The radius of a soap bubble is $r$. The surface tension of the soap solution is $T$. Keeping the temperature constant,the radius of the soap bubble is doubled. The energy necessary for this process will be: (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:

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 ........... $J$.

The work done in doubling the radius of a soap bubble of radius $2\,cm$ will be ( Surface Tension of soap solution $= 30\,dyne/cm$ ).

Surface tension of a soap solution is $2 \times 10^{-2} \, N/m$. The work done in producing a soap bubble of radius $2 \, cm$ is:

$A$ soap bubble of radius $R$ is blown. After heating the solution,a second bubble of radius $2R$ is blown. The work required to blow the second bubble in comparison to that required for the first bubble is