$A$ spherical drop of radius $r$ is divided into $8$ equal droplets. If the surface tension is $S$, then the work done in the process will be

The pressure inside two soap bubbles, $A$ and $B$, is $1.01 \,atm$ and $1.02 \,atm$ respectively. The ratio of their respective radii $(r_A : r_B)$ is (outside pressure $= 1 \,atm$).

The excess pressure inside a bubble in water is $P_1$. The excess pressure inside a drop of the same radius is $P_2$. Then:

Two small drops of mercury,each of radius $R$,coalesce to form a single large drop. The ratio of the total surface energies before and after the change is:

Two drops of equal radius coalesce to form a bigger drop. What is the ratio of the surface energy of the bigger drop to that of a smaller one?

The surface tension of a soap bubble is $0.03 \text{ N/m}$. The work done in increasing the diameter of the bubble from $2 \text{ cm}$ to $6 \text{ cm}$ is $\alpha \times 10^{-4} \text{ J}$. The value of $\alpha$ is . . . . . . (Take $\pi = 3.14$).