PhysicsCBSE Class 12 · Board exam 2026, set 55/5/1 · 5 marks · Long answer

CBSE Class 12 Physics 2026: question 31(b)

Set in the CBSE Class 12 board exam of 2026, paper 55/5/1, section E, question 31(b), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q31(b) · 5 marks · Long answerCurrent Electricity

(i) Consider a cylindrical conductor of length $l$ and area of cross-section A. Current I is maintained in the conductor and electrons drift with velocity $v_{d}$ ( $∣ v_{d} ∣ = \frac{e ∣ E ∣}{m} τ$ ), (where symbols have their usual meanings). Show that the conductivity $σ$ of the material of the conductor is given by $σ = \frac{n e ^{2}}{m} τ$ .
(ii) The resistance of a metal wire at 20°C is 1·05 $Ω$ and at 100°C is 1·38 $Ω$ . Determine the temperature coefficient of resistivity of this metal.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2026 board paper, set 55/5/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

Count the charge crossing a cross-section in time Δt to relate I with v_d.

Hint 2

Write V = IR and compare with R = ρl/A.

Hint 3

For (ii) use R_T = R₀[1 + α(T − T₀)] with the 20 °C value as R₀.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

(i) In time Δt, electrons within distance $v_{d} Δ t$ cross a section: charge $= n e A v_{d} Δ t$ , so $I = n e A v_{d}$ .
1 mark
Substituting $v_{d} = \frac{e E τ}{m}$ and $E = \frac{V}{l}$ : $I = n e A \cdot \frac{e V τ}{m l} = \frac{n e ^{2} τ A}{m l} V$
1 mark
So $R = \frac{V}{I} = \frac{m}{n e ^{2} τ} \cdot \frac{l}{A}$ . Comparing with $R = ρ \frac{l}{A}$ : $ρ = \frac{m}{n e ^{2} τ}$ , hence $σ = \frac{1}{ρ} = \frac{n e ^{2} τ}{m}$ .
1 mark
(ii) $R_{T} = R_{0} [1 + α (T - T_{0})]$ with $T_{0} = 2 0^{\circ}$ C: $α = \frac{R _{100} - R _{20}}{R _{20} ( 100 - 20 )}$
1 mark
$α = \frac{1.38 - 1.05}{1.05 \times 80} = \frac{0.33}{84} \approx 3.9 \times 1 0^{- 3}^{\circ} C^{- 1}$
1 mark

Answer: (i) σ = ne²τ/m; (ii) α ≈ 3.9 × 10^-3 °C^-1

Where the marks are

Current I = neAv_d (1 mark)
Substitution of v_d and E = V/l (1 mark)
Obtaining ρ = m/(ne²τ), hence σ = ne²τ/m (1 mark)
Formula for temperature coefficient (1 mark)
α ≈ 3.9 × 10⁻³ °C⁻¹ (1 mark)

Revise the chapter

Chapter: Current Electricity (notes, formulas, where marks are lost and every past question)
Topic: Drift Velocity, Mobility and Resistivity

Practise this chapterBoard-style questions from Current Electricity, one at a time, marked against the answer key and marking points.Back to the full paper2026 board paper, set 55/5/1: every question, by section.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.

CBSE Class 12 Physics 2026: question 31(b)

Hints

Step-by-step solution

Revise the chapter

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