The surface tension of a soap solution is $3.5 \times 10^{-2} \, N m^{-1}$. The amount of work done required to increase the radius of a soap bubble from $10 \, cm$ to $20 \, cm$ is $..... \times 10^{-4} \, J$.

$A$ 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 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 $\pi R^2 T$)

$A$ spherical drop of liquid splits into $1000$ identical spherical drops. If $E_1$ is the surface energy of the original drop and $E_2$ is the total surface energy of the resulting drops,then $\frac{E_1}{E_2} = \frac{x}{10}$. The value of $x$ is:

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 $= 0.03\, Nm^{-1}$)

Fill in the blanks:
$(i)$ As the surface area is large,the potential energy of the molecules is ...... (more / less).
$(ii)$ As the surface area is small,the potential energy of its molecules is ...... (more / less).