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

The work done in doubling the radius of a soap bubble of radius $2\,cm$ will be ( Surface Tension of soap solution $= 30\,dyne/cm$ ).

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

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

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

$A$ spherical drop of liquid splits into $1000$ identical spherical drops. If $E_1$ is the surface energy of the original drop and $E_2$ is the total surface energy of the resulting drops,then $\frac{E_1}{E_2} = \frac{x}{10}$. The value of $x$ is: