An air bubble doubles in radius when it rises from the bottom of the sea to the surface. If the atmospheric pressure is equal to the pressure exerted by a $10 \, m$ column of water,then the depth of the sea is $... \, m$. (Assume surface tension is negligible.)

The radius of a soap bubble is $r$ and the surface tension of the soap solution is $S$. The electric potential to which the soap bubble must be raised by charging it so that the pressure inside the bubble becomes equal to the pressure outside the bubble is $(\varepsilon_0 = \text{permittivity of the free space})$

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

Two water drops each of radius $r$ coalesce to form a bigger drop. If $T$ is the surface tension,the surface energy released in this process is

$1000$ droplets of water,each having a diameter of $2 \ mm$,coalesce to form a single drop. Given the surface tension of water is $0.072 \ N/m$. The energy loss in the process is

Two spherical soap bubbles of radii $r_1$ and $r_2$ in vacuum combine under isothermal conditions. The resulting bubble has a radius equal to: