If the excess pressure inside a soap bubble of radius 5 mm is balanced by a 1.5 mm column of oil of specific gravity 0.76, find the surface tension of the soap solution.

Explanation

Given, `r=5 mm =5 xx 10^(-3) m`
`h=1.5 mm =1.5 xx 10^(-3) m`
`p_("oil")=0.76 xx 10^(3)" kg/m"^(3)`
Excess pressure inside a soap bubble with two free surface is `p=(4 sigma)/R` which is equal to the presure exerted by 1.5 mm column of oi.
`p=hpg`
`rArr (4 sigma)/(R)=hpg`
`or sigma=(hpgR)/(4)`
`=(1.5 xx 10^(-3) xx 0.76 xx 10^(3) xx 9.8 xx 5 xx 10^(-3))/(4)`
`=13.96 xx 10^(-3) N//m`