Question

A drop of liquid of radius $R = {10^{ - 2\,}}\,m$ having surface tension $S = \dfrac{{0.1}}{{4\pi }}\,N{m^{ - 1}}$ divides itself into $K$ identical drops. In this process the total change in the surface energy $\Delta U = {10^{ - 3}}\,J$. If $K = \,{10^\alpha }$ then the value of the $\alpha$ is

Answer 1

From the given radius of the large raindrop, calculate the relation for the area of the small and the large raindrop. Using the formula of loss of energy, and the relation of the areas of the small and the large raindrop, find the value of the $K$. From it, find $\alpha$.

Useful formula:
(1) The volume of the sphere is given by
$V = \dfrac{4}{3}\pi {R^3}$
Where $V$ is the volume of the sphere and $R$ is the radius of the sphere.
(2) The loss in the energy is given by
$\Delta U = S \times dA$
Where $\Delta U$ is the loss in the energy, $S$ is the surface tension of the raindrop and $dA$ is the increase in the area of the small rain drops.
Complete step by step solution:
It is given that
The large raindrop is divided into many small rain drops.

The radius of the large rain drops, $R = {10^{ - 2\,}}\,m$
The surface tension of the raindrop, $S = \dfrac{{0.1}}{{4\pi }}\,N{m^{ - 1}}$
The change in the energy, $\Delta U = {10^{ - 3}}\,J$
From the question, it is clear that the large volume of rain drops breaks into $K$ small volume of rain drops.
$V = Kv$
$\dfrac{4}{3}\pi {R^3} = K\left( {\dfrac{4}{3}\pi {r^3}} \right)$
By simplifying the above equation,
${R^3} = K{r^3}$
$r = {K^{\dfrac{{ - 1}}{3}}}R$ -------------------(1)
The difference in the area of the water drop is obtained by subtracting the area of the small water drop from the large.
$dA = A - a$
$dA = K4\pi {r^2} - 4\pi {R^2}$
By simplifying the above equation and also substituting (1) in above equation.
$dA = 4\pi {R^2}\left( {{K^{\dfrac{1}{3}}} - 1} \right)$
By using the formula for loss in energy,
$\Delta U = S \times dA$
${10^{ - 3}} = \dfrac{{0.1}}{{4\pi }} \times 4\pi {R^2}\left( {{K^{\dfrac{1}{3}}} - 1} \right)$
By further simplification of the above equation.
${10^2} = {K^{\dfrac{1}{3}}} - 1$
$100 = {K^{\dfrac{1}{3}}} - 1$
By doing simple arithmetic operation,
$K = {10^6}$

It is given that the $K = \,{10^\alpha }$, from that the value of the $\alpha$ is obtained as $6$.

Note: The shape of the rain drop is larger at the bottom and the smaller at the top. This is mainly due to the surface tension of the rain water. This forms the spherical skin for the raindrop. Remember the formula of the loss in the energy.