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

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 . . . . . . . $(\pi = 22/7)$

$A$ wire of length $10 \, cm$ is used in a soap film experiment. What is the work done to move the wire by $1 \, mm$? (Surface tension $T = 7.2 \times 10^{-2} \, N/m$)

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 $W_1: W_2=$

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

$A$ mercury drop of radius $1 \,cm$ is sprayed into $10^6$ drops of equal size. The energy expended in joules is (Surface tension of mercury is $460 \times 10^{-3} \,N/m$)