If the current sensitivity of a moving coil galvanometer is increased by 20%, its resistance also increases by 1.5 times.How will the voltage sensitivity of galvanometer be affected?

If there is an increase in current sensitivity then there is also change in voltage sensitivity. So, using data in the equation for voltage sensitivity we get,
 $$\begin{gathered} \frac{\alpha \text{'}}{V}=\frac{N\text{'}B\text{'}A\text{'}}{kR\text{'}} \\ \frac{\alpha \text{'}}{V}=\left(\frac{20}{100}+1\right)\frac{NBA}{k\left(1.5R\right)} \\ \frac{\alpha \text{'}}{V}=\left(1.2\right)\frac{NBA}{k\left(1.5R\right)} \\ \frac{\alpha \text{'}}{V}=0.8\frac{\alpha }{V} \\ \frac{\alpha \text{'}}{V}=\left(80\%\right)\times \frac{\alpha }{V} \end{gathered}$$
So, there is an increase in 80% of original value in voltage sensitivity.