Question

A thin tube of uniform cross section is sealed at both ends. It lies horizontally, the middle $5$cm containing mercury and the two equal ends containing air at the same pressure P. When the tube is held at an angle ${60}^o$ with the vertical direction, the length of the air column 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 ${30}^o$C)

• A. $85.4$ cm of Hg column.
• B. $75.4$ cm of Hg column.
• C. $65.4$ cm of Hg column.
• D. $95.4$ cm of Hg column.

Answer 1

The correct option is B $75.4$ cm of Hg column.

Solution:-

Let the length of the air column initially in the tube be $lcm$.

Since the total length of the tube remains the same before and after tilting,

$\therefore 2l+5=46+44.5$

$\Rightarrow l=45.25cm$

When the tube is held vertically at an angle of $60º$, the $Hg$ column will be displaced to the lower end. If the pressure of top and bottom ends be $P_A$ and $P_B$ respectively, then

$P_B=P_A+5\cos 60°$

$\Rightarrow P_B-P_A=5\times \frac{1}{2}=2.5.....\left(1\right)$

As $PV=\text{constant}$ holds for each air column.

Therefore, for end A,

$P_A=\frac{l}{l_A}{P}_0$

$\Rightarrow P_A=\frac{45.25}{46}{P}_0.....\left(2\right)$

Now, for end B,

$P_B=\frac{l}{l_B}{P}_0$

$P_B=\frac{45.25}{44.5}{P}_0.....\left(3\right)$

Substituting the values of $P_A$ and $P_B$ in ${eq}^n\left(1\right)$, we have

$\frac{45.25}{44.5}{P}_0-\frac{45.25}{46}{P}_0=2.5$

$\Rightarrow P_0=\frac{2.5\times 46\times 44.5}{45.25}=75.4\text{cm of Hg}$

Hence the pressure $P$ is $75.4\text{cm of Hg}$.