paper-1 (b.e./b.tech.) jee main 2019 · jee main 12 april 2019 shift-1 physics 1. find the distance...

PAPER-1 (B.E./B.TECH.)

JEE MAIN 2019 Computer Based Test

Solutions of Memory Based Questions

Date: 12th April 2019 (Shift-1)

Time: 09:30 A.M. to 12:30 P.M.

Durations: 3 Hours | Max. Marks: 360

Subject: Physics

www.embibe.com

JEE MAIN 12 APRIL 2019 SHIFT-1 PHYSICS

1. Find the distance of image of an object (O) placed on the surface of water from the surface w.r.t

observer as shown in figure.

(A) 6.7 cm

(B) 13.8 cm

(C) 5.0 cm

(D) 8.5 cm

Solution: (A)

U = −45 cm

f = −20 cm

From mirror formula 1−45

+ 1ν

= 1−20

1ν

=1

−20+

145

⇒ ν = −36 cm

distance from surface: 45 − 36 = 9 cm

distance from surface w.r.t observer in air d = 91.33

= 6.75 cm

2. What is the dimension of electrical resistance

(A) �μ0ε0

(B) �ε0μ0

(C) � μ0ε0

(D) None

Solution: (A)

[Resistance]=[Volt]

[Ampere] =[ML2T−3A−1]

[A] = [ML2T−3A−2]

Option (A)

μ0 = [MLT−2A−2] ε0 = M−1L−3T4A2


�μ0ε0

= �[μ0][ε0] = �

[MLT−2A−2][M−1L−3T4A2]�

1 2⁄

= [M2L4T−6A−4]1 2⁄ = ML2T−3A−2

Option (B)

�ε0μ0

= M−1L−3T3A2

Option (C)

�μ0ε0 =1c

= ML−1T

3. A square frame is moving out of a magnetic field of field strength 1T, as shown in figure. Find current

i in the figure.

(A) 25 × 10−4A

(B) 9.35 × 10−4A

(C) 18.75 × 10−4A

(D) 27 × 10−4A

Option (C)

EMF generated ε = Bvl (∵ B ⊥ v ⊥ l)

= (1T)(5 cm/s)(5 cm)

= 1 × 5 × 10−2 × 5 × 10−2

= 25 × 10−4 V

As given circuit forms whetstone bridge, current through 3 Ω is zero


Equivalent resistance = 4×24+2

= 43Ω

Current i, in the circuit i = εReq

= 25×10−4

(4 3⁄ )

=754

× 10−4A

= 18.75 × 10−4A

4. The spaces between two identical capacitors of plate area A and plate distance ‘d’ are filed with 3

dielectrics of dielectric constant k1, k2 , k3 as shown in figure. If they are connected to battery of EMF V

then find the ratio of energy stored in two capacitors

(A) 9 k1,k2,k3(k1+k2+k3) (k1k2+k2k3+k3k4)

(B) k1,k2,k39 (k1+k2+k3) (k1k2+k2k3+k3k1)

(C) (k1k2+k2k3+k3k1) (k1+k2+k3)k1,k2,k3

(D) (k1k2+k2k3+k3k1) (k1+k2+k3)9 k1,k2,k3

Solution: (A)

Energy stored in a capacitor E = 12

CV2


⇒E1E2

=C1C2

v

C1 = �d

3 ε0k1A+

d3ε0Ak2

+d

3ε0Ak3�−1

= �d

3Aε0�−1

�1k1

+1

k2+

1k3�−1

= �d

3Aε0�−1

�k1k2 + k2k3 + k3k1

k1k2k3�−1

=3Aε0

d�

k1k2k3k1k2 + k2k3 + k3k1

�

C2 =ε0 �

A3�k1d

+ε0 �

A3�k2d

+ε0 �

A3�k3d

=Aε03d

(k1 + k2 + k3)

∴C1C2

=E1E2

=9 k1k2k3

(k1 + k2 + k3)(k1k2 + k2k3 + k3k1)

5. A 2m long ladder is leaning against a wall. The bottom of ladder is pulled with speed 25 cm/sec away

from the wall. How fast is the top of the ladder moving down when it is 1 m above the ground?

(A) 25 cm/sec

(B) 25√3

cm/sec


(C) 25 √3 cm/sec

(D) 256 cm/sec

Solution: (C)

From constrain motion 25 cos 30° = V0 cos 60°

25�√32� = V0 �

12�

V0 = 25√3 cm/sec

6. In the given circuit the power dissipated by resistance R is 4 watt, then find value of R.

(A) 4 Ω

(B) 8 Ω

(C) 12 Ω

(D) 1 Ω

Solution: (A)

In given circuit Req = 4R

Power dissipated by resistance R = i2R


4 = �164R�2

R

4 =16R

R = 4 Ω

7. A projectile is thrown at an angle θ and with velocity v0 from the ground. If the equation of trajectory

is y = 2x − 9x2 find the value of θ.

(A) θ = sin−1 � 2√5�

(B) θ = tan−1 � 2√5�

(C) θ = sin−1 � 1√2�

(D) θ = cos−1 �12�

Solution: (A)

y = 2x − 9x2

Comparing above equation with general equation of trajectory of projectile 𝑦𝑦 = 𝑥𝑥 𝑡𝑡𝑡𝑡𝑡𝑡 𝜃𝜃 �1− �𝑥𝑥𝑅𝑅��

tan θ = 2

⇒ sin θ =2√5

θ = sin−1 �2√5�


8. For the given logic circuit which of the following truth table is correct?

(A)

A B C

0 0 1

0 1 1

1 0 0

1 1 0

(B)

A B C

0 0 1

0 1 0

1 0 1

1 1 1

(C)

A B C

0 0 1

0 1 0

1 0 1

1 1 0

(D) None of these

Solution: (A)

For the given logic circuit


C = (A + B). A��

= (A + B)�� + A�

= A�. B� + A�

A B C

0 0 1

0 1 1

1 0 0

1 1 0

9. An annular disc of inner radius ‘a’ and outer radius ‘b’ has surface mass density σ0r

. Where σ0 is a

positive constant and r is distance form center of annular disc. Find the radius of gyration of the annular

disc.

(A) �a2+b2+ab3

(B) a + b

(C) √a + b

(D) �a2+b2−ab3

Solution: (A)

dm = σ 2nrdr

=σ0r

2 nrdr

m = 2nσ0 � drb

a

= 2n σ0(b − a)


I = 2π σ0 � r2drb

a=

23πσ0(b3 − a3)

Radius of gyration = �I m⁄ = �23π σ0(b3−a3)

2π σ0(b−a)

= �b3 − a3

3(b − a)= �

(b − a)(b2 + a2 + ab)3 (b − a)

= �b2 + a2 + ab3

10. In YDSE, the wavelength of light used is λ. When a glass slab of thickness t and refractive index μ is

placed in front of one slit the central maxima shifts by one fringe width. Find the thickness of glass slab.

(A) λμ−1

(B) λDd

(C) (μ − 1)λ

(D) (μ − 1)λ Dd

Solution: (A)

Shift = (μ − 1) t Dd

Fringe width (β) = λDd

(μ − 1) tDd

= λDd


t =λ

μ − 1

11. A dipole p�⃗ = pi is placed at origin. Find electric filed and potential due to dipole at

(0, d, 0). �k = 14nε0

�

(A) kpd3

i, 0

(B) −kpd3

i, 0

(C) k(2p)d3

i, 0

(D) −k(2p)d3

i, 0

Solution: (B)

E��⃗ = −kp�⃗r3

= −kpd3ı̂

V = 0

12. A and of mass m and length l is being rotated about one of its ends. Which of the following graph is

correct representation of tension (T) with distance (X) from hinge point?

(A)

(B)


(C)

(D)

Solution: (D)

dm =ML

. dx

T = (dm)ω2x

= �MLω2x dx

L

X

=MLω2 �

x2

2�X

L

M2L

ω2[L2 − X2]

Downward opening parabola.

13. For the given cyclic process A → B → C, find the work done on gas. given QAB = 250 J, QBC = 60 J, and QCA = −180 J


(A) 130 J

(B) 490 J

(C) 10 J

(D) 100 J

Solution: (A)

In a cyclic process, ΔU = 0

From 1st law of thermodynamics

Q = ΔU + W

Q = W

250 + 60 − 180 = W

130 = W

W = 130 J

14. Two projectiles with time of flight t1 & t2 have same range R. Find t1t2

(A) R2g

(B) Rg

(C) 2Rg

(D) None

Solution: (C)

We know for complementary angles range is same for same projection speed.


t12usin(θ)

g ….. (i)

t22usin(90−θ)

g ….. (ii)

R = 2u2sincosθg

…..(iii)

From equation (i) and (ii)

t1t2 =4 U2sinθ cosθ

g2

(Using the value of R from equation iii)

t1t2 = 2Rg

15. The graph between stopping potential and frequency of photon is given in figure. Find the work

function of metal.

(× 1014 Hz)

(A) 0.83 eV

(B) 0.95 eV

(C) 0.72 eV

(D) 0.65 eV

Solution: (A)

Work function (ϕ) = (hϑ)e

(in eV)

=6.63 × 10−34 × (2 × 1014)

1.6 × 10−19


= 0.83 eV

16. A wave is propagating in +x direction. At time t = 0, the shape of wave is as shown in figure. If the

equation of wave is given as y = A sin(ωt− kx +ϕ),then find the value of ϕ.

(A) ϕ = π

(B) ϕ = π2

(C) ϕ = 0

(D) ϕ = −π2

Solution: (C)

At t = 0 & x = 0 the particle is at origin as shown in figure

y = A sinϕ = 0

⇒ sinϕ = 0 ..............(i)

For the velocity of particle at x = 0 & t = 0

v =dydt

= A ω cosϕ

From the graph, we can say the particle is going up

v = A ω cosϕ > 0 ...........(ii)

From eq (i) & (ii)

ϕ = 0

17. A man and his child are resting on a frictionless surface. The mass of man is 50 kg and the mass of

child is 20 kg. If man pushes his child with relative velocity 0.7 m s⁄ then find the velocity of man just

after he pushes his child.


(A) 0.2 m s⁄

(B) 0.5 m s⁄

(C) 1.0 m s⁄

(D) 2.0 m s⁄

Solution: (A)

Let the velocity of man be V m s⁄ from conservation of linear momentum.

0 = −20(0.7− V) + 50 V

= −14 + 70 V

V =1470

=15

V = 0.2 m s⁄

18. For a common emitter n − p− n transistor the graph between output currrent (IC) and input

current (IB) is given in figure. If the input resistance is 100 Ω and output resistance is 100 KΩ, then find

the value voltage and power gain.

(A) 2 × 104 , 4 × 105

(B) 4 × 104 , 9 × 105


(C) 104 , 105

(D) 8 × 104 , 3 × 105

Solution: (A)

Current gain (β) = ΔICΔIB

= 200−10010−5

= 20

Voltage gain(AV) = β �RoutRin

� = 20 × 100×103

100= 2 × 104

Power gain = AV × β = 20 × 2 × 104 = 4 × 105

19. A point charge ‘q’ is moving on a circular path of radius 10 cm with angular frequency 40π rad sec⁄ .

The magnetic field produced by it at the centre of circle is 3.8 × 10−10 T, then find the value of charge.

(A) 3μc

(B) 4 μc

(C) 5 μc

(D) 1 μc

Solution: (A)

i =qt

=q

2π ω⁄=

qω2π

B =μ0i2R

=μ0qω4πR

4π × 10−7 × q × 10π4π × (0.1)

= 3.8 × 10−10

q = (40π) = 3.8 × 10−4

q =3.840π

× 10−4

= 3 × 10−6C

20. Two moles of He gas and three moles of H2 gas are mixed at constant volume. Find the Cv of mixture.

(A) 2110

R


(B) 3110

R

(C) 32

R

(D) 52

R

Solution: (A)

(Cv)mix =n1C1 + n2C2

n1 + n2=

2 �32 R�+ 3 �5

2 R�

2 + 3

=3R + 15

2 R5

=2110

R

21. m1 gm of ice at −10oC and m2 gm of water at 50oC are mixed in a calorimeter. If after long time we get only water at 0oC then latent heat of ice is: (Given Sice = 0.5 cal gm⁄ −℃ and Swater = 1 cal gm⁄ −℃)

(A) 50m2m1

(B) 50m2m1

− 5

(C) 50m2m1

+ 5

(D) 50 m1m2

Solution: (B)

From the principle of calorimetry

m1 Sice�0− (−10)�+ m1Lf = m2Swater(50 − 0)

m1(0.5)(10) + m1Lf = m2(50)

5m1 + m1Lf = m2(50)

Lf = 50m2

m1− 5

22. He+ in nth state emits two successive photon of wavelength 103.7 nm and 30.7 nm to come to ground state. Find the value of n.

(A) 3

(B) 5

(C) 7


(D) 9

Solution: (B)

E1 =1240λ1

=1240103.7

≈ 11.96 eV

E2 =1240λ2

=124030.7

≈ 40.40 eV

ETotal = E1 + E1 = 52.35 eV

52.35 = 13.6(2)2 �1 −1

n2�

52.35 = 54.4 �1−1

n2�

0.96 = 1 −1

n2

1n2

= 0.04

n2 =1

0.04=

1004

= 25

n = 5

23. Aperture diameter of the lens of a telescope is 1.25 m. If the wavelength of light used is 5000 Å then find its resolving power.

(A) 2.1 × 106

(B) 2.5 × 106

(C) 3.1 × 106

(D) 4.2 × 106

Solution: (A)

Resolving power= d1.22λ

= 1.251.22×5000×10−10

= 2.05 × 106

24. A submarine A is going with speed 18 km h⁄ . Another submarine B is chasing A with speed of 27 km h⁄ . Submarine B sends frequency of 500 Hz and hears after reflection from A. Find the frequency of reflected wave as heard by submarine B. (Given speed of sound in water 1500 m s⁄ )

(A) 500 Hz


(B) 498 Hz

(C) 502 Hz

(D) 333.33 Hz

Solution: (C)

𝑉𝑉𝐴𝐴 = 18 𝑘𝑘𝑘𝑘 ℎ⁄ = 5 𝑘𝑘 𝑠𝑠⁄ ; 𝑉𝑉𝐵𝐵 = 27 𝑘𝑘𝑘𝑘 ℎ = 7.5𝑘𝑘 𝑠𝑠⁄⁄

Frequency received by A, 𝑓𝑓𝐴𝐴 = 𝑓𝑓0 �1500−51500−7.5

�

Frequency of reflected wave as heard by B.

𝑓𝑓′𝐵𝐵 = 𝑓𝑓𝐴𝐴 �1500 + 7.51500 + 5

�

= 𝑓𝑓0 �1500 − 5

1500 − 7.5� �

1500 + 7.51500 + 5

�

= 500 �1495

1492.5� �

1507.51505

�

= 501.67 Hz

paper-1 (b.e./b.tech.) jee main 2019 · jee main 12 april 2019 shift-1 physics 1. find the distance...

Documents