The work done in blowing a soap bubble of radius $0.2 \, m$,given that the surface tension of the soap solution is $60 \times 10^{-3} \, N/m$,is:

In an isothermal process,$2$ water drops of radius $1 \, mm$ are combined to form a bigger drop. Find the energy change (in $\mu J$) in this process if the surface tension $T = 0.1 \, N/m$.

$A$ soap bubble of surface tension $0.04 \ N/m$ is blown to a diameter of $7 \ cm$. If $(15000 - x) \ \mu J$ of work is done in blowing it further to make its diameter $14 \ cm$, then the value of $x$ is . . . . . . . (Take $\pi = 22/7$)

The amount of work done in blowing a soap bubble such that its diameter increases from $d_1$ to $d_2$ is ($T=$ surface tension of soap solution).

$A$ liquid film is formed in a loop of area $0.05 \, m^2$. The increase in its potential energy will be $(T = 0.2 \, N/m)$.

$A$ 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: