$A$ drop of mercury of radius $2\, mm$ is split into $8$ identical droplets. Find the increase in surface energy in $\mu J$. (Surface tension of mercury is $0.465\;J/m^2$)

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

$A$ 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:

The surface tension of soap solution is $0.03 \,N/m$. The work done in blowing to form a soap bubble of surface area $40 \,cm^2$ (in $J$) is:

Find out the work done (in $\times 10^{-3} \; J$) to expand the soap bubble to radius $R = 5 \; cm$ (Surface tension of water $= 0.1 \; N/m$).

The work done in blowing a soap bubble of diameter $3 \ cm$ is (Surface tension of soap solution $= 0.035 \ N/m$). (in $\mu J$)