Question

A thin tube of uniform cross-section is sealed at both ends. It lies horizontally, the middle $5cm$ containing $Hg$ and the two equal ends containing air at the same pressure $P_0$. When the tube is held at an angle ${60}^{\circ }$ with the vertical, the lengths of the air column above and below the mercury are $46cm$ and $44.5cm$ respectively. Calculate pressure $P_0$ in cm of $Hg$. (The temperature of the system is kept at ${30}^{\circ }C$).

Answer 1

At horizontal position, let the length of air column tube be $Lcm$.
$\therefore 2L+5=46+5+44.5cm$
$L=45.25cm$
When the tube is held at ${60}^{\circ }$ with the vertical, the mercury column will slip down.
$P_Y=5\cos {60}^{\circ }=P_X$
$P_X-P_Y=\frac{5}{2}=3.5cmHg$
From end $X,P_0\times 45.25=P_X\times 44.5$
$P_X=\frac{45.25}{44.5}{P}_0$
From end $Y,P_0\times 45.25=P_Y\times 46$
$P_Y=\frac{45.25}{46}{P}_0$
Substituting the values of $P_X$ and $P_Y$ in equation (i) we get
$P_0=75.4$.