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

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 of radius $R$ is formed by $729$ small drops of water of radius $r$. Then the radius of each small drop will be:

The pressure inside a soap bubble $A$ is $1.01 \text{ atm}$ and that in a soap bubble $B$ is $1.02 \text{ atm}$. The ratio of the volume of $A$ to that of $B$ is

Pressure inside two soap bubbles are $1.01$ and $1.02$ atmosphere, respectively. The ratio of their volumes is (in $ : 1$)

$1000$ small water drops,each of radius $r$ and charge $q$,coalesce together to form one large spherical drop. The potential of the big drop is larger than that of the smaller drop by a factor of: