Search any question & find its solution
Question:
Answered & Verified by Expert
Two spherical soap bubbles of radii ' $a$ ' and ' $b$ ' in vacuum coalesce under isothermal conditions. The resulting bubble has a radius equal to
Options:
Solution:
2061 Upvotes
Verified Answer
The correct answer is:
$\sqrt{a^2+b^2}$
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
$\begin{aligned}
& \therefore \quad \frac{4 \mathrm{~T}}{\mathrm{a}}\left(\frac{4}{3} \pi \mathrm{a}^3\right)+\frac{4 \mathrm{~T}}{\mathrm{~b}}\left(\frac{4}{3} \pi \mathrm{b}^3\right)=\frac{4 \mathrm{~T}}{\mathrm{c}}\left(\frac{4}{3} \pi \mathrm{c}^3\right) \\
& \mathrm{a}^2+\mathrm{b}^2=\mathrm{c}^2 \\
& \mathrm{c}=\sqrt{\mathrm{a}^2+\mathrm{b}^2}
\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.