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Q.
The radius of a soap bubble is increased from $\frac{1}{\sqrt{\pi}} cm$ to $\frac{2}{\sqrt{\pi}} cm$. If the surface tension of water is $30$ dynes per $cm$, then the work done will be :-
Solution:
$W =8 \pi T \left( r _{2}^{2}- r _{1}^{2}\right)$
$=8 \pi T \left[\left(\frac{2}{\sqrt{\pi}}\right)^{2}-\left(\frac{1}{\sqrt{\pi}}\right)^{2}\right]$
$\therefore W =8 \times \pi \times 30 \times \frac{3}{\pi}=720\, erg$