Find out the work done (in $\times 10^{-3} \; J$) to expand the soap bubble to radius $R = 5 \; cm$ (Surface tension of water $= 0.1 \; N/m$).

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 $J$)

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

$A$ liquid drop having surface energy $E$ is spread into $512$ droplets of the same size. The final surface energy of the droplets is (in $E$)

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

The work done in blowing a soap bubble of radius $0.2\, m$ is (the surface tension of soap solution being $0.06\, N/m$).