Question

A liquid drop of radius r is divided into 64 similar tiny droplets if the surface tension of liquid is T calculate the increase in energy

Answer 1

The increase in energy from liquid droplet to tiny droplets is 192 T$\pi r^{2}$.

Explanation:

Increase in energy or gain in energy is the energy which is increased due to the increase in surface area of divided tiny droplets. It is the product of surface tension and increase of surface area of droplets.

Given:

              Let the radius of liquid drop = R

              and the radius of tiny droplets = r

              Surface tension of liquid   = T

We know, surface area of spherical liquid droplet = $4\pi R^{2}$

                 surface area of spherical tiny droplets = $64\times4\pi r^{2}$

So,                Increase in surface area = $64\times4\pi r^{2} - 4\pi R^{2}$

                                                              = $4\pi (64r^{2} - R^{2} )$ .........   (1)

                               

               We know that the volume of liquid droplet is equal to the volume of tiny droplets.

So,                         ⇒  $4\pi R^{3} = 64\times4\pi r^{3}$

Then,                      ⇒  $R= 4r$

       

      Put above value in equation ......(1).

                             ⇒  Increase in energy = $T\times 4\pi \times48r^{2}$

                               ⇒  Increase in energy = $192T\pi r^{2}$