Question

Two spherical rain drops with radii in the ratio $1:2$ fall from a great height through the atmosphere. The ratio of their momenta after they have attained terminal velocity is:

• A. $1:8$
• B. $2:1$
• C. $1:32$
• D. $1:2$

Answer 1

The correct option is C $1:32$

Given, the ratio of radii of the drops, $r_1:r_2=1:2$
The terminal velocity and mass of a body is:
$v_T=\frac{2}{9}{r}^{2}g\frac{(\sigma -\rho )}{\eta }$
where $\sigma$ is the density of the body i.e rain drop and $\rho$ is density of fluid (air) through which the body is falling.
$\Rightarrow v_T\propto r^2...\left(i\right)$
Similarly, mass of rain drop will be,
$m=\sigma \times \frac{4}{3}\pi r^3$
$\Rightarrow m\propto r^3...\left(ii\right)$
From Eq.$\left(i\right)\&\left(ii\right)$,
$m{v}_T\propto r^5$
Hence, when rain drops attain $v_T$, the ratio of their momenta will be:
$\frac{p_1}{p_2}=\frac{m_1{v}_1}{m_2{v}_2}={\left(\frac{r_1}{r_2}\right)}^5$
$\Rightarrow \frac{p_1}{p_2}={\left(\frac{1}{2}\right)}^5$
$\therefore p_1:p_2=1:32$