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

Total Questions: 50

41. In a house, a TV set rated as 150 W is operated for 4 hours, and a 1500 W electric heater is operated for 25 minutes. The energy consumption values per day for the TV set and the electric heater are: [RRC Group D 16/09/2022 (Afternoon)]

Correct Answer: (d) 600 W h and 625 W h, respectively
Solution:

Energy consumption = Power × Time. For the TV set, Energy consumption = 150 W × 4 hours = 600 Wh. For the electric heater, Energy
consumption = 1500 W  × 25/60 hours = 625 wh.

42. calculate the current in a wire if 1200 C of charge passes through it in 10 minutes. [RRC Group D 16/09/2022 (Afternoon)]

Correct Answer: (c) 2A
Solution:

Charge (q) = 1200C, t = 10 minutes = 10 × 60 = 600 seconds.
Current in wire (I) = q/t = 1200/600 = 2A

43. Consider two circuits, A and B, [RRC Group D 16/09/2022 (Evening)]

Correct Answer: (c)
Solution:

44. Two wires are of the same length and the same resistance, but the ratio of their cross-sectional areas is 1 : 8. The ratio of their resistivities will be: [RRC Group D 17/09/2022 (Morning)]

Correct Answer: (b) 1 : 8
Solution:

Resistivity of a material is given by (ρ) = RA/𝑙  where R - resistance, A - Cross sectional area, 𝑙 = Length. Since the length and Resistance are given the same and the area of cross sectional is 1 : 8. So, Resistivity directly depends on the cross sectional area. ρ₁ = 1, , ρ₂ = 8. ρ1/ρ2 = 1/8

45. If three resistors of 3 Ω, 2 Ω and 6 Ω are connected in series combination with a 9V battery, then the potential difference across the 6Ω resistor will be: [RRC Group D 17/09/2022 (Morning)]

Correct Answer: (d) 4.9 V
Solution:

46. Ashutosh plotted Voltage against Current for two Ohmic conductors and found that the slopes of these two conductors are 10 and 50, respectively. If the same 20 V battery is connected across these two Ohmic wires separately, what will be the ratio of the current flow in these two wires? [RRC Group D 17/09/2022 (Afternoon)]

Correct Answer: (b) 5 : 1
Solution:According to ohm's law V = IR . According to the question, Voltage is
same for both conductors So ratio of current flowing in these wires is given by I = V/R. Given R₁ = 10 Ω, R₂ = 50 Ω Then ratio of current I₁/I₂ = R₂/R₁1 = 50:10 = 5:1

47. The heat produced in a resistor of 100 Ω when a current of 5A passes through it for 15 minutes is: [RRC Group D 18/09/2022 (Morning)]

Correct Answer: (b) 2.25 × 10⁶ J
Solution:

Heat produced (H) = I²Rt Where H = heat, I (Current) = 5A, R (resistance) = 100Ω , t (time) = 15 minutes.
H = (5)² ×100 × 15 × 60 = 2250000 J or 2.25 × 10⁶ J.

48. An electric lamp is connected to a 240 V DC source. The current flowing through the lamp is 0.25 A. The power of the Jamp is: [RRC Group D 18/09/2022 (Morning)]

Correct Answer: (a) 60W
Solution:

P = VI, Where P = Power, V= voltage and I = Current
P = 240 × 0.25 = 60 Watt.

49. In a house, bulbs A, B and C of rating 30 W, 60 W and 75 W, respectively, are connected in parallel with an electric source. Which of the following is true? [RRC Group D 18/09/2022 (Afternoon)]

Correct Answer: (d) Brightness of bulb C is the maximum
Solution:

50. Consider two coils A and B. Suppose coil A is replaced with coil B that has the number of loops 2 times as that of loop A and the rate of change of magnetic flux is constant. Determine the ratio of the initial to the final induced EMF. [RRC Group D 18/09/2022 (Evening)]

Correct Answer: (c) 1 : 2
Solution: