JEE Physics Current Electricity MCQs Set 29

Answer: (c) \( R=r \)

Explanation:
1. Series: I = nε/(R + nr). Parallel: I = ε/(R + r/n) = nε/(nR + r).
2. Equating gives R + nr = nR + r, so R = r.

Teacher's Note:
1. Set the two currents equal.

Answer: (b) decreases, the thermal velocity of the electrons increases

Explanation:
1. Higher temperature means more frequent collisions, so τ and the drift velocity v_d = eEτ/m decrease.
2. The thermal speed rises with temperature.

Teacher's Note:
1. τ falls with T.

Answer: (b) 3 Ω

Explanation:
1. The two ends of a diameter divide the ring into two halves of 6 Ω each in parallel.
2. R = 6 × 6/12 = 3 Ω.

Teacher's Note:
1. Two equal halves in parallel.

Answer: (a) charge

Explanation:
1. The current entering a junction equals the current leaving it.
2. This is conservation of charge.

Teacher's Note:
1. Junction rule: charge.

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

Explanation:
1. Momentum per unit length = (nA)(m v_d), and I = nAev_d.
2. So the momentum per unit length = Im/e = I/s.

Teacher's Note:
1. p = I m/e.

Answer: (c) tin-lead alloy

Explanation:
1. A fuse wire needs a low melting point and a high resistivity.
2. A tin-lead alloy has both.

Teacher's Note:
1. Low melting point alloy.

Answer: (d) \( v_d\propto E \)

Explanation:
1. Ohm's law J = σE with J = nev_d requires v_d ∝ E.

Teacher's Note:
1. Linear v_d gives Ohm's law.

Answer: (a) \( \varepsilon,\ \frac rn \)

Explanation:
1. In parallel the emf stays ε.
2. The internal resistances are in parallel, giving r/n.

Teacher's Note:
1. Parallel cells: same emf, smaller r.

Answer: (d) copper strip decreases and that of germanium increases

Explanation:
1. The resistance of a metal falls on cooling.
2. A semiconductor has fewer charge carriers at low temperature, so its resistance rises.

Teacher's Note:
1. Metal: R falls.
2. Semiconductor: R rises.

Answer: (a) it measures potential in open circuit

Explanation:
1. At balance the potentiometer draws no current from the source.
2. So it reads the true emf without the internal-resistance drop.

Teacher's Note:
1. Open-circuit measurement.

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

Explanation:
1. The junction rule expresses charge conservation.
2. The loop rule follows from conservation of energy, not momentum, so the reason is false.

Teacher's Note:
1. Loop rule: energy.

Answer: (d) all of the above

Explanation:
1. Thermistors are made of semiconducting metal oxides such as those of nickel, iron, copper and manganese.

Teacher's Note:
1. Oxides of transition metals.

Answer: (c) I and II

Explanation:
1. In parallel 1/R = Σ1/Rᵢ makes R smaller than the least resistance, and I ∝ 1/R.
2. In series the same current flows through all, so III is false.

Teacher's Note:
1. Series: same current.

Answer: (c) \( \frac{n}{n+1} \)

Explanation:
1. I = ε/(nr + r) and V = I·nr.
2. V/ε = n/(n + 1).

Teacher's Note:
1. V = IR.

Answer: (b) increasing the length of the potentiometer wire

Explanation:
1. A longer wire gives a smaller potential gradient, so a small emf difference corresponds to a larger shift of the balance length.

Teacher's Note:
1. Smaller gradient, higher sensitivity.

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

Explanation:
1. A higher temperature reduces τ and the conductivity, so v_d = eEτ/m decreases at fixed E.
2. The reason explains the assertion.

Teacher's Note:
1. v_d ∝ τ.

Answer: (a) the maximum potential difference between the terminals of a cell when no current is drawn from the cell

Explanation:
1. The emf is the potential difference between the terminals in open circuit.
2. It is the maximum value, greater than the terminal voltage when current flows.

Teacher's Note:
1. Emf is the open-circuit voltage.

Answer: (d) III only

Explanation:
1. The wire must melt easily when the current exceeds the safe value, so the melting point is low.
2. A high specific resistance makes it heat quickly, so III is correct.

Teacher's Note:
1. Low melting point, high resistivity.

Answer: (b) galvanometer circuit

Explanation:
1. The null deflection means no current flows through the galvanometer branch.

Teacher's Note:
1. Zero current in the galvanometer.

Answer: (c) zero

Explanation:
1. The positive ions and the free electrons carry equal and opposite charges.
2. The conductor is electrically neutral even when a current flows.

Teacher's Note:
1. A current-carrying wire is neutral.

Answer: (d) \( (n\varepsilon-6\varepsilon),\ nr \)

Explanation:
1. Each reversed cell subtracts ε instead of adding it, a net loss of 2ε per cell.
2. With three reversed, emf = (n − 6)ε and the internal resistance stays nr.

Teacher's Note:
1. Reversed cell: −2ε.

Answer: (d) it is a null method

Explanation:
1. At balance no current flows through the galvanometer.
2. Being a null method, it does not depend on the accuracy of meter readings.

Teacher's Note:
1. Null method.

Answer: (b) parallel

Explanation:
1. Joining all the first ends together and all the second ends together is a parallel connection.

Teacher's Note:
1. Common ends: parallel.

Answer: (b) \( \frac PQ=\frac{R(S_1+S_2)}{S_1S_2} \)

Explanation:
1. The fourth arm is S = S₁S₂/(S₁ + S₂).
2. Balance: P/Q = R/S = R(S₁ + S₂)/(S₁S₂).

Teacher's Note:
1. Replace the parallel pair by one resistance.

Answer: (d) \( v_d/4 \)

Explanation:
1. v_d = I/(nπr²e), so v_d ∝ 1/r².
2. For radius 2r it becomes v_d/4.

Teacher's Note:
1. v_d ∝ 1/r².