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)Nm^(-1)`. The atmospheric pressure is `1.01xx10^(5)` Pa. Also given the excess pressure inside the drop.

Explanation

Here, `r=3.0mm=3xx10^(-3)m`, ,br> `S=4.65xx10^(-1)Nm^(-1)`,
`P=1.01xx10^(5)` Pa
Excess of pressure inside the drop of mercury is given by
`P=(2S)/(r)=(2xx4.65xx10^(-1))/(3xx10^(-3))=310`Pa
Total pressure inside the drop =` P+rho`
`=1.01xx10^(5)+310`
`=1.013xx10^(5)` Pa