The work done in blowing a soap bubble of volume $V$ is $W$. The work required to blow a soap bubble of volume $2V$ is [where $T$ is the surface tension of the soap solution].

$A$ liquid drop having surface energy $E$ is spread into $512$ droplets of the same size. The final surface energy of the droplets is (in $E$)

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

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

Two mercury drops,each with same radius $r$,merge to form a bigger drop. If $T$ is the surface tension of mercury,then the surface energy of the bigger drop is given by

$A$ 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: