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

Total Questions: 50

11. Two resistors, A of 8Ω and B of 12Ω , are connected in series with a battery of 6 V. If V₁ and V₂ are the potential drops across A and B and 𝐼₁ and 𝐼₂ are the currents through them, respectively, then (𝑉₁/𝑉₂ ) and (𝐼₁/𝐼₂) are respectively________. [RRC Group D 01/09/2022 (Morning)]

Correct Answer: (c)
Solution:

12. The resistance of a wire of length L and area of cross-section A is 1.0 Ω. The resistance of a wire of the same material, but of length 4L and area of cross-section 5A will be: [RRC Group D 01/09/2022 (Afternoon)]

Correct Answer: (b) 0.8 Ω
Solution:

13. When a 24 V potential is applied across a conductor and 96 mA current is flowing through it then the resistance of the conductor is: [RRC Group D 01/09/2022 (Evening)]

Correct Answer: (b) 250 Ω
Solution:

Given that V= 24 V and I = 96mA or 96 × 10⁻³A ⇒ Find, R= resistance
As per ohm’s law, we have V = IR
⇒ 24 = 96 × 10⁻³ × R
⇒ R = 250 ohm.

14. If the current passing through a conductor is doubled and the potential difference is tripled then the power will increase: [RRC Group D 02 /09/2022 (Morning)]

Correct Answer: (b) six fold
Solution:

As Power is equal to product of potential difference and Current. P = VI
New Power P’ = (3)V × (2)I = (6)VI = 6P. So, the power will increase 6 fold.

15. If the resistance of a conductor is doubled then the heat produced is: [RRC Group D 02/09/2022 (Afternoon)]

Correct Answer: (b) Two times
Solution:

Heat generated (H₁) in a current carrying conductor with Resistance (R₁) is expressed as: H₁=I²R₁t. Since, Resistance is doubled so new resistance (R₂) = 2R₁. Then, Heat produced (H₂) = I² (2R₁)t H₂= 2 I²R₁t ⇒ H₂ = 2H₁
Hence, heat generated will be two times.

16. How many resistors of 12 Ω must be connected in parallel combination to obtain 4 Ω resistance? [RRC Group D 02/09/2022 (Evening)]

Correct Answer: (c) 3 resistors
Solution:

17. Two resistors, A of 10 Ω and B of 20 Ω are connected in series with a battery of 6 V. If V₁ and V₂ are the potential drops across A and B and I₁ and I₂ are the currents through them, then (V₁/V₂) and (I₁/I₁) are respectively. [RRC Group D 05/09/2022 (Afternoon)]

Correct Answer: (c)
Solution:

18. The values of potential drops across, and corresponding currents in, a resistor are given in the following table. If the resistor obeys Ohm's law, then the values of X and Y are respectively: [RRC Group D 05/09/2022 (Evening)]

Correct Answer: (d) 2.4 and 1.2
Solution:

As per Ohm’s Law, V = IR
⇒ 0.8 = 0.2 × R ⇒ R = 4 ohm
Now, X = 0.6 × 4 = 2.4
And, 4.8 = Y × 4 ⇒ Y = 1.2

19. A wire of length L and R resistance is reshaped in such a way that its length is increased by 50%, and there is no change in its volume. The resistance of the new wire will be _______. [RRC Group D 06/09/2022 (Morning)]

Correct Answer: (d) (9/4) R
Solution:

20. Two identical resistances are connected in parallel with a 12 V battery. The total power dissipated in the circuit is 6 W. There will be current in each resistance. [RRC Group D 06/09/2022 (Afternoon)]

Correct Answer: (c) 0.25 A
Solution:

Given: Power, P = 6 W and Voltage, V = 12V.
Now, P = VI

⇒ 6 = 12 × I ⇒ I = 6/12 = 1/2 A = 0.5 A
Since 2 resistances are connected in parallel and are of equal value, this current will get divided equally between them. So, the current through each resistance
= 0.5 = 0.25 A.