JEE Physics Current Electricity MCQs Set 11

Answer: (b) 2.50 × 10⁻³ m/s

Explanation:
1. \( I=neAv_d\Rightarrow v_d=\frac{I}{neA} \).
2. \( v_d=\frac{40}{10^{29}\times1.6\times10^{-19}\times10^{-6}}=\frac{40}{1.6\times10^4}=2.5\times10^{-3} \) m/s.

Teacher's Note:
1. Drift velocity depends on current, electron density and area.
2. Keep all quantities in SI units.

Answer: (b) 10⁻² cm/s

Explanation:
1. For typical currents in copper wires, \( v_d\sim10^{-4} \) m/s.
2. This is about \( 10^{-2} \) cm/s.

Teacher's Note:
1. Drift velocities are tiny; the signal still travels fast because the field is set up almost instantly.
2. Thermal speeds are about 10⁵ m/s, far larger than drift speeds.

Answer: (b) 0.1 mm/s

Explanation:
1. \( n=\frac{9\times10^3}{63\times10^{-3}}\times6.02\times10^{23}=8.6\times10^{28} \) m⁻³.
2. Area \( A=\pi(0.5\times10^{-3})^2=7.85\times10^{-7} \) m².
3. \( v_d=\frac{I}{neA}=\frac{1.1}{8.6\times10^{28}\times1.6\times10^{-19}\times7.85\times10^{-7}}\approx1\times10^{-4} \) m/s = 0.1 mm/s.

Teacher's Note:
1. The number density follows from density, molar mass and Avogadro's number.
2. Atomic weight 63 means a molar mass of 63 g = 0.063 kg.

Answer: (a) \( v_d\propto E \)

Explanation:
1. \( v_d=\frac{eE\tau}{m} \), where the relaxation time τ is constant for a given temperature.
2. So \( v_d\propto E \).

Teacher's Note:
1. This proportionality leads directly to Ohm's law.
2. It holds as long as τ does not change with the field.

Answer: (b) 0.25R

Explanation:
1. \( R=\frac{\rho l}{A} \) and \( A\propto d^2 \).
2. Doubling the diameter makes \( A \) four times larger, so the resistance becomes \( 0.25R \).

Teacher's Note:
1. Resistance varies as 1/d² for a fixed length and material.
2. Do not confuse diameter with area.

Answer: (a) \( \frac{R}{n^4} \)

Explanation:
1. Volume is constant: area becomes \( n^2A \) so length becomes \( \frac{l}{n^2} \).
2. \( R'=\frac{\rho(l/n^2)}{n^2A}=\frac{R}{n^4} \).

Teacher's Note:
1. When a wire is reshaped, volume stays fixed so A × l is constant.
2. Result: R ∝ l²/V and R ∝ 1/A².

Answer: (b) 2R

Explanation:
1. Each parallel pair: \( \frac R2 \).
2. Four pairs in series: \( 4\times\frac R2=2R \).

Teacher's Note:
1. Parallel pairs of equal resistors halve the resistance.
2. Series simply adds the resistances.

Answer: (a) \( \frac{\rho_1l_1+\rho_2l_2}{l_1+l_2} \)

Explanation:
1. Total resistance \( R=\frac{\rho_1l_1}{A}+\frac{\rho_2l_2}{A} \).
2. A single wire of length \( l_1+l_2 \) and the same area has \( R=\frac{\rho(l_1+l_2)}{A} \), so \( \rho=\frac{\rho_1l_1+\rho_2l_2}{l_1+l_2} \).

Teacher's Note:
1. The equivalent resistivity is a length-weighted average.
2. This works because both wires have the same cross-section.

Answer: (c) Zero

Explanation:
1. The four arms are equal, so \( \frac{R_{AB}}{R_{BC}}=\frac{R_{AD}}{R_{DC}} \): the bridge is balanced.
2. Points B and D are at the same potential, so no current flows through the 60 Ω resistor.

Teacher's Note:
1. A balanced Wheatstone bridge carries no current in the middle arm, whatever its resistance.
2. The value of the bridge resistor does not matter in the balanced case.

Answer: (b) 100 Ω

Explanation:
1. Zero deflection means no current from the 2 V cell, so the potential drop across X equals 2 V.
2. Current in the main loop: \( I=\frac{12-2}{500}=0.02 \) A.
3. \( X=\frac{2}{0.02}=100 \) Ω.

Teacher's Note:
1. With no current through the galvanometer, the 2 V cell is balanced exactly against the voltage drop across X.
2. The 500 Ω resistor must then carry the remaining 10 V.

Answer: (a) B will be zero

Explanation:
1. Total emf 4 V and total resistance \( 0.1+0.3+0.2=0.6 \) Ω, so \( I=\frac{4}{0.6}=6.67 \) A.
2. Terminal voltage of A: \( 2-6.67\times0.1=1.33 \) V. Terminal voltage of B: \( 2-6.67\times0.3=0 \).

Teacher's Note:
1. Terminal voltage is V = E − Ir.
2. A terminal voltage of zero happens when I = E/r.

Answer: (a) 1.5 A

Explanation:
1. Each row: emf \( 5000\times0.15=750 \) V; resistance \( 5000\times0.25=1250 \) Ω.
2. 100 identical rows in parallel: emf 750 V and internal resistance \( \frac{1250}{100}=12.5 \) Ω.
3. \( I=\frac{750}{500+12.5}=1.46\approx1.5 \) A.

Teacher's Note:
1. Cells in series add emf and resistance; identical rows in parallel divide the resistance.
2. The emf of a parallel group of identical rows is that of one row.

Answer: (a) 1.7 A

Explanation:
1. Kirchhoff's junction rule: current in equals current out.
2. \( 2+2=1+1.3+i\Rightarrow i=1.7 \) A.

Teacher's Note:
1. Charge cannot accumulate at a junction.
2. Sum all arriving currents and subtract the leaving ones.

Answer: (c) 14 volt

Explanation:
1. Current circulating between the cells: \( I=\frac{18-12}{2+1}=2 \) A.
2. Terminal voltage across the 18 V cell: \( V=18-2\times2=14 \) V (check: \( 12+2\times1=14 \) V).

Teacher's Note:
1. Two cells joined in parallel drive a current around the loop if their emfs differ.
2. The common terminal voltage must be the same through either cell.

Answer: (a) End correction

Explanation:
1. Interchanging the known and unknown resistances and averaging the results cancels the effect of the end resistances (the unmeasured lengths at the ends of the wire).
2. So the error removed is the end correction.

Teacher's Note:
1. End resistances are the extra resistance at the ends of the bridge wire.
2. A second balance after interchange eliminates them.

Answer: (c) Low resistance in parallel

Explanation:
1. A shunt (small resistance) in parallel diverts most of the current, so the galvanometer carries only its safe current.
2. The combination has a very low resistance, as an ammeter should.

Teacher's Note:
1. Ammeter: low resistance in parallel (shunt). Voltmeter: high resistance in series.
2. An ideal ammeter has zero resistance.

Answer: (c) 4V

Explanation:
1. \( I=neAv\Rightarrow v\propto\frac1A \).
2. Halving the diameter makes the area \( \frac14 \) as large, so the drift velocity must be \( 4V \).

Teacher's Note:
1. For the same material (same n) and current, v ∝ 1/A.
2. Area scales with the square of the diameter.

Answer: (c) inversely proportional to area of cross-section

Explanation:
1. Current is the same at every section; \( J=\frac IA \).
2. So J is inversely proportional to the cross-sectional area.

Teacher's Note:
1. Current is constant along a conductor in steady state; current density is not.
2. Narrow sections carry a larger current density.

Answer: (d) 180 Ω

Explanation:
1. Volume constant: length ×3 means area ÷3, so \( R'=\rho\frac{3l}{A/3}=9R \).
2. \( R'=9\times20=180 \) Ω.

Teacher's Note:
1. R ∝ l² for a wire stretched at constant volume.
2. Stretching to n times the length gives n² times the resistance.

Answer: (b) 1/8

Explanation:
1. Let a fraction f be stretched to length \( L'=(\frac12+f)l \) so that the total length is \( 1.5l \).
2. Resistance: \( (1-f)R+R\frac{(L'/l)^2}{f}=R\left[(1-f)+\frac{(0.5+f)^2}{f}\right] \).
3. Setting this equal to \( 4R \): \( f-f^2+0.25+f+f^2=4f\Rightarrow f=\frac18 \).

Teacher's Note:
1. The stretched part keeps its volume, so its resistance grows as the square of its length ratio.
2. Verify by substituting f = 1/8: unstretched 7/8 R plus stretched (5/8)²/(1/8) R = 25/8 R, total 4R.