In $S.I.$ system,the total energy of the free surface of a liquid drop is $2 \pi$ times the surface tension of the liquid. The diameter of the drop is . . . . . . . (in $m$)

The work done in blowing a soap bubble of radius $R$ is $W_1$ and that of radius $2R$ is $W_2$. The ratio of $W_1$ to $W_2$ is

Two mercury drops,each with same radius $r$,merge to form a bigger drop. If $T$ is the surface tension of mercury,then the surface energy of the bigger drop is given by

What will be the work done in increasing the radius of a soap bubble from $\frac{r}{2}$ to $2r$, if the surface tension of the soap solution is $T$ (in $\pi r^2 T$)?

$A$ soap bubble of radius $\frac{1}{\sqrt{\pi}} \text{ cm}$ is expanded to radius $\frac{3}{\sqrt{\pi}} \text{ cm}$. The surface tension of the soap solution is $25 \text{ dyne/cm}$. The work done during expansion in $\text{erg}$ is:

The surface tension of soap solution is $0.03 \,N/m$. The work done in blowing to form a soap bubble of surface area $40 \,cm^2$ (in $J$) is: