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 $60^{\circ}$ 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 $30^{\circ}C$.

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

Answer 1

The Correct Option is A

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 $60^{\circ}$ 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 $P_{1}-P_{2} = 5\cos\,60^{\circ} = 5 \times \frac{1}{2} = \frac{5}{2}\, cm$ of $Hg$ Using Boyle?? law for constant temperature, $PV = P_{1}V_{1} = P_{2}V_{2}$ $P \times A \times 45.25 = P_{1} \times A \times 44.5 = P_{2} \times A \times 46$ $\therefore\quad \frac{P\times 45.25}{44.5} - \frac{P\times 45.25}{46} = \frac{5}{2}$ or $\quad P = \frac{5\times 44.5\times 46}{2 \times 45.25 \times 1.5 } = 75.4\,cm$ of $Hg$