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.

What is the pressure inside the drop of mercury of radius 3.00 mm at r

1.	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.
Answer» Solution :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`