Question

If ratio of terminal velocity of two drops falling in air is $3:4$, then what is the ratio of their surface areas?

• A. $\frac{2}{3}$
• B. $\frac{3}{4}$
• C. $\frac{4}{3}$
• D. $\frac{3}{2}$

Answer 1

Answer: (B)

Terminal velocity is given by,
$v_T=\frac{2}{9}.\frac{r^2\left(\rho -\sigma \right)g}{\eta }$
That means,
$v_T\infty r^2$
Surface area of a drop, A $=4\pi r^2$
So, $v_T\infty A$
Given, $\frac{v_{T_1}}{v_{T_2}}=\frac{3}{4}$
So, $\frac{v_{T_1}}{v_{T_2}}=\frac{A_1}{A_2}=\frac{3}{4}$