The excess pressure inside an air bubble of radius r just below the surface of the water is P1 . The pressure inside a drop of the same radius just outside the surface is P2 . If T is the surface tension then

Question

The excess pressure inside an air bubble of radius r just below the surface of the water is ​ P1 . The pressure inside a drop of the same radius just outside the surface is P2 . If T is the surface tension then (A) P1=2P2 (B) P 1=P⁡2 (C) P2=2P1 (D) P2=0 text,P1≠ 0

Tardigrade Admin · Accepted Answer

Excess pressure inside a bubble just below the surface of water is P 1=(2 T â¡/r â¡) and excess pressure inside a drop just outside the surface is P2=(2 T/r) . Hence, P 1=Pâ¡2

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

$P_{1} = 2 P_{2}$

$P_{1} = P_{2}$

$P_{2} = 2 P_{1}$

$P_{2} = 0, P_{1} \neq = 0$

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

Q. The excess pressure inside an air bubble of radius r just below the surface of the water is ​ P1​ . The pressure inside a drop of the same radius just outside the surface is P2​ . If T is the surface tension then

P1​=2P2​

P1​=P⁡2​

P2​=2P1​

P2​=0,P1​=0