If the work done in blowing a soap bubble of radius $R$ is $W$, then the work done in blowing a soap bubble of radius $2R$ is: (in $W$)

Two mercury drops of radii $r$ and $2r$ merge to form a bigger drop. The surface energy released in the process is nearly (Surface tension of mercury is $S$ and take $9^{2/3} = 4.326$). (in $\pi r^2 S$)

Surface tension of a soap bubble is $2.0 \times 10^{-2} \; N m^{-1}$. Work done to increase the radius of the soap bubble from $3.5 \; cm$ to $7 \; cm$ will be $......... \times 10^{-4} \; J$ [Take $\pi = \frac{22}{7}$]

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

$n$ drops of a liquid,each with surface energy $E$,join to form a single drop. Which of the following statements is correct?

$A$ 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 $0.465\;J/m^2$)