The current sensitivity of a moving coil galvanometer is doubled by making the number of turns double. Then its voltage sensitivity will be:

Remain unchanged

The correct answer is Option (4) → Remain unchanged

Current sensitivity of galvanometer is,

$\frac{\phi}{I}=\frac{N+B}{K}$   ...(1)

Shy, Voltage sensitivity is,

$\frac{\phi}{V}=\frac{NAB}{K}\frac{I}{V}=\frac{NAB}{K}\frac{1}{R}$

⇒ Voltage sensitivity = Current sensitivity × $\frac{1}{R}$

If the current sensitivity is double, resistance also becomes twice.

∴ Voltage sensitivity = $2×\frac{\phi}{I}×\frac{1}{2R}=\frac{\phi}{IR}$

∴ it remain unchanged.