The excess pressure inside one soap bubble is four times that of another. What is the ratio of the volume of the first bubble to the second bubble?

Two soap bubbles of radii $R$ and $r$ come in contact. $R$ is more than $r$. The radius of curvature of the common surface is

$A$ long cylindrical glass vessel has a pinhole of diameter $0.2 \,mm$ at its bottom. The depth to which the vessel can be lowered vertically in a deep water bath without the water entering into the vessel is (surface tension of water,$T=0.07 \,N/m$,acceleration due to gravity,$g=10 \,m/s^2$) (in $\,cm$)

The radii of two soap bubbles are $r_1$ and $r_2$. In isothermal conditions,they meet together in a vacuum. Then the radius of the resultant bubble is given by

The excess pressure inside the first soap bubble is three times that inside the second soap bubble. The ratio of the volume of the first soap bubble to the volume of the second soap bubble is

When $n$ identical mercury drops combine to form a single big drop,