PhysicsCBSE Class 12 · Sample paper, session 2025-26 · 1 mark · MCQ

CBSE Class 12 Physics sample paper 2025-26: question 2

From the official CBSE sample question paper for 2025-26, section A, question 2, for 1 mark. Try it first; the hints and our step-by-step solution are below.

Q2 · 1 mark · MCQCurrent Electricity

A coil contains N turns of insulated copper wire of diameter d and resistivity ρ wound on a cylinder of diameter D. What is the total resistance between the two ends of the coil of copper wire?(given: D>>d)

(A) $\frac{4 ρN D}{d ^{2}}$
(B) $\frac{8 ρN D}{d ^{2}}$
(C) $\frac{2 ρN D}{d ^{2}}$
(D) $\frac{12 ρN D}{d ^{2}}$

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2025-26.

Hints

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

Hint 1

Each turn has a length equal to the circumference of the cylinder.

Hint 2

Use $R = ρ L / A$ with the area of the wire's cross-section.

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.

Length of wire: $L = N π D$ . Cross-sectional area: $A = π d^{2} /4$ .
$R = \frac{ρ L}{A} = \frac{ρN π D}{π d ^{2} /4} = \frac{4 ρN D}{d ^{2}}$ . Option (A).
1 mark

Answer: $R = \frac{4 ρN D}{d ^{2}}$ , option (A)

Where the marks are

Takes $L = N π D$ , $A = π d^{2} /4$ and obtains $4 ρN D / d^{2}$ (1 mark)

Revise the chapter

Chapter: Current Electricity (notes, formulas, where marks are lost and every past question)

Practise this chapterBoard-style questions from Current Electricity, one at a time, marked against the answer key and marking points.Back to the full paperSample paper 2025-26: 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 sample paper 2025-26: question 2

Hints

Step-by-step solution

Revise the chapter

More questions from Current Electricity