The area of a liquid film is $60\, cm^2$ and its surface tension is $T = 20\, dyne/cm$. What is the work done to increase its area to $120\, cm^2$ (in $, erg$)?

$A$ mercury drop of radius $1\, cm$ is sprayed into $10^{6}$ drops of equal size. The energy expressed in joule is (surface tension of mercury is $460 \times 10^{-3}\, N/m$)

$A$ mercury drop of $1\, cm$ radius is broken into ${10^6}$ small drops. The energy used will be (surface tension of mercury is $35 \times {10^{ - 3}} \, N/cm$).

The radius of a soap bubble is increased from $\frac{1}{\sqrt{\pi}} \text{ cm}$ to $\frac{2}{\sqrt{\pi}} \text{ cm}$. If the surface tension of water is $30 \text{ dynes/cm}$,then the work done will be ....... $ergs$.

If the work done in blowing a soap bubble of volume $V$ is $W$,then the work done in blowing a bubble of volume $2V$ from the same soap solution is:

$A$ soap bubble of radius $R$ is blown. After heating the solution,a second bubble of radius $2R$ is blown. The work required to blow the second bubble in comparison to that required for the first bubble is