Properties Of Solids And Liquids Question 323

Question: 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}}N{{m}^{-1}} $ and atmospheric pressure = $ 1.013\times 10^{5}N{{m}^{-2}} $ ] [AMU (Med.) 2002]

Options:

A) $ 2.054\times 10^{3}Pa $

B)$ 1.027\times 10^{3}Pa $

C) $ 1.027\times 10^{5}Pa $

D)$ 2.054\times 10^{5}Pa $

Show Answer

Answer:

Correct Answer: C

Solution:

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

$ \Rightarrow P _{in}-P _{out}=\frac{2T}{r}=\frac{2\times 70\times {{10}^{-3}}}{0.1\times {{10}^{-3}}}=1400Pa $

therefore $ P _{in}=1400+1.013\times 10^{5} $

$ =0.014\times 10^{5}+1.013\times 10^{5}=1.027\times 10^{5}Pa $



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