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Observation show that rate of change of atmosphere pressure P with attitude h is proportional to pressure. Assume that pressure at 6000 meter is half of its value $P_0$ at sea level, find formula for pressure at any height.

$(a)\;P=p. \frac{h}{6000} \\ (b)\;P= 3000 p.h \\ (c)\;P= P.(\frac{1}{2})h \\ (d)\;P=P. (\frac{1}{2})^{h/6000} $
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given $\large\frac{dp}{dh}$$=KP$
$\large\frac{dP}{P}$$=K dh$
Integrate $ \ln P \bigg]_{P_0}^{P/o/2}=k h \bigg]_0^{6000}$
$\log \large\frac{1}{2}$$=6000 K$
Now $ \log P \bigg]_{P_0}^{P}=\large\frac{\log (1/2)}{6000} \times h \bigg]_0^h$
$ \log \frac{P}{P_0} =\large\frac{h \log (1/2)}{6000}$
$P=P_0 \bigg(\large\frac{1}{2}\bigg) ^{\Large\frac{h}{6000}}$
Hence d is the correct answer.


answered Feb 6, 2014 by meena.p

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