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

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 $T = 0.1 \, N/m$.

The surface energy of a liquid drop is $V$. It is sprayed into $1000$ equal droplets. The surface energy of all the droplets is

The amount of energy required to form a soap bubble of radius $2\,cm$ from a soap solution is nearly $..........\,\times 10^{-4}\,J$: (surface tension of soap solution $=0.03\,N\,m^{-1}$)

Consider a soap film on a rectangular frame of wire of area $4 \times 4 \text{ cm}^2$. If the area of the soap film is increased to $4 \times 5 \text{ cm}^2$,the work done in the process will be (The surface tension of the soap film is $3 \times 10^{-2} \text{ N/m}$).

Energy needed in breaking a liquid drop of radius $R$ into $n$ smaller drops,each of radius $r$,is [where $T$ is the surface tension of the liquid].