1. Recall the definitions

For a moving coil galvanometer:

• Current sensitivity is $S_i = \frac{\theta}{I} = \frac{NAB}{k}$ where $N$ = number of turns, $A$ = area, $B$ = magnetic field, and $k$ = torsional constant.

So, $S_i \propto N$

• Voltage sensitivity is $S_v = \frac{\theta}{V}$ But since $V = IG$ where $G$ is galvanometer resistance, $S_v = \frac{\theta}{IG} = \frac{S_i}{G}$

Also, the resistance of the coil is proportional to the length of wire used, hence proportional to number of turns: $G \propto N$

2. Current sensitivity increased by $50\%$

If current sensitivity increases by $50\%$, then $S_i' = 1.5 S_i$

Since $S_i \propto N$, $N' = 1.5N$

3. Effect on galvanometer resistance

Since resistance $G \propto N$, $G' = 1.5G$

4. Effect on voltage sensitivity

Now, $S_v = \frac{S_i}{G}$

Therefore, $S_v' = \frac{S_i'}{G'} = \frac{1.5S_i}{1.5G} = \frac{S_i}{G} = S_v$

So voltage sensitivity remains unchanged.

5. Percentage change in voltage sensitivity

$\frac{S_v' - S_v}{S_v} \times 100 = 0\%$

Hence, the correct option is:

A: $0\%$