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

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 $\pi = 3.14$).

$1000$ droplets of water, each having a diameter of $2 \ mm$, coalesce to form a single drop. Given the surface tension of water is $0.072 \ N/m$. The energy loss in the process is

$A$ spherical drop of water has a radius of $1\, mm$. If the surface tension of water is $70 \times 10^{-3} \, N/m$,then the difference of pressure between the inside and outside of the spherical drop is:

If the surface tension of a soap solution is $0.03 \, N/m$,then the excess pressure inside a soap bubble of diameter $6 \, mm$ over the atmospheric pressure will be:

Two bubbles $A$ and $B$ $(r_A > r_B)$ are joined through a narrow tube. Then