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Standard XII

Physics

Coefficient of Viscosity

Two soap bubb...

Question

Two soap bubbles with radii $r_{1}$ and $r_{2}$ $(r_{1} > r_{2})$ come in contact. Their common surface has a radius of curvature $r =$

(R_{1} + R_{2}) / 2

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(R_{1} R_{2}) / (R_{1} - R_{2})

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(R_{1} R_{2}) / (R_{1} + R_{2})

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\sqrt{R_{1} R_{2}}

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Solution

The correct option is B $(R_{1} R_{2}) / (R_{1} - R_{2})$
Since pressure difference = $\frac{4 T}{R}$
So, $\frac{4 T}{R_{n e t}} = \frac{4 T}{R_{2}} - \frac{4 T}{R_{1}}$
$R_{n e t} = \frac{R_{1} R_{2}}{R_{1} - R_{2}}$

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Two soap bubbles with radii $r_{1}$ and $r_{2}$ $(r_{1} > r_{2})$ come in contact. Their common surface has a radius of curvature $r =$

The correct option is B $(R_{1} R_{2}) / (R_{1} - R_{2})$
Since pressure difference = $\frac{4 T}{R}$
So, $\frac{4 T}{R_{n e t}} = \frac{4 T}{R_{2}} - \frac{4 T}{R_{1}}$
$R_{n e t} = \frac{R_{1} R_{2}}{R_{1} - R_{2}}$

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Two soap bubbles with radii r1 and r2 (r1>r2) come in contact. Their common surface has a radius of curvature r=

The correct option is B (R1R2)/(R1−R2)Since pressure difference = 4TRSo, 4TRnet=4TR2−4TR1 Rnet=R1R2R1−R2

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FREE TEXTBOOK SOLUTIONS

STATE BOARDS

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Two soap bubbles with radii $r_{1}$ and $r_{2}$ $(r_{1} > r_{2})$ come in contact. Their common surface has a radius of curvature $r =$

The correct option is B $(R_{1} R_{2}) / (R_{1} - R_{2})$
Since pressure difference = $\frac{4 T}{R}$
So, $\frac{4 T}{R_{n e t}} = \frac{4 T}{R_{2}} - \frac{4 T}{R_{1}}$
$R_{n e t} = \frac{R_{1} R_{2}}{R_{1} - R_{2}}$