Fluid Mechanics Viscosity Question 245
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 $ Þ $ P_{in}=1400+1.013\times 10^{5} $ $ =0.014\times 10^{5}+1.013\times 10^{5}=1.027\times 10^{5}Pa $
BETA
