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 $30^{{}^{\circ }}\text{C}$ ). Report answer after rounding off to the nearest integer value.

Answer 1

Answer: 75

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'
${\text{P}}_{\text{B}}-{\text{P}}_{\text{A}}=5\text{ cm}\times \text{cos }60^{{}^{\circ }}=2\cdot 5\text{ cm}$ ...(I)
$\frac{{\text{P}}_{\text{A}}\times 46\times \text{a}}{\text{RT}}=\frac{{\text{P}}_0\times 45\cdot 25\times \text{a}}{\text{RT}}={\text{n}}_{\text{I}}\text{ or }{\text{n}}_{{\text{I}}^{{}^{\prime }}}$
${\text{P}}_{\text{A}}={\text{P}}_0\frac{45\cdot 25}{46}$ ...(II)
$\frac{{\text{P}}_{\text{B}}\times 44\cdot 5\times \text{a}}{\text{RT}}=\frac{{\text{P}}_0\times 45\cdot 25\times \text{a}}{\text{RT}}{\text{n}}_{\text{II}}\text{ or }{\text{n}}_{{\text{II}}^{{}^{\prime }}}$
${\text{P}}_{\text{B}}={\text{P}}_0\frac{45\cdot 25}{44\cdot 5}$ ...(III)
$\text{III}-\text{II}\to {\text{P}}_{\text{B}}-{\text{P}}_{\text{A}}={\text{P}}_{0}45\cdot 25\left[\frac{1}{44\cdot 5}-\frac{1}{46\cdot 0}\right]=2\cdot 5$
$\therefore$ ${\text{P}}_0\frac{45\cdot 25\times 1\cdot 5}{44\cdot 5\times 46\cdot 0}=2\cdot 5$
${\text{P}}_0=\frac{2\cdot 5\times 44\cdot 5\times 46\cdot 0}{45\cdot 25\times 1\cdot 5}=75\text{ cm}={\text{P}}_0$