PhysicsCBSE Class 12 · Sample paper, session 2025-26 · 5 marks · Long answer

CBSE Class 12 Physics sample paper 2025-26: question 33(I)

From the official CBSE sample question paper for 2025-26, section E, question 33(I), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q33(I) · 5 marks · Long answerMoving Charges and Magnetism

(A) State the working principle of a moving coil galvanometer? What modification is required in the galvanometer to make its scale linear?
(B) If a galvanometer of resistance 49.5Ω has range of 0.05A. What will be the value of resistance needed to convert it in ammeter of range 5A?
(C) How these two resistors should be connected to galvanometer in both cases?

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

Think about what happens to a current loop placed between the poles of a magnet.

Hint 2

For the ammeter, the extra current $5 - 0.05$ A must bypass the galvanometer through a resistor that has the same voltage across it as the galvanometer.

Hint 3

An ammeter must have low total resistance and a voltmeter high total resistance; decide series or parallel from that.

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.

(A) Principle. A current-carrying coil placed in a magnetic field experiences a torque, $τ = N I A B sin θ$ . A spring provides a restoring torque $k ϕ$ . At equilibrium the deflection $ϕ$ is proportional to the current: $ϕ = \frac{N A B}{k} I$ .
1 mark
Modification for a linear scale. The magnetic field is made radial. To do this:
- the pole pieces of the magnet are made concave (cylindrical), and
- a soft-iron cylindrical core is placed inside the coil.
The plane of the coil then always stays parallel to B ( $sin θ = 1$ ), so $ϕ \propto I$ and the scale is linear.
1 mark
(B) Given $G = 49.5 Ω$ , $I_{g} = 0.05$ A and $I = 5$ A. The shunt carries $I - I_{g}$ , and the potential difference across the shunt equals that across the galvanometer:
$I_{g} G = (I - I_{g}) S$
1 mark
Solving:
$S = \frac{I _{g} G}{I - I _{g}} = \frac{0.05 \times 49.5}{5 - 0.05} = \frac{2.475}{4.95} = 0.5 Ω$
1 mark
(C)
- To make an ammeter, the low resistance (shunt, here 0.5 Ω) is connected in parallel with the galvanometer.
- To make a voltmeter, a high resistance is connected in series with the galvanometer.
1 mark

Answer: (A) Torque on a current loop in a magnetic field; use a radial field (concave pole pieces + soft-iron core); (B) 0.5 Ω shunt; (C) shunt in parallel for ammeter, high resistance in series for voltmeter

Where the marks are

States the principle: torque on a current-carrying coil in a magnetic field, giving $ϕ \propto I$ (1 mark)
Radial field using concave pole pieces and a soft-iron core (1 mark)
Uses $I_{g} G = (I - I_{g}) S$ (1 mark)
Gets S = 0.5 Ω (1 mark)
Ammeter: shunt in parallel; voltmeter: high resistance in series (1 mark)

Revise the chapter

Chapter: Moving Charges and Magnetism (notes, formulas, where marks are lost and every past question)

Practise this chapterBoard-style questions from Moving Charges and Magnetism, one at a time, marked against the answer key and marking points.Back to the full paperSample paper 2025-26: every question, by section.

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CBSE Class 12 Physics sample paper 2025-26: question 33(I)

Hints

Step-by-step solution

Revise the chapter

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