Railway Science (Physics-Miscellaneous) (Part-II)

Total Questions: 20

11. The density of an object of mass 300 kg and volume 150 m³ is ___________ [RRB Group D 29/10/2018 (Evening)]

Correct Answer: (d) 2 kgm⁻³
Solution:

Given that : Mass of the object (m) = 300 kg,
Volume of the object (V) = 150 m³
Density = 𝑚/𝑉 = 300/150 = 2 kg m⁻³

12. The relative density of silver is 10.8. The density of water is 10³ 𝑘𝑔𝑚⁻³ . What is the density of silver ? [RRB Group D 30/10/2018 (Morning)]

Correct Answer: (a) 10.8 × 10³ kg m⁻³
Solution:

13. 1 kWh = _________ | [RRB Group D 2/11/2018 (Morning)]

Correct Answer: (c) 3.6 × 10⁶ J
Solution:

1 kWh = 1 kilowatt × 1 hour
= 1000 watts × 3600 seconds (1 hour = 3600 seconds)
= 3,600,000 J = 3.6 × 10⁶ J.

14. A rocket is launched to travel vertically upward with a constant velocity of 20 m/s. After traveling for 35 seconds, the rocket develops a snag and its fuel supply is cut off. The rocket then travels like a free body. The height achieved by it is: [RRB ALP Tier - I (14/08/2018) Morning]

Correct Answer: (c) 720 m
Solution:

Given: A rocket is launched to travel vertically upward with a constant velocity of 20 m/s.
we know that S = ut, when it is moving at a constant velocity.
After 35 seconds the distance will be (S) = 20 × 35 = 700 m
Now from 700 m above the ground now it travels in a free fall.
The initial velocity = 20 m/s
Final velocity = 0
Acceleration due to gravity = 10
Now, V² = U² - 2gs ⇒ 0 = 400 - 20S
⇒ 20s = 400 ⇒ S = 400/20 ⇒ S = 20 m
The total height achieved = 700 + 20 = 720 m

15. If the initial velocity of a car is 5 m/s, and the final velocity is 10 m/s in 5 s, then the acceleration is _________. [RRB ALP Tier - I (17/08/2018) Evening]

Correct Answer: (a) 1 m/s²
Solution:

Acceleration the rate of change of velocity with respect to time. Acceleration is a vector quantity as it has both magnitude and direction. Initial Velocity (u) = 5 m/s , Final Velocity (v) = 10 m/s, Time = 5 s.
Acceleration= 𝑣 − 𝑢/𝑇
Acceleration = (10 - 5)/5
Acceleration = 1 m/s² .

16. A ball, thrown vertically upward, rises to a height of 80 m and returns to its original position. The magnitude of its displacement after 7 s of motion will be _____. (take g = 10 m/s²) [RRB ALP Tier - I (20/08/2018) Morning]

Correct Answer: (d) 35 m
Solution:

17. If a ball is thrown vertically upwards with a velocity of 40 m/s, then what will be the magnitude of its displacement after 6 s? (Take g = 10 m/𝑠²) [RRB ALP Tier - I (20/08/2018) Evening]

Correct Answer: (a) 60m
Solution:

18. A particle experiences constant acceleration for 20 s after starting from rest. If it travels a distance X₁ in the first 10 s and distance X₂ in the remaining 10 s, then which of the following is true? [RRB ALP Tier - I (21/08/2018) Afternoon]

Correct Answer: (d) X₂ = 3X₁
Solution:

A particle experiences constant acceleration for 20 s after starting from rest. If it travels a distance X₁ , in the first 10 s and the distance X₂ , in the remaining 10s.
We know that, s = ut + 1/2 at²,
[s = distance, u = initial velocity,
a = acceleration and t = time]
Let "v" be the velocity at t = 10 sec.
Then for first 10 seconds, as the particle
starts moving from rest then u = 0
⇒ X₁ = 0 × 10 + 1/2 a × (10)² ⇒ 50a.
Now, v = u + at = 0 + a × 10 = 10a. Now,
after 10 seconds the velocity becomes 10a.
⇒ X₂ = 10a × 10 + 1/2 a × 10² = 150a.
X₂ = 3X₁.

19. An object, starting from rest, moves with constant acceleration of 4 m/s² . After 8 s, its speed is: [RRB ALP Tier - I (21/08/2018) Evening]

Correct Answer: (a) 32 m/s
Solution:

Given: Acceleration of the object (a) = 4 m/s².
Time taken (t) = 8 s
speed (v) = u + at
u = 0 {u (initial velocity) Because the object is at rest}.
v = 0 + 8 × 4 ⇒ v = 32 m/s.

20. If the initial velocity of an object thrown upwards is 14 m/s, then the time taken for the object to reach its highest point will be______. (a = 9.8m/s²) [RRB ALP Tier - I (29/08/2018) Evening]

Correct Answer: (c) 1.43 s
Solution:

Initial Velocity of an object (u) = 14 m/s , Acceleration
against gravity (a) = - 9.81 m/s² .
When the object reaches maximum
height while thrown upwards its final
velocity (v) = 0
v = u + at
0 = 14 + [ - 9.81 (t) ] ⇒ 14 = 9.81(t)
t = 14/9.8 ⇒ t = 1.43 s.