Question

The excess pressure inside an air bubble of radius $r$ just below (inside) the surface of the water is ​ $P_{1}$ . The pressure inside a drop of the same size (radius) which is just outside the water surface is $P_{2}$ . If $T$ is the surface tension then

• A. $P_{1}=2P_{2}$
• B. $P _{1}=P⁡_{2}$
• C. $P_{2}=2P_{1}$
• D. $P_{2}=0\text{,}P_{1}\neq 0$

Answer 1

Answer: (B)

Excess pressure inside a bubble just below (inside) the surface of water is $P _{1}=\frac{2 T ⁡}{r ⁡}$ and excess pressure inside a drop just outside the water surface is $P_{2}=\frac{2 T}{r}$ .
Hence, $P _{1}=P⁡_{2}$