JEE Physics Current Electricity MCQs Set 18

Answer: (b) 3 and 4 only

Explanation:
1. Free electrons are in continuous random thermal motion, so statements 1 and 2 are false (speed is never zero).
2. Random motion in all directions makes the average velocity of one electron over a long time zero, and the average over all electrons at an instant also zero, so 3 and 4 are true.

Teacher's Note:
1. Speed is a magnitude and cannot average to zero; velocity is a vector and can.
2. Zero net current means zero average velocity, not zero motion.

Answer: (c) A is true, but R is false.

Explanation:
1. The field is set up through the wire at nearly the speed of light, so all electrons start drifting at once and the bulb glows immediately: A is true.
2. Drift velocity is only about 10⁻⁴ m/s, so R is false.

Teacher's Note:
1. Signal speed and drift speed are very different.
2. The bulb lights as soon as the field reaches the filament.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Copper has a low resistivity, so a wire of copper loses very little energy as heat.
2. This high conductivity is exactly why copper is used for connecting wires.

Teacher's Note:
1. Low resistance means little voltage drop along the connections.
2. Silver is better but too costly.

Answer: (c) A is true, but R is false.

Explanation:
1. Charge cannot accumulate at a junction, so the sum of currents in equals the sum out: A is true.
2. The loop rule follows from conservation of energy (the electrostatic field is conservative), not momentum: R is false.

Teacher's Note:
1. Junction rule: charge. Loop rule: energy.
2. Learn the pairing.

Answer: (c) A is true, but R is false.

Explanation:
1. The torque on the coil reverses with the current, so its average is zero and the pointer cannot settle: A is true.
2. AC does produce heat in a resistor (I² R is positive for either direction), so R is false.

Teacher's Note:
1. Moving-coil meters respond to the average current, which is zero for pure AC.
2. Hot-wire instruments use the heating effect to measure AC.

Answer: (d) A is false, but R is true.

Explanation:
1. Heating the metal resistor raises its resistance, so to keep the ratio the same the standard resistance must be increased, not decreased: A is false.
2. R is a correct statement.

Teacher's Note:
1. Null point stays fixed if the two resistances keep the same ratio.
2. Both gap resistances must change in the same direction.

Answer: (a) 400 C

Explanation:
1. \( q=\int_0^{10}(20+4t)\,dt=\left[20t+2t^2\right]_0^{10} \).
2. \( =200+200=400\ \text{C} \).

Teacher's Note:
1. Charge is the integral of current over time.
2. The average current is 40 A over 10 s.

Answer: (c) 4 Ω

Explanation:
1. Series: \( I_1=\frac{2E}{R+2} \). Parallel: \( I_2=\frac{E}{R+0.5} \).
2. \( \frac{J_1}{J_2}=\frac{I_1^2}{I_2^2}=\frac{4(R+0.5)^2}{(R+2)^2}=2.25 \), so \( \frac{2(R+0.5)}{R+2}=1.5 \).
3. \( 2R+1=1.5R+3\Rightarrow R=4\ \Omega \).

Teacher's Note:
1. Both arrangements have the same load R, so heat is proportional to current squared.
2. Two identical cells in parallel have emf E and internal resistance 0.5 Ω.

Answer: (a) 0.4 V

Explanation:
1. The voltmeter reading is \( V=iR+z \), where z is the zero error.
2. Subtracting the two readings: \( R=\frac{22.4-14.4}{2.75-1.75}=8\ \Omega \).
3. \( z=14.4-1.75\times8=0.4\ \text{V} \).

Teacher's Note:
1. The slope of the V–i line gives the true resistance.
2. The intercept on the V axis is the zero error.

Answer: (a) 0.07 Ω

Explanation:
1. Open-circuit reading is the emf, 1.52 V.
2. Lost volts \( =1.52-1.45=0.07\ \text{V}=Ir \) with I = 1.0 A, so r = 0.07 Ω.

Teacher's Note:
1. A high-resistance voltmeter on open circuit reads the emf.
2. Divide the lost voltage by the current.

Answer: (a) 2 : 1

Explanation:
1. The half coil has resistance R/2.
2. \( \frac{H'}{H}=\frac{V^2/(R/2)}{V^2/R}=2 \).

Teacher's Note:
1. At fixed voltage, power is inversely proportional to resistance.
2. Halving the length halves the resistance.

Answer: (c) its length

Explanation:
1. Heat generated per unit length is \( \frac{i^2\rho}{\pi r^2} \) and heat lost per unit length depends on the radius, so the melting current depends on ρ and r only.
2. The length cancels out, so it is immaterial.

Teacher's Note:
1. Both heat generation and loss are proportional to length.
2. Fuse rating is chosen by material and thickness.

Answer: (d) 1 : 4

Explanation:
1. Series resistance 2R, parallel resistance R/2.
2. \( \frac{H_s}{H_p}=\frac{V^2/2R}{V^2/(R/2)}=\frac{1}{4} \).

Teacher's Note:
1. At fixed voltage, more resistance means less heat.
2. The parallel combination dissipates four times as much as the series one.

Answer: (b) \( P=\left(\frac{V}{V_0}\right)^2P_0 \)

Explanation:
1. \( R=\frac{V_0^2}{P_0} \) is constant.
2. \( P=\frac{V^2}{R}=\frac{V^2}{V_0^2}P_0 \).

Teacher's Note:
1. For a fixed resistance power varies as the square of voltage.
2. Rated power and voltage give the filament resistance.

Answer: (b) \( R_A<R_B \)

Explanation:
1. In parallel the voltage is the same, so \( P=\frac{V^2}{R} \).
2. The brighter bulb has the greater power and hence the smaller resistance: \( R_A<R_B \).

Teacher's Note:
1. Parallel: brightness is inversely proportional to resistance.
2. Series: brightness is proportional to resistance.

Answer: (a) \( \frac14 \) amp

Explanation:
1. \( R=\frac{200^2}{100}=400\ \Omega \).
2. \( i=\frac{100}{400}=\frac14\ \text{A} \).

Teacher's Note:
1. Use the rating to find the resistance, then Ohm's law at the actual voltage.
2. The rated current of 0.5 A flows only at 200 V.

Answer: (a) \( \frac12KT_0 \)

Explanation:
1. Thermoelectric power \( S=\frac{dE}{dT} \).
2. \( S=K\left[T_0-\frac{T+T_r}{2}-\frac{T-T_r}{2}\right]=K(T_0-T) \).
3. At \( T=\frac{T_0}{2} \): \( S=\frac{KT_0}{2} \).

Teacher's Note:
1. Differentiate the product with respect to T, holding Tr fixed.
2. The thermoelectric power vanishes at the neutral temperature T = T0.

Answer: (c) \( R_2=R_3\ \text{and}\ R_1=\frac14R_2 \)

Explanation:
1. Equal dissipation in R2 and R3 (same voltage) requires \( R_2=R_3 \). If the current in R1 is i, each of R2 and R3 carries i/2.
2. \( i^2R_1=\left(\frac i2\right)^2R_2\Rightarrow R_1=\frac{R_2}{4} \).

Teacher's Note:
1. Equal power in parallel resistors needs equal resistances.
2. The current in R1 is the sum of the currents in the other two.

Answer: (a) 3P

Explanation:
1. Each resistor can safely take voltage \( V_m=\sqrt{PR} \), or current \( I_m=\sqrt{P/R} \).
2. In a parallel combination each resistor receives full voltage Vm and dissipates P, so the total is 3P (any series arrangement gives less).

Teacher's Note:
1. The limit is on each resistor, so the best arrangement keeps every resistor at its limit.
2. Parallel keeps the same voltage across all three.

Answer: (b) 275 °C and 540 °C

Explanation:
1. The neutral temperature depends only on the two metals, not on the cold junction temperature, so TN stays 275 °C.
2. Inversion occurs symmetrically about TN: \( T_N=\frac{T_0+T_i}{2}\Rightarrow T_i=2T_N-T_0=550-10=540\ ^\circ\text{C} \).

Teacher's Note:
1. Neutral temperature is a property of the metal pair.
2. Raising the cold junction temperature lowers the inversion temperature.

Answer: (c) half the atomic weight of Cu to the atomic weight of Ag

Explanation:
1. The same current flows, so by Faraday's law the masses deposited per second are proportional to the equivalent weights.
2. Cu has equivalent weight A/2 (valency 2) and Ag has A/1 (valency 1).

Teacher's Note:
1. Mass deposited is proportional to atomic weight divided by valency.
2. Density has nothing to do with the mass deposited.

Answer: (a) 19.8 × 10⁻⁷ kg

Explanation:
1. \( m=Zit=3.3\times10^{-7}\times3\times2 \).
2. \( =19.8\times10^{-7}\ \text{kg} \).

Teacher's Note:
1. Faraday's first law: m = Zq = Zit.
2. Charge passed is 6 C.

Answer: (b) from Sb to Bi at the cold junction

Explanation:
1. In a thermocouple the current flows at the cold junction from the metal that comes earlier in the Seebeck series to the later one.
2. So the current flows from Sb to Bi at the cold junction (and from Bi to Sb at the hot junction).

Teacher's Note:
1. Rule: earlier metal to later metal at the cold junction.
2. At the hot junction the direction is reversed.

Answer: (d) 0.126 g

Explanation:
1. The same charge passes at both poles, so \( \frac{m_{Cu}}{m_{Zn}}=\frac{Z_{Cu}}{Z_{Zn}} \).
2. \( m_{Cu}=0.13\times\frac{31.5}{32.5}=0.126\ \text{g} \).

Teacher's Note:
1. Masses liberated by the same charge are in the ratio of the electrochemical equivalents.
2. Time is not needed since the same current flows in the cell.

Answer: (d) 1 mJ

Explanation:
1. Heat absorbed \( =\sigma\,\Delta T\,q \).
2. \( =(10\times10^{-6})\times10\times10=10^{-3}\ \text{J}=1\ \text{mJ} \).

Teacher's Note:
1. Thomson heat is proportional to both the temperature difference and the charge.
2. Temperature difference in kelvin equals that in degrees Celsius.