Physics - Fluid Question with Solution | TestHub

Let the radius of the initial drop be $R$ and the surface tension of the liquid be $T$.

The initial surface energy is given by $E=4\pi R^{2}T$.

Let the radius of each of the $512$ smaller droplets be $r$. Since the total volume remains constant during the process:

$\frac{4}{3}\pi R^3=512\times \frac{4}{3}\pi r^3$

$R^3=512r^3$

$R=8r\Rightarrow r=\frac{R}{8}$

The final surface energy of the $512$ droplets is:

$E^{\prime }=512\times 4\pi r^{2}T$

Substituting $r=\frac{R}{8}$:

$E^{\prime }=512\times 4\pi {\left(\frac{R}{8}\right)}^{2}T$

$E^{\prime }=512\times \frac{4\pi R^{2}T}{64}$

$E^{\prime }=8\times (4\pi R^{2}T)$

$E^{\prime }=8E$

Answer: $8E$