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Question

Two soap bubbles of radii $$r_{1}$$ and $$r_{2}$$ equal to 4cm and 5 cm are touching each other over a common surface $$S_{1}S_{2}$$ (shown in figure). Its radius will be :

282372.png


A
4 cm
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B
20 cm
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C
5 cm
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D
4.5 cm
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Solution

The correct option is A 20 cm
The excess pressure inside any soap bubble is given by the expression,
$$ P_{excess} = \frac {4T}{r} $$

$$ P_1 = \frac {4T}{4} $$
$$ P_2 = \frac {4T}{5} $$

The difference between the 2 pressures will account for the radius of the new surface

$$ P_1 - P_2 = \frac {4T}{r} $$

$$ \frac {4T}{4} - \frac {4T}{5} = \frac {4T}{r} $$

Solving, gives r = 20 cm

Physics

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