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

$A$ 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, the energy spent in the blowing is: (in $\pi T r^2$)

The amount of work done in forming a soap film of size $10\,cm \times 10\,cm$ is (Surface tension $T = 3 \times 10^{-2}\,N/m$).

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:

The radius of a soap bubble is $r$, and the surface tension of the soap solution is $T$. Without increasing the temperature, how much energy is required to double its radius (in $\pi r^2 T$)?

$A$ mercury drop of radius $1 \ cm$ is divided into $10^6$ droplets of equal size. If the surface tension of mercury is $35 \times 10^{-3} \ N/m$,then the change in surface energy in the process is: