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
$v=\frac{2 g r^{2}(\rho-\sigma)}{9 \eta}$
where, $\rho \rightarrow$ density of the rain drop
$\sigma \rightarrow$ density of air
$\eta \rightarrow$ 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{2 g r^{2}(\rho-\sigma)}{9 \eta}\right)^{2} \left[\because \rho=\frac{m}{V}\right]$
$k =\frac{8 \pi \rho g^{2}(\rho-\sigma)^{2} r^{7}}{243 \eta^{2}} \Rightarrow W=k r^{7} \text { or } W \propto r^{7}$