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Electric current, drift velocity, mobility and Ohm’s law: NEET MCQs with answers

50 questions on this topic of Current Electricity (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. In a hydrogen discharge tube, 3.0 × 10¹⁸ protons move to the right through a cross-section per second, while 2.0 × 10¹⁸ electrons move to the left through the same cross-section per second. The net electric current passing through this cross-section is:

  1. 0.16 A towards the left
  2. 0.16 A towards the right
  3. 0.80 A towards the right
  4. 0.80 A towards the left
Show answer and explanation

Answer: (C) 0.80 A towards the right

Current due to positive charges moving to the right is I₁ = n₁e directed to the right. Current due to negative charges moving to the left is I₂ = n₂e, also directed to the right. Thus, the total current is I = (n₁ + n₂)e = (3.0 × 10¹⁸ + 2.0 × 10¹⁸)(1.6 × 10⁻¹⁹ C) = 5.0 × 10¹⁸ × 1.6 × 10⁻¹⁹ = 0.80 A towards the right.

medium

2. Mobility of charge carriers is defined as:

  1. Drift velocity per unit electric field, μ = vd/E
  2. Current per unit area
  3. Charge per unit time
  4. Resistance per unit length
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Answer: (A) Drift velocity per unit electric field, μ = vd/E

Mobility μ = vd/E has SI unit m² V⁻¹ s⁻¹. It indicates how readily carriers drift in a field.

NCERT Class 12

hard

3. A metal wire of length L and uniform cross-sectional area A has a non-uniform relaxation time τ(x) = τ₀ (1 + x/L) along its length, where x is measured from one end (0 ≤ x ≤ L). If the electron density n is uniform throughout, the total resistance of the wire is:

  1. (m L) / (n e² A τ₀ ln 2)
  2. (m L) / (2 n e² A τ₀)
  3. (2 m L) / (n e² A τ₀)
  4. (m L ln 2) / (n e² A τ₀)
Show answer and explanation

Answer: (D) (m L ln 2) / (n e² A τ₀)

The local resistivity is ρ(x) = m / [n e² τ(x)] = m / [n e² τ₀ (1 + x/L)]. The resistance of an infinitesimal slice dx is dR = ρ(x) dx / A = [m / (n e² A τ₀)] · [dx / (1 + x/L)]. Integrating from x = 0 to x = L: ∫ dR = [m / (n e² A τ₀)] · L · [ln(1 + x/L)] from 0 to L = (m L ln 2) / (n e² A τ₀).

medium

4. A wire of resistance 16 Ω is drawn out uniformly through a die so that its radius is reduced to half its initial value. Assuming the density and resistivity of the material remain unchanged, what is the new resistance of the wire?

  1. 64 Ω
  2. 256 Ω
  3. 128 Ω
  4. 32 Ω
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Answer: (B) 256 Ω

When the wire is stretched, its volume V = A × L remains constant. Since radius r' = r/2, area A' = A/4. Conservation of volume gives L' = 4L. The new resistance is R' = ρL'/A' = ρ(4L)/(A/4) = 16(ρL/A) = 16R. Therefore, R' = 16 × 16 Ω = 256 Ω.

easy

5. A wire of resistance R is stretched uniformly such that its length increases by 0.5%. What is the approximate percentage increase in its resistance?

  1. 1.0%
  2. 0.5%
  3. 0.25%
  4. 2.0%
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Answer: (A) 1.0%

When a wire is stretched uniformly, its volume V = A × l remains constant, so R = ρ l / A = ρ l² / V. For small fractional changes, ΔR/R ≈ 2(Δl/l) = 2 × 0.5% = 1.0%.

hard

6. A potential difference V is maintained across a cylindrical conductor of length L and uniform cross-sectional area A. If the mass of an electron is m, its charge is e, and the relaxation time is τ, the total rate of heat energy dissipated in the conductor is:

  1. (n² e² τ A V) / (m L²)
  2. (n e² τ A V²) / (2 m L)
  3. (m L V²) / (n e² τ A)
  4. (n e² τ A V²) / (m L)
Show answer and explanation

Answer: (D) (n e² τ A V²) / (m L)

The rate of heat dissipation is electric power P = V²/R. The resistance of the wire is R = ρL/A. From the microscopic theory of conduction, resistivity is ρ = m / (n e² τ), so R = m L / (n e² τ A). Thus, P = V² / R = (n e² τ A V²) / (m L).

medium

7. An electric field of magnitude 150 V/m is applied across an n-type semiconductor crystal. If the electron mobility is 0.36 m² V⁻¹ s⁻¹ and electron density is 5.0 × 10¹⁹ m⁻³, what is the current density due to the electrons?

  1. 432 A m⁻²
  2. 288 A m⁻²
  3. 864 A m⁻²
  4. 144 A m⁻²
Show answer and explanation

Answer: (A) 432 A m⁻²

Drift speed v_d = μ_e E = 0.36 × 150 = 54 m/s. The current density is J = n e v_d = (5.0 × 10¹⁹ m⁻³) × (1.6 × 10⁻¹⁹ C) × (54 m/s) = 8.0 × 54 = 432 A m⁻².

easy

8. The charge flowing through a conductor varies with time as q(t) = αt - βt² + γt³, where α, β, and γ are positive constants. The initial value of the electric current at t = 0 is:

  1. Zero
  2. α
  3. α - 2β
  4. γ
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Answer: (B) α

Current is defined as I(t) = dq/dt = α - 2βt + 3γt². Substituting t = 0 gives I(0) = α.

medium

9. A wire of initial resistance R₀ is subjected to a uniform tensile stress so that its length increases by 2%. Assuming the volume and temperature of the wire remain constant, the percentage change in its electrical resistance is:

  1. 2.0%
  2. 4.04%
  3. 1.0%
  4. 8.16%
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Answer: (B) 4.04%

Since volume V = A·L is constant, L' = 1.02 L gives A' = A / 1.02. The new resistance is R' = ρ L' / A' = ρ (1.02 L) / (A / 1.02) = (1.02)² R₀ = 1.0404 R₀. Thus, the percentage increase is (1.0404 - 1) × 100% = 4.04%.

medium

10. The current I in a conductor of cross-section A, with n free electrons per unit volume moving at drift velocity vd, is:

  1. I = nAvd/e
  2. I = neA/vd
  3. I = neAvd
  4. I = ne/Avd
Show answer and explanation

Answer: (C) I = neAvd

In time t, the charge crossing the section is n(Avd t)e, so I = neAvd.

NCERT Class 12

easy

11. A potential difference V is applied across a copper wire of length L and diameter d. If the potential difference V is doubled and the length L is halved while the diameter remains unchanged, the drift velocity of free electrons in the wire will become:

  1. Four times its initial value
  2. Twice its initial value
  3. Half of its initial value
  4. Unchanged
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Answer: (A) Four times its initial value

The drift velocity is given by vd = eEτ/m = eVτ/(mL). Therefore, vd is directly proportional to V/L. When V is doubled (2V) and L is halved (L/2), the ratio (V/L) increases by a factor of 2 / (1/2) = 4. Hence, the drift velocity becomes four times its original value.

easy

12. The SI unit of electrical mobility of charge carriers is:

  1. m² V⁻² s⁻¹
  2. m² V s⁻¹
  3. m V⁻¹ s⁻²
  4. m² V⁻¹ s⁻¹
Show answer and explanation

Answer: (D) m² V⁻¹ s⁻¹

Mobility is defined as drift speed per unit electric field (μ = v_d / E). In SI units, this is (m s⁻¹) / (V m⁻¹) = m² V⁻¹ s⁻¹.

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