Physics - Fluid Question with Solution | TestHub

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

Since the volume remains constant during the splitting process:

$\frac{4}{3}\pi R^3=64\times \frac{4}{3}\pi r^3$

$R^3=64r^3\Rightarrow r=\frac{R}{4}$

Initial surface area of the large drop:

$A_i=4\pi R^2$

Final surface area of the 64 small droplets:

$A_f=64\times 4\pi r^2=64\times 4\pi {\left(\frac{R}{4}\right)}^2=16\pi R^2$

Change in surface area:

$\Delta A=A_f-A_i=16\pi R^2-4\pi R^2=12\pi R^2$

Work done in splitting the drop is equal to the product of surface tension and the change in surface area:

$W=\mathrm{T}\Delta A=12\pi \mathrm{T}{R}^2$

Answer: $12\pi {\mathrm{TR}}^2$