The amount of work done to break a big water drop of radius $R$ into $27$ small drops of equal radius is $10 \ J$. The work done required to break the same big drop into $64$ small drops of equal radius will be (in $J$)

$A$ soap bubble of radius $r$ is blown up to form a bubble of radius $2r$ under isothermal conditions. If $T$ is the surface tension of the soap solution, the energy spent in the blowing is: (in $\pi T r^2$)

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).

Consider a soap film on a rectangular frame of wire of area $4 \times 4 \text{ cm}^2$. If the area of the soap film is increased to $4 \times 5 \text{ cm}^2$,the work done in the process will be (The surface tension of the soap film is $3 \times 10^{-2} \text{ N/m}$).

The surface energy required to create a liquid surface of area $0.04 \text{ m}^2$ for a liquid of surface tension $75 \text{ N/m}$ will be ....... $\text{J}$

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