The surface energy of a liquid film on a ring of area $0.15\;m^2$ is ....... $J$ (Surface tension of the liquid $= 5\;N/m$).

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 $3 \times 10^{-2} \text{ N/m}$).

$A$ liquid drop of diameter $D$ splits into $3375$ small identical drops. If $S$ is the surface tension of the liquid,then the change in the surface energy in the process is

$A$ mercury drop of $1\, cm$ radius is broken into ${10^6}$ small drops. The energy used will be (surface tension of mercury is $35 \times {10^{ - 3}} \, N/cm$).

The work done in blowing a soap bubble of radius $R$ is $W_1$ at room temperature. Now the soap solution is heated. From the heated solution another soap bubble of radius $2R$ is blown and the work done is $W_2$. Then:

$A$ big drop is formed by coalescing $1000$ small droplets of water. The ratio of surface energy of $1000$ droplets to that of the energy of the big drop is $\frac{10}{x}$. The value of $x$ is . . . . . . .