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

The surface energy of a liquid drop is $U$. It splits up into $512$ equal droplets. The surface energy becomes (in $U$)

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 $\pi R^2 T$)

$A$ frame made of metallic wire enclosing a surface area $A$ is covered with a soap film. If the area of the frame of metallic wire is reduced by $50 \%$,the energy of the soap film will be changed by (in $\%$)

The work done in blowing a soap bubble of radius $R$ is $W_1$ at room temperature. Now the soap solution is heated. From the heated solution another soap bubble of radius $2R$ is blown and the work done is $W_2$. Then:

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).