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Two spherical soap bubbles of radii $r_1$ and $r_2$ in vacuum combine under isothermal conditions. The resulting bubble has a radius equal to:
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Verified Answer
The correct answer is:
$\sqrt{r_1^2+r_2^2}$
Excess of pressure, inside the first bubble
$p_1=\frac{4 T}{r_1}$
Similarly, $\quad p_2=\frac{4 T}{r_2}$
Let the radius of the large bubble be $R$.
Then, excess of pressure inside the large bubble,
$p=\frac{4 T}{R}$
Under isothermal condition, temperature remains constant.
So, $\quad P V=p_1 V_1+p_2 V_2$
$\frac{4 T}{R}\left(\frac{4}{3} \pi R^3\right)=\frac{4 T}{r_1}\left(\frac{4}{3} \pi r_1^3\right)+\frac{4 T}{r_2}\left(\frac{4}{3} \pi r_2^3\right)$
$R^2=r_1^2+r_2^2$
$\Rightarrow \quad R=\sqrt{r_1^2+r_2^2}$
$p_1=\frac{4 T}{r_1}$
Similarly, $\quad p_2=\frac{4 T}{r_2}$
Let the radius of the large bubble be $R$.
Then, excess of pressure inside the large bubble,
$p=\frac{4 T}{R}$
Under isothermal condition, temperature remains constant.
So, $\quad P V=p_1 V_1+p_2 V_2$
$\frac{4 T}{R}\left(\frac{4}{3} \pi R^3\right)=\frac{4 T}{r_1}\left(\frac{4}{3} \pi r_1^3\right)+\frac{4 T}{r_2}\left(\frac{4}{3} \pi r_2^3\right)$
$R^2=r_1^2+r_2^2$
$\Rightarrow \quad R=\sqrt{r_1^2+r_2^2}$
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