Railway Science (Physics-Force and Pressure) (Part-V)

Total Questions: 23

1. An object of mass 20 kg moves with an acceleration of 4 m/s² . Calculate the amount of force acting on it. [RRB Group D 26/10/2018 (Afternoon)]

Correct Answer: (a) 80 N
Solution:

Given : Mass of the body (m) = 20 kg,
Acceleration (a) = 4m/s²
For a body of constant mass,
F = ma = 20 × 4 = 80N.

2. The momentum of a body is 50 kg.ms⁻¹ and its velocity is 5 ms⁻¹. What is the mass of the body ? [RRB Group D 29/10/2018 (Evening)]

Correct Answer: (a) 10 kg
Solution:

Given that : Momentum of the body (p) =
50 kg m s⁻¹ , Velocity of the body (v) = 5 ms⁻¹
p = m × v
⇒ Mass (m) =  𝑝/v = 50/5 = 10 kg.

3. Which one of the following is the equation for Velocity-Time relation ? [RRB Group D 29/10/2018 (Evening)]

Correct Answer: (c) v = u + at
Solution:

The second equation of motion is s = ut + 1/2 at²,can be called the position - time relation. Third equation of motion is v² - u² + 2as, which can be called position - velocity relation.

4. A ball thrown vertically upward returns to the ground in 13.5 seconds. At what speed did it go up? (g =10 ms⁻²) [RRB Group D 31/10/2018 (Morning)]

Correct Answer: (d) 67.5 m
Solution:

Given : Time (t) = 13.5 s (for going up and coming down)
So for going up = t = 13.5/2 = 6/75 s
For going up initial velocity u = ?
At highest point, velocity = v = 0
∵ v = u - gt
0 = u - (10 × 6.75)
u = 67.5 m.

5. Which of the following statements is true? [RRB Group D 31/10/2018 (Afternoon)]

A. The value of 'g' at Mount Everest is > g.
B. The value of 'g' at Mount Everest is < g.
C. A ball thrown up vertically returns to the ground after 15 second. Its velocity is 75 m/s.
D. A ball thrown up vertically returns to the ground after 15 second. Its velocity is 150 m/s.

Correct Answer: (d) Only statements B and C are true
Solution:

The value of 'g' on Mount Everest is < g, meaning it is less than the standard acceleration due to gravity (9.8 m/s²).
Time to reach Maximum height: t = 15/2 = 7.5 s
u = 0 (at the maximum height).
⇒ a = g = 9.8 ms⁻² or 10 ms⁻²
∵ v = u + at
⇒ v = 0 + 10 × 7.5 ⇒ v = 75 ms⁻¹.

6. An object of mass 50 kg is moving with a constant velocity of 6 ms⁻¹. Calculate the momentum of the object. [RRB Group D 2/11/2018 (Morning)]

Correct Answer: (b) 300 kg ms⁻¹
Solution:

Given : Mass (m) = 50 kg, Velocity (v) = 6 m/s
Momentum (p) = mass (m) × velocity (v) = 50 kg × 6 m/s = 300 kg m/s.

7. The weight of an object on the surface of the earth is 20 N. What would be its mass when measured on the surface of the moon ? [RRB Group D 5/11/2018 (Afternoon)]

Correct Answer: (b) 3.33 N
Solution:

The acceleration due to gravity on the moon is about 1/6ₜₕ of that on Earth.
The weight of an object on the surface of the moon = 1/6 × 20 = 3.33 N.

8. What will be the momentum of a body of mass 50 kg moving with a velocity of 20 ms⁻¹? [RRB Group D 12/11/2018 (Evening)]

Correct Answer: (d) 1000 kg ms⁻¹
Solution:

Given that : Mass of the body (m) = 50 kg, Velocity of the body (v) = 20 m/s Momentum of the body (p) = Mass (m) × Velocity (v) = 50 kg × 20 m/s = 1000 kg ms⁻¹.

9. A car falls from an outcrop and hits the ground in 1.0 seconds. (Assume g = 10 ms ⁻²) upon hitting the ground. What is its speed? [RRB Group D 26/11/2018 (Evening)]

Correct Answer: (a) 10 ms⁻¹
Solution:

Given that : initial speed (u) = zero,
Acceleration due to gravity (g) = 10 ms⁻² ,
Time (t) = 1 s.
Using the equation of motion, v = u + gt
⇒ v = 0 + 10 × 1 = 10 ms⁻¹

10. A car is moving with a velocity of 72 km/h. It takes 4 s to stop it after the brakes are applied. If the mass of the car is 1000 kg, the magnitude of the force exerted by the brakes is: [RRB Group D 27/11/2018 (Morning)]

Correct Answer: (d) 5.0 × 10³ N
Solution:

Given velocity (v) = 72 km/h = 72 × 5/18 = 20 m/s,
final velocity (v) = Initial velocity (u) + acceleration (a) × time (t).
⇒ 0 = 20 + a × 4
a = - 5 (here negative sign indicates that acceleration is in the opposite direction).
Then, Force (F) = mass (m) × acceleration (a) = 1000 × 5 ⇒ 5000 N.