(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) ¼

3Ra

2= (2) a R / 3=

(3) a = 8–1/4R (4) a = 2–1/4 R


Sol. E 4pa2 =

a 2

0

0

kr 4 r drp

eò

E = 4

0

k 4 a4 4

p´ pe

2Q = R 2

0kr 4 r drpò

k = 4

2Q

Rp

QE = 0

14pe 2

QQ(2a)

R = a81/4

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13. 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 = 2 h1h2 (2) R2 = 16h1h2

(3) R2 = 4 h1h2 (4) R2 = h1h2


Sol.q 90–q

u u

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

R = 2u 2sin cos

gq q

h1 = 2 2u sin

gq

h2 = 2 2u sin (90 )

g- q

2

1 2

R16

h h=

14. A spring whose unstretched length is l has aforce constant k. The spring is cut into twopieces of unstretched lengths l1 and l2 where,l1 = nl2 and n is an integer. The ratio k1/k2 ofthe corresponding force constants, k1 and k2

will be :

(1) 2

1n

(2) n2 (3) 1n

(4) n


Sol. k1 = 1

Cl

k2 = 2

Cl

1 2

2 1

k Ck C

= l

ll2 = 2

2

1n n

=l

l

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, asshown in the figures. The coefficient of frictionbetween the block and floor is m = 0.2. Thedifference between the accelerations of theblock, in case (B) and case (A) will be :(g = 10 ms–2)

F = 20 N

F = 20 N

30º30º

(A) (B)

(1) 0 ms–2 (2) 0.8 ms–2

(3) 0.4 ms–2 (4) 3.2 ms–2


Sol.30º

N1

5010

20

fk1

10 330º

N2

50

20

fk2

10 3

10

N1 = 60 N2 = 40

a1 = 10 3 0.2 60

5- ´

a2 = 10 3 0.2 40

5- ´

a1 – a2 = 0.816. Consider the LR circuit shown in the figure. If

the switch S is closed at t = 0 then the amountof charge that passes through the battery

between t = 0 and t = LR

is :

E S

RL

i

(1) 2

EL7.3R (2) 2

EL2.7R

(3) 2

7.3ELR

(4) 2

2.7ELR


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Sol. q I dt= ò

RtL / RL

0

Eq 1 e dt

R

-é ù= -ê ú

ê úë ûò

2

EL 1q

eR=

2

ELq

2.7R=

17. 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) 16.00 (2) 1.80

(3) 5.33 (4) 10.67


Sol. Since unpolarised light falls on P1 Þ intensity

of light transmitted from P1 = 0I2

Pass axis of P2 will be at an angle of 30º withP1

\ Intensity of light transmitted from

P2 = 0I2

cos2 30º = 03I8

Pass axis of P3 is at an angle of 60º with P2

\ Intensity of light transmitted from

P3 = 03I8

cos2 60º = 03I32

\ 0I 3210.67

I 3æ ö = =ç ÷è ø

18. Half lives of two radioactive nuclei A and B are10 minutes and 20 minutes, respectively. If,initially a sample has equal number of nuclei,then after 60 minutes, the ratio of decayednumbers of nuclei A and B will be :(1) 9 : 8 (2) 1 : 8(3) 8 : 1 (4) 3 : 8Official Ans. by NTA (1)

Sol.1/2

t OA OAA OA t / t 6

N NN N e

22-l= = =

\ Number of nuclei decayed

= OA OAOA 6

N 63NN

642- =

1/2

t OB OBB OB t / t 3

N NN N e

22-l= = =

\ Number of nuclei decayed

= OB OBOB 3

N 7NN

82- =

Since NOA = NOB

\ Ratio of decayed numbers of nuclei

A & B = OA

OB

63N 8 964 7N 8

´=

´

19. A solid sphere, of radius R acquires a terminalvelocity n1 when falling (due to gravity) througha viscous fluid having a coefficient of viscosityh. The sphere is broken into 27 identical solidspheres. If each of these spheres acquires aterminal velocity, n2, when falling through thesame fluid, the ratio (n1/n2) equals :(1) 1/27 (2) 1/9(3) 27 (4) 9Official Ans. by NTA (4)

Sol. We have

VT = 29

2

0r

( )gr -rh l Þ vT µ r2

since mass of the sphere will be same

\ r43

pR3 = 27.43

pr3r Þ r = R3

\2

12

2

v R9

v r= =

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20. The ratio of the weights of a body on the Earth'ssurface to that on the surface of a planet is

9 : 4. The mass of the planet is 19

th of that of

the Earth. If 'R' is the radius of the Earth, whatis the radius of the planet ? (Take the planetsto have the same mass density)

(1) R3

(2) R2

(3) R4

(4) R9

Official Ans. by NTA (2)Sol. Since mass of the object remains same

\ Weight of object will be proportional to 'g'(acceleration due to gravity)Given

earth earth

planet planet

W 9 gW 4 g

= =

Also, gsurface = 2

GMR

(M is mass planet, G is

universal gravitational constant, R is radius ofplanet)

\2

earth planet2

planet earth

GM R94 GM R

= = 2 2

earth Planet Planet2 2

planet earth earth

M R R9

M R R´ =

\ Rplanet = earthR R2 2

=

21. Figure shown a DC voltage regulator circuit,with a Zener diode of breakdown voltage = 6V.If the unregulated input voltage varies between10 V to 16 V, then what is the maximum Zenercurrent ?

R = 4 kL W

IL

IZ

IS

R = 2 kS W

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


Sol. Maximum current will flow from zener if input

voltage is maximum.

R = 4 kL W

IL

IZ

IS

R = 2 kS W

16V

When zener diode works in breakdown state,voltage across the zener will remain same.

\ Vacross 4kW = 6V

\ Current through 4KW = 6 6

A mA4000 4

=

Since input voltage = 16V

\ Potential difference across 2KW = 10V

\ Current through 2kW = 10

5mA2000

=

\ Current through zener diode

= (IS – IL) = 3.5 mA

22. 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) 124ºC, 62ºC (2) 37ºC, 99ºC

(3) 62ºC, 124ºC (4) 99ºC, 37ºC


Sol. Efficiency of Carnot engine = 1 – sink

source

TT

Given,

sink

source

1 T1

6 T= - Þ

sink

source

T 5T 6

= .....(1)

Also,

sink

source

2 T 621

6 T-

= - Þ source

62 1T 6

= .....(2)

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\ Tsource = 372 K = 99ºC

Also, Tsink = 56

× 372 = 310 K = 37ºC

(Note :- Temperature of source is more thantemperature of sink)

23. 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


Sol. For a diatomic gas, Cp = 72

R

Since gas undergoes isobaric process

Þ DQ = nCPDT

Also, DW = nRDT = 10J(given)

\ DQ = n72

RDT = 72

(nRDT) = 35 J

24. 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 10–12W/m2]

(1) 10 cm (2) 30 cm

(3) 40 cm (4) 20 cm


Sol. Loudness of sound is given by

dB = 10 log 0

II

0

I is intensity of soundI is reference intensity of sound

æ öç ÷è ø

\ 120 = 10 log0

II

æ öç ÷è ø

Þ I = 1 W/m2

Also I = 2 2

P 2

4 r 4 r=

p p

\ r = 2 1

4 2=

p pm = 0.399 m » 40 cm

25. 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) equals to :

(1) F/(3pr2YT) (2) 3F/(pr2YT)

(3) 6F/(pr2YT) (4) 9F/(pr2YT)


Sol.

L

r

Q Length of cylinder remains unchanged

so Compressive Thermal

F FA A

æ ö æ ö=ç ÷ ç ÷è ø è ø

2

FY T

r= a

p

( is linear coefficient of expansion)a

\ a = 2

F

YT rp

\The coefficient of volume expansion g = 3a

\ g = 3 2

F

Y T rp

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26. 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 :

Y

m = 150 g3

m = 100 g2

1.0m0.5m60º

050 g = m1

(1) 7 3

m, m12 8

æ öç ÷ç ÷è ø

(2) 3 5

m, m4 12


(3) 7 3

m, m12 4


(4) 3 7

m, m8 12



Sol.

Y

150 g (1/2, /2)

m = 100 g2

1.0m0.5m60º

050 g = m1

3

The co-ordinates of the centre of mass

cm

1 3ˆ ˆ ˆ0 150 i j 100 i2 2

r300

æ ö+ ´ + + ´ç ÷ç ÷

è ø=r

cm7 3ˆ ˆr i j

12 4= +

r

\ Co-ordinate 7 3

,12 4


m

27. An electron, moving along the x-axis with

an initial energy of 100 eV, enters a region

of magnetic field 3 ˆB (1.5 10 T)k-= ´r

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 × 10–19 C, mass of

electron = 9.1 × 10–31 kg)

2cm

8 cm

P

Q

d

S

(1) 12.87 cm (2) 1.22 cm

(3) 11.65 cm (4) 2.25 cm


Sol. R = mvqB

= 2m (K.E.)

qB

R = 31 19

19 3

2 9.1 10 (100 1.6 10 )1.6 10 1.5 10

- -

- -

´ ´ ´ ´ ´´ ´ ´

R = 2.248 cm

2cm

8 cm

P

Q

Sq q U

T

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sin q = 2

2.248

tan q = QUTU

21.026

= QU6

QU = 11.69PU = R(1 – cos q)

= 1.22d = QU + PU

28. A plane electromagnetic wave having afrequency n = 23.9 GHz propagates along thepositive z-direction in free space. The peakvalue of the electric field is 60 V/m. Whichamong the following is the acceptablemagnetic field component in theelectromagnetic wave ?

(1) 7 3 11 ˆB 2 10 sin(0.5 10 z 1.5 10 t)i= ´ ´ + ´r

(2) 7 2 11 ˆB 2 10 sin(1.5 10 x 0.5 10 t) j-= ´ ´ + ´r

(3) 7 3 11 ˆB 2 10 sin(0.5 10 z 1.5 10 t)i-= ´ ´ - ´r

(4) 3 11 ˆB 60sin (0.5 10 x 1.5 10 t)k= ´ + ´r

Official Ans. by NTA (3)Sol. Magnetic field when electromagnetic wave

propagates in +z directionB = B0 sin (kz – wt)where

B0 = 8

60

3 10´ = 2 × 10–7

k = 320.5 10

p= ´

l

w = 2pf = 1.5 × 1011

29. In an amplitude modulator circuit, the carrierwave is given by,C(t) = 4 sin (20000 pt) while modulating signalis given by, m(t) = 2 sin (200 pt). The valuesof modulation index and lower side bandfrequency are :(1) 0.5 and 9 kHz (2) 0.5 and 10 kHz(3) 0.3 and 9 kHz (4) 0.4 and 10 kHzOfficial Ans. by NTA (1)

Sol. Modulation index is given by

m

c

A 2m 0.5

A 4= = =

& (a) carrier wave frequency is given by

= 2pfc = 2 × 104 pfc = 10 kHz

(b) modulating wave frequency (fm)

2pfm = 2000 pÞ fm = 1 kHz

lower side band frequency Þ fc – fm

Þ 10 kHz – 1 kHz = 9 kHz

30. In the given circuit, the charge on 4 mFcapacitor will be :

1 Fm4 Fm

3 Fm

5 Fm

10 V

(1) 5.4 mC (2) 24 mC

(3) 13.4 mC (4) 9.6 mC


Sol.

6 Fm4 Fm

3 Fm

10 V

Þ

2.4 Fm

3 Fm

10 V

q1

q2

q

So total charge flow = q = 5.4 mF × 10V

= 54 mC

The charge will be distributed in the ratio of

capacitance

Þ1

2

q 2.4 4q 3 5

= =

\ 9X = 54 mC Þ X = 6 mC\ charge on 4 mF capacitor

will be = 4X = 4 × 6 mC

= 24 mC

Page No: 10

Documents

FINAL JEE–MAIN EXAMINATION – APRIL,2019...2019/12/04 · 1 m 2 90–C C sin c = 1 2 m m m 1 sin q = m 2 sin (90º – C) sin q = 2 1 2 2 1 1 m m-m m q = sin–1 22 21 2 1 m -m