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

T= surface tension of mercury
r = radius of the drop
The excess pressure inside the mercury drop `=(2T)/(r)`
`=(2 xx 4.65 xx 10^(-1))/(3 xx 10^(-3))`
`p=2 xx 1.55 xx 10^(2) Pa`
Atmospheric pressure, `P=1.01 xx 10^5 Pa`
Total pressure inside the drop =atmospheric pressure (P)+ excess pressure
`=1.01 xx 10^(5) Pa+0.00310 xx 10^(5)Pa`
`=1.0131 xx 10^(5)Pa`