Voltage sensitivity of a moving coil galvanometer doesn't depend on:

The correct answer is Option (2) → Number of turns of the coil

The voltage sensitivity of a moving coil galvanometer is the deflection per unit voltage applied across the coil, that is the current sensitivity divided by the coil resistance.

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

Doubling the number of turns doubles the current sensitivity \( NAB/k \). But the coil is wound from the same wire, so doubling \( N \) also doubles the length of wire and hence the coil resistance \( R \). With \( R \propto N \), the factor \( N/R \) stays the same, so the voltage sensitivity does not change when the number of turns is changed.

The other three quantities do change it: a stronger magnetic field \( B \) raises it, while a larger resistance \( R \) (for the same turns) or a stiffer spring (larger torsional constant \( k \)) lowers it.

NCERT (2026-27):
The Moving Coil Galvanometer – “… We define the voltage sensitivity as the deflection per unit voltage. From Eq. (4.26),

    \( \displaystyle \frac{\phi}{V} = \left(\frac{NAB}{k}\right)\frac{I}{V} = \left(\frac{NAB}{k}\right)\frac{1}{R} \qquad (4.28) \)”

The Moving Coil Galvanometer – “… 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.”