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

Answer

`1.013 xx 10^(5) Pa`

Explanation

Total pressure `= 101000 +( 2 xx 4.65 xx 10^(-1) // 3 xx 10^(-3) ) = 1.013 xx 10^(5)Pa`.