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

Total Questions: 50

41. How much current will an electric bulb draw a 220 V source if the resistance of the bulb filament is 500 Ohm ? [RRB Group D 09/10/2018 (Evening)]

Correct Answer: (c) 0.44 A
Solution:

Given that,Voltage(V) = 220 V,
Resistance(R) = 500 Ω
Using Ohm's law
Voltage(V ) = Current (I) × Resistance (R)
I = V/R = 220/500 = 0.44 A.

42. How much current will flow through the electric heater coil from a 220 V source if the resistance of the coil is 750 ohms? [RRB Group D 10/10/2018 (Morning)]

Correct Answer: (a) 0.29 A
Solution:

Given: V = 220 V, R = 750 Ω, I is the current (in amperes)
According to Ohm's Law, I = 𝑉/R
⇒ I = 220 𝑉/750 Ω ⇒ I = 220/750 A
I = 0.293 A

43. How much work is required to transfer a charge of 3C across two points with a potential difference of 10V? [RRB Group D 10/10/2018 (Morning)]

Correct Answer: (d) 30 J
Solution:

Given : Charge (Q) = 3 C, Potential difference (V) = 10 V.
By the formula : Work (W) = Charge (Q) × Potential Difference (V)
W = 3 × 10 = 30 J (The answer is positive because the work is done by the electric field on the charge, transferring potential energy to the charge).

44. The amount of electric charge transported through an electric circuit in 10 minutes is 300 C. Find the electric current used by the filament of the electric bulb. [RRB Group D 11/10/2018 (Evening)]

Correct Answer: (a) 0.5 A
Solution:

Given, Charge (Q)= 300 C, Time (t) = 10 min = 10 × 60 = 600 sec.
Charge (Q) = Current (I) × Time (t)
Current (I) = charge(q)/time(t) = 300/600 = 0.5 A.

45. When 0.5 A current flows in a circuit, the amount of electric charge is 300 C. Electricity flows through the filament of the bulb. Find out the time. [RRB Group D 12/10/2018 (Morning)]

Correct Answer: (a) 600 s
Solution:

Given that : The electric charge (in coulombs), Q = 300 C, The electric current (in amperes), I = 0.5 A, The time (in seconds) t = ?
Formula : t = 𝑄/l = t = 300 C/0.5 A = 600 s.

46. An electric fan of 300 W is used for 8 hours per day. The amount of energy used will be ______unit. [RRB Group D 12/10/2018 (Morning)]

Correct Answer: (b) 2.4
Solution:

Given : Power of the electric fan (P) = 300 W, Time the fan is used per day (t) = 8 hours.
Substituting the given values into the formula : Energy = Power × Time = 300 W × 8 h = 2400 Wh = 2.4 kWh or 2.4 units.

47. The potential difference is 12 V and the work done is 60 J. Find the electric charge that flows through the circuit. [RRB Group D 12/10/2018 (Evening)]

Correct Answer: (d) 5C
Solution:

Work = Potential Difference × Charge.
⇒ Charge = Work / Potential Difference
= 60 J / 12 V = 5 C.

48. An electric iron of 1250 W is used for 3 h per day. The energy consumed in one day by the electric iron is __________ units. [RRB Group D 12/10/2018 (Evening)]

Correct Answer: (c) 3.75
Solution:

Given that, Power = 1250 W = 1.25 kW, time = 3 h, electric energy = ?
We know that, Electric energy = Power × Time = 1.25 kW × 3 h = 3.75 kWh or 3.75 units.

49. An electric bulb of 200 W is used for 5 hours per day. Calculate the energy consumed by the bulb in a week. [RRB Group D 15/10/2018 (Morning)]

Correct Answer: (c) 7 unit
Solution:

Given that, Power of the bulb = 200 W,
Time = 5 hours
Energy = Power × Time
Energy consumed = 200 W × 5 h = 0.2 kW × 5 h = 1 unit (in one day). Energy consumed in a week = 1 unit per day × 7 day = 7 units per week.

50. How much work is done in moving a charge of 4C across two points having a P.D of 10 V ? [RRB Group D 15/10/2018 (Afternoon)]

Correct Answer: (a) 40 J
Solution:

W = Q × V, where W represents work done in moving a charge Q, V is the potential difference.
Given : Q = 4 C, V = 10 V
W = 4 × 10 = 40 J.