JEE Physics Current Electricity MCQs Set 14

Answer: (b) A and C only

Explanation:
1. Sensitivity means the smallest potential difference that can be measured, which is the potential gradient multiplied by the smallest measurable length.
2. A smaller potential gradient \( k=\frac{V}{L} \) gives a more sensitive potentiometer, so sensitivity is inversely proportional to the potential gradient (C).
3. A longer wire (for the same voltage) has a smaller gradient, so sensitivity is directly proportional to the length (A).

Teacher's Note:
1. A smaller potential gradient lets small differences in emf produce large changes in balance length.
2. Using a longer wire is the standard way to improve sensitivity.

Answer: (a) 40 V

Explanation:
1. The voltmeter in parallel with one resistor: \( \frac{100\times400}{500}=80\ \Omega \).
2. Total resistance \( =80+100=180\ \Omega \), so \( I=\frac{90}{180}=0.5\ \text{A} \).
3. Reading \( =0.5\times80=40\ \text{V} \).

Teacher's Note:
1. A real voltmeter loads the circuit, reducing the voltage it measures.
2. An ideal voltmeter would read 45 V.

Answer: (b) 10 m

Explanation:
1. \( R=\frac VI=\frac{3.4}{2}=1.7\ \Omega \).
2. Volume \( Al=\frac{m}{d}=10^{-6}\ \text{m}^3 \), so \( A=\frac{10^{-6}}{l} \).
3. \( R=\frac{\rho l}{A}=\frac{\rho l^2}{10^{-6}}\Rightarrow l^2=\frac{1.7\times10^{-6}}{1.7\times10^{-8}}=100\Rightarrow l=10\ \text{m} \).

Teacher's Note:
1. Express the area using the mass and density.
2. Both R and the volume are known, which fixes l.

Answer: (c) 125 Ω

Explanation:
1. At constant volume \( R\propto l^2 \).
2. \( R'=5^2\times5=125\ \Omega \).

Teacher's Note:
1. Stretching a wire to n times its length multiplies its resistance by n².
2. Length increases and area decreases by the same factor.

Answer: (a) 3 : 1

Explanation:
1. In parallel the voltage V is the same, so \( H=\frac{V^2}{R}t\propto\frac1R \).
2. \( H_R:H_{3R}=\frac1R:\frac1{3R}=3:1 \).

Teacher's Note:
1. For the same voltage, the smaller resistance dissipates more heat.
2. In series (same current) the ratio would be reversed.

Answer: (a) 30 Ω²

Explanation:
1. Series: \( \frac{R_1+R_2}{10}=\frac{60}{40}\Rightarrow R_1+R_2=15 \).
2. Parallel: \( \frac{R_1R_2/(R_1+R_2)}{3}=\frac{40}{60}\Rightarrow\frac{R_1R_2}{R_1+R_2}=2 \).
3. \( R_1R_2=2\times15=30\ \Omega^2 \).

Teacher's Note:
1. Apply the balance condition twice and combine the results.
2. Product = (parallel resistance) x (series resistance).

Answer: (a) 5 Ω

Explanation:
1. Terminal voltage with shunt R: \( V=\frac{ER}{R+r}\propto l \).
2. \( \frac{200}{300}=\frac{5/(5+r)}{15/(15+r)}=\frac{15+r}{3(5+r)} \).
3. \( 2(5+r)=15+r\Rightarrow r=5\ \Omega \).

Teacher's Note:
1. A cell shunted by R delivers a terminal voltage E R/(R + r).
2. The null length is proportional to the terminal voltage.

Answer: (a) \( \frac23\ \Omega \)

Explanation:
1. The left vertical wire joins the two left corners, so both are at the potential of A; the right vertical wire makes both right corners the potential of B. The centre of the cross is a junction.
2. Between the A-side and B-side: top arc \( 1.5+0.5=2\ \Omega \), 12 Ω, bottom 6 Ω, lower arc \( 1+1=2\ \Omega \).
3. Through the centre: \( (8\parallel2)+(6\parallel4)=1.6+2.4=4\ \Omega \).
4. Conductances add: \( \frac12+\frac1{12}+\frac16+\frac12+\frac14=\frac32\ \text{S} \), so \( R=\frac23\ \Omega \).

Teacher's Note:
1. Redraw the circuit with all the points joined by plain wires merged into single nodes.
2. Then every branch is simply in parallel between A and B.

Answer: (a) 5 Ω

Explanation:
1. Series: \( I=\frac{2E}{R+2r} \). Parallel: \( I=\frac{E}{R+r/2} \).
2. \( \frac{2}{R+2r}=\frac{1}{R+r/2}\Rightarrow2R+r=R+2r\Rightarrow r=R=5\ \Omega \).

Teacher's Note:
1. The two currents are equal only if the internal resistance equals the external resistance.
2. This is the condition at which the two groupings are equally effective.

Answer: (d) \( \frac83R \)

Explanation:
1. Take 2R and 6R in the lower path: from A, 2R goes to a node X, from which 9R goes to node Y (where the top R also ends) and 6R goes to B. Y connects to B through 3R.
2. Check the bridge balance: the potentials of X and Y are equal (ratio of the 2R:6R arm to the R:3R arm is 1:3 in both), so the 9R carries no current.
3. Path 1: \( 2R+6R=8R \). Path 2: \( R+3R=4R \). \( R_{AB}=\frac{8R\times4R}{8R+4R}=\frac83R \).

Teacher's Note:
1. This is a balanced Wheatstone bridge: a resistor between two equipotential points can be ignored.
2. The condition is 2R/6R = R/3R.

Answer: (a) \( \frac{(n-1)R}{n^2} \)

Explanation:
1. Between adjacent corners there are two paths: one side of resistance \( \frac Rn \) and the remaining \( n-1 \) sides of resistance \( \frac{(n-1)R}{n} \).
2. \( R_{eq}=\frac{\frac Rn\cdot\frac{(n-1)R}{n}}{R}=\frac{(n-1)R}{n^2} \).

Teacher's Note:
1. Each side has resistance R/n and the two paths are in parallel.
2. For n = 2 this gives R/4, the parallel combination of two R/2 halves.

Answer: (a) 2

Explanation:
1. The 10 V cell fixes the potential difference between L and P: \( |V_L-V_P|=10\ \text{V} \).
2. Each parallel resistor then carries \( I_1=I_2=\frac{|V|}{10}=1\ \text{A} \).
3. In branch (iii) the current is \( I_3=\frac{V_L-V_P\pm20}{10} \). Writing \( v=V_L-V_P \), \( I_1=I_2=-\frac v{10} \): \( \frac{I_1+I_3}{I_2}=\frac{-v+v\pm20}{-v}=\mp\frac{20}{v} \).
4. Since \( |v|=10\ \text{V} \), the magnitude is 2.

Teacher's Note:
1. Cells connected directly across two junctions fix the voltage between them.
2. The 20 V cell adds its emf to the branch, and the ratio is independent of the polarity of each cell.