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 $W$)

On the surface of the liquid in equilibrium,molecules of the liquid possess

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 $\pi r^2 S$)

$A$ 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 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 surface tension of a soap solution is $3.5 \times 10^{-2} \text{ N/m}$. The work required to increase the radius of a soap bubble from $1 \text{ cm}$ to $2 \text{ cm}$ is $\alpha \times 10^{-6} \text{ J}$. The value of $\alpha$ is . . . . . . . $(\pi = 22/7)$