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# Top-15 RRB NTPC Physics Problems PDF

Download Top-15 Expected Physics Problems for RRB NTPC Stage-1 exam. Go through the video of Repeatedly asked and most important RRB NTPC physics problems questions. These questions are based on previous year questions in Railways and other Govt exams.

Practice:

Practice 4500+ Solved Questions for RRB NTPC ## Read this Post in Telugu

Question 1: If a person moves a trolley for a distance of 10 m with a force of 50 N, then the work done is:

a) 0.2 J

b) 5 J

c) 20 J

d) 500 J

Question 2: A mass of 20 kg is at a height of 8 m above the ground. Then the potential energy possessed by the body is: [Given g = 9.8$ms^{-2}$]

a) 1568 J

b) 1568 C

c) 1568 W

d) 1568 N

Question 3: The focal length of a lens is 50 centimetre. Its power is:

a) 50 dioptre

b) 1 dioptre

c) 10 dioptre

d) 2 dioptre

Question 4: The potential energy (P.E.) of a body at a certain height is 200 J. The kinetic energy possessed by it when it just touches the surface of the earth is:

a) zero

b) = P.E.

c) <P.E.

d) >P.E.

Question 5: A sound wave has a frequency of 3.5 kHz and wave length 0.1 m. How long will it take to travel 700 m?

a) 3.0 s

b) 1.5 s

c) 2.0 s

d) 1 s

Question 6: A ball thrown up vertically returns to the ground after 10 second. Find the velocity with which it was thrown up? (if g = 10 m/s2).

a) 120 m/s

b) 50 m/s

c) 600 m/s

d) 60 m/s

Question 7: An object is placed 30 cm before a concave mirror of focal length of 20 cm to get a real
image. What will be the distance of the image from the mirror?

a) 60 cm

b) 20 cm

c) 30 cm

d) 40 cm

Question 8: A certain house hold has consumed 320 units of energy during a month. How much energy is this in joules?

a) $9 \times 10^{8} J$

b) $5 \times 10^{8} J$

c) $10 \times 10^{5} J$

d) $1152 \times 10^{6} J$

Question 9: Calculate the work done by the force of gravity when satellite moves in an orbit of
radius 40,000 km around the earth.

a) 8,000 J

b) 4,00,000 J

c) 0 J

d) 4,000 J

Question 10: An object of 1.2 cm height is placed 30 cm before a concave mirror of focal length of 20 cm to get a real image at a distance of 60 cm from the mirror. What is the height of the image formed?

a) -3.6 cm

b) -2.4 cm

c) 1.2 cm

d) 2.4 cm

Question 11: The radius of curvature of a concave mirror is 30 cm. Following Cartesian Sign Convention, its focal length is expressed as:

a) -30 cm

b) -15 cm

c) +30 cm

d) +15 cm

Question 12: Calculate the current flowing through a resistor of 10 ohms when potential difference of 140V is applied across it.

a) 14 Amperes

b) 140 Amperes

c) 1400 Amperes

d) 1.4 Amperes

Question 13: A transformer has 1000 primary turns. It is connected to 250 volts A.C. supply. Find the number of secondary turns to get secondary voltage of 400 volts.

a) 1600

b) 625

c) 100

d) 1250

Question 14: The time period of a vibrating body is 0.04s, then the frequency of the wave is……

a) 25HZ

b) 20Hz

c) 250Hz

d) 200Hz

Question 15: The power of a lens is —2.5 D. The type of lens and its focal length are respectively:

a) Convex, -0.40 m

b) Concave, -0.40 m

c) Concave, 0.40 m

d) Convex, 0.40 m

The velocity of the soundwave is wavelength*frequency = 3500*0.1 = 350 m/s

Hence, the amount of time it takes to travel 700m is 700/350 = 2 seconds.

The ball comes to a stop in the air after 10/2 = 5 seconds.

Hence, the initial velocity of the ball is 5*10 = 50 m/s

Given, height of the object = 1.2 cm
Focal length f = -20 cm (focal length is negative for a concave mirror)
Distance of the image v = -60 cm
We know that
$\dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}$
Here, u is object distance, v is image distance, f is focal length
$\dfrac{-1}{60} + \dfrac{1}{u} = \dfrac{-1}{20}$

$\dfrac{1}{u} = \dfrac{1}{60} – \dfrac{1}{20}$

$\dfrac{1}{u} = \dfrac{-1}{30}$

=> u = -30 cm
We know that,
$\dfrac{\text{Height of the image}}{\text{Height of the object}} = \dfrac{-v}{u}$

$\dfrac{\text{Height of the image}}{1.2} = \dfrac{-(-60)}{-30}$

Height of the image = -2.4 cm

Focal length is half of radius.
Then, f = 30/2 = 15 cm
For a concave mirror, focal length will be negative. Hence, f = -15 cm