In $S.I.$ system,the total energy of the free surface of a liquid drop is $2 \pi$ times the surface tension of the liquid. The diameter of the drop is . . . . . . . (in $m$)

Let $W_1$ be the work done in blowing a soap bubble of radius $r$ from a soap solution at room temperature. The soap solution is now heated and a second soap bubble of radius $2r$ is blown from the heated soap solution. If $W_2$ is the work done in forming this bubble,then:

Two small droplets combine to form a single large drop. What is the ratio of the surface energy of the small droplets to the surface energy of the large drop?

Two mercury drops (each of radius $r$) merge to form a bigger drop. The surface energy of the bigger drop, if $T$ is the surface tension, is

If $T$ is the surface tension of a soap solution, the amount of work done in blowing a soap bubble from a diameter $D$ to $2D$ is (in $\pi D^2 T$)

The amount of work done in blowing a soap bubble such that its diameter increases from $d_1$ to $d_2$ is ($T=$ surface tension of soap solution).