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(B) Given,radius of rain drop $= r$. Since the rain drop starts falling from rest,its initial speed $u = 0$. The terminal velocity $v$ of the rain dro

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$r^7$

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(B) Given,radius of rain drop $= r$. Since the rain drop starts falling from rest,its initial speed $u = 0$. The terminal velocity $v$ of the rain drop is given by $v = \frac{2 g r^2(\rho - \sigma)}{9 \eta}$,where $\rho$ is the density of the rain drop,$\sigma$ is the density of air,and $\eta$ is the coefficient of viscosity. According to the work-energy theorem,the work done by all forces on the drop is equal to the change in its kinetic energy: $W = \Delta K = \frac{1}{2} m v^2 - \frac{1}{2} m u^2 = \frac{1}{2} m v^2$. Substituting $m = \frac{4}{3} \pi r^3 \rho$ and $v = \frac{2 g r^2(\rho - \sigma)}{9 \eta}$: $W = \frac{1}{2} \left( \frac{4}{3} \pi r^3 \rho \right) \left( \frac{2 g r^2(\rho - \sigma)}{9 \eta} \right)^2$. Simplifying the expression: $W = \frac{2}{3} \pi r^3 \rho \cdot \frac{4 g^2 r^4(\rho - \sigma)^2}{81 \eta^2} = \frac{8 \pi \rho g^2(\rho - \sigma)^2}{243 \eta^2} r^7$. Since all other terms are constant,$W \propto r^7$.

$A$ 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

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