What is the excess pressure inside a bubble of soap solution of radius 5.00 mm , given that the surface of soap solution at the temperature `(20^(@)C)` is `2.50xx10^(-2)Nm^(-1)` ? If an air bubble of the same dimension were formed at depth of 40.0 cm inside a container containing the soap solution (of relative density 1.20) , what would be the pressure is `1.01xx10^(5)Pa)`.

Explanation

Excess pressure inside soap bubble ,
`P_(i)-P_(o)=(4S)/(R)=(4xx2.5xx10^(-2))/(5xx10^(-3))=20Pa`
Excess pressure inside the soap bubble in air ,
`P_(i)-P_(o)=(2S)/(R)=(2xx2.5xx10^(-2))/(5xx10^(-3))=10Pa`
Inside pressure air bubble at depth of 40 cm ,
`P_(i)=P_(0)+(2S)/(r)`
`=1.01xx10^(5)+hrhog+(2S)/(r)`
`=1.01xx10^(5)+0.4xx1.2xx10^(3)xx9.8+10`
`=1.01xx10^(5)+4.704xx10^(3)+10`
`1.01xx10^(5)+0.04704xx10^(5)+10`
`=1.05704xx10^(5)Pa+10Pa`
`=1.06xx10^(5)Pa+0.00010xx10^(5)Pa`
`=1.0601xx10^(5)Pa`
`=1.06xx10^(5)Pa`