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Question: The pressure inside a small air bubble of radius 0.1 mm situated just below the surface of water wil...

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 70×10−3Nm−170 \times 10 ^ { - 3 } \mathrm { Nm } ^ { - 1 } and atmospheric pressure = 1.013×105Nm−21.013 \times 10 ^ { 5 } \mathrm { Nm } ^ { - 2 }]

A

B

1.027×103 Pa1.027 \times 10 ^ { 3 } \mathrm {~Pa}

C

1.027×105 Pa1.027 \times 10 ^ { 5 } \mathrm {~Pa}

D

2.054×105 Pa2.054 \times 10 ^ { 5 } \mathrm {~Pa}

Answer

1.027×105 Pa1.027 \times 10 ^ { 5 } \mathrm {~Pa}

Explanation

Solution

Excess pressure inside the air bubble =2Tr= \frac { 2 T } { r }

⇒ =1.027×105 Pa= 1.027 \times 10 ^ { 5 } \mathrm {~Pa}