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, $R^{\prime }=2R,$
therefore $W=T\times 2\times 4\pi (2R{)}^2$
$W=32\pi R^{2}T$