CBSE Class 10 Science Electricity MCQs Set 14

Answer: (c) coulomb. Charge is measured in coulomb (C); 1 C is the charge carried by a current of 1 A in 1 s.

Teacher's Note:
1. Ampere is the unit of current, not of charge.
2. Remember that Q = I × t.

Answer: (c) 600 C. Q = I × t = 5 × 120 = 600 C.

Teacher's Note:
1. Convert minutes to seconds before substituting.
2. This is a common source of lost marks.

Answer: (a) 6.25 × 10¹⁸. n = Q/e = 1/(1.6 × 10⁻¹⁹) = 6.25 × 10¹⁸.

Teacher's Note:
1. Divide total charge by the charge on one electron.
2. Check the power of ten at the end.

Answer: (b) 1 J / 1 C. Potential difference is work done per unit charge, so 1 V = 1 J/C.

Teacher's Note:
1. Define it in words as well as by the formula V = W/Q.
2. Both forms are asked.

Answer: (b) in series and in parallel. An ammeter has a low resistance and is placed in series; a voltmeter has a high resistance and is placed across the component.

Teacher's Note:
1. A wrongly connected ammeter in parallel can damage it.
2. Remember that the voltmeter is connected across the points being measured.

Answer: (b) directly proportional to the potential difference. V ∝ I, or V = IR, provided the temperature and other physical conditions stay the same.

Teacher's Note:
1. The graph of V against I is a straight line through the origin.
2. The slope of this line gives the resistance.

Answer: (c) its nature and temperature. Resistivity is a property of the material and changes only with temperature, not with the size of the wire.

Teacher's Note:
1. Resistance depends on length and area, but resistivity does not.
2. Its SI unit is ohm metre (Ω m).

Answer: (c) 4R. Area becomes one-fourth, and as R = ρl/A, the resistance becomes 4R.

Teacher's Note:
1. Area depends on the square of the radius.
2. Halving the radius therefore reduces the area four times.

Answer: (b) 5 Ω. Each part is 10 Ω; two 10 Ω in parallel give 10 × 10/(10 + 10) = 5 Ω.

Teacher's Note:
1. Cutting the wire halves each part's resistance.
2. Parallel combination then lowers it further.

Answer: (b) 10 Ω. In series, R = 2 + 3 + 5 = 10 Ω.

Teacher's Note:
1. The equivalent resistance in series is always larger than the largest resistor.
2. Use this as a quick check.

Answer: (c) 2 Ω. R = (6 × 3)/(6 + 3) = 2 Ω.

Teacher's Note:
1. In parallel, the equivalent resistance is less than the smallest resistor.
2. Product over sum works only for two resistors.

Answer: (c) 4 A. R = (4 × 12)/16 = 3 Ω, so I = V/R = 12/3 = 4 A.

Teacher's Note:
1. You can also add the branch currents, 3 A and 1 A.
2. Both methods must give the same answer.

Answer: (c) 4 V. I = 6/(2 + 4) = 1 A, so V = IR = 1 × 4 = 4 V.

Teacher's Note:
1. In series the current is the same in all resistors.
2. The larger resistor takes the larger share of the voltage.

Answer: (b) All three in parallel. Three 6 Ω in parallel give 6/3 = 2 Ω, the least possible value.

Teacher's Note:
1. Parallel connection always reduces the resistance.
2. The series connection gives the maximum value, 18 Ω.

Answer: (b) resistance. Since V = IR, the ratio V/I, which is the slope, equals the resistance.

Teacher's Note:
1. This is valid for an ohmic conductor, where the graph is a straight line.
2. Check which quantity is on which axis.

Answer: (b) I². By Joule's law, H = I²Rt, so the heat produced varies with the square of the current.

Teacher's Note:
1. Doubling the current makes the heat four times.
2. Remember that R and t are kept constant.

Answer: (b) 200 J. H = I²Rt = 2² × 10 × 5 = 200 J.

Teacher's Note:
1. Square the current first and then multiply.
2. Write the unit as joule.

Answer: (b) high resistivity and high melting point. A high resistivity produces much heat, and a high melting point lets the element glow without melting; it also does not oxidise easily.

Teacher's Note:
1. Both properties must be stated together.
2. Alloys generally have a higher resistivity than the metals they are made of.

Answer: (b) has a very high melting point. Tungsten melts at about 3380 °C, so it stays solid at the high temperature at which the filament glows.

Teacher's Note:
1. Bulbs are filled with an inert gas to prolong filament life.
2. The filament still has high resistance.

Answer: (b) low melting point. A fuse is made of an alloy of tin and lead with a low melting point, so it melts and breaks the circuit when excess current flows.

Teacher's Note:
1. A fuse is connected in series with the live wire.
2. It protects appliances from overloading and short circuits.

Answer: (b) 484 Ω. R = V²/P = (220)²/100 = 484 Ω.

Teacher's Note:
1. Use the form of the power formula that contains voltage and power.
2. Do not forget to square the voltage.

Answer: (c) 25 W. Resistance stays 484 Ω, so P = (110)²/484 = 25 W.

Teacher's Note:
1. Power is proportional to the square of the voltage.
2. Halving the voltage reduces the power to one-fourth.

Answer: (b) 3.6 × 10⁶ J. 1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J.

Teacher's Note:
1. The kilowatt-hour is the commercial unit of energy.
2. One such unit is called one unit of electricity.

Answer: (b) ₹216. Energy = (4 × 60 W × 5 h × 30)/1000 = 36 kWh, so cost = 36 × 6 = ₹216.

Teacher's Note:
1. Convert watts into kilowatts before multiplying by hours.
2. Then multiply the number of units by the rate.

Answer: (a) 2 kWh. E = P × t = 1 kW × 2 h = 2 kWh.

Teacher's Note:
1. Use kilowatt for power and hour for time to get kWh directly.
2. No conversion to joule is needed here.

Answer: (a) 1.6 Ω. A = 1 mm² = 10⁻⁶ m², so R = ρl/A = 1.6 × 10⁻⁸ × 100 / 10⁻⁶ = 1.6 Ω.

Teacher's Note:
1. Convert mm² to m² by multiplying by 10⁻⁶.
2. Mistakes in this conversion are common.

Answer: (b) each appliance gets the full supply voltage and works independently. In parallel every appliance gets the same voltage, and switching one off does not affect the others.

Teacher's Note:
1. In series, one failure would break the whole circuit.
2. Mention both advantages in a long answer.

Answer: (a) Both Assertion (A) and Reason (R) are true and R is the correct explanation of A. Both statements are true, and the high resistivity and resistance to oxidation are why alloys are chosen.

Teacher's Note:
1. The reason directly explains the assertion.
2. Remember nichrome as the standard example.