CBSE Class 10 Physics Electricity MCQs Set 20

Answer: (a) high melting point and high specific resistance

Explanation:
1. Heat produced is \( I^2Rt \), so a large resistivity gives large heating.
2. The element glows red hot, so its melting point must be very high.
3. Hence a high-melting, high-resistivity alloy such as nichrome is used.

Teacher's Note:
1. Heating elements are always alloys like nichrome.
2. Copper is unsuitable because its resistivity is low.

Answer: (c) 1 joule per coulomb

Explanation:
1. Potential difference is work done per unit charge: \( V = W/q \).
2. So 1 V = 1 J/C.

Teacher's Note:
1. Newton per coulomb is the unit of electric field, not potential.

Answer: (a) 806.7 Ω, 0.27 A

Explanation:
1. \( R = V^2/P = 220^2/60 \approx 806.7\ \Omega \).
2. \( I = P/V = 60/220 \approx 0.27 \) A.

Teacher's Note:
1. The rated values give the resistance of a filament directly through \( V^2/P \).
2. Check: \( 806.7 \times 0.27 \approx 220 \) V.

Answer: (a) Rs. 49.5

Explanation:
1. Power = 220 × 15 = 3300 W = 3.3 kW.
2. Time = 0.5 h × 15 = 7.5 h.
3. Energy = 3.3 × 7.5 = 24.75 kWh.
4. Cost = 24.75 × 2 = Rs. 49.5.

Teacher's Note:
1. Convert power to kW and time to hours before multiplying.
2. Energy in kWh times the rate gives the bill.

Answer: (c) n²H

Explanation:
1. Each part has resistance R/n.
2. The parallel combination has \( R_{eq} = \dfrac{R}{n}\cdot\dfrac{1}{n} = \dfrac{R}{n^2} \).
3. Heat per second \( = V^2/R_{eq} = n^2 V^2/R = n^2H \).

Teacher's Note:
1. Cutting a wire into n pieces and joining them in parallel divides the resistance by n².
2. Heat per second varies inversely with R at fixed voltage.

Answer: (d) one fourth

Explanation:
1. Area of cross-section \( A \propto d^2 \), so doubling d makes A four times.
2. \( R = \rho l/A \) so R becomes one fourth.

Teacher's Note:
1. Resistance varies inversely with the area of cross-section.
2. The diameter enters the area as a square.

Answer: (c) iron, nickel, aluminium, titanium

Explanation:
1. Nipermag is a permanent-magnet alloy.
2. Its constituents are iron, nickel, aluminium and titanium.

Teacher's Note:
1. Alnico (not Nipermag) contains cobalt.
2. Learn alloy compositions as pairs of name and metals.

Answer: (d) 16 × 10¹⁹

Explanation:
1. Charge \( q = It = 6.4 \times 4 = 25.6 \) C.
2. Number \( n = q/e = 25.6/(1.6\times10^{-19}) = 1.6 \times 10^{20} \).
3. \( 1.6\times10^{20} = 16\times10^{19} \).

Teacher's Note:
1. Number of electrons equals total charge divided by charge of one electron.
2. Check the power of ten against the options.

Answer: (a) 0 Hz

Explanation:
1. Direct current does not change direction.
2. Hence it has no oscillation, i.e. zero frequency.

Teacher's Note:
1. Mains alternating current in India has frequency 50 Hz.

Answer: (a) 1 V

Explanation:
1. \( V = W/q = 1/1 = 1 \) V.

Teacher's Note:
1. This is the definition of the volt.

Answer: (d) 7.2 × 10⁸

Explanation:
1. 1 unit = 1 kWh = 3.6 × 10⁶ J.
2. 200 units = 200 × 3.6 × 10⁶ = 7.2 × 10⁸ J.

Teacher's Note:
1. Memorise 1 kWh = 3.6 × 10⁶ J.

Answer: (c) the atoms of the paper get polarised by the charged comb

Explanation:
1. Rubbing charges the comb.
2. The charged comb polarises the atoms of the paper, so the nearer end of each atom shows opposite charge.
3. The attraction on this nearer charge is stronger than the repulsion on the farther charge.

Teacher's Note:
1. Charged bodies attract neutral light objects by induced polarisation.
2. Plastic combs are insulators, not conductors.

Answer: (a) 4R

Explanation:
1. Volume is constant, so doubling length halves the area.
2. \( R = \rho l/A \) becomes \( \rho (2l)/(A/2) = 4R \).

Teacher's Note:
1. On stretching, resistance changes as the square of the length factor.
2. Compare with melting and recasting problems.

Answer: (b) The 60 W bulb glows brighter because it has the higher resistance

Explanation:
1. \( R = V^2/P \): the 60 W bulb has the larger resistance.
2. In series the same current flows, so power \( I^2R \) is larger for the larger resistance.
3. So the 60 W bulb glows brighter.

Teacher's Note:
1. In series, the larger resistance dissipates more power.
2. In parallel it is the other way round.

Answer: (d) \( \dfrac{r}{R+r} \)

Explanation:
1. True voltage \( V_0 = Ir \).
2. With the voltmeter, the parallel resistance is \( rR/(r+R) \), so reading \( V = I\,rR/(r+R) \).
3. Error \( = (V_0-V)/V_0 = 1-\dfrac{R}{r+R} = \dfrac{r}{R+r} \).

Teacher's Note:
1. A voltmeter should have a very large resistance to keep this error small.
2. If R is much larger than r, the error tends to zero.

Answer: (c) 16 V

Explanation:
1. Net emf in the loop \( = 20-14 = 6 \) V, total resistance \( = 3\ \Omega \).
2. Current \( I = 6/3 = 2 \) A.
3. Terminal voltage \( = 20 - 2\times2 = 16 \) V (check: \( 14 + 2\times1 = 16 \) V).

Teacher's Note:
1. The stronger cell drives current into the weaker one.
2. Terminal voltage = emf minus the drop on the internal resistance.

Answer: (b) electric field intensity

Explanation:
1. Electric field \( E = V/d \), measured in volt per metre.
2. It is also equal to newton per coulomb.

Teacher's Note:
1. Field intensity can be written as V/m or N/C.

Answer: (a) 2.5 kW

Explanation:
1. \( P = V^2/R = 250^2/25 = 2500 \) W = 2.5 kW.

Teacher's Note:
1. Use \( P=V^2/R \) when voltage and resistance are given.

Answer: (a) 1.6 × 10⁻¹⁹ J

Explanation:
1. 1 eV is the energy gained by one electron crossing 1 V.
2. \( W = qV = 1.6\times10^{-19}\times1 \) J.

Teacher's Note:
1. eV is a unit of energy, not of potential.

Answer: (a) a straight line through the origin

Explanation:
1. Ohm's law: \( V = IR \) for constant R.
2. This is a straight line through the origin with slope R.

Teacher's Note:
1. Non-ohmic devices (diode, filament) give curved graphs.

Answer: (c) the same current flows through each and the effective resistance is greater than each individual resistance

Explanation:
1. There is only one path, so the current is the same in every resistor.
2. \( R_s = R_1+R_2+\dots \) is larger than any one of them.

Teacher's Note:
1. Series: same current, voltage divides.
2. Parallel: same voltage, current divides.

Answer: (c) 15 units

Explanation:
1. Energy = 1.5 kW × 10 h = 15 kWh.
2. 1 unit = 1 kWh, so 15 units.

Teacher's Note:
1. Convert watts to kilowatts first.

Answer: (d) 1 Ω

Explanation:
1. \( R = \rho l/A \).
2. New \( R' = \rho (l/2)/(2A) = R/4 = 1\ \Omega \).

Teacher's Note:
1. Length and area both enter in a ratio, so multiply the factors.

Answer: (b) live (phase) wire

Explanation:
1. The fuse must break the live wire so that the appliance is disconnected from the high-potential line.
2. If it were in the neutral, the appliance would stay at live potential after the fuse blows.

Teacher's Note:
1. The switch and fuse go in the live wire.

Answer: (a) zero

Explanation:
1. Without an electric field, electron motion is random.
2. So the average (drift) velocity is zero.

Teacher's Note:
1. Drift velocity appears only when a field is applied.