Question

The excess pressure inside an air bubble of radius $r$ just below the surface of water is $p_1$. The excess pressure inside a drop of the same radius just outside the surface is $p_2$. If $T$ is surface tension, then:-

• A. $p_1=2p_2$
• B. $p_1=p_2$
• C. $p_2=2p_1$
• D. $p_2=0,p_1\ne 0$

Answer 1

The correct option is A $p_1=2p_2$

Since, an air bubble has 2 air-liquid interface, therefore the excess pressure inside a bubble is a result of 2 surfaces rather than just 1 in case of drop.

Excess pressure -

Bubble: $\frac{4T}{R}$

Drop : $\frac{2T}{R}$

Therefore, $p_1=2p_2$