The work done in blowing a soap bubble of $10\, cm$ radius is (Surface tension of the soap solution is $\frac{3}{100}\,N/m$).

If $\sigma$ is the surface tension,the work done in breaking a big drop of radius $R$ into $n$ drops of equal radius is:

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 ....... $ergs$.

The work done in doubling the radius of a soap bubble of radius $2\,cm$ will be ( Surface Tension of soap solution $= 30\,dyne/cm$ ).

The radius $R$ of the soap bubble is doubled under isothermal conditions. If $T$ is the surface tension of the soap bubble, the work done in doing so is given by (in $\pi R^2 T$)

$8000$ identical water drops are combined to form a big drop. Then the ratio of the final surface energy to the initial surface energy of all the drops together is