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 $T = 0.465\,J/m^2$)

$A$ mercury drop of radius $1 \ cm$ is divided into $10^6$ droplets of equal size. If the surface tension of mercury is $35 \times 10^{-3} \ N/m$,then the change in surface energy in the process is:

The area of a liquid film is $60\, cm^2$ and its surface tension is $T = 20\, dyne/cm$. What is the work done to increase its area to $120\, cm^2$ (in $, erg$)?

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:

$A$ 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, what is the energy spent in blowing the bubble (in $\pi T r^2$)?