JEE Physics Current Electricity MCQs Set 15

Answer: (b) 4 : 1

Explanation:
1. The same current flows through both wires and they have the same area, and \( I = neAv_d \), so \( v_d \propto 1/n \).
2. \( \dfrac{v_1}{v_2} = \dfrac{n_2}{n_1} = \dfrac{4}{1} \).

Teacher's Note:
1. In a series connection the current is the same everywhere. The drift velocity is smaller where there are more carriers.

Answer: (b) \( v_P > v_Q \)

Explanation:
1. The current is the same at every cross-section: \( I = neAv_d \).
2. So \( v_d \propto 1/A \). P has the smaller area, so \( v_P > v_Q \).

Teacher's Note:
1. Think of water in a pipe: it flows faster where the pipe is narrower. The current is the same, but the current density is not.

Answer: (b) 8 Ω

Explanation:
1. On stretching, volume is constant, so the area becomes half when the length doubles.
2. \( R = \rho\dfrac{l}{A} \) becomes \( \rho\dfrac{2l}{A/2} = 4R = 8\ \Omega \).

Teacher's Note:
1. When a wire is stretched at constant volume, \( R \propto l^2 \). A length factor of n multiplies the resistance by \( n^2 \).

Answer: (d) \( \dfrac{\rho_1+\rho_2}{2} \)

Explanation:
1. Each wire has length l and area A. Series resistance: \( R = \dfrac{\rho_1l}{A} + \dfrac{\rho_2l}{A} \).
2. The combined wire has length 2l, so \( \rho_{eff}\dfrac{2l}{A} = \dfrac{(\rho_1+\rho_2)l}{A} \), giving \( \rho_{eff} = \dfrac{\rho_1+\rho_2}{2} \).

Teacher's Note:
1. Do not forget that the length of the combined wire is doubled. For wires in parallel the answer would be the harmonic-type expression.

Answer: (d) minimum for the wire of cross-sectional area 2A

Explanation:
1. With \( R_0 = \rho\ell/A \): the first wire has \( R_0 \), the second \( \rho(2\ell)/(A/2) = 4R_0 \), and the third \( \rho(\ell/2)/(2A) = R_0/4 \).
2. The third wire, of area 2A, has the minimum resistance.

Teacher's Note:
1. Resistance is proportional to length and inversely proportional to area, so compare the ratio l/A.

Answer: (d) 44%

Explanation:
1. At constant volume, \( R \propto l^2 \).
2. \( R'/R = (1.2)^2 = 1.44 \), an increase of 44%.

Teacher's Note:
1. Do not just add 20% twice (40%). The area also decreases by the same factor, giving the square.

Answer: (b) \( x > z > y \)

Explanation:
1. Across AB: length 4a, area \( 2a\times a = 2a^2 \), so \( x = \dfrac{\rho\,4a}{2a^2} = \dfrac{2\rho}{a} \).
2. Across CD: length a, area \( 4a\times2a = 8a^2 \), so \( y = \dfrac{\rho}{8a} \).
3. Across EF: length 2a, area \( 4a\times a = 4a^2 \), so \( z = \dfrac{\rho}{2a} \).
4. Hence \( x > z > y \).

Teacher's Note:
1. For each pair of faces, the length is the dimension between them and the area is the product of the other two.

Answer: (a) \( T_1 < T_2 \)

Explanation:
1. On an I-V graph, the slope is \( 1/R \). A larger slope means a smaller resistance.
2. The resistance of a metal increases with temperature, so the line with the smaller slope (T2) is at the higher temperature: \( T_1 < T_2 \).

Teacher's Note:
1. Be careful with the axes: with I on the vertical axis, a steeper line means lower resistance.

Answer: (c) \( \dfrac{\beta}{\alpha} \)

Explanation:
1. The change in total resistance is zero: \( R_1\alpha\Delta T + R_2(-\beta)\Delta T = 0 \).
2. So \( R_1\alpha = R_2\beta \), and \( \dfrac{R_1}{R_2} = \dfrac{\beta}{\alpha} \).

Teacher's Note:
1. This is how temperature-compensated resistors are made: one material with a positive and one with a negative coefficient.

Answer: (c) \( \dfrac{\alpha_1R_1+\alpha_2R_2}{R_1+R_2} \)

Explanation:
1. At temperature t: \( R_1(1+\alpha_1t) + R_2(1+\alpha_2t) = (R_1+R_2)(1+\alpha t) \).
2. Comparing the terms in t: \( \alpha = \dfrac{R_1\alpha_1+R_2\alpha_2}{R_1+R_2} \).

Teacher's Note:
1. The effective coefficient is a weighted average, weighted by the resistances at 0°C.

Answer: (c) 25R

Explanation:
1. At constant volume, \( R \propto l^2 \).
2. \( R' = 5^2R = 25R \).

Teacher's Note:
1. Stretching by a factor n multiplies the resistance by n², because length increases n times and area decreases n times.

Answer: (b) Yellow - Violet - Orange - Silver

Explanation:
1. 47 kΩ = 47 × 10³ Ω. Digits: 4 = yellow, 7 = violet. Multiplier \( 10^3 \) = orange.
2. The tolerance is \( 4.7/47 = 10\% \), which is silver.
3. So the sequence is yellow, violet, orange, silver.

Teacher's Note:
1. Colour code: black 0, brown 1, red 2, orange 3, yellow 4, green 5, blue 6, violet 7, grey 8, white 9. Gold is 5% and silver is 10%.

Answer: (b) 1.0 A

Explanation:
1. \( P/Q = 1/3 = R/S \), so the bridge is balanced and the galvanometer carries no current.
2. The two arms in series: \( P+Q = 4\ \Omega \) and \( R+S = 12\ \Omega \), in parallel: \( \dfrac{4\times12}{16} = 3\ \Omega \).
3. \( I = \dfrac{7}{3+4} = 1.0 \) A.

Teacher's Note:
1. Test for balance first. If it is balanced, the galvanometer branch can be removed from the calculation.

Answer: (c) \( \dfrac13 \) Ω

Explanation:
1. \( E = 4(2+r) = 1(9+r) \).
2. \( 8+4r = 9+r \Rightarrow r = \dfrac13\ \Omega \).

Teacher's Note:
1. Write E = I(R + r) for both cases and equate, since E and r are fixed for the battery.

Answer: (a) 10 Ω

Explanation:
1. \( I = \dfrac{E}{R+r} \Rightarrow 0.2 = \dfrac{2.1}{0.5+r} \).
2. \( 0.5 + r = 10.5 \Rightarrow r = 10\ \Omega \).

Teacher's Note:
1. A large internal resistance is typical of a weak or old cell. Check: the terminal voltage is only 0.1 V.

Answer: (d) 6 V

Explanation:
1. With R = r, the current is \( I = \dfrac{E}{2r} \).
2. Terminal voltage \( V = IR = \dfrac{E}{2} = 3 \), so \( E = 6 \) V.

Teacher's Note:
1. When the external resistance equals the internal resistance, the terminal voltage is exactly half the e.m.f.

Answer: (d) 20

Explanation:
1. First case: balance at 50 cm, so \( R_1+10 = R_2 \).
2. Second case: \( \dfrac{R_1}{R_2} = \dfrac{40}{60} = \dfrac23 \Rightarrow R_1 = \dfrac23(R_1+10) \).
3. \( \dfrac{R_1}{3} = \dfrac{20}{3} \Rightarrow R_1 = 20\ \Omega \).

Teacher's Note:
1. In a metre bridge the ratio of the resistances equals the ratio of the lengths of the wire on the two sides of the balance point.

Answer: (c) 1.5 A

Explanation:
1. One row: \( E_{row} = 5000\times0.15 = 750 \) V, \( r_{row} = 5000\times0.25 = 1250\ \Omega \).
2. 100 rows in parallel: \( E = 750 \) V and \( r = 1250/100 = 12.5\ \Omega \).
3. \( I = \dfrac{750}{500+12.5} \approx 1.46 \approx 1.5 \) A.

Teacher's Note:
1. Identical cells in series add their e.m.f. and resistance. Identical rows in parallel keep the e.m.f. the same and divide the resistance.

Answer: (a) +11 V

Explanation:
1. Going from A to B in the direction of the current, the potential falls by \( 2\times2 = 4 \) V across the first resistor.
2. Entering the cell at the positive terminal and leaving at the negative one, it falls by a further 3 V.
3. It falls by 4 V across the second resistor. \( V_A - V_B = 4+3+4 = 11 \) V.

Teacher's Note:
1. Moving through a cell from + to − is a drop of E. Moving through a resistor in the direction of the current is a drop of IR.

Answer: (a) 6

Explanation:
1. Series: \( I = \dfrac{E}{nR+R} = \dfrac{E}{(n+1)R} \).
2. Parallel: \( I' = \dfrac{E}{R/n+R} = \dfrac{nE}{(n+1)R} = nI \).
3. Given \( I' = 6I \), so n = 6.

Teacher's Note:
1. Note that the internal resistance is the same as R, so it appears in both expressions in the same form.

Answer: (b) I remains constant (a horizontal line)

Explanation:
1. For n cells in series, \( E_{total} = nE \) and \( r_{total} = nr \).
2. On short-circuit, \( I = \dfrac{nE}{nr} = \dfrac{E}{r} \), independent of n.
3. So the graph is a horizontal line.

Teacher's Note:
1. A series arrangement increases e.m.f. and internal resistance in the same proportion, so the short-circuit current does not change.

Answer: (d) 2 volts

Explanation:
1. The conductances add: \( \dfrac{1}{48}+\dfrac{1}{48}+\dfrac{1}{24}+\dfrac16+\dfrac14+\dfrac12 = \dfrac{1+1+2+8+12+24}{48} = 1 \) S.
2. So the parallel combination is 1 Ω and carries 2 A, giving a potential difference \( V = 2\times1 = 2 \) V.
3. The same potential difference appears across each resistor of the combination, including the 24 Ω one.

Teacher's Note:
1. The potential difference is common to all branches in parallel. Only the 3 Ω and 5 Ω resistors are extra distractors here.