\({Mg}=6 \pi \eta \mathrm{rv}\)
\(\mathrm{V} \propto \frac{1}{\mathrm{r}} \quad\) (as mass is constant)
Now, \(\frac{\mathrm{v}}{\mathrm{v}^{\prime}}=\frac{\mathrm{r}^{\prime}}{\mathrm{r}}\)
\(r^{\prime}=2 \mathrm{r}\)
So, \(v^{\prime}=\frac{v}{2}\)