CBSE Class 10 Physics Electricity MCQs Set 13

Answer: (a) volt-ampere

Explanation:
1. Power P = VI, so its unit may be written as volt × ampere (volt-ampere).
2. Kilowatt-hour and watt-second are units of energy, and joule-second is not a unit of power.

Teacher's Note:
1. 1 watt = 1 volt × 1 ampere = 1 joule per second.
2. Do not confuse units of power (watt) with units of energy (kWh, watt-second).

Answer: (b) electric power

Explanation:
1. Electric power is defined as the rate at which electrical energy is consumed or delivered in a circuit.
2. P = E/t = VI.

Teacher's Note:
1. Power is energy per unit time.
2. Its SI unit is the watt.

Answer: (b) 3.6 × 10⁶ J

Explanation:
1. 1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J.

Teacher's Note:
1. Kilowatt-hour is the commercial unit of electrical energy and is called one unit.
2. Convert kW to W and hour to seconds.

Answer: (b) IR²

Explanation:
1. Electric power P = VI.
2. Using V = IR gives P = I²R, and using I = V/R gives P = V²/R.
3. IR² does not represent power.

Teacher's Note:
1. Remember the three forms of power: VI, I²R and V²/R.
2. Check by substituting Ohm’s law in P = VI.

Answer: (c) Its length

Explanation:
1. A fuse wire melts when the rate of heat produced by the current equals the rate of heat lost from its surface.
2. Heat produced is I²ρl/πr² and heat lost is proportional to 2πrl, so the condition becomes I²ρ ∝ r³.
3. The length l cancels out, so the length of the fuse wire is immaterial.

Teacher's Note:
1. A fuse wire is selected on the basis of its material, thickness and the rated current.
2. Its length does not decide the current at which it melts.

Answer: (c) High resistance, low melting point

Explanation:
1. The fuse wire must heat up quickly when excess current flows, so it needs a high resistance.
2. It must melt at once to break the circuit, so it needs a low melting point.

Teacher's Note:
1. A fuse is made of an alloy of lead and tin.
2. It is always connected in series with the live wire.

Answer: (c) the leading wire has less resistance than the filament

Explanation:
1. The same current flows through the filament and the leading wire, since they are in series.
2. H = I²Rt, so the heat is more where the resistance is more.
3. The leading wire has a much smaller resistance, so it does not get hot enough to glow.

Teacher's Note:
1. Connecting wires are made of copper, which has very low resistivity.
2. The filament is made of tungsten with high resistivity.

Answer: (a) High resistance, high melting point

Explanation:
1. A heating element must produce a large amount of heat, so it needs a high resistance.
2. It becomes red hot, so it must also have a high melting point.

Teacher's Note:
1. Nichrome has both high resistivity and a high melting point.
2. Heat produced is H = I²Rt, which is larger for a larger resistance.

Answer: (d) 5 A

Explanation:
1. Current drawn I = P/V = 1000/220 ≈ 4.5 A.
2. The fuse rating must be slightly higher than the working current, so a 5 A fuse is used.

Teacher's Note:
1. A fuse must not melt in normal working but must melt when current exceeds the safe limit.
2. Choose the nearest standard fuse rating above the calculated current.

Answer: (a) 75 mW

Explanation:
1. V = 0.75 V and I = 100 mA = 0.1 A.
2. P = VI = 0.75 × 0.1 = 0.075 W = 75 mW.

Teacher's Note:
1. Convert milliampere to ampere before multiplying.
2. 1 W = 1000 mW.

Answer: (c) 3600 J

Explanation:
1. E = Pt = 60 W × 60 s = 3600 J.

Teacher's Note:
1. Convert the time to seconds when the power is in watt.
2. 1 watt = 1 joule per second.

Answer: (d) 90 kJ

Explanation:
1. I = 5 A, R = 2 Ω, t = 30 min = 1800 s.
2. H = I²Rt = 25 × 2 × 1800 = 90 000 J = 90 kJ.

Teacher's Note:
1. Convert time to seconds before using H = I²Rt.
2. Divide by 1000 to convert joule to kilojoule.

Answer: (b) 3 × 10⁵ J

Explanation:
1. Heat produced = power × time.
2. H = 1000 W × (5 × 60) s = 1000 × 300 = 3 × 10⁵ J.

Teacher's Note:
1. Convert minutes to seconds before multiplying.
2. Power is the energy produced per second.

Answer: (a) doubled

Explanation:
1. H = I²Rt.
2. When I becomes 2I and t becomes t/2, H′ = (2I)² R (t/2) = 4 × 1/2 × I²Rt = 2H.

Teacher's Note:
1. Heat is proportional to the square of the current and to the time.
2. Factor 4 from the current and factor 1/2 from the time gives 2.

Answer: (c) 750 J

Explanation:
1. I = 0.5 A, R = 10 Ω, t = 5 min = 300 s.
2. H = I²Rt = 0.25 × 10 × 300 = 750 J.

Teacher's Note:
1. Convert the time into seconds.
2. Use Joule’s law H = I²Rt.

Answer: (a) two times

Explanation:
1. The potential difference V of the circuit is unchanged, so H = V²t/R.
2. When R becomes R/2, H becomes V²t/(R/2) = 2 V²t/R, that is twice the earlier value.

Teacher's Note:
1. With V fixed, heat produced is inversely proportional to resistance.
2. If current were fixed instead, halving R would halve the heat.

Answer: (d) 50 watts

Explanation:
1. Resistance of each lamp R = V²/P = 220²/100 = 484 Ω.
2. In series the total resistance = 968 Ω.
3. Power consumed P = V²/R = 220²/968 = 50 W.

Teacher's Note:
1. In series the supply voltage is divided between the two lamps, so each lamp gets 110 V.
2. Each lamp then consumes 25 W, and the total power is 50 W.

Answer: (d) 2 minutes

Explanation:
1. Heat required is the same, H = I²Rt.
2. With the current doubled, I² becomes 4 times, so the time becomes 8/4 = 2 minutes.

Teacher's Note:
1. At constant R, t ∝ 1/I² for the same heat.
2. Doubling the current cuts the time to one fourth.

Answer: (a) R₂/R₁

Explanation:
1. In parallel the potential difference V across each is the same.
2. H = V²t/R, so H₁/H₂ = R₂/R₁.

Teacher's Note:
1. For the same voltage and time, heat is inversely proportional to resistance.
2. For resistors in series (same current), H ∝ R instead.

Answer: (a) doubled

Explanation:
1. Resistance of one half = R/2.
2. For the same supply voltage, power P = V²/R becomes V²/(R/2) = 2V²/R, so it is doubled.

Teacher's Note:
1. The mains voltage is constant, so a smaller resistance means more heat.
2. A shorter coil has less resistance.

Answer: (b) Both 40 W and 50 W bulbs will fuse

Explanation:
1. Resistances: R₄₀ = 200²/40 = 1000 Ω, R₅₀ = 800 Ω, R₁₀₀ = 400 Ω; total = 2200 Ω.
2. Current in series I = 600/2200 = 0.2727 A.
3. Rated currents are 40/200 = 0.2 A, 50/200 = 0.25 A and 100/200 = 0.5 A.
4. 0.2727 A exceeds 0.2 A and 0.25 A, so the 40 W and 50 W bulbs fuse.

Teacher's Note:
1. In series, the same current flows through all the bulbs.
2. A bulb fuses when the current is more than its rated current.

Answer: (d) 25 W

Explanation:
1. Resistance of the bulb R = V²/P = 220²/100 = 484 Ω.
2. At 110 V, P′ = 110²/484 = 25 W.

Teacher's Note:
1. Power is proportional to the square of the voltage for a fixed resistance.
2. Voltage halved means power becomes one fourth.

Answer: (b) ₹ 24

Explanation:
1. Energy consumed = power × time = 2 kW × 3 h = 6 kWh.
2. Cost = 6 × 4 = ₹ 24.

Teacher's Note:
1. Use the power in kW and the time in hours to get energy in kWh.
2. Multiply the energy by the cost per unit.

Answer: (b) 36 kJ

Explanation:
1. Power P = VI = 5 × 0.5 = 2.5 W.
2. Time t = 4 × 3600 = 14400 s.
3. Energy = Pt = 2.5 × 14400 = 36000 J = 36 kJ.

Teacher's Note:
1. Convert mA to A and hours to seconds.
2. Energy = power × time.

Answer: (c) 11

Explanation:
1. Current drawn by each bulb I = P/V = 100/220 = 0.4545 A.
2. The maximum number = 5/0.4545 = 11.

Teacher's Note:
1. In a parallel circuit the currents add up.
2. The total current must not exceed the fuse rating.

Answer: (d) 12.5 A

Explanation:
1. Total power = 1500 + 1000 = 2500 W at 200 V.
2. Total current I = P/V = 2500/200 = 12.5 A.
3. So a fuse of 12.5 A rating is needed.

Teacher's Note:
1. Appliances in a household circuit are connected in parallel, so the currents add.
2. The fuse rating must be at least the total current.

Answer: (d) 4P

Explanation:
1. Let the resistance of the coil be R. Each half has R/2.
2. The two halves in parallel give (R/2)/2 = R/4.
3. At the same voltage P = V²/R, so the new power is 4 times, that is 4P.

Teacher's Note:
1. Cutting and joining in parallel reduces the resistance to one fourth.
2. At constant voltage, power is inversely proportional to resistance.

Answer: (c) n²H

Explanation:
1. Each part has resistance R/n.
2. The n parts in parallel have resistance (R/n)/n = R/n².
3. At the same voltage, P = V²/R, so the heat per second becomes n²H.

Teacher's Note:
1. More parallel branches mean less resistance and more heat at the same voltage.
2. The equivalent resistance falls n² times.

Answer: (d) 3 : 1

Explanation:
1. In parallel the voltage is the same, so P = V²/R, which means P ∝ 1/R.
2. P₁ : P₂ = 1/1 : 1/3 = 3 : 1.

Teacher's Note:
1. At the same voltage the smaller resistance dissipates more power.
2. The ratio of powers is the inverse of the ratio of resistances.

Answer: (c) 500 W

Explanation:
1. Resistance of each heater R = 240²/1000 = 57.6 Ω.
2. In series the total resistance = 115.2 Ω.
3. Combined power = 240²/115.2 = 500 W.

Teacher's Note:
1. Doubling the resistance at the same voltage halves the power.
2. The series combination consumes half of the power of one heater.

Answer: (d) 0.055 W

Explanation:
1. For the same resistor P = V²/R, so P ∝ V².
2. P′ = 0.5 × (1/3)² = 0.5/9 = 0.055 W.

Teacher's Note:
1. The resistance is the same in both cases.
2. Reducing the voltage to one third reduces the power to one ninth.

Answer: (c) A is true, but R is false.

Explanation:
1. A is true: appliances with a metallic body (iron, geyser) need the third earth pin, a bulb does not.
2. R is false: the earth pin provides a low resistance path to the ground so that a person does not get a shock; it does not reduce heating of wires.
3. So A is true but R is false.

Teacher's Note:
1. The earth wire protects the user from electric shock by taking leakage current to the ground.
2. The thickness and rating of the wire, not the number of pins, decides the heating of wires.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. A is true: high power appliances such as geysers and air coolers are connected to a separate 15 A circuit.
2. R is also true: the metallic body is connected to the earth wire of low resistance for safety.
3. But R does not explain why a 15 A circuit is used for such appliances, so R is not the correct explanation of A.

Teacher's Note:
1. Heavy-duty appliances draw a large current and are therefore put on a 15 A circuit.
2. The earth wire is a safety feature and is a separate fact.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. At the same voltage, resistance R = V²/P, so the 100 W bulb has more resistance than the 200 W bulb (R is true).
2. The 200 W bulb therefore draws more current and consumes more power, so it glows brighter (A is true).
3. R explains A.

Teacher's Note:
1. At a fixed voltage, a higher-wattage bulb has lower resistance and draws more current.
2. Brightness depends on the power consumed.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. A is true: a fuse wire is thin, has a high resistance and a low melting point so that it melts when too much current flows.
2. R is true: alloys do not oxidise readily at high temperature.
3. But R is not the explanation of why a fuse wire is thin and has high resistance.

Teacher's Note:
1. A fuse wire is made of an alloy of lead and tin with a low melting point.
2. Thinner wire means larger resistance and quicker heating.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. A is true: the heating elements of electric toasters are made of alloys such as nichrome.
2. R is true: alloys have high resistivity and do not oxidise (burn) readily at high temperature.
3. R is the correct explanation of A.

Teacher's Note:
1. Heating elements need high resistivity, to produce more heat, and resistance to oxidation.
2. Pure metals have low resistivity and oxidise quickly when red hot.

Answer: (d) A is false, but R is true.

Explanation:
1. A is false: the commercial unit of electrical energy is the kilowatt-hour (kWh), not the kilowatt-second.
2. R is true: 1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J.
3. So A is false but R is true.

Teacher's Note:
1. One kilowatt-hour is also known as one unit of electricity.
2. The electricity bill is calculated in kWh.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. For a transmitted power P = VI, a higher voltage V means a smaller current I.
2. The power loss in the transmission wires is I²R, so it becomes much smaller at high voltage (R is true).
3. This is the reason for transmitting at high voltage, so R explains A.

Teacher's Note:
1. Transmission lines have resistance, so a large current would waste a lot of energy as heat.
2. Step-up transformers raise the voltage before transmission.

Answer: (c) A is true, but R is false.

Explanation:
1. With an ideal battery the voltage V is constant, so P = V²/R. Decreasing R increases P, so A is true.
2. R is false: P is directly proportional to R only when the current is constant; here the current increases as R decreases.

Teacher's Note:
1. Choose the formula for power according to which quantity is kept constant.
2. With constant V, P ∝ 1/R.

Answer: (c) A is true, but R is false.

Explanation:
1. A is true: by Joule’s law H = I²Rt, so more current gives more heating.
2. R is false: heat depends on the square of the current and also on the resistance and the time, not on the current only.
3. So A is true but R is false.

Teacher's Note:
1. Joule’s law has three factors: I², R and t.
2. The word ‘only’ makes the Reason false.