What is the pressure inside the drop of mercury of radius 3.00 mm at room temperature `(20"^(@)C)` is `4.65xx10^(-1)Nm^(-1)`. The atmospheric pressure is `1.01xx10^(5)` Pa. Also give excess pressure inside the drop.

Explanation

Drop of mercury has one free surface,
`thereforeP_(i)-P_(o)=(2S)/(r)`
`P=(2S)/(r)`
`=(2xx4.65xx10^(-1))/(3xx10^(-3))`
`=310Pa`
`thereforeP_(i)-P_(o)=P`
`thereforeP_(i)=P_(o)+P`
`=1.01xx10^(5)+310`
`=1.01xx10^(5)+0.00310xx10^(5)`
`=1.01310xx10^(5)`
Correct upto three significant figures,
`P_(i)=1.01xx10^(5)Pa`