Question

A rain drop of radius $r$ is falling through air, starting from rest. The work done by all the forces on the drop, when it attains terminal velocity, is proportional to

• A. $r^3$
• B. $r^7$
• C. $r^5$
• D. $r^4$

Answer 1

Answer: (B)

Given, radius of rain drop $=r$
Since, rain drop starts falling from rest, hence its initial speed, $u=0$
Final velocity of rain drop is equal to the terminal velocity $v$, which is given by
$v=\frac{2g{r}^2(\rho -\sigma )}{9\eta }$
where, $\rho \to$ density of the rain drop
$\sigma \to$ density of air
$\eta \to$ coefficient of viscosity
$\therefore$ According to the work-energy theorem,
work done by all the forces on the drop
= change in its kinetic energy
$W=\frac{1}{2}m{v}^2-\frac{1}{2}m{u}^2=\frac{1}{2}m{v}^2[\because u=0]$
$=\frac{1}{2}\times \frac{4}{3}\pi r^3\cdot \rho {\left(\frac{2g{r}^2(\rho -\sigma )}{9\eta }\right)}^2\left[\because \rho =\frac{m}{V}\right]$
$k=\frac{8\pi \rho g^2(\rho -\sigma {)}^2{r}^7}{243{\eta }^2}\Rightarrow W=k{r}^7\text{ or }W\propto r^7$