The volume of an air bubble becomes three times as it rises from the bottom of a lake to its surface. Assuming atmospheric pressure to be $75\, cm$ of $Hg$ and the density of water to be $1/10  $ of the density of mercury, the depth of the lake is ....... $m$

In capillary pressure below the curved surface of water will be

The excess of pressure inside a soap bubble is twice the excess pressure inside a second soap bubble. The volume of the first bubble is $n$ times the volume of the second where $n$ is

What is the pressure inside the drop of mercury of radius $3.00 \;mm$ at room temperature? Surface tension of mercury at that temperature $\left(20\,^{\circ} C \right)$ is $4.65 \times 10^{-1}\; N m ^{-1} .$ The atmospheric pressure is $1.01 \times 10^{5}\; Pa$. Also give the excess pressure inside the drop.

There are two liquid drops of different radii. The excess pressure inside over the outside is

If the radius of a soap bubble is four times that of another, then the ratio of their excess pressures will be