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

Total Questions: 50

1. Select the most appropriate option. 1KWh = _______ . [RRB Group D 18/09/2018 (Afternoon)]

Correct Answer: (c) 36,00,000 J
Solution:

Energy (kWh) = Power (kW) × Time (h) The kW above is 1000 W (1000 J/s). 1 hour is 3,600 seconds. So, 1 kWh = 1,000 J/s × 3,600 s = 36,00,000 J.

2. In a house, 150 units of energy has been used during a month. What will be the amount of this energy in joules ? [RRB Group D 18/09/2018 (Evening)]

Correct Answer: (b) 5.4 × 10⁸ J
Solution:

Energy consumed (E) = 150 units = 150
kWh = 150 × 3.6 × 10⁶ J ( ∵ 1 kWh = 3.6 × 10⁶ J) = 5.4 × 10⁸  J.
Hence, the household has consumed 5.4 × 10⁸  J of energy during the month.

3. A source of 6 V maintains a current of 0.5 A in a resistor. The power supplied by the source to the resistor will be ______ . [RRB Group D 19/09/2018 (Afternoon)]

Correct Answer: (c) 3.0 W
Solution:

Given, Potential Difference = 6 V, Current (I) = 0.5 A
∴ Power supplied (P) = V × I = 6 × 0.5 = 3.0 W

4. If the three resistors of 10Ω, 8Ω, and 7Ω are connected in a series, the effective resistance in the circuit will be _________. [RRB Group D 19/09/2018 (Evening)]

Correct Answer: (d) 25 Ω
Solution:

To find the effective resistance of resistors connected in a series, we simply add up the individual resistances. Given : R₁ =10 Ω, R₂ = 8 Ω, R₃ = 7 Ω Rₜₒₜₐₗ= R₁ + R₂ + R₃  = 10 Ω + 8 Ω + 7 Ω = 25 Ω.

5. 2A of electric current is placed across a resistance of 5 Ω. The amount of charge flowing through the resistance in one minute will be: [RRB Group D 20/09/2018 (Morning)]

Correct Answer: (d) 120 C
Solution:

Given, 𝐼 = 2 A , 𝑡 = 1 min = 60 sec
Now, 𝑞 = 𝐼 × 𝑡 ⇒ 𝑞 = 2 × 60 = 120 C.

6. Calculate the current drawn from a generator having a voltage output of 220 V w.hen connected to a motor having a power of 1100 W [RRB Group D 20/09/2018 (Afternoon)]

Correct Answer: (a) 5 A
Solution:

As we know, Power = Voltage × Current
According to question,
Current = 𝑃𝑜𝑤𝑒r / voltage  = 1100/220 = 5 A.

7. A constant current of 1.0 A is maintained in a resistor of 12 Ω. The amount of charge that flows through this resistor in one minute will be ________. [RRB Group D 20/09/2018 (Evening)]

Correct Answer: (c) 60 C
Solution:

Given, Current (I) = 1.0 A, Resistance (R) = 12 Ω, Time (t) = 1 minute = 60 sec. Electric charge flown (Q) = I × t = 1.0 × 60 = 60 C.

8. One kWh is equal to ___________ . [RRB Group D 20/09/2018 (Evening)]

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

1 kWh = 1 kW × 1 h ⇒ 1000 W × 3600 s = 3.6 × 10⁶  W s = 3.6 × 10⁶ J

9. A resistor of length L and area of cross-section A has a resistance R. Another resistors, of the same material, of length 𝐿/2 and area of cross-section, 2A will have a resistance of __________ . [RRB Group D 20/09/2018 (Evening)]

Correct Answer: (c) R/4
Solution:

10. A current of 0.5A is maintained in a resistor of 10Ω. The amount of charge flowing throwing the resistor in one minute is : [RRB Group D 22/09/2018 (Morning)]

Correct Answer: (d) 30 C
Solution:

Given that, Current (I) = 0.5 A, Time (t) = 1 minute = 1 × 60 = 60 seconds, Resistance (R) = 10 Ω .
Total charge (Q) = I × t = 0.5 × 60 = 30 C.