Question

A mecury drop of radius 1 cm is sprayed into 10^(5) droplets of equal size. Calculate the increase in surface energy if surface tension of mercury is 35xx10^(-3)N//m.

Answer 1

The increase in surface energy is:

• Given, Radius, R = 1 cm = 0.01 m

        Number of droplets = 100000

        Surface Tension, T = 0.035 N/m

• As 100000 droplets are formed from 1 cm mercury drop, volume must be equal.

            $\frac{4}{3}\pi R^3 = 100000 (\frac{4}{3}\pi r^3)$

                                    where r = radius of droplets

            ⇒ r = 0.01 R = 0.0001 m

• Initial Area = $4\pi (0.01)^2$

        Final Area = $100000 * 4\pi (0.0001)^2$

        ΔA = $36 \pi * 10^{-4} m^2$

• Increase in surface energy = T * ΔA = $3956.4 * 10^{-7} N m$