Correct answer = 1.628 x 10-7 J
Explaination ::
Given :: r=0.1mm =0.1 x 10-3m , 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
