Railway Science (Physics-Electric Current and its Effects) (Part-IX)

Total Questions: 31

11. A potential of 17 V exists between the two points, how much work must be done to move a charge of 5 C between these two points? [RRB Group D 23/10/2018 (Evening)]

Correct Answer: (b) 85 J
Solution:

Given : Potential Difference = 17 V,
Charge = 5 C
Work Done = Charge × Potential Difference = 17 × 5 = 85 J.

12. What is the work done in transferring a charge of 5C between two points having a potential difference of 11 V? [RRB Group D 24/10/2018 (Afternoon)]

Correct Answer: (c) 55 J
Solution:

Given : Charge (Q) = 5 C, Potential difference (V) = 11 V
W = Q × V
W = 5 × 11 = 55 J.

13. A current of 0.5A is drawn by a filament of an electric bulb for 8 minutes. Find the amount of electric charge that flows through the current. [RRB Group D 25/10/2018 (Afternoon)]

Correct Answer: (a) 240 C
Solution:

Given : Current (I) = 0.5 A, Time (t) = 8
min = 60 × 8 = 480 s.
∵ Q = I × t.
Q = 0.5 × 480 ⇒ Q = 240 C.

14. An electric current of 0.75 A flows in a filament of an electric bulb in 10 minutes. Find the amount of electric charge flowing through the current. [RRB Group D 30/10/2018 (Afternoon)]

Correct Answer: (a) 450 C
Solution:

Given that : Electric current (I) = 0.75 A
and time (t) = 10 min = 10 × 60 = 600 s.
∵ Electric charge (Q)= Electric current (I) × time (t) = 0.75 A × 600 = 450 C.

15. A current of 0.75A is drawn by a filament of an electric bulb in 1 minute. Find the amount of electric charge flowing through the current. [RRB Group D 30/10/2018 (Evening)]

Correct Answer: (b) 45 C
Solution:

Given that : Electric current (I) = 0.75 A and time (t) = 1 min = 1 × 60 = 60 s. Electric charge (Q) = Electric current (I) × Time (t) = 0.75 A × 60 = 45 C.

16. The potential difference between the terminals of an electric heater is 60 V when it draws a current of 4 A from the source. What current will the heater draw if the potential difference is increased to 150 V? [RRB Group D 31/10/2018 (Afternoon)]

Correct Answer: (c) 10 A
Solution:

17. What will be the resultant resistance of a circuit consisting of three resistors of 25 Ω each in series? [RRB Group D 31/10/2018 (Evening)]

Correct Answer: (b) 75Ω
Solution:

Total resistance (in series) = R₁ + R₂ + R₃ = 25 + 25 + 25 = 75 Ω.

18. A 9 Ω resistance wire is doubled on it. Calculate the new resistance of the wire. [RRB Group D 2/11/2018 (Evening)]

Correct Answer: (c) 2.25 Ω
Solution:

19. A 10 Ω resistance wire is doubled on it. Calculate the new resistance of the wire. [RRB Group D 5/11/2018 (Evening)]

Correct Answer: (b) 2.5 Ω
Solution:

20. Electric lamps of 100 W are used for 8 hours per day. Calculate the units of energy consumed by the lamp in 3 days. [RRB Group D 3/12/2018 (Afternoon)]

Correct Answer: (d) 2.4 unit
Solution:

Given : Power (P) = 100 W, Time (t) = 8 hours per day Energy consumed by bulb in one day = 8 × 100 = 800 Wh
∵ 1 kWh = 1000 Wh = 1 unit
Hence 800 Wh = 0.8 unit for 1 day
For 3 days = 0.8 × 3 = 2.4 units.