Two soap bubbles of radii $3r$ and $4r$ are in contact with each other. The radius of curvature of the interface between the bubbles is:

Let $R_1, R_2$ and $R_3$ be the radii of three mercury drops. $A$ big mercury drop is formed from them under isothermal conditions. The radius of the resultant drop is

When a large bubble rises from the bottom of a lake to the surface,its radius doubles. If atmospheric pressure is equal to that of a column of water of height $H$,then the depth of the lake is:

An air bubble of radius $r$ in water is at depth $h$ below the water surface. If $P$ is the atmospheric pressure, and $d$ and $T$ are the density and surface tension of water respectively, then the pressure inside the bubble will be:

Assume that a drop of liquid evaporates by a decrease in its surface energy,such that its temperature remains unchanged. What should be the minimum radius of the drop for this to be possible? The surface tension is $T$,the density of the liquid is $\rho$,and $L$ is its latent heat of vaporization.

Pressure inside two soap bubbles is $1.01 \, atm$ and $1.03 \, atm$. The ratio between their volumes is (Pressure outside the soap bubble is $1 \, atm$). (in $ : 1$)