A column of mercury is in the middle of a horizontal tube of uniform bore closed at both ends. The pressure of the enclosed air column on either side of mercury being 75 cm (of mercury). When the tube is placed vertical the length of the air column below mercury column is $3 / 5^{t h}$ of that above it. Find the length of the mercury column(in cm).

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Solution

Given $P_{0}$ =75 cm of Hg

Hence, $P_{0} + ρ g h = \frac{3 P_{0}}{5}$
$75 c m o f H g + h = 45 c m o f H g$
$h = 30 c m o f H g$

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Given P0=75 cm of HgHence, P0+ρgh=3P05 75cmofHg+h=45cmofHg h=30cmofHg

Given $P_{0}$ =75 cm of Hg

Hence, $P_{0} + ρ g h = \frac{3 P_{0}}{5}$
$75 c m o f H g + h = 45 c m o f H g$
$h = 30 c m o f H g$