Question

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 ${60}^o$ 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 ${30}^o$C.

• A. $75.4$
• B. $45.8$
• C. $67.5$
• D. $89.3$

Answer 1

Answer: (A)

$75.4$

Let A be the area of cross-section of the tube. When the tube is horizontal, the $5$cm column of 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 ${60}^o$ 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 $P_1-P_2=5\cos {60}^o=5\times \frac{1}{2}=\frac{5}{2}$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