Find the pressure in a bubble of radius 0.2 cm formed at the depth of 5cm from the free surface of water . The surface tension of water is 70 dyne `cm^(-1)` and its density is `1gcm^(-3)` . Atmospheric pressure is `10^(6)` dyne `cm^(-2)` .The gravitational acceleration is `980cms^(-2)`.

Explanation

`{:(h=5cm,R=0.2cm),(T=70 "dyne " cm ^(-1),rho=1gcm^(-3)),(P=10^(6) "dyne" cm^(-2),g=980 cm-s^(-2)):}`
If the pressures inside and outside of the air bubble formed in water are `P_(i)andP_(o)` respectively ,
`P_(i)-P_(o)=(2T)/(R)`
(A bubble in water has one free surface only.)
`thereforeP_(i)=P_(o)+(2T)/(R)` ...(1)
But `P_(o)=` atmmospheric pressure + pressure due to watar column of height h.
`thereforeP_(o)=P+hrhog` ...(2)
From equations (1) and (2),
`P_(i)=P+hrhog+(2T)/(R)`
`=10^(6)+(5xx1xx980)+(2xx70)/(0.2)`
`=10^(6)+4900+700`
`P_(i)=1.0056xx10^(6)"dyne " cm^(-2)