Question

A thin tube of uniform cross section is sealed at both the ends. It lies horizontally, the middle $5\text{cm}$ 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 $46\text{cm}$ and $44.5\text{cm}$ respectively. Calculate the pressure $P,$ if the temperature of the system is kept at ${30}^{\circ }C$.

• A. $75.4\text{cm}$ of Hg
• B. $55.4\text{cm}$ of Hg
• C. $65.4\text{cm}$ of Hg
• D. $70.4\text{cm}$ of Hg

Answer 1

The correct option is A $75.4\text{cm}$ of Hg

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

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

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,

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

$\Rightarrow \begin{array}{c}{P}_1=\left(\frac{P{l}_0}{l_1}\right)\end{array}....\left(i\right)$.

Similarly, for the lower part,

$\begin{array}{c}{P}_2=\left(\frac{P{l}_0}{l_2}\right)\end{array}....\left(ii\right)$.

Thus,

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

Putting from (i) and (ii),

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

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

$\Rightarrow P=\left[\frac{mg}{2A{l}_0(\frac{1}{l_2}-\frac{1}{l_1})}\right]$.

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

$P=\rho gh$.

Also mas of the mercury pellet is $m=\left(5\text{cm}\right)A\rho$

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

$\Rightarrow h=\left[\frac{\left(5\right)}{2\times \left(45.25\right)\times (\frac{1}{44.5}-\frac{1}{46})}\right]=75.4\text{cm}$.

The pressure $P$ is equal to $75.4\text{cm}$ of Hg.