Energy required to form a soap bubble of diameter $20 \, cm$ will be (Surface tension for soap solution is $30 \, dynes/cm$)

$A$ spherical liquid drop of radius $R$ is divided into $8$ equal droplets. If surface tension is $S$,then the work done in this process will be

If $8$ identical droplets are formed from a single drop of radius $2\,mm$, what is the change in surface energy in $\mu J$? (Surface tension $T = 0.465\,J/m^2$)

The potential energy of a molecule on the surface of a liquid compared to the molecules inside the liquid is:

$A$ 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:

$A$ film of soap solution is formed between two straight parallel wires of length $10 \,cm$ each, separated by $0.5 \,cm$. If their separation is increased by $1 \,mm$ while still maintaining their parallelism, how much work will have to be done? (Surface tension of solution $= 65 \times 10^{-3} \,N/m$)