A thin tube of uniform cross-section is sealed at both ends. When it lies horizontally, the middle $5 cm$ length contains mercury and the two equal ends contain air at the same pressure P. When the tube is held at an angle of $60°$ with the vertical, then the lengths of the air columns above and below the mercury column are $46 cm$ and $44.5 cm$ respectively. Calculate the pressure P in cm of mercury. The temperature of the system is kept at $30°C$.

Let $A$be the area of cross-section of the tube. When the tube is horizontal, the $5 cm$ column of $\mathrm{Hg}$ is in the middle, so length of air column on either side at pressure $P$

$=\frac{46+44.5}{2}=45.25 cm$

When the tube is held at $60°$ with the vertical, the lengths of air columns at the bottom and the top are $44.5 cm$ and $46 cm$ respectively. If $P_1$ and $P_2$ are their pressures, then

$P_1-P_2=5\cos 60°=5\times \frac{1}{2}=\frac{5}{2}cm$of $Hg$

Using Boyle's law for constant temperature,

$PV = P_1{V}_1= P_2{V}_2$

$P\times A\times 45.25=P_1\times A\times 44.5=P_2\times A\times 46$

Therefore , $\frac{P\times 45.25}{44.5}-\frac{P\times 45.25}{46}=\frac{5}{2}$

or $P=\frac{5\times 44.5\times 46}{2\times 45.25\times 1.5}=75.4$ $cm$ of $Hg$