If the work done in blowing a soap bubble of volume $V$ is $W$,then the work done in blowing a soap bubble of volume $2V$ will be

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

Find out the work done (in $\times 10^{-3} \; J$) to expand the soap bubble to radius $R = 5 \; cm$ (Surface tension of water $= 0.1 \; N/m$).

The work done in blowing a soap bubble of $10 \,cm$ radius is ...... $J$ (surface tension of soap solution is $0.03 \,N/m$).

Consider a soap film on a rectangular frame of wire of area $4 \times 4 \text{ cm}^2$. If the area of the soap film is increased to $4 \times 5 \text{ cm}^2$,the work done in the process will be (The surface tension of the soap film is $3 \times 10^{-2} \text{ N/m}$).

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 $\pi R^2 T$)