Q149 / 155 · WBJEE 2021 · Physics

Under isothermal conditions, two soap bubbles of radii a and b coalesce to form a single bubble of radius c. If the external pressure is $\mathrm{P}$, then surface tension of the bubbles is

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

Solution

Total number of moles $=$ constant $$ \begin{array}{l} \frac{P_{1} V_{1}}{R T}+\frac{P_{2} V_{2}}{R T}=\frac{P V}{R T} \Rightarrow P_{1} V_{1}+P_{2} V_{2}=P V \Rightarrow\left(P+\frac{4 T}{a}\right) \frac{4}{3} \pi a^{3}+\left(P+\frac{4 T}{b}\right) \frac{4}{3} \pi b^{3}=\left(P+\frac{4 T}{c}\right) \frac{4}{3} \pi c^{3} \\ \Rightarrow P\left(a^{3}+b^{3}-c^{3}\right)=4 T\left(c^{2}-a^{2}-b^{2}\right) \Rightarrow T=\frac{P\left(a^{3}+b^{3}-c^{3}\right)}{4\left(c^{2}-a^{2}-b^{2}\right)} \Rightarrow T=\frac{P}{4} \frac{\left(c^{3}-a^{3}-b^{3}\right)}{\left(a^{2}+b^{2}-c^{2}\right)} \end{array} $$

Hence, the correct answer is option: 'B'.

Relevant Questions