Due to surface tension, the excess pressure inside a smaller drop is $9$ units. If $27$ smaller drops combine, then the excess pressure inside the bigger drop is:

The surface tension of a soap solution is $2 \times 10^{-2} \ N/m$. To blow a bubble of radius $1 \ cm$,the work done is:

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.)

$A$ spherical liquid drop splits into $729$ identical spherical drops. If $E$ is the surface energy of the original drop and $U$ is the total surface energy of the resulting drops,then $\frac{E}{U} = \frac{1}{x}$. The value of $x$ is

Two bubbles $A$ and $B$ $(r_A > r_B)$ are joined through a narrow tube. Then

When two soap bubbles of radius $r_1$ and $r_2$ $(r_2 > r_1)$ coalesce,the radius of curvature of the common surface is: