The current sensitivity of a galvanometer increases by 10%. If its resistance also increases by 20%, the voltage sensitivity will

decrease by 8.3 %

The correct answer is Option (1) → decrease by 8.3 %

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

Voltage sensitivity $S_v = \frac{\theta}{V} = \frac{\theta}{IR} = \frac{S_c}{R}$

Initial: $S_{v1} = \frac{S_c}{R}$

Final: $S_{v2} = \frac{1.1 S_c}{1.2 R} = \frac{1.1}{1.2} \cdot \frac{S_c}{R} = \frac{11}{12} S_{v1}$

$\frac{S_{v2}}{S_{v1}} = \frac{11}{12} \approx 0.9167$

Hence, voltage sensitivity decreases by about 8.3%.