The excess pressure due to surface tension inside a spherical drop is 6 units. If eight such drops coalesce, the excess pressure inside the new drop is:

3 units

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6 units

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12 units

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48 units

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Solution

The correct option is A 3 units

$G i v e n : P_{i n} - P_{o r t} = \frac{4 T}{r} = 6 u n i t s$
If eight such drops coalesce
$\Rightarrow 8 (\frac{4}{3} π r^{3}) = \frac{4}{3} π r_{1}^{3}$
$r_{1} = 2 r$
$∴ e x c e s s P r e s s u r e : P_{i n} - P_{o r t} = \frac{4 T}{r_{1}} = \frac{4 T}{2 r} = \frac{6}{2}$ $=$ 3 units

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r

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σ

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