Pressure inside two soap bubbles are $1.01$ and $1.02$ atmosphere, respectively. The ratio of their volumes 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 is an air bubble of radius $1.0\,mm$ in a liquid of surface tension $0.075\,Nm ^{-1}$ and density $1000\,kg$ $m ^{-3}$ at a depth of $10\,cm$ below the free surface. The amount by which the pressure inside the bubble is greater than the atmospheric pressure is $....Pa \left( g =10\,ms ^{-2}\right)$

The diameter of rain-drop is $0.02 \,cm$. If surface tension of water be $72 \times {10^{ - 3}}\,newton$ per metre, then the pressure difference of external and internal surfaces of the drop will be

The excess pressure in a soap bubble is thrice that in other one. Then the ratio of their volume is

A soap bubble in vacuum has a radius of $3 \,cm$ and another soap bubble in vacuum has a radius of $4 \,cm$. If the two bubbles coalesce under isothermal condition, then the radius of the new bubble is ....... $cm$