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

Total Questions: 50

11. If an iron marked 1,000 W is operated for 3 hours a day, the total energy produced by it in 30 days in KWh will be - [RRB Group D 22/09/2018 (Morning)]

Correct Answer: (c) 90
Solution:

Given that, Power (P) = 1000 W = 1 kWh, Time (t) = 3 hours/day × 30 days = 3 × 30 = 90 h.
Energy = Power × Time = 90 × 1= 90 kWh.

12. 5.6 kWh = ? [RRB Group D 22/09/2018 (Afternoon)]

Correct Answer: (c) 20.16 × 10⁶ J
Solution:

∵ 1 kWh = 1 kW × 1 h
⇒ 1 kWh = 1000 W × 3600 s
⇒ 1 kWh = 3.6 × 10⁶ W s
Hence, 5.6 kWh = 5.6 × 3.6 × 10⁶ J
= 20.16 × 10⁶ J.

13. If a 50 W bulb consumes 1000 J of energy, then the time taken by the bulb is: [RRB Group D 23/09/2018 (Morning)]

Correct Answer: (d) 20 s
Solution:

Given : P = 50 W, Energy (E) = 1000J
∵ Power = 𝐸𝑛𝑒𝑟𝑔𝑦 / 𝑇𝑖𝑚e
∴ Time = 𝐸𝑛𝑒𝑟𝑔𝑦 / 𝑝𝑜𝑤𝑒r = 1000/50 = 20 s.

14. An electric washing machine of 750 W is used for 4 hours a day. The energy consumed by the machine in one day will be ________ [RRB Group D 23/09/2018 (Morning)]

Correct Answer: (d) 3 units
Solution:

Given: Time = 4 hours, Power = 750 W.
Power =  𝐸𝑛𝑒𝑟𝑔𝑦 / 𝑇𝑖𝑚e  ⇒ Energy = Power  × time = 750 × 4 = 3000 watt-hours = 3 units {1 kilowatt-hour (units) = 1000 watt-hours}.

15. A person has five resistances, each of which has a value of 1/5 Ω. Find the value of maximum resistance obtained by connecting them. [RRB Group D 24/09/2018 (Morning)]

Correct Answer: (b) 1 Ω
Solution:

The resistance of each individual resistor is R = 1/5 Ω.
We know that in series, Rₜₒₜₐₗ = R₁ + R₂ + R₃ + R₄ + R₅ = 5R = 5 × 1/5 = 1 Ω.

16. If a current of 0.3 A is passed for 1 minute in a coil having resistance of 418 Ω, then what will be the thermal energy produced ? [RRB Group D 25/09/2018 (Morning)]

Correct Answer: (d) 540 cal
Solution:

Given I = 0.3 A, R = 418 Ω , t = 60 seconds.
Q = I² × R × t. Calculations: Q = (0.3)² ×  418 × 60 = 2257.2 J = 2257.2/4.184
≈ 540 cal.

17. A battery of 12 V is connected in series with resistors of 0.2 ohm, 0.3.ohm, 0.4.ohm, 0.5 ohm and 12 ohm. How much current would flow through the 0.3 ohm resistor ? [RRB Group D 25/09/2018 (Afternoon)]

Correct Answer: (b) 0.895 A
Solution:

18. When the length of the wire is doubled the reading of the ammeter decreases by _________. [RRB Group D 26/09/2018 (Morning)]

Correct Answer: (b) half
Solution:

Resistance (R) is directly proportional to wire length; doubling length doubles resistance.
According to Ohm's Law, current (I) = 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 (𝑉)/𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 (𝑅) ; current is inversely  proportional to resistance. Thus, doubling wire length halves the current.

19. A shop uses 250 units of energy in a month. How much energy was used in joules ? [RRB Group D 26/09/2018 (Afternoon)]

Correct Answer: (b) 9 × 10⁸ J
Solution:

1 unit of energy is equal to 1 kilowatt hour (kWh) i.e. 1 unit = 1 kWh.
1 kWh = 3.6 10⁶ × J.
∴ 250 units of energy = 250 × 3.6 × 10⁶ J
= 9 × 10⁸ × J.

20. If a bulb uses 100 J of energy and remains on for 20 seconds, the power consumed by the bulb is: [RRB Group D 27/09/2018 (Afternoon)]

Correct Answer: (b) 5W
Solution:

Given: Work = Energy ⇒ 100J and Time (t) = 20s
By formula,
Power = 𝐸𝑛𝑒𝑟𝑔𝑦 / 𝑇𝑖𝑚e ⇒ Power = 100/20 = 5W,