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

If the surface tension of a soap solution is $3 \times 10^{-2} \,N/m$, then the work done in forming a soap film of $20 \,cm \times 5 \,cm$ will be

$A$ water drop of diameter $2\,cm$ is broken into $64$ equal droplets. The surface tension of water is $0.075\,N/m$. In this process,the gain in surface energy will be ...........$J$.

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

$A$ mercury drop of radius $1 \,cm$ is sprayed into $10^6$ droplets of equal size. Calculate the energy expended if surface tension of mercury is $435 \times 10^{-3} \,N/m$.

$A$ soap bubble of initial radius $R$ is to be blown up. The surface tension of the soap film is $T$. The surface energy needed to double the diameter of the bubble is (in $\pi R^2 T$)