MHT CET · Physics · Mechanical Properties of Fluids
Two spherical soap bubbles of radii ' \(a\) ' and ' \(b\) ' in vacuum coalesce under isothermal conditions. The resulting bubble has a radius equal to
- A \(a+b\)
- B \(\frac {a+b}{2}\)
- C \(\sqrt{a^2+b^2}\)
- D \(\frac {a+b}{ab}\)
Answer & Solution
Correct Answer
(C) \(\sqrt{a^2+b^2}\)
Step-by-step Solution
Detailed explanation
Number of moles is conserved, so
\(\mathrm{P}_1 \mathrm{~V}_1+\mathrm{P}_2 \mathrm{~V}_2=\mathrm{P}_3 \mathrm{~V}\)
But, \(\mathrm{P}=\frac{4 \mathrm{~T}}{\mathrm{r}}\) where, \(\mathrm{r}\) is the radius of the bubble
\(\begin{aligned}
& \therefore \quad \frac{4 \mathrm{~T}}{\mathrm{a}}\left(\frac{4}{3} \pi \mathrm{a}^3\right)+\frac{4 \mathrm{~T}}{\mathrm{~b}}\left(\frac{4}{3} \pi \mathrm{b}^3\right)=\frac{4 \mathrm{~T}}{\mathrm{c}}\left(\frac{4}{3} \pi \mathrm{c}^3\right) \\
& \mathrm{a}^2+\mathrm{b}^2=\mathrm{c}^2 \\
& \mathrm{c}=\sqrt{\mathrm{a}^2+\mathrm{b}^2}
\end{aligned}\)
\(\mathrm{P}_1 \mathrm{~V}_1+\mathrm{P}_2 \mathrm{~V}_2=\mathrm{P}_3 \mathrm{~V}\)
But, \(\mathrm{P}=\frac{4 \mathrm{~T}}{\mathrm{r}}\) where, \(\mathrm{r}\) is the radius of the bubble
\(\begin{aligned}
& \therefore \quad \frac{4 \mathrm{~T}}{\mathrm{a}}\left(\frac{4}{3} \pi \mathrm{a}^3\right)+\frac{4 \mathrm{~T}}{\mathrm{~b}}\left(\frac{4}{3} \pi \mathrm{b}^3\right)=\frac{4 \mathrm{~T}}{\mathrm{c}}\left(\frac{4}{3} \pi \mathrm{c}^3\right) \\
& \mathrm{a}^2+\mathrm{b}^2=\mathrm{c}^2 \\
& \mathrm{c}=\sqrt{\mathrm{a}^2+\mathrm{b}^2}
\end{aligned}\)
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