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

Energy needed in the breaking of a drop of radius $R$ into $n$ drops of radius $r$ is given by : (With $S$ surface tension and $P$ atmospheric pressure)

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 $\pi R^2 T$)

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

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 $W_1: W_2=$

The surface tension of soap solution is $0.03 \,N/m$. The work done in blowing to form a soap bubble of surface area $40 \,cm^2$ (in $J$) is: