The number of turns in a moving coil galvanometer is doubled (keeping…

The correct answer is Option (4) → doubled, same

The deflection of a moving coil galvanometer is \( \phi = (NAB/k)\,I \), where N is the number of turns, A the area of the coil, B the magnetic field and k the torsional constant of the spring.

The current sensitivity is the deflection per unit current:

    \( \displaystyle \frac{\phi}{I} = \frac{NAB}{k} \)

The voltage sensitivity is the deflection per unit voltage. With G the resistance of the coil, V = IG:

    \( \displaystyle \frac{\phi}{V} = \frac{NAB}{kG} \)

Doubling N with A unchanged doubles the current sensitivity:

    \( \displaystyle \left(\frac{\phi}{I}\right)' = \frac{(2N)AB}{k} = 2\,\frac{NAB}{k} \)

The extra turns are wound from the same wire, so the length of wire, and hence the resistance of the coil, also doubles: G′ = 2G.

    \( \displaystyle \left(\frac{\phi}{V}\right)' = \frac{(2N)AB}{k\,(2G)} = \frac{NAB}{kG} \)

So the current sensitivity is doubled and the voltage sensitivity stays the same.

Note: The question does not say how the resistance of the coil changes. The usual reading, used above, is that the extra turns are wound from the same wire, so the resistance doubles; this gives the answer given by NTA. If the resistance of the coil is taken as unchanged, the voltage sensitivity \( NAB/kG \) also doubles, and Option (1), doubled, doubled, would be correct.

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. Let us take Eq. (4.27) which provides a measure of current sensitivity. If \( N \rightarrow 2N \), 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), \( N \rightarrow 2N \), and \( R \rightarrow 2R \), thus the voltage sensitivity,

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

remains unchanged.”