JEE Physics Current Electricity MCQs Set 13

Answer: (b) 0.6 A

Explanation:
1. Current = charge crossing a point per second: \( I=qf=2\times10^{-2}\times30=0.6\ \text{A} \).

Teacher's Note:
1. The diameter is not needed.
2. Current equals charge times frequency for a circulating charge.

Answer: (a) n is unaltered but v is increased

Explanation:
1. \( I=neAv_d \). The number density n is a property of the material and does not change.
2. A larger current therefore needs a larger drift velocity.

Teacher's Note:
1. n depends only on the metal.
2. Current is directly proportional to the drift velocity.

Answer: (d) 1 : 3

Explanation:
1. \( R=\frac{\rho L}{A} \). Area of the solid wire \( \pi r^2 \); area of the tube \( \pi(4r^2-r^2)=3\pi r^2 \).
2. \( \frac{R_{tube}}{R_{solid}}=\frac{\pi r^2}{3\pi r^2}=\frac13 \).

Teacher's Note:
1. Resistance varies inversely with the area of cross-section.
2. The area of a hollow tube is the difference of two circles.

Answer: (c) 0.4%

Explanation:
1. Volume is constant, so \( L\propto\frac1{r^2} \) and \( R=\frac{\rho L}{\pi r^2}\propto\frac1{r^4} \).
2. A 0.1% decrease in r gives \( 4\times0.1\%=0.4\% \) increase in R.

Teacher's Note:
1. Stretching keeps the volume constant.
2. R varies as the inverse fourth power of the radius.

Answer: (b) 10

Explanation:
1. \( I=\frac{n\varepsilon}{R+nr}\Rightarrow0.6=\frac{1.5n}{20+0.5n} \).
2. \( 12+0.3n=1.5n\Rightarrow n=10 \).

Teacher's Note:
1. Cells in series add both the emf and the internal resistance.
2. Solve the linear equation for n.

Answer: (a) 4

Explanation:
1. Maximum resistance: current along the longest side (4 cm) through the smallest face \( 2\times3=6\ \text{cm}^2 \): \( R_{max}=\rho\frac46 \).
2. Minimum resistance: along 2 cm through the face \( 3\times4=12\ \text{cm}^2 \): \( R_{min}=\rho\frac2{12} \).
3. \( \frac{R_{max}}{R_{min}}=\frac{4/6}{2/12}=4 \).

Teacher's Note:
1. R = ρL/A is largest for the longest length and smallest area.
2. Pair each length with the face perpendicular to it.

Answer: (b) \( 1:n^2 \)

Explanation:
1. Each part has resistance r. Series: \( R_s=nr \). Parallel: \( R_p=\frac rn \).
2. Heat per second \( =\frac{V^2}{R} \): \( \frac{H_s}{H_p}=\frac{R_p}{R_s}=\frac{r/n}{nr}=\frac1{n^2} \).

Teacher's Note:
1. Same supply voltage: power is inversely proportional to resistance.
2. Series combination has the larger resistance, so less heat.

Answer: (c) both bulbs will be fused

Explanation:
1. Each bulb is designed for 250 V.
2. In parallel, both are subjected to the full line voltage of 500 V, twice their rating, so each draws four times its rated power.
3. Both bulbs will be fused.

Teacher's Note:
1. Parallel connection gives each branch the full supply voltage.
2. Power rises as the square of the voltage.

Answer: (a) The lamp having lower resistance

Explanation:
1. In parallel, the voltage V across each lamp is the same.
2. Power \( P=\frac{V^2}{R} \) is larger for the lower resistance.

Teacher's Note:
1. Brightness depends on the power dissipated.
2. Same voltage means P ∝ 1/R.

Answer: (b) The lamp having higher resistance

Explanation:
1. In series the same current I flows through both lamps.
2. Power \( P=I^2R \) is larger for the higher resistance.

Teacher's Note:
1. Brightness depends on the power dissipated.
2. Same current means P ∝ R.

Answer: (d) \( T_1\lt T_2 \)

Explanation:
1. The slope of the I-V graph is \( \frac1R \). The T1 line is steeper, so the wire has a lower resistance at T1.
2. The resistance of a metal increases with temperature, so T1 is the lower temperature: \( T_1\lt T_2 \).

Teacher's Note:
1. Slope of I against V is 1/R, not R.
2. Metals have a positive temperature coefficient of resistance.

Answer: (c) 12°C

Explanation:
1. Heat produced \( H=I^2Rt \), so the temperature rise is proportional to \( I^2 \).
2. Doubling the current gives \( 4\times3=12^\circ\text{C} \).

Teacher's Note:
1. Joule heating varies as the square of the current.
2. Assume the same time and constant resistance.

Answer: (b) \( \frac{\pi}{\pi+4}\ \Omega \)

Explanation:
1. Each semicircle has length \( \pi\ \text{m} \), so resistance \( \frac{\pi}{2}\ \Omega \) each.
2. The diameter wire is 2 m long: \( 1\ \Omega \).
3. Three resistors in parallel: \( \frac1R=\frac2\pi+\frac2\pi+1=\frac{4+\pi}{\pi} \), so \( R=\frac{\pi}{\pi+4}\ \Omega \).

Teacher's Note:
1. The two semicircles and the diameter are three parallel paths between A and B.
2. Add reciprocals for parallel resistors.

Answer: (a) 26.4 Ω

Explanation:
1. Balance condition \( \frac PQ=\frac RS \) requires \( S'=\frac{RQ}{P}=\frac{4\times11}{9}=\frac{44}{9}\ \Omega \), which is less than 6 Ω, so a resistor X in parallel is needed.
2. \( \frac1X=\frac9{44}-\frac16=\frac{10}{264}\Rightarrow X=26.4\ \Omega \).

Teacher's Note:
1. Parallel resistance is always smaller than the smallest branch.
2. Find the required S first, then the parallel resistor.

Answer: (d) be doubled if the length of the wire is doubled

Explanation:
1. The slope of the V-I graph is the resistance \( R=\frac{\rho L}{A} \).
2. Doubling L doubles R, so the slope doubles.
3. Higher temperature increases R, and silver has a lower resistivity, so the other options are wrong.

Teacher's Note:
1. Slope of V against I is R.
2. R is proportional to length.

Answer: (c) 1°C

Explanation:
1. Heat per minute \( =I^2Rt=49\times60\times60=176400\ \text{J} \).
2. \( \Delta T=\frac{176400}{42\times4200}=1^\circ\text{C} \).

Teacher's Note:
1. Joule heating H = I²Rt.
2. Divide by mass × specific heat.

Answer: (a) 11.9 kΩ

Explanation:
1. Total resistance for 12 V at 1 mA: \( \frac{12}{10^{-3}}=12000\ \Omega \).
2. Series resistance \( =12000-100=11900\ \Omega=11.9\ \text{k}\Omega \).

Teacher's Note:
1. The galvanometer and the series resistor form the voltmeter.
2. Subtract the coil's own resistance.

Answer: (b) 1.09, 2.18

Explanation:
1. Minimum gradient (rheostat at maximum): \( I=\frac{22}{21.8+0.2+22}=0.5\ \text{A} \), \( V_{wire}=0.5\times21.8=10.9\ \text{V} \), gradient \( 1.09\ \text{V/m} \).
2. Maximum gradient (rheostat at zero): \( I=\frac{22}{22}=1\ \text{A} \), \( V_{wire}=21.8\ \text{V} \), gradient \( 2.18\ \text{V/m} \).

Teacher's Note:
1. Gradient = (current × wire resistance)/length.
2. Rheostat resistance changes the current.

Answer: (a) 11 Ω

Explanation:
1. The galvanometer gets 10% of the current and the shunt carries 90%.
2. Equal voltages: \( 0.1I\times99=0.9I\times S \), so \( S=\frac{99\times0.1}{0.9}=11\ \Omega \).

Teacher's Note:
1. The shunt is in parallel, so the voltage across it equals that across the galvanometer.
2. Currents divide inversely as the resistances.

Answer: (d) \( 5\times10^{-3}\ \text{V/m} \)

Explanation:
1. Resistance of the wire \( =100\times0.005=0.5\ \Omega \).
2. Current \( I=\frac{2}{0.5+1.5}=1\ \text{A} \).
3. Gradient \( =I\times0.005=5\times10^{-3}\ \text{V/m} \).

Teacher's Note:
1. Include the cell's internal resistance when finding the current.
2. Gradient = current × resistance per unit length.

Answer: (c) 4 m

Explanation:
1. The accumulator is connected directly across the wire (no other resistance), so the potential difference across the wire is always 2 V.
2. The potential gradient is unchanged by changing the diameter, so the null point stays at 4 m.

Teacher's Note:
1. Potential gradient = V/L when nothing else is in series with the wire.
2. The wire's resistance does not matter if the supply has no internal resistance.

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

Explanation:
1. Resistance \( R=\frac{\rho\ell}{A}=\frac{4\rho\ell}{\pi d^2} \): it varies directly as the length but inversely as the square of the diameter, so A is false.
2. R is true.

Teacher's Note:
1. R is inversely proportional to the area of cross-section.
2. The area depends on the square of the diameter.

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

Explanation:
1. A bulb of lower power has a higher resistance (\( R=\frac{V^2}{P} \)).
2. In series the same current flows through both, so \( P=I^2R \) is larger for the higher resistance: A is true.
3. R is false, since the current is the same.

Teacher's Note:
1. In series, the current is the same in all bulbs.
2. The lower rated power means the higher resistance.