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?

When $n$ identical mercury drops combine to form a single big drop,

$A$ spherical soap bubble of radius $3\,cm$ is formed inside another spherical soap bubble of radius $6\,cm$. If the internal pressure of the smaller bubble of radius $3\,cm$ in the above system is equal to the internal pressure of another single soap bubble of radius $r\,cm$,then the value of $r$ is:

In a cylinder provided with a piston, air is under pressure $P_1$ at a constant temperature $t$. $A$ soap bubble with radius $r$ and surface tension $T$ is lying inside the cylinder. To reduce the radius of the soap bubble to half, the required air pressure inside the cylinder is

Let $R_1$ and $R_2$ be the radii of two mercury drops. $A$ big mercury drop is formed from them under isothermal conditions. The radius of the resultant drop is

$A$ cylinder with a movable piston contains air under a pressure $p_1$ and a soap bubble of radius $r$. The pressure $p_2$ to which the air should be compressed by slowly pushing the piston into the cylinder for the soap bubble to reduce its size by half will be: (The surface tension is $\sigma$, and the temperature $T$ is maintained constant)