Question

The radius $ R $ of the soap bubble is doubled under isothermal condition. If $ T $ be the surface tension of soap bubble, the work done in doing so is given by:

• A. $ 32\pi {{R}^{2}}\,T $
• B. $ 24\pi {{R}^{2}}\,T $
• C. $ 8\pi {{R}^{2}}\,T $
• D. $ 4\pi {{R}^{2}}\,T $

Answer 1

Answer: (A)

The surface tension $(T)$ of a liquid is equal to the work $(W) $ required to increase the surface area of the liquid film by unity at constant temperature.
As per key idea, tension $=\frac{\text { work done }}{\text { surface area }}$
or $T=\frac{W}{\Delta A}$
Since, soap bubble has two surfaces and surface area of soap bubble is $4 \pi R^{2}$.
where $R$ is radius of bubble.
then $W=T \times 2 \times 4 \pi R^{2}$
Given, $ R'=2 R, $
therefore $W =T \times 2 \times 4 \pi(2 R)^{2} $
$W=32 \pi R^{2} T $