A thin tube of uniform cross section is sealed at both ends. It lies horizontally, the middle 5 cm containing mercury and the parts on its two sides containing air at the same pressure P. When the tube is held at an angle of ${60}^{\circ }$ with the vertical, the length of the air column above and below the mercury pellet are 46cm and 44.5cm respectively. Calculate the pressure P in centimeters of mercury. The temperature of the system is kept at ${30}^{\circ }C$.

The correct option is Bb 75.39 cms Hg.

When the tube is kept inclined to the vertical, the length of the upper part is $l_1=46cm,$ and that of the lower part is $l_2=44.5cm.$ When the tube lies horizontally, the length on each side is -

$l_0=\left[\frac{l_1+l_2}{2}\right]=\left[\frac{46cm+44.5cm}{2}\right]=45.25cm.$

Let $P_1$ and $P_2$ be the pressures in the upper and the lower parts when the tube is kept inclined. As the temperature is constant throughout, we can apply Boyle's law. For the upper part,

$P_1{l}_{1}A=P{l}_{0}A$

$P_1=\left(\frac{P{l}_0}{l_1}\right)$.................(i)

Similarly, for the lower part,

$P_2=\left(\frac{P{l}_0}{l_2}\right)$................(ii)

Thus,

$P_2=P_1+\frac{mg}{A}cos{60}^{\circ }$.

Putting from (i) and (ii),

$\left(\frac{P{l}_0}{l_2}\right)=\left(\frac{P{l}_0}{l_1}\right)+\left(\frac{mg}{2A}\right)$

$\Rightarrow P{l}_0(\frac{1}{l_2}-\frac{1}{l_1})=\frac{mg}{2A}$

$\Rightarrow P=\left[\frac{mg}{2A{l}_0(\frac{l}{l_2}-\frac{l}{l_1})}\right]$.

If the pressure P is equal to a height h of mercury,

$P=\rho gh$.

Also, m = (5 cm) $A\rho$

$\Rightarrow \rho gh=\left[\frac{\left(5cm\right)A\rho g}{2A{l}_0(\frac{1}{l_2}-\frac{1}{l_1})}\right]$

$\Rightarrow h=\left[\frac{\left(5cm\right)}{2\times \left(45.25cm\right)\times (\frac{1}{44.5}-\frac{1}{46cm})}\right]=75.39cm$.

The pressure P is equal to 75.39 cm of Hg.