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

The surface tension of a soap solution is $3.5 \times 10^{-2} \, N m^{-1}$. The amount of work done required to increase the radius of a soap bubble from $10 \, cm$ to $20 \, cm$ is $..... \times 10^{-4} \, J$.

The surface energy of a liquid drop is $U$. It splits up into $512$ equal droplets. The surface energy becomes (in $U$)

$A$ liquid drop having surface energy $E$ is spread into $216$ droplets of the same size. The final surface energy of the droplets is (in $E$)

$A$ mercury drop of radius $10^{-3} \ m$ is broken into $125$ equal size droplets. Surface tension of mercury is $0.45 \ Nm^{-1}$. The gain in surface energy is $...... \times 10^{-5} \ J$.

$A$ spherical drop of liquid splits into $1000$ identical spherical drops. If $E_1$ is the surface energy of the original drop and $E_2$ is the total surface energy of the resulting drops,then $\frac{E_1}{E_2} = \frac{x}{10}$. The value of $x$ is: