Physics - Fluid Question with Solution | TestHub

Let $n$ be the number of small drops of radius $r$.

Equating the initial and final volumes:

$n\left(\frac{4}{3}\pi r^3\right)=\frac{4}{3}\pi R^3=V$

The initial surface area is $A_i=n(4\pi r^2)$.

Using $n=\frac{V}{\frac{4}{3}\pi r^3}$, we get $A_i=\frac{3V}{r}$.

The final surface area is $A_f=4\pi R^2$.

Using $V=\frac{4}{3}\pi R^3$, we get $A_f=\frac{3V}{R}$.

The decrease in surface area is $\Delta A=A_i-A_f=3V\left(\frac{1}{r}-\frac{1}{R}\right)$.

The energy released is given by $E=T\Delta A=3VT\left(\frac{1}{r}-\frac{1}{R}\right)$.

Answer: $3VT\left(\frac{1}{r}-\frac{1}{R}\right)$