Question

A thin tube of uniform cross-section is sealed at both ends. When it lies horizontally, the middle $5\text{cm}$ 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 length of the air columns above and below the mercury column are $46\text{cm}$ and $44.5\text{cm}$ respectively. Calculate the pressure $P$ in $\text{centimetres}$ of mercury.
(The temperature of the system is kept at $30{}^{\circ }\text{C})$

• A. $45.8$
• B. $75.4$
• C. $67.5$
• D. $79.3$

Answer 1

The correct option is B $75.4$

Let $a$ be the area of cross-section of the tube.

When the tube is exactly horizontal $5\text{cm Hg}$ is in the middle.

Length $L$ of the air column in tube will be,

$2L+5=44.5+5+46$

$\therefore L=45.25\text{cm}$

Now making the tube at ${60}^{\circ }$ with the vertical, the pressure difference across mercury will be,

$\Delta P=P_1-P_2=5\cos {60}^{\circ }$

$\therefore \Delta P=2.5\text{cm of Hg}$

Using Boyle's Law,

$PV=P_1{V}_1=P_2{V}_2$

$P_1=\frac{PV}{V_1}\&P_2=\frac{PV}{V_2}$

$\therefore \frac{P\times 45.25a}{44.5a}-\frac{P\times 45.25a}{46a}=2.5$

$\therefore P=75.4\text{cm of Hg}$

Hence, option $\left(\text{B}\right)$ is correct.