The radius $R$ of the soap bubble is doubled under isothermal conditions. If $T$ is the surface tension of the soap bubble, the work done in doing so is given by (in $\pi R^2 T$)

$A$ 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 $\pi = \frac{22}{7}$.

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 the work done in blowing a soap bubble of volume $V$ is $W$,then the work done in blowing a soap bubble of volume $2V$ will be

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).

$A$ soap bubble of surface tension $0.04 \ N/m$ is blown to a diameter of $7 \ cm$. If $(15000 - x) \ \mu J$ of work is done in blowing it further to make its diameter $14 \ cm$, then the value of $x$ is . . . . . . . (Take $\pi = 22/7$)