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, $\frac{v_{T_1}}{v_{T_2}}=\frac{A_{1}}{A_{2}}=\frac{3}{4}$