Q 11-09-034JEE MainJEE Main 2026 (2 Apr, Shift 1)Medium
A liquid drop of diameter $2$ mm breaks into $512$ droplets. The change in surface energy is $\alpha\times10^{-6}$ J. The value of $\alpha$ is ______. (Take surface tension of liquid $=0.08$ N/m)
Answer: (B) $7$
Volume is conserved: $\tfrac43\pi R^3=512\times\tfrac43\pi r^3\Rightarrow r=\dfrac{R}{8}$, with $R=1$ mm.
Increase in surface area:
$$\Delta A=512\times4\pi r^2-4\pi R^2=4\pi R^2\left(\frac{512}{64}-1\right)=28\pi R^2=28\pi\times10^{-6}\ \text{m}^2$$
$\Delta E=T\,\Delta A=0.08\times28\pi\times10^{-6}\approx7.0\times10^{-6}$ J, so $\alpha=7$.
Solution by Sreeraj P, M.Sc Physics