The surface energy required to create a liquid surface of area $0.04 \text{ m}^2$ for a liquid of surface tension $75 \text{ N/m}$ will be ....... $\text{J}$

The surface tension of a liquid is $5 \,N/m$. If a film is held on a ring of area $0.02 \,m^2$,its surface energy is about ........... $J$.

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

If the work done in blowing a soap bubble of volume $V$ is $W$,then the work done in blowing a bubble of volume $2V$ from the same soap solution is:

Two mercury drops (each of radius $r$) merge to form a bigger drop. The surface energy of the bigger drop, if $T$ is the surface tension, is

$A$ soap bubble of radius $r$ is blown up to form a bubble of radius $2r$ under isothermal conditions. If $T$ is the surface tension of the soap solution, what is the energy spent in blowing the bubble (in $\pi T r^2$)?