Question

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 :-

• A. 180 ergs
• B. 360 ergs
• C. 720 ergs
• D. 960 ergs

Answer 1

Answer: (C)

$W=8\pi T\left(r_2^2-r_1^2\right)$
$=8\pi T\left[{\left(\frac{2}{\sqrt{\pi }}\right)}^2-{\left(\frac{1}{\sqrt{\pi }}\right)}^2\right]$
$\therefore W=8\times \pi \times 30\times \frac{3}{\pi }=720erg$