Physics - Fluid Question with Solution | TestHub

The terminal velocity can be given by 

$v_T=\frac{2(\rho -\sigma )r^{2}g}{9\eta }$.

Clearly, $v_T\propto r^2$

Let the radius of the small drops be $r_1$ and the radius of the big drop be $r_2$.

So,

 $$\begin{gathered} 8\left(\frac{4\pi }{3}\right)r_1^3=\left(\frac{4\pi }{3}\right)r_2^3 \\ \Rightarrow r_1=\frac{r_2}{2} \end{gathered}$$

Hence, taking the ratio
$\frac{{\left(v_T\right)}_1}{{\left(v_T\right)}_2}={\left(\frac{r_1}{r_2}\right)}^2$
$$$\Rightarrow \frac{10}{{\left(v_T\right)}_2}=\frac{1}{4}$
$\Rightarrow {\left(v_T\right)}_2=40 \mathrm{cm} {\mathrm{s}}^{-1}$