An air bubble of radius 1 mm is formed inside water at a depth 10 m below free surface (where, air pressure is 10⁵ Nm^-2). The pressure inside the bubble is (surface tension of water = 7 X 10^-2 Nm^-1)

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$2.0014 \times 10^{5} {Nm}^{- 2}$

$2.14 \times 10^{5} {Nm}^{- 2}$

$2.28 \times 10^{5} {Nm}^{- 2}$

$2.0028 \times 10^{5} {Nm}^{- 2}$

answer is D.

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Detailed Solution

Pressure inside the bubble

= Pressure outside $+ \frac{2 T}{r} = (p_{0} + ρ g h) + \frac{2 T}{r}$

$= 10^{5} + (10^{3} \times 10 \times 10) + \frac{2 \times 7 \times 10^{- 2}}{10^{- 3}}$

$= 2.0014 \times 10^{5} {Nm}^{- 2}$

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