The surface energy of a liquid film on a ring of area $0.15\;m^2$ is ....... $J$ (Surface tension of the liquid $= 5\;N/m$).

The work done in blowing a soap bubble of radius $R$ is $W_1$ at room temperature. Now the soap solution is heated. From the heated solution another soap bubble of radius $2R$ is blown and the work done is $W_2$. Then:

$A$ film of water is formed between two straight parallel wires of length $10 \ cm$ each,separated by $0.5 \ cm$. If their separation is increased by $1 \ cm$ while still maintaining their parallelism,how much work will have to be done? (Surface tension of water $= 7.2 \times 10^{-2} \ N/m$)

$A$ liquid drop having surface energy $E$ is spread into $216$ droplets of the same size. The final surface energy of the droplets is (in $E$)

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

The surface tension of soap solution is $0.03 \,N/m$. The work done in blowing to form a soap bubble of surface area $40 \,cm^2$ (in $J$) is: