Question

A thin tube of uniform cross section is sealed at both ends. It lies horizontally. The middle $5cm$ containing $Hg$ and the two equal parts containing air at the pressure $P_{0}$ . When the tube is held at an angle $60^\circ $ with the vertical, the length of air column above and below the mercury are $46$ and $44.5cm$ respectively. Calculate the pressure $P_{0}$ in $cm$ of $Hg$ (the temperature of the system is kept at $3 0^{^\circ } \text{C}$ ). Report answer after rounding off to the nearest integer value.

Answer 1

Let a is cross-sectional area I, II, I' , II' are sections of the tube.
Let P₀ is initial pressure in both sections
P_A and P_B are pressures in section I' and II'
$\text{P}_{\text{B}} - \text{P}_{\text{A}} = 5 \text{ cm} \times \text{cos } 6 0^{^\circ } = 2 \cdot 5 \text{ cm}$ ...(I)
$\frac{\text{P}_{\text{A}} \times 4 6 \times \text{a}}{\text{RT}} = \frac{\text{P}_{0} \times 4 5 \cdot 2 5 \times \text{a}}{\text{RT}} = \text{n}_{\text{I}} \text{ or } \text{n}_{\text{I}^{'}}$
$\text{P}_{\text{A}} = \text{P}_{0} \frac{4 5 \cdot 2 5}{4 6}$ ...(II)
$\frac{\text{P}_{\text{B}} \times 4 4 \cdot 5 \times \text{a}}{\text{RT}} = \frac{\text{P}_{0} \times 4 5 \cdot 2 5 \times \text{a}}{\text{RT}} \text{n}_{\text{II}} \text{ or } \text{n}_{\text{II}^{'}}$
$\text{P}_{\text{B}} = \text{P}_{0} \frac{4 5 \cdot 2 5}{4 4 \cdot 5}$ ...(III)
$\text{III} - \text{II} \rightarrow \text{P}_{\text{B}} - \text{P}_{\text{A}} = \text{P}_{0} 4 5 \cdot 2 5 \left[\frac{1}{4 4 \cdot 5} - \frac{1}{4 6 \cdot 0}\right] = 2 \cdot 5$
$\therefore $ $\text{P}_{0} \frac{4 5 \cdot 2 5 \times 1 \cdot 5}{4 4 \cdot 5 \times 4 6 \cdot 0} = 2 \cdot 5$
$\text{P}_{0}=\frac{2 \cdot 5 \times 4 4 \cdot 5 \times 4 6 \cdot 0}{4 5 \cdot 2 5 \times 1 \cdot 5}=75\text{ cm}=\text{P}_{0}$