If $W_1$ is the work done in increasing the radius of a soap bubble from $r$ to $2r$ and $W_2$ is the work done in increasing the radius of the soap bubble from $2r$ to $3r$,then $W_1: W_2=$

What is the ratio of the surface energy of $1$ small drop to $1$ large drop,if $1000$ small drops combine to form $1$ large drop?

If $1000$ droplets of water of surface tension $0.07\,N/m$,each having the same radius $1\,mm$,combine to form a single drop,the released surface energy in the process is: (Take $\pi = \frac{22}{7}$)

$A$ liquid drop of diameter $2 \text{ mm}$ breaks into $512$ droplets. The change in surface energy is $\alpha \times 10^{-6} \text{ J}$. The value of $\alpha$ is . . . . . . . (Take surface tension of liquid = $0.08 \text{ N/m}$)

If $T$ is the surface tension of a liquid,the energy needed to break a liquid drop of radius $R$ into $64$ drops is:

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