$A$ soap bubble is blown to a diameter of $7 \ cm$. $36960 \ erg$ of work is done in blowing it further. If the surface tension of the soap solution is $40 \ dyne/cm$,then the new radius is . . . . . . $cm$. Take $\pi = \frac{22}{7}$.

On the surface of a liquid in equilibrium,molecules of the liquid possess

$A$ water film is formed between two parallel wires of $10 \text{ cm}$ length. The distance of $0.5 \text{ cm}$ between the wires is increased by $1 \text{ mm}$. The work done in the process is (surface tension of water $= 72 \text{ mN/m}$).

What is the ratio of the surface energy of $1$ small drop to $1$ large drop,if $1000$ small drops combine to form $1$ large drop?

$A$ drop of liquid of radius $R=10^{-2} \,m$ having surface tension $S=\frac{0.1}{4 \pi} \,Nm^{-1}$ divides itself into $K$ identical drops. In this process, the total change in the surface energy is $\Delta U=10^{-3} \,J$. If $K=10^\alpha$, then the value of $\alpha$ is:

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