The surface tension of a soap solution is $T$. Work done in blowing a soap bubble of diameter $2d$ is (in $\pi d^2 T$)

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?

When a mercury drop of radius $R$ splits up into $1000$ droplets of radius $r$,the change in surface energy is ($\pi R^2 T$). ($T=$ surface tension of mercury).

$A$ water film is formed between two straight parallel wires,each of length $10 \text{ cm}$,kept at a separation of $0.5 \text{ cm}$. Now,the separation between them is increased by $1 \text{ mm}$ without breaking the water film. The work done for this is (surface tension of water $= 7.2 \times 10^{-2} \text{ N/m}$)

The amount of work done in blowing a soap bubble such that its diameter increases from $d_1$ to $d_2$ is ($T=$ surface tension of soap solution).

$A$ 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 $\pi = \frac{22}{7}$.