JEE Physics Current Electricity MCQs Set 27

Multiple Choice Questions (MCQs) for JEE Physics: Current Electricity

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Multiple Choice Questions

Question 1: Which of the following statements are incorrect? I. The order of magnitude of the current flowing in household appliances is one ampere. II. The order of magnitude of the current in lightning is about one ampere. III. The order of magnitude of the current in nerves in the human body is one ampere.
(a) II and III
(b) I and II
(c) I and III
(d) I, II and III
Show Answer & Explanation

Answer: (a) II and III

Explanation:
1. Household appliances draw about an ampere, so I is correct.
2. Lightning carries about 10⁴ A and nerve currents are of the order of microamperes, so II and III are incorrect.

Teacher's Note:
1. Know the orders of magnitude.

Question 2: The current in the primary circuit of a potentiometer is 0.5 A, ρ for the wire is 4 × 10⁻⁷ Ω m and the area of cross-section of the wire is 8 × 10⁻⁶ m². The potential gradient in the wire would be
(a) 25 mV/m
(b) 2.5 mV/m
(c) 25 V/m
(d) 10 V/m
Show Answer & Explanation

Answer: (a) 25 mV/m

Explanation:
1. The gradient is Iρ/A = 0.5 × 4 × 10⁻⁷/(8 × 10⁻⁶).
2. = 0.025 V/m = 25 mV/m.

Teacher's Note:
1. k = Iρ/A.

Question 3: When no electric field is present, which of the following are correct? I. The electrons move due to thermal motion during which they collide with the fixed ions. II. An electron colliding with an ion emerges with the same speed as before the collision, but the direction of its velocity after the collision is completely random. III. At a given time there is no preferential direction for the velocities of the electrons, so on average the number of electrons travelling in any direction equals the number travelling in the opposite direction and there is no net current.
(a) I and II
(b) II and III
(c) I and III
(d) I, II and III
Show Answer & Explanation

Answer: (d) I, II and III

Explanation:
1. All three statements describe the random thermal motion of free electrons in the Drude model.

Teacher's Note:
1. No field, no net current.

Question 4: Consider the following statements and select the correct statement(s). I. Current is the time rate of flow of charge through any cross-section. II. For a given conductor, the current does not change with change in cross-sectional area. III. The net charge in a current carrying conductor is infinite.
(a) I and II
(b) II and III
(c) I and III
(d) I, II and III
Show Answer & Explanation

Answer: (a) I and II

Explanation:
1. Current is dq/dt, and for a steady current it is the same at every cross-section.
2. The net charge in a current-carrying conductor is zero, so III is false.

Teacher's Note:
1. Steady current is the same everywhere.

Question 5:
Assertion (A): Current is a vector quantity.
Reason (R): Current has magnitude as well as direction.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

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

Explanation:
1. Current is a scalar because it does not obey the vector addition law, so the assertion is false.
2. It does have magnitude and direction, so the reason is true.

Teacher's Note:
1. Direction alone does not make a vector.

Question 6: The amount of charge Q passed in time t through a cross-section of a wire is Q = 5t² + 3t + 1. The value of the current at time t = 5 s is
(a) 9 A
(b) 49 A
(c) 53 A
(d) none of these
Show Answer & Explanation

Answer: (c) 53 A

Explanation:
1. I = dQ/dt = 10t + 3.
2. At t = 5 s, I = 53 A.

Teacher's Note:
1. I = dQ/dt.

Question 7: In the equation AB = C, A is the current density and C is the electric field. Then B is
(a) resistivity
(b) conductivity
(c) potential difference
(d) resistance
Show Answer & Explanation

Answer: (b) conductivity

Explanation:
1. J = σE.
2. So B is the conductivity σ.

Teacher's Note:
1. J = σE.

Question 8: The example of a non-ohmic resistance is
(a) diode
(b) copper wire
(c) filament lamp
(d) carbon resistor
Show Answer & Explanation

Answer: (a) diode

Explanation:
1. A diode has a strongly non-linear I–V curve and does not obey Ohm's law.
2. A copper wire and a carbon resistor are ohmic at a fixed temperature.

Teacher's Note:
1. Diode is non-ohmic.

Question 9:
Assertion (A): E = ρj is the statement of Ohm's law.
Reason (R): If the resistivity of the conducting material is independent of the direction and magnitude of the applied field, then the material obeys Ohm's law.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

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

Explanation:
1. E = ρj is the vector form of Ohm's law, valid when ρ is constant.
2. The reason gives exactly this condition.

Teacher's Note:
1. E = ρj.

Question 10: The internal resistance of a 2.1 V cell which gives a current of 0.2 A through a resistance of 10 Ω is
(a) 0.5 Ω
(b) 0.8 Ω
(c) 1.0 Ω
(d) 0.2 Ω
Show Answer & Explanation

Answer: (a) 0.5 Ω

Explanation:
1. ε = I(R + r), so r = ε/I − R.
2. r = 2.1/0.2 − 10 = 0.5 Ω.

Teacher's Note:
1. r = ε/I − R.

Question 11: To minimise the power loss in the transmission cables connecting the power stations to homes and factories, the transmission cables carry current
(a) at a very low voltage
(b) at a very high voltage
(c) at 220 volt
(d) neither at a very high voltage nor at a very low voltage
Show Answer & Explanation

Answer: (b) at a very high voltage

Explanation:
1. For a given power P = VI, a high voltage means a small current.
2. The loss I²R is then small.

Teacher's Note:
1. Loss = I²R.

Question 12: The current which is assumed to be flowing in a circuit from the positive terminal to the negative is called
(a) direct current
(b) pulsating current
(c) conventional current
(d) alternating current
Show Answer & Explanation

Answer: (c) conventional current

Explanation:
1. By convention current is taken in the direction of flow of positive charge, from the positive to the negative terminal.
2. This is the conventional current.

Teacher's Note:
1. Conventional current: + to −.

Question 13: The resistances of the four arms P, Q, R and S in a Wheatstone's bridge are 10 ohm, 30 ohm, 30 ohm and 90 ohm respectively (P and Q in one branch, R and S in the other). The e.m.f. and internal resistance of the cell are 7 volt and 5 ohm respectively. If the galvanometer resistance is 50 ohm, the current drawn from the cell will be
(a) 0.2 A
(b) 0.1 A
(c) 2.0 A
(d) 1.0 A
Show Answer & Explanation

Answer: (a) 0.2 A

Explanation:
1. P/Q = 10/30 = R/S = 30/90, so the bridge is balanced and the galvanometer carries no current.
2. The branches are 40 Ω and 120 Ω in parallel = 30 Ω, so I = 7/(30 + 5) = 0.2 A.

Teacher's Note:
1. Balanced bridge: ignore the galvanometer.

Question 14: Match the physical quantities in Column I with their mathematical expressions in Column II.
Column IColumn II
(A) Current(1) ne²τ/m
(B) Conductivity(2) (1/ρ)(dρ/dT)
(C) Current density(3) j·ΔS
(D) Temperature coefficient of resistivity(4) n q v_d

(a) A-3, B-4, C-1, D-2
(b) A-3, B-1, C-2, D-4
(c) A-3, B-2, C-4, D-1
(d) A-3, B-1, C-4, D-2
Show Answer & Explanation

Answer: (d) A-3, B-1, C-4, D-2

Explanation:
1. I = j·ΔS (A-3).
2. σ = ne²τ/m (B-1).
3. j = nqv_d (C-4).
4. α = (1/ρ)(dρ/dT) (D-2).

Teacher's Note:
1. Standard expressions.

Question 15: If negligibly small current is passed through a wire of length 15 m and resistance of 5 Ω, having uniform cross-section of 6 × 10⁻⁷ m², then the coefficient of resistivity of the material is
(a) \( 1\times10^{-7}\ \Omega\text{m} \)
(b) \( 2\times10^{-7}\ \Omega\text{m} \)
(c) \( 3\times10^{-7}\ \Omega\text{m} \)
(d) \( 4\times10^{-7}\ \Omega\text{m} \)
Show Answer & Explanation

Answer: (b) \( 2\times10^{-7}\ \Omega\text{m} \)

Explanation:
1. ρ = RA/l = 5 × 6 × 10⁻⁷/15.
2. = 2 × 10⁻⁷ Ω m.

Teacher's Note:
1. ρ = RA/l.

Question 16:
Assertion (A): Two non-ideal batteries are connected in parallel. The equivalent emf is smaller than either of the two emfs.
Reason (R): The equivalent internal resistance is smaller than either of the two internal resistances.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

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

Explanation:
1. The equivalent emf lies between the two emfs, so the assertion is false.
2. The parallel internal resistance is smaller than either, so the reason is true.

Teacher's Note:
1. Equivalent emf lies in between.

Question 17: A conductor carries a current of 50 μA. If the area of cross-section is 50 mm², the value of the current density in A m⁻² is
(a) 0.5
(b) 1
(c) \( 10^{-3} \)
(d) \( 10^{-6} \)
Show Answer & Explanation

Answer: (b) 1

Explanation:
1. J = I/A = 50 × 10⁻⁶/(50 × 10⁻⁶) = 1 A/m².

Teacher's Note:
1. J = I/A.

Question 18:
Assertion (A): Kirchhoff's junction rule can be applied to a junction of several lines or to a point in a line.
Reason (R): When a steady current is flowing, there is no accumulation of charge at any junction or at any point in a line.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

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

Explanation:
1. In steady flow the charge entering a point equals that leaving it.
2. So the rule applies to any junction or point.

Teacher's Note:
1. Steady current, no accumulation.

Question 19: A cell, when balanced with a potentiometer, gave a balance length of 50 cm. When an external resistance of 4.5 Ω is introduced in the circuit (across the cell), it is balanced at 45 cm. The internal resistance of the cell is
(a) 0.25 Ω
(b) 0.5 Ω
(c) 1.0 Ω
(d) 1.5 Ω
Show Answer & Explanation

Answer: (b) 0.5 Ω

Explanation:
1. r = R(l₁/l₂ − 1) = 4.5 × (50/45 − 1).
2. = 4.5 × 1/9 = 0.5 Ω.

Teacher's Note:
1. r = R(l₁ − l₂)/l₂.

Question 20: A primary cell has an e.m.f. of 1.5 volt. When short-circuited, it gives a current of 3 ampere. The internal resistance of the cell is
(a) 4.5 ohm
(b) 2 ohm
(c) 0.5 ohm
(d) \( (1/4.5)\ \text{ohm} \)
Show Answer & Explanation

Answer: (c) 0.5 ohm

Explanation:
1. When short-circuited, I = ε/r.
2. r = 1.5/3 = 0.5 Ω.

Teacher's Note:
1. I_sc = ε/r.

Question 21: The relaxation time in conductors
(a) increases with the increase of temperature
(b) decreases with the increase of temperature
(c) does not depend on temperature
(d) changes suddenly at 400 K
Show Answer & Explanation

Answer: (b) decreases with the increase of temperature

Explanation:
1. With increasing temperature the ions vibrate more and the electrons collide more often.
2. So the relaxation time decreases.

Teacher's Note:
1. τ decreases as T rises.

Question 22: A battery of e.m.f. 10 V and internal resistance 0.5 Ω is connected across a variable resistance R. The value of R for which the power delivered in it is maximum is given by
(a) 0.5 Ω
(b) 1.0 Ω
(c) 2.0 Ω
(d) 0.25 Ω
Show Answer & Explanation

Answer: (a) 0.5 Ω

Explanation:
1. The power in the external resistance is maximum when R equals the internal resistance.
2. So R = 0.5 Ω.

Teacher's Note:
1. Maximum power transfer.

Question 23:
Assertion (A): A potentiometer of longer length is used for accurate measurement.
Reason (R): The potential gradient for a potentiometer of longer length with a given source of e.m.f. becomes small.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

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

Explanation:
1. A longer wire gives a smaller potential gradient for the same source.
2. A smaller gradient makes each division of length correspond to a smaller voltage, so the accuracy improves.

Teacher's Note:
1. Smaller gradient, better accuracy.

Question 24: The electric resistance of a certain wire of iron is R. If its length and radius are both doubled, then
(a) the resistance and the specific resistance will both remain unchanged
(b) the resistance will be doubled and the specific resistance will be halved
(c) the resistance will be halved and the specific resistance will remain unchanged
(d) the resistance will be halved and the specific resistance will be doubled
Show Answer & Explanation

Answer: (c) the resistance will be halved and the specific resistance will remain unchanged

Explanation:
1. R = ρl/(πr²), so doubling l and r gives R′ = R × 2/4 = R/2.
2. The specific resistance is a property of the material and does not change.

Teacher's Note:
1. ρ depends on the material only.

Question 25: Match Column I with Column II.
Column IColumn II
(A) Junction rule(1) Another statement of Ohm's law
(B) Loop rule(2) Magnitude of drift velocity per unit electric field
(C) j = σE(3) Based on the law of conservation of charge
(D) Mobility(4) Based on the law of conservation of energy

(a) A-1, B-4, C-3, D-2
(b) A-3, B-2, C-1, D-4
(c) A-4, B-3, C-1, D-2
(d) A-3, B-4, C-1, D-2
Show Answer & Explanation

Answer: (d) A-3, B-4, C-1, D-2

Explanation:
1. The junction rule expresses conservation of charge (A-3).
2. The loop rule expresses conservation of energy (B-4).
3. j = σE is Ohm's law in vector form (C-1).
4. Mobility is v_d/E (D-2).

Teacher's Note:
1. Conservation laws behind the rules.

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