Physics - Fluid Question with Solution | TestHub

Let the radius of the initial drop be $R$ and the radius of each small droplet be $r$.

Initial surface energy of the drop is given by:
$E=T\times 4\pi R^2$

By conservation of volume:
$\frac{4}{3}\pi R^3=512\times \frac{4}{3}\pi r^3$

$R^3=512r^3\Rightarrow R=8r\Rightarrow r=\frac{R}{8}$

Final surface energy of the 512 droplets is:
$E^{\prime }=512\times T\times 4\pi r^2$

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

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

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

Answer: 8 E