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 $70 \times 10^{- 3} {Nm}^{- 1}$ and atmospheric pressure = $1 .013 \times 10^{5} {Nm}^{- 2}$ ]

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$2 .054 \times 10^{3} Pa$

$1.027 \times 10^{3} Pa$

$1.027 \times 10^{5} Pa$

$2.054 \times 10^{5} Pa$

answer is C.

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Detailed Solution

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

$\begin{array}{l} \Rightarrow P_{in} - P_{out} = \frac{2 T}{r} = \frac{2 \times 70 \times 10^{- 3}}{0 .1 \times 10^{- 3}} = 1400 Pa \\ \Rightarrow P_{in} = 1400 + 1 .013 \times 10^{5} = 1 .027 \times 10^{5} Pa \end{array}$

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