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

The correct option is A $32\pi R^{2}T$

The surface tension(T) of a liquid is equal to the work (W) required to increase the surface area of the liquid file by unity at constant temperature.
As per key ides, tension
$=\frac{workdone}{surfacearea}$
or $T=\frac{W}{\Delta A}$
Since soa bubble has two surface 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 $R=2R$
therefore $W=T\times 2\times 4\pi \times (2R{)}^2$
$W=32\pi R^{2}T$.