$A$ 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 $460 \times 10^{-3} \,N/m$)

The work done in blowing a soap bubble of diameter $3 \ cm$ is (Surface tension of soap solution $= 0.035 \ N/m$). (in $\mu J$)

$A$ soap bubble of initial radius $R$ is to be blown up. The surface tension of the soap film is $T$. The surface energy needed to double the diameter of the bubble is (in $\pi R^2 T$)

If the work done in blowing a soap bubble of volume $V$ is $W$,then the work done in blowing a bubble of volume $2V$ from the same soap solution is:

$A$ water drop of radius $1\,cm$ is broken into $729$ equal droplets. If the surface tension of water is $75\,dyne/cm$,then the gain in surface energy up to the first decimal place will be $...\times 10^{-4}\,J$.

$A$ soap bubble is blown to a diameter of $7 \ cm$. $36960 \ erg$ of work is done in blowing it further. If the surface tension of the soap solution is $40 \ dyne/cm$,then the new radius is . . . . . . $cm$. Take $\pi = \frac{22}{7}$.