Railway Science (Physics-Electric Current and its Effects) (Part-VII)Total Questions: 5041. Keeping the potential difference constant, the resistance of a circuit is halved. What happens to the current? [RRB JE 26/05/2019 (Afternoon)](a) It becomes four times(b) It gets halved(c) It becomes one-fourth(d) It gets doubledCorrect Answer: (d) It gets doubledSolution:42. If the potential difference across the ends of a conductor is halved, what happens to the current flowing through it? [RRB JE 27/05/2019 (Morning)](a) It gets increased(b) It gets doubled(c) It gets decreased(d) It gets halvedCorrect Answer: (d) It gets halvedSolution:According to Ohm's law, potential difference is directly proportional to current in the circuit. (V ∝ I) ⇒ V = IR (where R is resistance), if potential difference is doubled across the ends of a conductor then the current flowing through the conductor also gets doubled.43. How much work is done in moving 2 Coulombs of charge across two points having a potential difference of 12 V? [RRB JE 27/05/2019 (Morning)](a) 6 Joule(b) 24 Joule(c) 18 Joule(d) 12 JouleCorrect Answer: (b) 24 JouleSolution:The potential difference across the two points is directly proportional to the work done per unit charge. ⇒ V = 𝑊/𝑄 where V is the potential difference, W is work done and Q is the charge. ⇒ W = Q × V Given Q = 2 Coulombs, Voltage (V) = 12 V. ⇒ W = 2 × 12 = 24 Joule.44. If the potential difference across the ends of a conductor is doubled, what happens to the current flowing through it ? [RRB JE 29/05/2019 (Morning)](a) It gets doubled(b) It gets quadrupled(c) It gets halved(d) It gets decreasedCorrect Answer: (a) It gets doubledSolution:By Ohm's law, - Voltage, V = current(I) x resistance (R). V = IR.45. A current of 5 Amperes flows around a circuit for 10 seconds. How much charge flows past a point in the circuit in this time? [RRB JE 29/05/2019 (Afternoon)](a) 2 Columbs(b) 0.5 Columbs(c) 25 Columbs(d) 50 ColumbsCorrect Answer: (d) 50 ColumbsSolution:Given that, Current (I) = 5A, Time (T) = 10s, Charge (Q) = ? We know that charge (Q) = Current(I) × Time(T) ⇒ Q = 5 × 10 ⇒ 50 Columbs.46. Three 2 V cells are connected in series and used as a battery in a circuit. What is the potential difference at the terminals of the battery ? [RRB JE 31/05/2019 (Morning)](a) 6 volt(b) 4 volt(c) 2 volt(d) 1 voltCorrect Answer: (a) 6 voltSolution:Given one cell potential difference (p.d) = 2V In series connection Potential differences across terminal is sum of all cell potential difference So, Total p.d = 2 + 2 + 2 = 6V.47. If the length of a wire is doubled by taking more of the wire, what happens to its resistance? [RRB JE 01/06/2019 (Evening)](a) Gets decreased(b) Gets halved(c) Gets doubled(d) Remains unaffectedCorrect Answer: (c) Gets doubledSolution:Resistance : A measure of the opposition to current flow in an electrical circuit. Measured in ohms (Ω). Resistance (R) = ρ𝑙/𝐴 Where ρ is the resistivity of a conductor, l is the length of the conductor and A is the cross-sectional area. If the length of the wire will get doubled then new resistance (Rⁱ ) will be Rⁱ = ρ 2𝑙/A ⇒ Rⁱ = 2 R.48. _______ is the amount of work done in carrying a charge of 4C across two points having a potential difference of 18V. [RRB Group D 17/09/2018 (Morning)](a) 24 J(b) 72 J(c) 7.2 J(d) 4.5 JCorrect Answer: (b) 72 JSolution:Given, Charge (q) = 4 C, potential difference (V) = 18 V Since, Work done to carry a charge (W) = Charge (q) × Potential difference (V), ⇒ W = 4 × 18 ⇒ W = 72 J.49. If the current flowing through a circuit is 0.6 A for 6 mins, the amount of electric charge flowing through it is _____. [RRB Group D 17/09/2018 (Afternoon)](a) 360 C(b) 216 C(c) 60 C(d) 36 CCorrect Answer: (b) 216 CSolution:Given : Current (I) = 0.6 A, Time (t) = 6 min, Charge (Q) = ? ∵ Charge = I × t = 0.6 × (6 × 60) = 216 ampere - second = 216 C.50. An electric lamp of 120 W is used for 8 hours per day. Calculate the units of energy consumed by the lamp in one day. [RRB Group D 18/09/2018 (Morning)](a) 16.00 unit(b) 1.50 unit(c) 0.96 uni(d) 2.00 unitCorrect Answer: (c) 0.96 uniSolution:Given : Electric lamp (power) = 120 W; Total time taken by lamp = 8 hours/day. Power (P) = E/t 120 = E/8 E = 960 watt- hours = 0.96 kWh or 0.96 units.Submit Quiz« Previous12345