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

If $T$ is the surface tension of a soap solution, then the work done in blowing a soap bubble from diameter $D$ to diameter $2D$ is (in $\pi TD^{2}$)

The work done in doubling the radius of a soap bubble of radius $2\,cm$ will be ( Surface Tension of soap solution $= 30\,dyne/cm$ ).

$A$ 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:

One thousand small water drops of equal radii combine to form a big drop. The ratio of final surface energy to total initial surface energy is

The surface tension of a soap solution is $3.5 \times 10^{-2} \text{ N/m}$. The work required to increase the radius of a soap bubble from $1 \text{ cm}$ to $2 \text{ cm}$ is $\alpha \times 10^{-6} \text{ J}$. The value of $\alpha$ is . . . . . . . $(\pi = 22/7)$