$A$ mercury drop of radius $1\, cm$ is sprayed into $10^{6}$ drops of equal size. The energy expressed in joule is (surface tension of mercury is $460 \times 10^{-3}\, N/m$)

Two small droplets combine to form a single large drop. What is the ratio of the surface energy of the small droplets to the surface energy of the large drop?

If $T$ is the surface tension of a soap solution, then the amount of work done in increasing the diameter of a soap bubble from $D$ to $2D$ is: (in $\pi D^2 T$)

$A$ 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:

$A$ soap bubble of radius $r$ is blown up to form a bubble of radius $2r$ under isothermal conditions. If $T$ is the surface tension of the soap solution, the energy spent in the blowing is: (in $\pi T r^2$)

The work done in increasing the diameter of a soap bubble from $2 \ cm$ to $4 \ cm$ is (Surface tension of soap solution $= 3.5 \times 10^{-2} \ N/m$)