What is the pressure inside a drop of mercury of radius 3.0mm at room temperature? Surface tension of mercury at that temperature `(20^(@)C)` is `4.65xx10^(-1)Nm^(-1)`. The atmospheric pressure is `1.01xx10^(5)Pa`. Also give the excess pressure inside the drop.

Explanation

Excess pressure `=(2sigma)/(R)=(2xx465xx10^(-1))/(3xx10^(-3))=310Pa`
Total pressure `=1.01xx10^(5)+(2sigma)/(R)`
`=1.01xx10^(5)+310=1.0131xx10^(5)Pa`
Since data is correct up to three significant figures, we should write total pressure inside the drop as `1.01xx10^(5)Pa`.