To increase the current sensitivity of a moving coil galvanometer by 50%, its resistance is increased so that the new resistance becomes twice its initial resistance. By what percentage does its voltage sensitivity change?

It decreases by 25%

The correct answer is Option (4) → It decreases by 25%

Definitions:

Current sensitivity $S_I = \frac{\theta}{I}$

Voltage sensitivity $S_V = \frac{\theta}{V} = \frac{S_I}{R}$

Let initial current sensitivity = $S_I$, initial resistance = $R$

Initial voltage sensitivity: $S_V = \frac{S_I}{R}$

New current sensitivity (increased by 50%): $S_I' = 1.5 S_I$

New resistance: $R' = 2R$

New voltage sensitivity: $S_V' = \frac{S_I'}{R'} = \frac{1.5 S_I}{2R} = 0.75 \frac{S_I}{R}$

$S_V' = 0.75 S_V$

Thus, voltage sensitivity decreases by $25\%$.