Two soap bubbles each with radius $r_{1}$ and $r_{2}$ coalesce in vacuum under isothermal conditions to form a bigger bubble of radius R.Then, R is equal to

\sqrt{r_{1}^{2} + r_{2}^{2}}

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\sqrt{r_{1}^{2} - r_{2}^{2}}

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r_{1}^{2} + r_{2}^{2}

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\frac{\sqrt{r_{1}^{2} + r_{2}^{2}}}{2}

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2 \sqrt{r_{1}^{2} + r_{2}^{2}}

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Solution

The correct option is A $\sqrt{r_{1}^{2} + r_{2}^{2}}$
By Boyle's law
pV= constant
So, $p_{1} V_{1} + p_{2} V_{2} = p V$

where $p_{1} = \frac{T}{r_{1}}$

$V_{1} = \frac{4}{3} π r_{1}^{3}$

$p_{2} = \frac{2 T}{r_{2}}$

$v_{2} = \frac{4}{3} π r_{2}^{3}$

$p = \frac{2 T}{R}$

$V = \frac{4}{3} π R^{3}$

$\frac{2 T}{r_{1}} \times \frac{4}{3} π r_{1}^{3} + \frac{2 T}{r_{2}} \times \frac{4}{3} = \frac{2 T}{R} \times \frac{4}{3} π r^{3}$

$2 T \times \frac{4}{3} π (\frac{1}{r_{1}} \times r^{3} + \frac{1}{r_{2}} \times r_{2}^{3}) = 2 T \times \frac{4}{3} π (\frac{1}{R} \times R^{3})$

$r_{1}^{2} + r_{2}^{2} = R^{2}$

$R = \sqrt{r_{1}^{2} + r_{2}^{2}}$

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Similar questions

If the sum of the areas of two circles with radii R₁ and R₂ is equal to the area of a circle of radius R, then

If the sum of the areas of two circles with radii $R_{1}$ and $R_{2}$ is equal to the area of a circle of radius R then

(a) $R_{1} + R_{2} = R$

(b) $R_{1} + R_{2} > R$

(d) $R_{1}^{2} + R_{2}^{2} = R^{2}$

R_{1}

R_{2}

R_{1} + R_{2} = R

R_{1}^{2} + R_{2}^{2} = R^{2}

R_{1} + R_{2} < R

R_{1}^{2} + R_{2}^{2} < R_{2}

\frac{{R_{1}}^{2}}{{R_{2}}^{2}} = \frac{9}{576}, \frac{R_{1}}{R_{2}} = ?

r_{1}^{2} + r_{2}^{2} + r_{3}^{2} + r^{2} =

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