Work done in increasing the size of a soap bubble from a radius of $3\, cm$ to $5\, cm$ is (Surface tension of soap solution $= 0.03\, Nm^{-1}$)

$A$ 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 $= 72 \text{ mN/m}$).

Explain surface energy and surface tension.

The work done in blowing a soap bubble of radius $0.2\, m$ is (the surface tension of soap solution being $0.06\, N/m$).

Two small drops of mercury each of radius $R$ coalesce to form a single large drop. The ratio of total surface energy before and after the change is -

Two mercury drops of radii $r$ and $2r$ merge to form a bigger drop. The surface energy released in the process is nearly (Surface tension of mercury is $S$ and take $9^{2/3} = 4.326$). (in $\pi r^2 S$)