JEE Main 2022 — Mechanical Properties of Fluids Question with Solution

The volume of the bigger drop will be, $V=\frac{4}{3}\pi R^3$. If the radius of smaller drops is $r$, using the conservation of volume,

$$\begin{gathered} n\frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3 \\ \Rightarrow 729\times \frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3 \\ \Rightarrow r=\frac{R}{9} \end{gathered}$$

The initial surface energy will be,

$E_i=\left(4\pi R^2\right)T$ 

And final surface energy will be,

$E_f=n\left(4\pi r^2\right)T=729\times 4\pi {\left(\frac{R}{9}\right)}^{2}T$$=36\pi R^{2}T$

Therefore, the change in the surface energy will be,

$$\begin{gathered} \Rightarrow \Delta E=E_f-E_i=32\pi R^{2}T=32\times 3.14\times {\left(1\times {10}^{-2}\right)}^2\times 75\times {10}^{-3} \\ \Rightarrow \Delta E=7.5\times {10}^{-4} \mathrm{J} \end{gathered}$$