Under isothermal conditions,two soap bubbles of radii $r_{1}$ and $r_{2}$ coalesce to form a single larger bubble. The radius of the larger bubble is

Write the equation of excess pressure (pressure difference) for a bubble in air and a bubble in water.

Eight mercury drops, each of radius $r$, coalesce to form a bigger drop. The surface energy released in this process is . . . . . . . ($S$ is the surface tension of mercury). (in $\pi r^2 S$)

$A$ 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 $\left(\frac{T_2}{T_1}\right)^{\frac{5}{2}} = \frac{P_{a2} + \frac{4S}{r_2}}{P_{a1} + \frac{4S}{r_1}}$

The excess pressure in a soap bubble is double that in another one. The ratio of their volumes is .............

Work done in splitting a drop of water of $1 \, mm$ radius into $10^6$ droplets is (Surface tension of water $= 72 \times 10^{-3} \, J/m^2$).