JEE Physics Current Electricity MCQs Set 24

Answer: (a) proportional to \( \sqrt T \)

Explanation:
1. The thermal (random) speed comes from the kinetic energy of electrons: \( \tfrac12mv^2\propto T \).
2. So \( v_{th}\propto\sqrt T \), independent of the applied potential difference.

Teacher's Note:
1. The applied field only adds a very small drift to the large random motion.
2. Thermal speed is about 10⁵ m/s while drift speed is about 10⁻⁴ m/s.

Answer: (c) \( I/2 \)

Explanation:
1. \( I=neAv \) with \( A\propto r^2 \).
2. Area becomes one-fourth and the speed doubles: \( I'=I\times\tfrac14\times2=I/2 \).

Teacher's Note:
1. Same material means the same carrier density n.
2. Area scales as the square of the radius.

Answer: (b) \( 1\times10^{6}\pm5\% \)

Explanation:
1. Brown = 1, black = 0, green = multiplier \( 10^5 \), gold = tolerance 5%.
2. Value \( 10\times10^5=1\times10^6\ \Omega\pm5\% \).

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.
2. Gold tolerance is 5%, silver 10%.

Answer: (a) \( 5.7\times10^{-5}\ \Omega \)

Explanation:
1. Outer radius 5 cm, inner radius 4.5 cm: \( A=\pi(0.05^2-0.045^2)=1.49\times10^{-3}\ \text{m}^2 \).
2. \( R=\dfrac{\rho L}{A}=\dfrac{1.7\times10^{-8}\times5}{1.49\times10^{-3}}=5.7\times10^{-5}\ \Omega \).

Teacher's Note:
1. The cross-section is the annulus, not a full disc.
2. Convert cm and mm to metres.

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

Explanation:
1. With \( R=\rho L/A \): \( x=\dfrac{\rho\,4a}{2a\cdot a}=\dfrac{2\rho}{a} \), \( y=\dfrac{\rho\,2a}{4a\cdot a}=\dfrac{\rho}{2a} \), \( z=\dfrac{\rho\,a}{4a\cdot2a}=\dfrac{\rho}{8a} \).
2. So \( x>y>z \).

Teacher's Note:
1. The longest path through the smallest area gives the largest resistance.
2. Use the same formula for each pair of faces.

Answer: (a) 2.4 V

Explanation:
1. Resistances: \( R_1=\dfrac{\rho\,2L}{A} \) and \( R_2=\dfrac{\rho L}{2A}=\dfrac{R_1}{4} \).
2. The 7 V drop divides in the ratio 4 : 1: the long wire takes \( \tfrac45\times7=5.6 \) V.
3. Junction potential \( 8.0-5.6=2.4\ \text{V} \).

Teacher's Note:
1. Same material and same current: potential drop is proportional to resistance.
2. Check your answer lies between 1 V and 8 V.

Answer: (d) \( y/x \)

Explanation:
1. \( V=E-Ir \): at \( I=0 \), \( V=E=y \); at \( V=0 \), \( I=E/r=x \).
2. So \( r=y/x \).

Teacher's Note:
1. The slope magnitude of the V–I line is the internal resistance.
2. The intercepts give emf and short-circuit current.

Answer: (d) 1/2

Explanation:
1. Halving per section means the shunt combination (R₂ in parallel with the rest of the ladder) equals R₁, so the whole ladder has \( R_{eq}=2R_1 \).
2. Then \( \dfrac{R_2\cdot2R_1}{R_2+2R_1}=R_1 \) gives \( R_2=2R_1 \).
3. So \( R_1/R_2=1/2 \).

Teacher's Note:
1. An infinite ladder looks the same after removing one section.
2. Equal voltage division requires the parallel part to equal the series resistor.

Answer: (c) 89%

Explanation:
1. Line current \( I=\dfrac{15000}{240}=62.5\ \text{A} \), loss \( I^2R=62.5^2\times0.5=1953\ \text{W} \).
2. Efficiency \( \dfrac{15000}{15000+1953}=0.885\approx89\% \).

Teacher's Note:
1. The input is output plus line loss.
2. Current is delivered power over delivered voltage.

Answer: (b) both the length and the radius of the wire are doubled

Explanation:
1. \( H=\dfrac{V^2t}{R} \) with \( R\propto\dfrac{L}{r^2} \).
2. Doubling both L and r gives \( R'=\dfrac{2L}{4r^2}\rho\ldots=R/2 \), so H doubles.
3. Halving both makes R double and H halve; doubling only r makes H four times; doubling only L halves H.

Teacher's Note:
1. Heat at fixed voltage is inversely proportional to resistance.
2. Check each option's effect on R.

Answer: (b) proportional to the square of their marked voltage

Explanation:
1. For a bulb \( R=V^2/P \), so at equal rated power \( R\propto V^2 \).
2. In series the current is the same, so \( P_{actual}=I^2R\propto V^2 \).

Teacher's Note:
1. Higher marked voltage means higher resistance for equal power rating.
2. In series, power is proportional to resistance.

Answer: (a) \( \dfrac R6\ \text{joule} \)

Explanation:
1. Current \( i=dQ/dt=2-16t \), which vanishes at t = 1/8 s.
2. Heat \( =\int_0^{1/8}i^2R\,dt=R\int_0^{1/8}4(1-8t)^2dt=\dfrac{4R}{24}=\dfrac R6 \).

Teacher's Note:
1. Differentiate the charge to get the current.
2. Integrate i squared times R over the interval.

Answer: (a) increase by about 11%

Explanation:
1. \( P=V^2/R \) and \( R\propto L \): \( P'/P=\dfrac{1}{0.9}=1.11 \).

Teacher's Note:
1. At fixed supply voltage, power is inversely proportional to resistance.
2. Reducing resistance by 10% raises power by 1/0.9.

Answer: (c) 200 W

Explanation:
1. \( R_A=\dfrac{200^2}{300}=133.3\ \Omega \), \( R_B=\dfrac{200^2}{600}=66.7\ \Omega \).
2. In series \( R=200\ \Omega \) and \( P=\dfrac{200^2}{200}=200\ \text{W} \).

Teacher's Note:
1. Find each resistance from the rated power at 200 V.
2. In series, resistances add.

Answer: (d) 48 V

Explanation:
1. The voltmeter in parallel with R₂ gives \( \dfrac{60\times120}{60+120}=40\ \Omega \).
2. Total resistance 60 + 40 = 100 Ω; current 1.2 A.
3. Reading \( 1.2\times40=48\ \text{V} \).

Teacher's Note:
1. A real voltmeter loads the circuit and reads less than the 60 V that R₂ would have without it.
2. Parallel resistance first, then series.

Answer: (a) 12 V

Explanation:
1. The ideal voltmeter has very large resistance, so the current is practically zero.
2. No drop occurs in the cell's internal resistance or the 4 Ω resistor, so the voltmeter reads the full emf.

Teacher's Note:
1. A very high-resistance meter in series blocks the current.
2. The voltage across the meter is then the full emf.

Answer: (a) \( \dfrac23\ \text{m} \)

Explanation:
1. The potential gradient of the wire is \( 6\ \text{V per m} \).
2. Zero current means the 4 V cell is balanced by the potential across AC: \( 6\cdot AC=4 \), so \( AC=\dfrac23\ \text{m} \).

Teacher's Note:
1. No current in the cell branch means its internal resistance has no drop.
2. This is the potentiometer principle.

Answer: (d) 20 cm

Explanation:
1. Initially \( AJ=\dfrac{12}{12+18}\times100=40\ \text{cm} \).
2. With R = 8 Ω: \( AJ'=\dfrac{12}{12+8}\times100=60\ \text{cm} \).
3. The jockey moves 20 cm.

Teacher's Note:
1. Balance length from A is 100 X over (X + R).
2. Smaller R in the right gap shifts the balance point to the right.

Answer: (c) 16 : 1

Explanation:
1. \( v_d=\dfrac{I}{neA} \).
2. \( \dfrac{v_1}{v_2}=\dfrac{I_1}{I_2}\cdot\dfrac{A_2}{A_1}=\dfrac41\cdot\dfrac41=16 \).

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

Answer: (a) \( \dfrac{i}{psq} \)

Explanation:
1. \( i=pAvq \) with \( A=s \) gives \( v=\dfrac{i}{psq} \).

Teacher's Note:
1. The length of the wire does not appear in the drift velocity.
2. Carrier density is per unit volume.

Answer: (c) \( 0.2\times10^{-6}\ \Omega\,\text{m} \)

Explanation:
1. \( \rho=\dfrac{RA}{L}=\dfrac{5\times6\times10^{-7}}{15}=2\times10^{-7}=0.2\times10^{-6}\ \Omega\,\text{m} \).

Teacher's Note:
1. Resistivity equals resistance times area over length.
2. A negligible current ensures no heating.

Answer: (d) 20 Ω

Explanation:
1. Stretching keeps the volume constant, so \( R\propto L^2 \).
2. Doubling the length multiplies R by 4: 20 Ω.

Teacher's Note:
1. Both length and area change on stretching.
2. Resistance varies as the square of length at constant volume.

Answer: (c) \( \left|\dfrac{\alpha_C\rho_C}{\alpha_B\rho_B}\right| \)

Explanation:
1. The total resistance \( R=\dfrac{1}{A}\left[\rho_Bt_B(1+\alpha_B\Delta T)+\rho_Ct_C(1+\alpha_C\Delta T)\right] \).
2. Independence of temperature requires \( \rho_Bt_B\alpha_B+\rho_Ct_C\alpha_C=0 \).
3. Hence \( \dfrac{t_B}{t_C}=\left|\dfrac{\alpha_C\rho_C}{\alpha_B\rho_B}\right| \) (the coefficients have opposite signs).

Teacher's Note:
1. Carbon has a negative coefficient, brass a positive one.
2. The two contributions to the temperature variation must cancel.