Railway Science (Physics-Work, Energy and Power) (Part-III)

Total Questions: 50

11. An object of mass 6 kg and having a potential energy of 480 J is placed at a height. Find the height of the object w.r.t the ground. (g = 10 m/s). [RRB Group D 18/09/2018 (Afternoon)]

Correct Answer: (b) 8 m
Solution:

P.E = mgh, where m is mass of the object, g is the acceleration due to gravity and h is the height.
Given, m = 6 kg, g = 10 m/sĀ² , P.E = 480 J
āˆ“ h = 480/6 Ɨ 10 = 8 m.

12. The average power required to lift an 80 kg object to a height of 40 m in approximately 50 seconds will be ____. (g =10 m/sĀ²) [RRB Group D 18/09/2018 (Evening)]

Correct Answer: (b) 640 J/s
Solution:

Power required, P = š‘Šš‘œš‘Ÿš‘˜ š‘‘š‘œš‘›š‘’ /Ā  š‘‡š‘–š‘šš‘’ š‘”š‘Žš‘˜š‘’n = š‘šš‘”ā„Ž/š‘”
= 80 Ɨ 10 Ɨ 40 /50 = 640 J/s.

13. If a girl weighing 400 N climbs a rope with a power of 160 W for 20 seconds, then what height will she be able to reach? [RRB Group D 18/09/2018 (Evening)]

Correct Answer: (c) 8 meters
Solution:

We know that, P = š‘Š/š‘” .
Given that, Power (P) = 160 W, Time (t) = 20 seconds.
W = P Ɨ t = 160 Ɨ 20 = 3200 J.
Now, Work (W) = Force (F) Ɨ Distance (d)
ā‡’ Distance (d) = š‘Šš‘œš‘Ÿš‘˜ (š‘Š) /š¹š‘œš‘Ÿš‘š‘’ (š¹) = 3200/400 = 8 m.

14. Fill in the blanks according to question . [RRB Group D 19/09/2018 (Afternoon)]

Correct Answer: (b) 2
Solution:

15. If a car uses 2000 J of energy and its output is 500 J, then the efficiency of the car is _______. [RRB Group D 20/09/2018 (Morning)]

Correct Answer: (b) 25%
Solution:

Efficiency, Ī· = š‘œš‘¢š‘”š‘š‘¢š‘”/š‘–š‘›š‘š‘¢š‘” = 500 š½/2000š½ = 0.25 = 25%

16. An object of 1 kg is raised to a height of 10 m. The work done by the force of gravity will be ______. (Assume g = 9.8 m/sĀ²) [RRB Group D 20/09/2018 (Morning)]

Correct Answer: (b) -98 J
Solution:

Given, mass (m) = 1 kg, height (h) = 10 m
āˆ“ work done, W = m(āˆ’g)h = 1 Ɨ (āˆ’ 9.8) Ɨ 10 = āˆ’ 98 J (minus sign indicates that the work is done against gravity).

17. A block of mass 10 kg accelerates uniformly from rest to a speed of 2 m/s in 20 s. The average power developed in time interval of 0 to 20 seconds is: [RRB Group D 20/09/2018 (Afternoon)]

Correct Answer: (d) 1 W
Solution:

18. A horizontal force of 10 N displaces a 5 kg object through a distance of 2 m in the direction of force. The work done by the force will be __________ . [RRB Group D 20/09/2018 (Evening)]

Correct Answer: (a) 20 J
Solution:

Given, Force = 10 N, Mass of object = 5 kg, Distance = 2 m
āˆµ Work Done = F.s cos Īø
Here, the angle between force and displacement is zero cos0ā° ā‡’ = 1
So, Work Done = F Ɨ s = 10 Ɨ 2 = 20 J.

19. The work done to increase the velocity of a car of 800 kg from 5 m/s to 10 m/s is [RRB Group D 22/09/2018 (Afternoon)]

Correct Answer: (d) 30 kJ
Solution:

Given, Mass (m) = 800 kg, Final velocity (v) = 10 m/s, initial velocity (u) = 5 m/s. According to the work-energy theorem, Work done = Change in K.E
ā‡’ W = Ī” K.E = 1/2 Ɨ m Ɨ (vĀ² - uĀ² ).
ā‡’ W = 1/2 Ɨ 800 Ɨ(10Ā² - 5Ā² )= 30000 J
= 30 kJ.

20. What is the maximum amount of work done in 10 s by a 20 kW engine? [RRB Group D 22/09/2018 (Evening)]

Correct Answer: (b) 200 kJ
Solution:

Given, Power (P) = 20 kW, Time (t) = 10 s.
Since, Work (W) = P Ɨ t
W = 20 kW Ɨ 10 s = 200 kJ.