$8000$ identical water drops are combined to form a big drop. Then the ratio of the final surface energy to the initial surface energy of all the drops together is

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

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 $3 \times 10^{-2} \text{ N/m}$).

Energy required to form a soap bubble of diameter $20 \, cm$ will be (Surface tension for soap solution is $30 \, dynes/cm$)

If $T$ is the surface tension of a soap solution, the amount of work done in blowing a soap bubble from a diameter $D$ to $2D$ is (in $\pi D^2 T$)

$A$ soap bubble of radius $R$ is blown. After heating the solution,a second bubble of radius $2R$ is blown. The work required to blow the second bubble in comparison to that required for the first bubble is