To increase the current sensitivity of a moving coil galvanometer by 75%, its resistance is increased so that the new resistance becomes twice its initial resistance. By what factor does its voltage sensitivity change?

decreases by 12.5%

The correct answer is Option (1) → decreases by 12.5%

$R=\frac{V}{I}$ [By Ohm's law]

if $i$ is increased by 75%

$I'=1.75I$

Also, $R_{new}=2R$

$∴V'=\frac{1.75I}{2R}$

$⇒V'=0.875V$

∴ it decreases by 12.5%