JEE Physics Current Electricity MCQs Set 22

Answer: (a) 1, 2, 3 and 4

Explanation:
1. The field accelerates the electrons to the right, so their speed and momentum increase from A to B (1).
2. Conventional current is opposite to the electron motion: right to left (2).
3. In steady flow the same charge crosses every section per second (3), and with a uniform cross-section \( J=\frac IA \) is the same (4).

Teacher's Note:
1. Test each statement separately against the physical law behind it.
2. Then choose the option that lists exactly the true statements.

Answer: (b) 1 and 4 only

Explanation:
1. The current is the same at every section, so the charge crossing per unit time does not depend on the cross-section (1).
2. The free-electron density is a property of the material (4).
3. Drift speed \( v_d=\frac{I}{neA} \) and current density \( \frac IA \) both vary with A (2 and 3 false).

Teacher's Note:
1. Test each statement separately against the physical law behind it.
2. Then choose the option that lists exactly the true statements.

Answer: (d) 1 and 4 only

Explanation:
1. From the far end: \( 2+4+2=8\ \Omega \) in parallel with the 8 \( \Omega \) rung gives 4 \( \Omega \); plus \( 2+2 \) gives 8 \( \Omega \), in parallel with the first 8 \( \Omega \) rung gives 4 \( \Omega \).
2. Total \( 1+3+2+4=10\ \Omega \) so the battery current is 1 A, which flows through the 3 \( \Omega \) resistor (1).
3. It splits equally (8 and 8 \( \Omega \)) at each rung: 0.5 A, then 0.25 A through the 4 \( \Omega \) resistor (4).

Teacher's Note:
1. Reduce the network or apply Kirchhoff's laws; check the answer with a power balance.
2. Remember that an ideal voltmeter has infinite resistance and an ideal ammeter has zero resistance.

Answer: (a) 1, 2 and 4

Explanation:
1. At junction D: \( \frac{70-V_D}{10}=\frac{V_D}{20}+\frac{V_D-10}{30} \) gives \( V_D=40 \) V (1).
2. Currents: \( \frac{30}{10}=3 \) A, \( \frac{40}{20}=2 \) A, \( \frac{30}{30}=1 \) A: ratio 3 : 2 : 1 (2).
3. Power \( 3^2\times10+2^2\times20+1^2\times30=200 \) W (4).

Teacher's Note:
1. Reduce the network or apply Kirchhoff's laws; check the answer with a power balance.
2. Remember that an ideal voltmeter has infinite resistance and an ideal ammeter has zero resistance.

Answer: (c) 1, 2 and 3

Explanation:
1. \( V=E-ir \) is a straight line with intercept E on the V axis: \( E=10 \) V (2).
2. Short-circuit current \( \frac Er=2 \) A gives \( r=5\ \Omega \) (1) and the maximum current is 2 A (3).
3. The graph is exactly of this form, so statement 4 is false.

Teacher's Note:
1. Reduce the network or apply Kirchhoff's laws; check the answer with a power balance.
2. Remember that an ideal voltmeter has infinite resistance and an ideal ammeter has zero resistance.

Answer: (d) 1, 2 and 4

Explanation:
1. \( V=E-ir \) when the battery discharges (current from - to + inside): \( V<E \), and \( V=0 \) if the external resistance is zero (1, 2).
2. When it is being charged (current from + to - inside) \( V=E+ir>E \) (4).
3. With no current \( V=E \), not zero (3 false).

Teacher's Note:
1. Reduce the network or apply Kirchhoff's laws; check the answer with a power balance.
2. Remember that an ideal voltmeter has infinite resistance and an ideal ammeter has zero resistance.

Answer: (a) 1, 3 and 4

Explanation:
1. The cell converts power \( \varepsilon i \) (1).
2. Heat in R is \( i^2R=i\cdot\frac{\varepsilon R}{R+r} \) (3) and in the internal resistance \( i^2r=\varepsilon i\frac{r}{R+r} \) (4); the two add to \( \varepsilon i \), so statement 2 is false.

Teacher's Note:
1. Reduce the network or apply Kirchhoff's laws; check the answer with a power balance.
2. Remember that an ideal voltmeter has infinite resistance and an ideal ammeter has zero resistance.

Answer: (c) 1, 2 and 3

Explanation:
1. Total resistance ranges from \( 5+75+120=200 \) to \( 5+75+170=250\ \Omega \), so the current is 0.10 A to 0.08 A.
2. The potential drop across the whole wire is \( 75I \): from 7.5 V (box 120 \( \Omega \)) down to 6.0 V (box 170 \( \Omega \)).
3. A voltage can be measured only if it does not exceed the wire drop, so 5, 6 and 7 V can be measured (7 V with a suitable box setting) but 8 V cannot.

Teacher's Note:
1. In a potentiometer or metre bridge, a null condition means no current in the galvanometer branch.
2. Write the balance condition as a ratio of lengths or resistances.

Answer: (a) 2 and 3 only

Explanation:
1. Voltmeter: \( R=\frac{V}{I_g}-G \): for 50 V, \( 10^6-100\approx1000\ \text{k}\Omega \) (1 false); for 10 V, \( 2\times10^5-100\approx200\ \text{k}\Omega \) (2 true).
2. Ammeter shunt \( S=\frac{I_gG}{I-I_g} \): for 5 mA \( \frac{50\times10^{-6}\times100}{4.95\times10^{-3}}=1.01\ \Omega \) (3 true); for 10 mA \( 0.50\ \Omega \) (4 false).

Teacher's Note:
1. A shunt goes in parallel and carries the excess current; a multiplier goes in series and drops the excess voltage.
2. Full-scale current of the galvanometer is the fixed quantity in every conversion.

Answer: (b) 1, 2, 3 and 4

Explanation:
1. The cell of larger emf drives a current through the other in the loop, charging it (1).
2. Both cells are connected between the same two points, so they have the same terminal voltage (2).
3. The charged cell has terminal voltage \( \varepsilon+ir>\varepsilon \) (3).
4. The loop current is \( \frac{\varepsilon_1-\varepsilon_2}{r_1+r_2} \) and \( V_{AB}=\varepsilon_1-ir_1=\frac{\varepsilon_1r_2+\varepsilon_2r_1}{r_1+r_2} \) (4).

Teacher's Note:
1. Reduce the network or apply Kirchhoff's laws; check the answer with a power balance.
2. Remember that an ideal voltmeter has infinite resistance and an ideal ammeter has zero resistance.

Answer: (c) 2 and 3 only

Explanation:
1. V1 and V2 are in series with different resistances, so the voltage divides unequally: \( V_1\ne V_2 \) (2).
2. The series pair is in parallel with V3, so \( V_1+V_2=V_3 \) (3).

Teacher's Note:
1. Reduce the network or apply Kirchhoff's laws; check the answer with a power balance.
2. Remember that an ideal voltmeter has infinite resistance and an ideal ammeter has zero resistance.

Answer: (d) 2 and 3 only

Explanation:
1. The potential difference across the whole wire is \( \varepsilon\frac{R_{AB}}{R_{AB}+r}<\varepsilon \).
2. At balance \( \varepsilon\frac{R_{AB}}{R_{AB}+r}\cdot\frac{\ell}{100}=\frac\varepsilon2 \), so \( \ell=50\left(1+\frac{r}{R_{AB}}\right)>50 \) cm (2).
3. Balance exists only if \( \ell\le100 \), i.e. \( R_{AB}\ge r \) (3).

Teacher's Note:
1. In a potentiometer or metre bridge, a null condition means no current in the galvanometer branch.
2. Write the balance condition as a ratio of lengths or resistances.

Answer: (b) 2, 3 and 4

Explanation:
1. On open circuit the two fields balance, but statement 1 says 'always': once a current is drawn they are unequal, so it is false.
2. The emf is the work per unit charge done by the non-electrostatic (chemical) force and the potential difference that of the electrostatic field (2).
3. When the cell is being charged the current flows + to - inside it (3), and when a load is connected the terminal voltage (electrostatic field) falls below the emf (4).

Teacher's Note:
1. Test each statement separately against the physical law behind it.
2. Then choose the option that lists exactly the true statements.

Answer: (c) 6 V

Explanation:
1. Battery 2 fixes the voltage across the parallel branches at \( E_2 \), so the current in battery 1 is \( \frac{E_1-E_2}{R_1} \): it decreases with \( E_2 \) and vanishes at \( E_2=E_1 \).
2. The falling line has slope \( \frac{-0.1-0.1}{8-4}=-0.05 \) A/V; from \( 0.1 \) A at 4 V it reaches zero at 6 V.
3. So \( E_1=6 \) V.

Teacher's Note:
1. Reduce the network or apply Kirchhoff's laws; check the answer with a power balance.
2. Remember that an ideal voltmeter has infinite resistance and an ideal ammeter has zero resistance.

Answer: (a) 40 \( \Omega \)

Explanation:
1. The current through battery 2 is \( \frac{E_2}{R_2}+\frac{E_2-E_1}{R_1} \): slope \( \frac1{R_2}+\frac1{R_1}=\frac{0.3}{4}=0.075 \) A/V.
2. With \( \frac1{R_1}=0.05 \): \( \frac1{R_2}=0.025 \), so \( R_2=40\ \Omega \). (Check: its current is zero at \( E_2=\frac{E_1/R_1}{0.075}=4 \) V.)

Teacher's Note:
1. Reduce the network or apply Kirchhoff's laws; check the answer with a power balance.
2. Remember that an ideal voltmeter has infinite resistance and an ideal ammeter has zero resistance.

Answer: (a) 0.067 div/\( \mu A \)

Explanation:
1. Same current through G and the wire, and equal voltages mean \( G=S=50\ \Omega \).
2. Current \( I=\frac{1.5}{10000+50+50}=1.485\times10^{-4} \) A = 148.5 \( \mu A \).
3. Sensitivity \( \frac{10}{148.5}=0.067 \) div/\( \mu A \).

Teacher's Note:
1. A shunt goes in parallel and carries the excess current; a multiplier goes in series and drops the excess voltage.
2. Full-scale current of the galvanometer is the fixed quantity in every conversion.

Answer: (b) \( R\gg G \)

Explanation:
1. Without the wire \( I_G=\frac{E}{R+G} \). With the wire in parallel, \( I_G'=\frac{ES}{R(S+G)+SG} \).
2. Setting \( I_G'=\frac12I_G \) gives \( R=\frac{SG}{G-S} \), which is very large when \( G=S=50\ \Omega \).
3. If \( R\gg G \) the supply current is fixed and the shunt S = G splits it equally, giving d/2.

Teacher's Note:
1. A shunt goes in parallel and carries the excess current; a multiplier goes in series and drops the excess voltage.
2. Full-scale current of the galvanometer is the fixed quantity in every conversion.

Answer: (c) 16

Explanation:
1. Current \( I=\frac{16\ \text{mV}}{100\ \Omega}=160\ \mu A \).
2. For 10 divisions: \( \frac{160}{10}=16 \) \( \mu A \) per division.

Teacher's Note:
1. A shunt goes in parallel and carries the excess current; a multiplier goes in series and drops the excess voltage.
2. Full-scale current of the galvanometer is the fixed quantity in every conversion.

Answer: (d) 75

Explanation:
1. The same current 160 \( \mu A \) flows through the galvanometer.
2. So \( G=\frac{12\ \text{mV}}{0.16\ \text{mA}}=75\ \Omega \).

Teacher's Note:
1. A shunt goes in parallel and carries the excess current; a multiplier goes in series and drops the excess voltage.
2. Full-scale current of the galvanometer is the fixed quantity in every conversion.

Answer: (b) 12

Explanation:
1. The meter branch has resistance \( 25075\ \Omega \), negligible compared with 24 \( \Omega \): current in the circuit \( \frac{5}{24+1}\approx0.2 \) A, voltage across the 24 \( \Omega \) about 4.8 V.
2. Meter current \( \frac{4.8}{25075}=191\ \mu A \).
3. Divisions \( \frac{191}{16}\approx12 \).

Teacher's Note:
1. A shunt goes in parallel and carries the excess current; a multiplier goes in series and drops the excess voltage.
2. Full-scale current of the galvanometer is the fixed quantity in every conversion.