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The surface tension of soap water is 0.04 Nm^-1. The excess of pressure inside a soap water bubble of diameter 10 mm is

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16 Pa

8 Pa

32 Pa

64 Pa

answer is C.

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Detailed Solution

$P = \frac{4 T}{R} = \frac{4 (0 .04)}{\frac{10}{2} \times 10^{- 3}} = 32 Pa$

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