Physics - Fluid Question with Solution | TestHub

Excess pressure inside a liquid drop

$∆p=\frac{2T}{R}$

Where T is surface tension and R is radius of liquid drop.

$\therefore \frac{∆p_1}{∆p_2}=\frac{R_2}{R_1}=\frac{0.75}{0.50}$

$⟹ \frac{∆p_1}{∆p_2}=\frac{3}{2}$