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The pressure inside a small air bubble of radius 0.1 mm situated just below the surface of water will be equal to [Take surface tension of water 70×10-3 Nm-1 and atmospheric pressure = 1.013×105 Nm-2]

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a

2.954 × 103 Pa

b

1.027 × 103 Pa

c

1.027×105 Pa

d

2.054×105 Pa

answer is C.

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Detailed Solution

Excess pressure inside the air bubble = 2Tr⇒Pin-Pout = 2Tr = 2×70×10-30.1×10-3 = 1400 Pa⇒Pin = 1400+1.013×105 = 0.014×105+1.013×105 = 1.027×105 Pa

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