Two small droplets combine to form a single large drop. What is the ratio of the surface energy of the small droplets to the surface energy of the large drop?

The surface tension of a soap solution is $3.5 \times 10^{-2} \, N m^{-1}$. The amount of work done required to increase the radius of a soap bubble from $10 \, cm$ to $20 \, cm$ is $..... \times 10^{-4} \, J$.

The radius of a soap bubble is increased from $\frac{1}{\sqrt{\pi}} \text{ cm}$ to $\frac{2}{\sqrt{\pi}} \text{ cm}$. If the surface tension of water is $30 \text{ dynes/cm}$,then the work done will be ....... $ergs$.

$A$ soap bubble of radius $R$ is blown. After heating a solution,a second bubble of radius $2R$ is blown. The work required to blow the $2^{nd}$ bubble in comparison to that required for the $1^{st}$ bubble is

The work done in blowing a soap bubble of $10\, cm$ radius is (Surface tension of the soap solution is $\frac{3}{100}\,N/m$).

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 $\pi D^2 T$)