A steel ball of mass 'm' falls in a viscous liquid with a terminal velocity 4 ms^-1. If another steel ball of mass '8 m' falls through the same liquid then its terminal velocity is

We know that

${\mathrm{V}}_{\mathrm{T}}=\frac{2}{9}\frac{{\mathrm{r}}^2\mathrm{g}(\mathrm{s}-\mathrm{\sigma })}{\mathrm{\eta }}$

$\Rightarrow {\mathrm{V}}_{\mathrm{T}}\propto {\mathrm{r}}^2$

Since the mass is made 8 times, volume of should have been made 8 times since the ball's density is constant.

$\Rightarrow \frac{4}{3}{\mathrm{\pi R}}^3=8\times \left(\frac{4}{3}{\mathrm{\pi r}}^3\right)$

$\Rightarrow {\mathrm{R}}^3 = 8{\mathrm{r}}^3\Rightarrow {\mathrm{R}}^2=4{\mathrm{r}}^2$

$\therefore {\mathrm{V}}_{\mathrm{T}}$ becomes 4 times the initial terminal velocity.

Hence the correct answer is $16{\mathrm{ms}}^{-1}.$