JEE Physics Current Electricity MCQs Set 26

Answer: (b) V is doubled

Explanation:
1. \( v_d = \frac{eE\tau}{m} = \frac{eV\tau}{m\ell} \), so \( v_d \propto V/\ell \).
2. Doubling V doubles the drift velocity; doubling ℓ would halve it; diameter does not matter; a higher temperature lowers τ.

Teacher's Note:
1. Drift velocity depends on the electric field V/ℓ in the wire.
2. Area of cross-section has no direct effect on drift velocity.

Answer: (a) 0.04%

Explanation:
1. Volume is constant, so \( \ell \propto 1/d^2 \) and \( R \propto \ell/A \propto 1/d^4 \).
2. \( \frac{\Delta R}{R} \approx 4\frac{\Delta d}{d} = 0.04\% \).

Teacher's Note:
1. Resistance of a stretched wire at constant volume varies as ℓ² or 1/d⁴.
2. Use small-change approximation.

Answer: (c) 4.5 A

Explanation:
1. Kirchhoff's current law at J with potential V: \( \frac{20-V}{2}+\frac{5-V}{4} = \frac V2 \).
2. \( 40-2V+5-V = 2V \Rightarrow V = 9 \) V, so \( I = \frac92 = 4.5 \) A.

Teacher's Note:
1. Apply node (junction) analysis with one unknown potential.
2. The current into the junction equals the current out.

Answer: (d) \( 2E\left(1-\frac2n\right) \)

Explanation:
1. Net emf \( = (n-2)E - 2E = (n-4)E \); total resistance \( nr \), so \( I = \frac{(n-4)E}{nr} \).
2. A reversed cell is being charged: \( V = E + Ir = E + \frac{(n-4)E}{n} = 2E\left(1-\frac2n\right) \).

Teacher's Note:
1. A cell connected against the current has terminal voltage E + Ir.
2. Count net emf with signs.

Answer: (b) 4 Ω

Explanation:
1. Let the total be R. Then \( R = 2R_1 + \frac{R_2R}{R_2+R} \), since the remaining ladder is again R.
2. \( R = 2 + \frac{4R}{4+R} \Rightarrow R^2 - 2R - 8 = 0 \Rightarrow R = 4 \) Ω.

Teacher's Note:
1. An infinite ladder repeats itself after removing one section.
2. Reject the negative root.

Answer: (a) 10

Explanation:
1. Series: \( P_s = \frac{V^2}{nR} \). Parallel: \( P_p = \frac{nV^2}{R} \).
2. \( \frac{P_p}{P_s} = n^2 = 100 \Rightarrow n = 10 \).

Teacher's Note:
1. Series resistance is nR; parallel resistance is R/n.
2. Power at fixed voltage is V²/R.

Answer: (b) the dim bulb

Explanation:
1. Household bulbs are in parallel, so each has the same voltage and \( P = V^2/R \).
2. The dim bulb has the smaller power, hence the larger resistance.

Teacher's Note:
1. Parallel connection means equal voltage.
2. Brightness follows power.

Answer: (a) The 60 W bulb

Explanation:
1. \( R_{60} = \frac{110^2}{60} \approx 201.7\ \Omega,\; R_{100} = \frac{110^2}{100} = 121\ \Omega \). Series current \( = \frac{220}{322.7} \approx 0.68 \) A.
2. Rated currents: 60 W bulb 0.545 A, 100 W bulb 0.909 A. Only the 60 W bulb carries more than its rated current.

Teacher's Note:
1. Compare the actual current with each bulb's rated current.
2. The lower-wattage bulb has the higher resistance, so it takes more voltage in series.

Answer: (c) \( \frac{10}{3}\ \Omega,\ 5\ \Omega \)

Explanation:
1. First balance: \( \frac{R_1}{R_2} = \frac{40}{60} = \frac23 \). Second: \( R_1 = R_2' \) (50 cm each side), where \( R_2' = \frac{10R_2}{10+R_2} \).
2. \( \frac23R_2 = \frac{10R_2}{10+R_2} \Rightarrow R_2 = 5\ \Omega,\ R_1 = \frac{10}{3}\ \Omega \).

Teacher's Note:
1. Balance at the midpoint means equal resistances in the two gaps.
2. Shunting lowers R₂, so the balance point moves toward the R₂ side.

Answer: (d) 5 : 1

Explanation:
1. \( E_1+E_2 \propto 60 \) and \( E_1-E_2 \propto 40 \), so \( \frac{E_1+E_2}{E_1-E_2} = \frac32 \).
2. \( 2E_1+2E_2 = 3E_1-3E_2 \Rightarrow E_1 = 5E_2 \).

Teacher's Note:
1. Potential difference is proportional to balancing length on a potentiometer wire.
2. Reversing a cell turns the sum into a difference.

Answer: (b) V_A = +1.5 V, V_D = −0.5 V

Explanation:
1. Current flows from higher to lower potential through resistors: \( V_A = V_B + 1\times1.5 = 1.5 \) V and \( V_C = V_B - 1\times2.5 = -2.5 \) V.
2. Crossing the cell from C to D (negative to positive terminal) raises the potential by 2 V: \( V_D = -2.5+2 = -0.5 \) V.

Teacher's Note:
1. Potential drops along the current through a resistor.
2. Going from the negative to the positive terminal of a cell raises the potential by its emf.