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Calculate pressure inside a small air bubble of radius 10^(-4)m radius found just below the surface of water (given S.T. of water =0.072 Nm^(-1) and" 1 atm "=1.013 xx 10^(5)Pa). |
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Answer» Solution :Given, `R=10^(-4)m""T=0.072Nm^(-1)""P_(o)=1.013 XX 10^(5)Pa` Excess of pressure inside an air BUBBLE `P-P_(o)=(2T)/(R)` i.e, `P=P_(o)+(2T)/(R)=(1.013 xx 10^(5))+(2 xx 0.072)/(10^(-4))=(1.013+0.0144)xx 10^(5)` `P=1.0274 xx 10^(5) NM^(-2) or Pa`. |
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