The work done in splitting a water drop of radius $R$ into $64$ droplets is ($T=$ surface tension of water). (in $\pi TR^2$)

Two mercury droplets of radii $0.1 \ cm$ and $0.2 \ cm$ coalesce into one single drop. What amount of energy is released? The surface tension of mercury is $T = 435.5 \times 10^{-3} \ N \ m^{-1}$.

$A$ bubble has surface tension $S$. The ideal gas inside the bubble has a ratio of specific heats $\gamma = \frac{5}{3}$. The bubble is exposed to the atmosphere and it always retains its spherical shape. When the atmospheric pressure is $P_{a1}$,the radius of the bubble is $r_1$ and the temperature of the enclosed gas is $T_1$. When the atmospheric pressure is $P_{a2}$,the radius of the bubble and the temperature of the enclosed gas are $r_2$ and $T_2$,respectively.
Which of the following statement$(s)$ is(are) correct?
$(A)$ If the surface of the bubble is a perfect heat insulator,then $\left(\frac{r_1}{r_2}\right)^5 = \frac{P_{a2} + \frac{4S}{r_2}}{P_{a1} + \frac{4S}{r_1}}$
$(B)$ If the surface of the bubble is a perfect heat insulator,then the total internal energy of the bubble including its surface energy does not change with the external atmospheric pressure.
$(C)$ If the surface of the bubble is a perfect heat conductor and the change in atmospheric temperature is negligible,then $\left(\frac{r_1}{r_2}\right)^3 = \frac{P_{a2} + \frac{4S}{r_2}}{P_{a1} + \frac{4S}{r_1}}$
$(D)$ If the surface of the bubble is a perfect heat insulator,then $\left(\frac{T_2}{T_1}\right)^{\frac{5}{2}} = \frac{P_{a2} + \frac{4S}{r_2}}{P_{a1} + \frac{4S}{r_1}}$

Two soap bubbles of radii $3r$ and $4r$ are in contact with each other. The radius of curvature of the interface between the bubbles is:

When two soap bubbles of radii $a$ and $b$ $(b > a)$ coalesce,the radius of curvature of the common surface is:

Under isothermal conditions, two soap bubbles of radii $a$ and $b$ coalesce to form a single bubble of radius $c$. If the external pressure is $P$, then the surface tension of the bubbles is: