JEE Physics Current Electricity MCQs Set 23

Answer: (d) \( \frac1{R_{100}}>\frac1{R_{60}}>\frac1{R_{40}} \)

Explanation:
1. For the same supply voltage \( R=\frac{V^2}{P} \), so the higher-power bulb has the smaller resistance: \( R_{100}<R_{60}<R_{40} \).
2. Taking reciprocals: \( \frac1{R_{100}}>\frac1{R_{60}}>\frac1{R_{40}} \).

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) 4

Explanation:
1. Series: \( I_1=\frac{2E}{R+2} \); parallel: \( I_2=\frac{E}{R+0.5} \).
2. Heat \( \propto I^2 \): \( \frac{I_1}{I_2}=\sqrt{2.25}=1.5=\frac{2(R+0.5)}{R+2} \).
3. So \( 2R+1=1.5R+3 \) and \( R=4\ \Omega \).

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, 3 and 4

Explanation:
1. By the symmetry of the network (left half 2 \( \Omega \) and 4 \( \Omega \) mirrored by the right half), solving the node equations gives \( V_P=V_Q=8 \) V and \( V_S=V_T=4 \) V for the left corner at 12 V and the right corner at 0.
2. So no current flows in PQ (1); \( I_1=\frac{12-8}{2}+\frac{12-8}{4}=3 \) A (2); \( V_S=4<V_Q=8 \) (3); \( I_2=\frac{8-4}{2}=2 \) A (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: (d) 2 and 4 only

Explanation:
1. A wire of diameter 2d has area four times as large, so its resistance is \( \frac{R_0}{4} \).
2. Series: \( \frac{R_0}{2} \); parallel: \( \frac{R_0}{8} \). For the same voltage, \( t=\frac{Q R}{V^2}\propto R \).
3. Series: \( \frac42=2 \) min (2); parallel: \( \frac48=0.5 \) min (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. With no current in \( R_2 \) the voltage between the two nodes is zero, so the two source branches must carry equal and opposite currents: \( \frac{V_1}{R_1}=\frac{V_2}{R_3} \).
2. 1: \( \frac{V}{R}=\frac VR \) yes. 2: \( R_1=R_3 \), \( V_1=V_2 \) yes. 3: \( \frac{2V_2}{R_1} \) versus \( \frac{V_2}{2R_1} \) no. 4: \( \frac{V_2/2}{R_1}=\frac{V_2}{2R_1}=\frac{V_2}{R_3} \) yes.

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) \( 60\pm0.25\ \Omega \)

Explanation:
1. \( \frac{R}{90}=\frac{40}{60} \) gives \( R=60\ \Omega \).
2. From \( R=\frac{90\,\ell}{100-\ell} \): \( \frac{dR}{d\ell}=\frac{9000}{(100-\ell)^2}=2.5 \) ohm per cm.
3. Uncertainty for 0.1 cm: \( 2.5\times0.1=0.25\ \Omega \).

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: (d) \( \frac{1875}{64}\ \mu\Omega \)

Explanation:
1. Aluminium area \( 49-4=45\ \text{mm}^2 \): \( R_{Al}=\frac{2.7\times10^{-8}\times0.05}{45\times10^{-6}}=30\ \mu\Omega \). Iron: \( R_{Fe}=\frac{10^{-7}\times0.05}{4\times10^{-6}}=1250\ \mu\Omega \).
2. In parallel: \( \frac{30\times1250}{1280}=\frac{1875}{64}\ \mu\Omega \).

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) 2 and 3 only

Explanation:
1. Voltage range: all in series gives \( I_g(2R+2R_C) \); arrangement 2 gives total current \( 2I_g \) through \( 2R+\frac{R_C}{2} \): \( I_g(4R+R_C) \), larger since \( 2R>R_C \) (statement 2 true).
2. Current range: all parallel gives \( 2I_g\left(1+\frac{R_C}{R}\right) \); arrangement 4 gives \( I_g\left(1+\frac{4R_C}{R}\right) \), smaller since \( R_C<\frac R2 \) (statement 3 true).

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) In a balanced Wheatstone bridge, if the cell and the galvanometer are exchanged, the null point is disturbed

Explanation:
1. The balance condition \( \frac PQ=\frac RS \) is symmetric: interchanging the cell and galvanometer does not disturb the null point, so that statement is false.
2. The other three statements are standard true statements.

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: (c) \( 1.985\times10^3\ \Omega \)

Explanation:
1. \( R=\frac{V}{I_g}-G=\frac{10}{5\times10^{-3}}-15=1985\ \Omega \).

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: (d) 11.5 V and 11.6 V

Explanation:
1. Millman's theorem: \( V=\frac{12/1+13/2}{1/1+1/2+1/10}=\frac{18.5}{1.6}=11.56 \) V.
2. This lies between 11.5 V and 11.6 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) \( 1.5\ \Omega \)

Explanation:
1. Open circuit: \( E\propto52 \). Shunted: terminal voltage \( V=\frac{5E}{5+r}\propto40 \).
2. So \( \frac{5}{5+r}=\frac{40}{52} \), giving \( 5+r=6.5 \) and \( r=1.5\ \Omega \).

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) \( 550\ \Omega \)

Explanation:
1. Let the balance point be at l cm: \( \frac XY=\frac{l}{100-l} \). After interchange \( \frac YX=\frac{l-10}{110-l} \).
2. Multiplying: \( (100-l)(110-l)=l(l-10) \) gives \( 11000=200l \), so \( l=55 \) cm.
3. Then \( X=\frac{55}{100}\times1000=550\ \Omega \).

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: (c) 27 k\( \Omega \), 10%

Explanation:
1. Red = 2, violet = 7, orange = multiplier \( 10^3 \): \( 27\times10^3=27\ \text{k}\Omega \).
2. Silver tolerance is 10%.

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) 0.3 V

Explanation:
1. First case: current \( \frac{6}{2+4}=1 \) A, so the wire drop is 4 V and \( \frac{AJ}{AB}=\frac{0.5}{4} \).
2. Second case: current \( \frac{6}{6+4}=0.6 \) A, wire drop 2.4 V.
3. Same fraction: \( \varepsilon_2=\frac{0.5}{4}\times2.4=0.3 \) V.

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: (d) \( 100\ \Omega \)

Explanation:
1. The null point means the potential drop along 100 cm of wire equals the emf, 5 V (no current in the cell, so its internal resistance is irrelevant).
2. Drop along 120 cm: \( 5\times\frac{120}{100}=6 \) V.
3. Resistance \( R=\frac{6}{0.060}=100\ \Omega \).

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) 9.9 k\( \Omega \)

Explanation:
1. \( R=\frac{10}{10^{-3}}-100=9900\ \Omega=9.9\ \text{k}\Omega \).

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) 45 V

Explanation:
1. With the meter across 400 ohm: 60 V is dropped across the 300 ohm resistor, so the current is 0.2 A and the 400 ohm resistor in parallel with the meter is \( \frac{60}{0.2}=300\ \Omega \), giving \( R_V=1200\ \Omega \).
2. Across 300 ohm: \( \frac{300\times1200}{1500}=240\ \Omega \), total \( 640\ \Omega \).
3. Reading \( 120\times\frac{240}{640}=45 \) V.

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) N = 6 and P = 30

Explanation:
1. Ammeter: total resistance \( 400+800+10=1210 \), so \( I=\frac{6}{1210} \) A and N = 6.
2. Voltmeter: \( 400\parallel1000=\frac{2000}{7}\ \Omega \), total \( 800+\frac{2000}{7}=\frac{7600}{7} \).
3. Reading \( 6\times\frac{2000/7}{7600/7}=\frac{12000}{7600}=\frac{30}{19} \) V, so P = 30.

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.