10 Students NCERT Questions Solution Electricity Questions solution

Answers to Science Questions

1. A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R’, then the ratio R/R’ is –

(d) 25  (c) 5  (b) 1/5  (a) 1/25

Answer: (d) 25

2. Which of the following terms does not represent electrical power in a circuit?

(d) V^2/R  (c) VI  (b) IR^2  (a) PR

Answer: (b) IR^2

3. An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be –

(d) 25 W  (c) 50 W  (b) 75 W  (a) 100 W

Answer: (c) 50 W

4. Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be–

(d) 4:1  (c) 1:4  (b) 2:1  (a) 1:2

Answer: (c) 1:4

5. How is a voltmeter connected in the circuit to measure the potential difference between two points?

Answer: A voltmeter is connected in parallel across the two points where the potential difference is to be measured.

6. A copper wire has diameter 0.5 mm and resistivity of 1.6 × 10^-8 Ω m. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?

Answer: Resistance R = ρL/A, where A = π(d/2)^2. For R = 10 Ω, L = (R × A) / ρ = (10 × π × (0.00025)^2) / (1.6 × 10^-8) ≈ 122.7 m. If diameter is doubled (d = 1 mm), A becomes 4 times, so R becomes 1/4 of original, i.e., 2.5 Ω.

7. The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below –

I (amperes) 0.5 1.0 2.0 3.0 4.0

V (volts) 1.6 3.4 6.7 10.2 13.2

Answer: The resistance R = V/I is approximately constant (~3.2 Ω), indicating Ohm’s law holds.

8. When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.

Answer: R = V/I = 12 / (2.5 × 10^-3) = 4800 Ω or 4.8 kΩ.

9. A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω, respectively. How much current would flow through the 12 Ω resistor?

Answer: Total resistance = 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 Ω. Current I = V/R = 9 / 13.4 ≈ 0.67 A, same through all in series.

10. How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?

Answer: Equivalent resistance R = V/I = 220 / 5 = 44 Ω. For n resistors of 176 Ω in parallel, 1/R = n/176, so n = 176 / 44 ≈ 4 resistors.

11. Show how you would connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of (i) 9 Ω, (ii) 4 Ω.

Answer: (i) Connect two in parallel (3 Ω) and one in series (3 + 6 = 9 Ω). (ii) Connect all three in parallel (6/3 = 2 Ω, but adjust to 4 Ω by series-parallel mix not possible exactly; closest is two in series (12 Ω) with one in parallel adjustment, but 4 Ω requires non-standard mix).

12. Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A?

Answer: Power per bulb = 10 W, current per bulb = 10/220 ≈ 0.0455 A. Max lamps = 5 / 0.0455 ≈ 110 lamps.

13. A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 Ω resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?

Answer: (i) Separately: I = V/R = 220/24 ≈ 9.17 A each. (ii) Series: R = 48 Ω, I = 220/48 ≈ 4.58 A. (iii) Parallel: R = 12 Ω, I = 220/12 ≈ 18.33 A.

14. Compare the power used in the 2 Ω resistor in each of the following circuits: (i) a 6 V battery in series with 1 Ω and 2 Ω resistors, and (ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.

Answer: (i) Total R = 3 Ω, I = 6/3 = 2 A, Power in 2 Ω = I^2R = 2^2 × 2 = 8 W. (ii) Voltage across 2 Ω = 4 V, Power = V^2/R = 4^2 / 2 = 8 W. Ratio = 1:1.

15. Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V?

Answer: Total power = 100 + 60 = 160 W, Current = P/V = 160/220 ≈ 0.727 A.

16. Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?

Answer: Energy = Power × Time. TV: 250 × 1 = 250 Wh. Toaster: 1200 × (10/60) = 200 Wh. TV uses more energy.

17. An electric heater of resistance 44 Ω draws 5 A from the service mains for 2 hours. Calculate the rate at which heat is developed in the heater.

Answer: Power = I^2R = 5^2 × 44 = 1100 W or 1.1 kW.

18. Explain the following.

(a) Why is the tungsten used almost exclusively for filament of electric lamps?

Answer: (a) Tungsten has a high melting point and can withstand high temperatures without melting, making it ideal for lamp filaments.

(b) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal?

Answer: (b) Alloys have higher resistivity and do not oxidize easily, providing better heating and durability.

(c) Why is the series arrangement not used for domestic circuits?

Answer: (c) Series circuits affect all appliances if one fails, and voltage divides, making it impractical.

(d) How does the resistance of a wire vary with its area of cross-section?

Answer: (d) Resistance decreases as the area of cross-section increases, as R = ρL/A.

(e) Why are copper and aluminium wires usually employed for electricity transmission?

Answer: (e) Copper and aluminium have low resistivity, are ductile, and cost-effective for efficient power transmission.