When two soap bubbles of radii $4 \, cm$ and $5 \, cm$ coalesce,what is the radius of the common interface surface in $cm$?

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.

$A$ big drop is formed by coalescing $1000$ small identical drops of water. If $E_1$ be the total surface energy of $1000$ small drops of water and $E_2$ be the surface energy of the single big drop of water,then the ratio $E_1 : E_2$ is $x : 1$,where $x = . . . . . . $.

$A$ soap bubble, blown by a mechanical pump at the mouth of a tube, increases in volume with time at a constant rate. The graph that correctly depicts the time dependence of pressure inside the bubble is given by

The radii of two soap bubbles are $r_1$ and $r_2$. In isothermal condition,they combine with each other to form a single bubble. The radius of the resultant bubble is

$A$ capillary type tube $AB$ is connected to a manometer $M$ as shown in the figure. Stopper $S$ controls the flow of air. $AB$ is dipped into a soap solution where surface tension is $T$. On opening the stopper for a while,a bubble is formed at $B$ end of the manometer and the level difference in manometer limbs is $h$. If $\rho$ is the density of the soap solution and $r$ is the radius of curvature of the bubble,then the surface tension $T$ of the liquid is given by: