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

$A$ 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 $1\, cm$ radius is broken into ${10^6}$ small drops. The energy used will be (surface tension of mercury is $35 \times {10^{ - 3}} \, N/cm$).

The amount of work done to break a big water drop of radius $R$ into $27$ small drops of equal radius is $10 \ J$. The work done required to break the same big drop into $64$ small drops of equal radius will be (in $J$)

How much energy is required to double the radius of a soap bubble of radius $R$ and surface tension $T$ (in $\pi R^2 T$)?

$A$ drop of liquid of radius $R=10^{-2} \,m$ having surface tension $S=\frac{0.1}{4 \pi} \,Nm^{-1}$ divides itself into $K$ identical drops. In this process, the total change in the surface energy is $\Delta U=10^{-3} \,J$. If $K=10^\alpha$, then the value of $\alpha$ is: