**Twenty seven droplets of water, each of radius 0.1 mm coalesce into a single drop. Find the change in surface energy. Surface tension of water is 0.072 N/m.**

**Correct answer = 1.628 x 10 ^{-7} J **

**Explaination ::**

Given :: r=0.1mm =0.1 x 10^{-3}m , T=0.072 N/m

Let R be the radius of the single drop formed due to the coalescence of 27 droplets of mercury.

Volume of 27 droplets = volume of the single drop as the volume of the liquid remains constant.

\(\therefore 27 \times \frac{4}{3} \pi r^3 =\frac{4}{3} \pi R^3\)

\(\therefore 27 r^3= R^3\)

\(\therefore 3r=R\) ---{taking cube root}

surface area of 27 droplets =\(27\times 4\pi r^2\)

surface area of single droplet =\(4\pi R^2\)

Decrease in surface area =\(27\times r^2 4\pi - 4\pi R^2\)

\( = 4\pi (27r^2-R^2)=4\pi[27r^2-(3r)^2]\)

=\(4\pi \times 18r^2\)

∴ The energy released = surface tension × decrease in surface area

=\(T\times 4\pi \times 18 r^2=\) \(0.072\times 4\times 3.142\times18 \) \(\times (1\times10^{-4})^2\)

= 1.628 x 10^{-7} J