If $8$ identical droplets are formed from a single drop of radius $2\,mm$,what is the change in energy in $\mu J$? (Surface tension $T = 0.465\,J/m^2$)

$A$ 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 initial radius $R$ is to be blown up. The surface tension of the soap film is $T$. The surface energy needed to double the diameter of the bubble is (in $\pi R^2 T$)

$A$ big water drop is formed by the combination of $n$ small water droplets of equal radii. The ratio of the surface energy of $n$ droplets to the surface energy of the big drop is

$A$ soap bubble is blown to a diameter of $7 \ cm$. $36960 \ erg$ of work is done in blowing it further. If the surface tension of the soap solution is $40 \ dyne/cm$,then the new radius is . . . . . . $cm$. Take $\pi = \frac{22}{7}$.

$A$ liquid drop having surface energy $E$ is spread into $512$ droplets of the same size. The final surface energy of the droplets is (in $E$)