CBSE Class 12 Physics Current Electricity MCQs Set 05

Answer: (a) 2 A

Explanation:
1. I = \( \frac{q}{t} \) and t = 1 min = 60 s
2. I = \( \frac{120}{60} \) = 2 A

Teacher's Note:
1. Always convert time to seconds before dividing.
2. 7200 A comes from multiplying instead of dividing, a common slip.

Answer: (c) currents add algebraically and do not obey the law of vector addition

Explanation:
1. Current has a direction (shown by an arrow in a wire), so (a) is wrong.
2. However, currents meeting at a junction simply add or subtract; they are not combined by the triangle or parallelogram law.
3. So current is a scalar because currents add algebraically and do not obey the law of vector addition.

Teacher's Note:
1. Having a direction is not enough to make a quantity a vector.
2. Current density j, on the other hand, is a vector.

Answer: (b) 4R

Explanation:
1. Volume constant: if l becomes 2l, the area A becomes A/2.
2. R′ = \( \frac{\rho (2l)}{A/2} \) = 4 \( \frac{\rho l}{A} \) = 4R

Teacher's Note:
1. Stretching changes both length and area, so the effect is n² times, not n times.
2. Resistivity ρ does not change on stretching.

Answer: (d) 1 × 10⁻⁴ m s⁻¹

Explanation:
1. v_d = \( \frac{I}{neA} \), with A = 1.0 mm² = 1.0 × 10⁻⁶ m²
2. v_d = \( \frac{1.36}{8.5 \times 10^{28} \times 1.6 \times 10^{-19} \times 1.0 \times 10^{-6}} \) = \( \frac{1.36}{13.6} \) m s⁻¹
3. v_d = 1 × 10⁻⁴ m s⁻¹ (0.1 mm s⁻¹)

Teacher's Note:
1. 1 mm² = 10⁻⁶ m², not 10⁻³ m².
2. Drift speeds are tiny, of the order of a fraction of a millimetre per second.

Answer: (b) the relaxation time of the free electrons decreases

Explanation:
1. ρ = \( \frac{m}{ne^2\tau} \)
2. In a metal, n is practically constant with temperature.
3. At higher temperature the ions vibrate more, collisions are more frequent and τ decreases, so ρ increases. Answer: the relaxation time decreases.

Teacher's Note:
1. An increase in n would lower ρ; that is what happens in semiconductors.
2. Thermal expansion changes R only slightly and does not explain the change in ρ.

Answer: (a) 2.5 × 10⁻⁴ m² V⁻¹ s⁻¹

Explanation:
1. μ = \( \frac{v_d}{E} \) = \( \frac{7.5 \times 10^{-4}}{3} \)
2. μ = 2.5 × 10⁻⁴ m² V⁻¹ s⁻¹

Teacher's Note:
1. Mobility is drift speed per unit field, so divide, do not multiply.
2. Check the unit: (m s⁻¹)/(V m⁻¹) = m² V⁻¹ s⁻¹.

Answer: (c) 2 × 10⁻³ °C⁻¹

Explanation:
1. α = \( \frac{R_T - R_0}{R_0 (T - T_0)} \), taking R₀ = 10 Ω at T₀ = 20 °C
2. α = \( \frac{12 - 10}{10 \times (120 - 20)} \) = \( \frac{2}{1000} \) = 2 × 10⁻³ °C⁻¹

Teacher's Note:
1. Divide by the resistance at the reference temperature, not by the final resistance (that gives 1.67 × 10⁻³).
2. A positive α tells you the material is a metal.

Answer: (d) T₁ < T₂

Explanation:
1. On an I–V graph the slope is \( \frac{I}{V} \) = \( \frac{1}{R} \).
2. The T₁ line is steeper, so the resistance at T₁ is smaller.
3. The resistance of a metal increases with temperature, so T₁ < T₂.

Teacher's Note:
1. Read the axes first: on a V–I graph the slope would be R itself.
2. Both lines are straight, so the wire is ohmic at both temperatures.

Answer: (b) 250 W

Explanation:
1. R = \( \frac{V^2}{P} \) = \( \frac{220^2}{1000} \) = 48.4 Ω
2. P′ = \( \frac{V'^2}{R} \) = \( \frac{110^2}{48.4} \) = 250 W

Teacher's Note:
1. P is proportional to V² for a fixed resistance, so halving V gives one-fourth the power.
2. 500 W is the trap answer from assuming P is proportional to V.

Answer: (a) 1.75 V

Explanation:
1. I = \( \frac{\varepsilon}{R + r} \) = \( \frac{2}{3.5 + 0.5} \) = 0.5 A
2. V = ε − Ir = 2 − 0.5 × 0.5 = 1.75 V

Teacher's Note:
1. The terminal voltage is less than the emf whenever the cell supplies current.
2. 0.25 V is only the drop across the internal resistance.

Answer: (c) 1.5 V, 0.1 Ω

Explanation:
1. For m identical cells in parallel, ε_eq = ε = 1.5 V.
2. r_eq = \( \frac{r}{m} \) = \( \frac{0.4}{4} \) = 0.1 Ω
3. Answer: 1.5 V, 0.1 Ω

Teacher's Note:
1. Parallel cells do not add their emfs; series cells do.
2. A parallel group is useful when a large current is needed through a small external resistance.

Answer: (d) energy

Explanation:
1. The loop rule says the sum of potential changes round a closed loop is zero.
2. A charge taken round a closed loop returns to the same potential, so no net work is done on it by the electrostatic field.
3. The rule therefore expresses conservation of energy.

Teacher's Note:
1. The junction rule comes from conservation of charge; do not mix the two.
2. The electrostatic force is conservative, which is why the loop rule holds.

Answer: (b) 1.5 A

Explanation:
1. Check balance: \( \frac{P}{Q} = \frac{4}{6} \) and \( \frac{R}{S} = \frac{8}{12} \), both equal to \( \frac{2}{3} \). The bridge is balanced, so no current flows through G.
2. Remove the galvanometer arm: ABC has 4 + 6 = 10 Ω and ADC has 8 + 12 = 20 Ω, in parallel.
3. R_eq = \( \frac{10 \times 20}{10 + 20} \) = \( \frac{20}{3} \) Ω, so I = \( \frac{10}{20/3} \) = 1.5 A

Teacher's Note:
1. Always test the balance condition first; it removes the middle arm.
2. 1.0 A is the current in arm ABC alone, not the total current.

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

Explanation:
1. Drift speeds are about 10⁻⁴ m s⁻¹, but the bulb lights at once. The Assertion is true.
2. The bulb does not wait for electrons from the switch to arrive; the field reaches all parts of the circuit very quickly and electrons already in the filament start moving. The Reason is true.
3. The Reason explains the Assertion, so the answer is (a).

Teacher's Note:
1. Compare a pipe already full of water: water flows out the moment the tap opens.
2. The signal travels fast; the electrons themselves move slowly.

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

Explanation:
1. For a metal α is positive, so ρ increases with temperature. The Assertion is true.
2. In a metal n is almost independent of temperature; it is τ that decreases. The Reason is false.

Teacher's Note:
1. A sharp rise in n with temperature is the behaviour of semiconductors.
2. Use ρ = m/(ne²τ) to reason out such statements.

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

Explanation:
1. Semiconductors have a negative temperature coefficient, so the Assertion is true.
2. More carriers become free as temperature rises, so n increases. The Reason is true.
3. The rise in n outweighs the fall in τ in ρ = m/(ne²τ), so the Reason explains the Assertion.

Teacher's Note:
1. Metals and semiconductors behave oppositely with temperature.
2. This is the basis of temperature sensors made from semiconductors.

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

Explanation:
1. Standard resistors are made of manganin, so the Assertion is true.
2. Manganin, an alloy, does have a higher resistivity than copper, so the Reason is true.
3. The real reason for using it is that its resistivity changes very little with temperature. The Reason does not explain the Assertion.

Teacher's Note:
1. A standard resistor must keep the same value as the room warms up.
2. High resistivity helps keep coils compact, but stability is the main requirement.

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

Explanation:
1. The loop rule follows from conservation of energy, not charge. The Assertion is false.
2. The Reason correctly states the loop rule, so it is true.

Teacher's Note:
1. Junction rule: charge. Loop rule: energy.
2. When the Assertion is false and the Reason true, choose (d).

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

Explanation:
1. Power lines carry power at very high voltages, so the Assertion is true.
2. I = P/V, so the loss P_c = I²R_c = P²R_c/V² falls as V rises. The Reason is true.
3. This is exactly why high voltage is used, so the Reason explains the Assertion.

Teacher's Note:
1. Doubling the voltage cuts the line loss to one-fourth.
2. Transformers step the voltage up for transmission and down for use.

Answer: (b) 1 Ω

Explanation:
1. ε = I(R + r) for both readings: 0.50(4 + r) = 0.25(9 + r)
2. 2 + 0.5r = 2.25 + 0.25r  ⇒  0.25r = 0.25
3. r = 1 Ω

Teacher's Note:
1. Two readings give two equations in two unknowns.
2. Equate the emf from both readings; ε is the same each time.

Answer: (c) 2.5 V

Explanation:
1. ε = I(R + r) = 0.50 × (4 + 1)
2. ε = 2.5 V (check: 0.25 × (9 + 1) = 2.5 V)

Teacher's Note:
1. Check your value with the second reading.
2. 2.0 V is the terminal voltage, not the emf.

Answer: (a) 2.0 V

Explanation:
1. V = IR = 0.50 × 4 = 2.0 V
2. Check: V = ε − Ir = 2.5 − 0.50 × 1 = 2.0 V

Teacher's Note:
1. The terminal voltage equals the drop across the external resistor.
2. The 0.5 V 'lost' is the drop across the internal resistance.

Answer: (d) 2.5 A

Explanation:
1. With R ≈ 0, I = \( \frac{\varepsilon}{r} \)
2. I = \( \frac{2.5}{1} \) = 2.5 A

Teacher's Note:
1. This is the maximum current the battery can give.
2. Short-circuiting heats the cell and drains it quickly.

Answer: (b) 0.56 W

Explanation:
1. P = I²R = (0.25)² × 9
2. P = 0.0625 × 9 ≈ 0.56 W

Teacher's Note:
1. 0.63 W is the total power ε × I, which also includes the loss inside the cell.
2. Square the current before multiplying by R.

Answer: (c) 0.17 Ω

Explanation:
1. R = \( \frac{\rho l}{A} \) = \( \frac{1.7 \times 10^{-8} \times 20}{2.0 \times 10^{-6}} \)
2. R = 0.17 Ω

Teacher's Note:
1. Convert 2.0 mm² to 2.0 × 10⁻⁶ m².
2. Copper wires have very small resistance, which is why they are used for wiring.

Answer: (a) 1.8 × 10⁻⁴ m s⁻¹

Explanation:
1. v_d = \( \frac{I}{neA} \) = \( \frac{5}{8.5 \times 10^{28} \times 1.6 \times 10^{-19} \times 2.0 \times 10^{-6}} \)
2. v_d = \( \frac{5}{27.2} \) × 10⁻³ ≈ 1.8 × 10⁻⁴ m s⁻¹

Teacher's Note:
1. At this speed an electron would take more than a day to travel the 20 m wire.
2. Drift speed does not depend on the length of the wire.

Answer: (d) 2.5 × 10⁶ A m⁻²

Explanation:
1. j = \( \frac{I}{A} \) = \( \frac{5}{2.0 \times 10^{-6}} \)
2. j = 2.5 × 10⁶ A m⁻²

Teacher's Note:
1. Current density is current per unit area, so divide I by A.
2. 2.5 × 10³ comes from using mm² as if it were 10⁻³ m².

Answer: (b) one-fourth

Explanation:
1. Doubling the diameter doubles the radius, so the area becomes 4 times.
2. R = ρl/A, so R becomes one-fourth (0.0425 Ω).

Teacher's Note:
1. Area depends on the square of the diameter.
2. A thicker wire wastes less power as heat.

Answer: (c) 4.25 W

Explanation:
1. P = I²R = 5² × 0.17
2. P = 4.25 W

Teacher's Note:
1. Use I²R since both the current and the wire resistance are known.
2. 0.85 W comes from using IR instead of I²R.