Physics - Fluid Question with Solution | TestHub

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

Since the volume remains constant when the drop is divided:

$\frac{4}{3}\pi R^3=8\times \frac{4}{3}\pi r^3$

$R^3=8r^3\Rightarrow R=2r$

The pressure difference (excess pressure) for a liquid drop is given by $\Delta P=\frac{2T}{R}$, where $T$ is the surface tension.

For the big drop, the pressure difference is:

$\Delta P_{\text{big}}=\frac{2T}{R}$

For the smaller droplet, the pressure difference is:

$\Delta P_{\text{small}}=\frac{2T}{r}$

Substituting $R=2r$ in the expression for the big drop:

$\Delta P_{\text{big}}=\frac{2T}{2r}=\frac{1}{2}\left(\frac{2T}{r}\right)=\frac{1}{2}\Delta P_{\text{small}}$

Thus, the pressure difference for the big drop is half of that for the smaller droplet.

Answer: half of that for smaller droplet