The surface energy of a liquid film on a ring of area $0.15\;m^2$ is ....... $J$ (Surface tension of the liquid $= 5\;N/m$).

The work done in blowing a soap bubble of radius $R$ is $W$. The work done in blowing a bubble of radius $2R$ of the same soap solution is:

Energy required to form a soap bubble of diameter $20 \, cm$ will be (Surface tension for soap solution is $30 \, dynes/cm$)

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 $\pi r^2 T$)?

$A$ liquid film is formed in a loop of area $0.05 \, m^2$. The increase in its potential energy will be $(T = 0.2 \, N/m)$.

$A$ liquid drop at temperature $t$,isolated from its surroundings,breaks into a number of droplets. The temperature of the droplets will be