JEE Physics Current Electricity MCQs Set 12

Answer: (a) 0.005 V/cm

Explanation:
1. Current \( I=\frac{2}{15+5}=0.1 \) A (ideal battery).
2. Potential drop across the wire \( =0.1\times5=0.5 \) V over 100 cm.
3. Gradient \( =0.005 \) V/cm.

Teacher's Note:
1. Potential gradient = voltage across the wire ÷ its length.
2. The 15 Ω series resistor lowers the current and hence the gradient.

Answer: (d) 19995 Ω

Explanation:
1. Required total resistance \( =\frac{100}{5\times10^{-3}}=20000 \) Ω.
2. Series resistance \( =20000-5=19995 \) Ω.

Teacher's Note:
1. Voltmeter conversion: R = V/I_g − G.
2. The series resistance is much bigger than the instrument resistance.

Answer: (c) 20 Ω

Explanation:
1. Balance condition: \( \frac{R}{80}=\frac{AC}{CB}=\frac{20}{80} \).
2. \( R=20 \) Ω.

Teacher's Note:
1. In a meter bridge R/S = l/(100 − l).
2. The unknown in the left gap goes with the length on the left.

Answer: (c) 4 Ω, 1.0 A

Explanation:
1. \( \frac X6=\frac{40}{60}\Rightarrow X=4 \) Ω.
2. The branch X + 6 = 10 Ω is in parallel with the bridge wire of \( 100\times0.1=10 \) Ω, giving 5 Ω in total.
3. \( I=\frac{5}{5}=1.0 \) A.

Teacher's Note:
1. The bridge wire is another parallel path across the cell.
2. Convert length to resistance using the resistance per cm.

Answer: (c) I/2

Explanation:
1. \( I=neAv \) with \( A\propto r^2 \).
2. New area \( =\frac A4 \) and velocity \( 2v \): \( I'=\frac14\times2\times I=\frac I2 \).

Teacher's Note:
1. Combine the effects of area and velocity as ratios.
2. Same material means the same electron density n.

Answer: (b) 954 Ω

Explanation:
1. \( R_1=600[1+0.001(30)]=618 \) Ω.
2. \( R_2=300[1+0.004(30)]=336 \) Ω.
3. Series: \( 618+336=954 \) Ω.

Teacher's Note:
1. Apply R = R₀(1 + αΔT) separately, then add in series.
2. The temperature rise is 50 − 20 = 30°C.

Answer: (c) 1/500 per degree

Explanation:
1. For series coils the increase in resistance per degree adds: \( R_0\alpha=600(0.001)+300(0.004)=0.6+1.2=1.8 \) Ω/°C.
2. \( \alpha=\frac{1.8}{900}=0.002=\frac1{500} \) per °C.

Teacher's Note:
1. The effective α is the resistance-weighted average of the individual coefficients.
2. For parallel coils the weighting would be different.

Answer: (b) 16 : 1

Explanation:
1. \( v=\frac{I}{neA}\propto\frac{I}{r^2} \).
2. \( \frac{v_1}{v_2}=\frac41\times\frac{4}{1}=16 \).

Teacher's Note:
1. Area is proportional to the square of the radius.
2. Same material means the same n.

Answer: (d) 60 Ω

Explanation:
1. Voltage across the parallel combination: \( V=0.3\times20=6 \) V.
2. Current through the 15 Ω resistor: \( \frac{6}{15}=0.4 \) A. Remaining current through R₁: \( 0.8-0.3-0.4=0.1 \) A.
3. \( R_1=\frac{6}{0.1}=60 \) Ω.

Teacher's Note:
1. Parallel branches share the same voltage.
2. Branch currents add up to the ammeter reading.

Answer: (c) 12 V

Explanation:
1. During charging \( V=E+Ir \).
2. \( 12.5=E+0.5\times1\Rightarrow E=12 \) V.

Teacher's Note:
1. A cell being charged has terminal voltage greater than its emf.
2. During discharge it is V = E − Ir instead.

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

Explanation:
1. Each part has resistance \( \frac Rn \); n of them in parallel give \( \frac{R}{n^2} \).
2. \( P=\frac{V^2}{R/n^2}=n^2\frac{V^2}{R}=n^2H \).

Teacher's Note:
1. Cutting into n parts and joining them in parallel cuts the resistance by n².
2. Power across a fixed voltage is inversely proportional to resistance.

Answer: (d) 4 Ω

Explanation:
1. Balance: \( \frac{6}{R}=\frac{AD}{DB}=\frac{30}{20} \).
2. \( R=4 \) Ω.

Teacher's Note:
1. On a 50 cm wire the right length is 50 − 30 = 20 cm.
2. The resistance ratio equals the ratio of the lengths.

Answer: (c) a high resistance in series

Explanation:
1. A voltmeter needs a very high resistance and must limit the current to the ammeter's range.
2. A large resistance connected in series does this.

Teacher's Note:
1. Voltage reading = I × (R_ammeter + R_series).
2. A series resistor converts the ammeter into a voltmeter.

Answer: (c) Linear increase for Cu, exponential decrease for Si

Explanation:
1. For a metal such as copper, resistance rises almost linearly with temperature in this range.
2. For intrinsic silicon the number of carriers grows exponentially with temperature, so resistance falls exponentially.

Teacher's Note:
1. Metals: positive temperature coefficient. Semiconductors: negative.
2. Carrier number controls semiconductor conduction.

Answer: (a) 0.01 Ω

Explanation:
1. Shunt: \( S=\frac{I_gG}{I-I_g}=\frac{10^{-3}\times100}{10-10^{-3}}\approx0.01 \) Ω.

Teacher's Note:
1. The shunt carries nearly all the current.
2. Connect it in parallel with the galvanometer.

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

Explanation:
1. (a), (b) and (c) are standard true statements.
2. In a balanced bridge the cell and galvanometer are interchangeable: the balance condition P/Q = R/S is unchanged, so the null point is not disturbed. Statement (d) is false.

Teacher's Note:
1. The balance condition of a Wheatstone bridge does not involve the cell or the galvanometer.
2. Sensitivity is highest when all four arms are of the same order.

Answer: (c) \( 1.985\times10^3 \) Ω

Explanation:
1. Total resistance needed \( =\frac{10}{5\times10^{-3}}=2000 \) Ω.
2. Series resistance \( =2000-15=1985 \) Ω \( =1.985\times10^3 \) Ω.

Teacher's Note:
1. Voltmeter range V = I_g (G + R).
2. Subtract the coil resistance from the total.

Answer: (a) 20 mA

Explanation:
1. Green (5), black (0), red (×10²): \( R=50\times10^2=5000 \) Ω (brown = 1% tolerance).
2. \( P=I^2R\Rightarrow I=\sqrt{\frac{2}{5000}}=0.02 \) A = 20 mA.

Teacher's Note:
1. The fourth band only gives tolerance.
2. Maximum current from the power rating: I = √(P/R).

Answer: (c) 1.0%

Explanation:
1. At constant volume \( R\propto l^2 \), so \( \frac{\Delta R}{R}=2\frac{\Delta l}{l}=2\times0.5\%=1\% \).

Teacher's Note:
1. For small changes the percentage change in R is twice that of l.
2. Resistance rises by about 1%.

Answer: (a) 0.02

Explanation:
1. \( v=\frac{I}{neA}=\frac{1.5}{9\times10^{28}\times1.6\times10^{-19}\times5\times10^{-6}}=2.1\times10^{-5} \) m/s.
2. That is about 0.02 mm/s.

Teacher's Note:
1. Convert mm² to m² (1 mm² = 10⁻⁶ m²).
2. Drift speeds are of the order of hundredths of a millimetre per second.

Answer: (a) \( 11\times10^{-5} \) W

Explanation:
1. \( R=\frac{P}{I^2}=\frac{4.4}{(2\times10^{-3})^2}=1.1\times10^6 \) Ω.
2. \( P'=\frac{V^2}{R}=\frac{121}{1.1\times10^6}=1.1\times10^{-4}=11\times10^{-5} \) W.

Teacher's Note:
1. First find the resistance from P = I²R.
2. Then apply P = V²/R at the new voltage.

Answer: (c) 500 Ω

Explanation:
1. Red, black, brown: \( 20\times10=200 \) Ω.
2. Green is 5: \( 50\times10=500 \) Ω.

Teacher's Note:
1. First band = first digit, second band = second digit, third band = multiplier.
2. Green stands for 5.