To increase the current sensitivity of a moving coil galvanometer by $25 \%$, its resistance is increased so that the new resistance becomes twice its initial resistance. By what factor does the voltage sensitivity change?

The current sensitivity of a moving coil galvanometer is given by $I_s = \dfrac{\theta}{I} = \dfrac{NAB}{k}$, where $N$ is the number of turns, $A$ is the area, $B$ is the magnetic field, and $k$ is the torsional constant of the spring.

The voltage sensitivity is given by $V_s = \dfrac{\theta}{V} = \dfrac{\theta}{IR} = \dfrac{I_s}{R}$, where $R$ is the resistance of the galvanometer.

Let the initial current sensitivity be $I_{s1}$ and initial resistance be $R_1$. The initial voltage sensitivity is $V_{s1} = \dfrac{I_{s1}}{R_1}$.

The new current sensitivity is $I_{s2} = I_{s1} + 0.25 I_{s1} = 1.25 I_{s1} = \dfrac{5}{4} I_{s1}$.

The new resistance is $R_2 = 2 R_1$.

The new voltage sensitivity is $V_{s2} = \dfrac{I_{s2}}{R_2} = \dfrac{1.25 I_{s1}}{2 R_1} = 0.625 \dfrac{I_{s1}}{R_1} = 0.625 V_{s1}$.

The change in voltage sensitivity is $V_{s2} - V_{s1} = 0.625 V_{s1} - V_{s1} = -0.375 V_{s1}$.

The percentage change is $-0.375 \times 100 \% = -37.5 \%$.

The negative sign indicates a decrease of $37.5 \%$.

Answer: Decreases by $37.5 \%$