The current sensitivity of a galvanometer which is used as an ammeter…

The correct answer is Option (3) → remain unchanged

The current sensitivity (deflection per unit current) and the voltage sensitivity (deflection per unit voltage) of a moving coil galvanometer are

    \( \displaystyle \frac{\theta}{I} = \frac{N A B}{k} \qquad \frac{\theta}{V} = \frac{N A B}{k R} \)

where \( R \) is the resistance of the coil.

Doubling \( N \) doubles the current sensitivity. But the coil now has twice the length of wire, so its resistance also doubles. In the voltage sensitivity the two factors of 2 cancel:

    \( \displaystyle \frac{\theta}{V}\Big|_{\mathrm{new}} = \frac{2N A B}{k\,(2R)} = \frac{N A B}{k R} \)

So the voltage sensitivity remains unchanged.

Why the other options are incorrect

• Option (1) be doubled — Incorrect. This would be true only if the resistance of the coil did not change; doubling the wire doubles \( R \).

• Option (2) become half — Incorrect. Both \( N \) and \( R \) double, so the ratio cannot fall to half.

• Option (4) become four times — Incorrect. \( N \) appears once in the numerator and \( R \) once in the denominator, so no factor of 4 arises.

Note: A similar question was also asked in CUET (UG) 2025 on 02-06-2025 (Slot 2), and in CUET (UG) 2024 on 19-07-2024 (Slot 1).

NCERT (2026-27):
The Moving Coil Galvanometer – “… An interesting point to note is that increasing the current sensitivity may not necessarily increase the voltage sensitivity. … If N → 2N, i.e., we double the number of turns, then

    \( \displaystyle \frac{\phi}{I} \rightarrow 2\frac{\phi}{I} \)

Thus, the current sensitivity doubles. However, the resistance of the galvanometer is also likely to double, since it is proportional to the length of the wire. In Eq. (4.28), N → 2N, and R → 2R, thus the voltage sensitivity,

    \( \displaystyle \frac{\phi}{V} \rightarrow \frac{\phi}{V} \)

remains unchanged.”