Question

Two soap bubbles of radii $a$ and $b$ combine to form a single bubble of radius $c$. If $P$ is the external pressure, then the surface tension of the soap solution is

• A. $\frac{P\left(c^{3}+a^{3}+b^{3}\right)}{4\left(a^{2}+b^{2}-c^{2}\right)}$
• B. $\frac{P\left(c^{3}-a^{3}-b^{3}\right)}{4\left(a^{2}+b^{2}-c^{2}\right)}$
• C. $P c^{3}-4 a^{2}-4 b^{2}$
• D. $P c^{2}-2 a^{2}-3 b^{2}$

Answer 1

Answer: (B)

Assuming isothermal conditions,
$\left(P+\frac{4 \sigma}{a}\right)\left(\frac{4}{3} \pi a^{3}\right)+\left(P+\frac{4 \sigma}{b}\right)\left(\frac{4}{3} \pi b^{3}\right)$
$=\left(P+\frac{4 \sigma}{c}\right)\left(\frac{4}{3} \pi c^{3}\right)$
or $P\left[a^{3}+b^{3}-c^{3}\right]=4 \sigma\left[c^{2}-a^{2}-b^{2}\right]$
or $\sigma=\frac{P\left(c^{3}-a^{3}-b^{3}\right)}{4\left(a^{2}+b^{2}-c^{2}\right)}$