Question

The heights of mercury surfaces in the two arms of the manometer shown in figure are 2 cm and 8 cm.
Atmospheric pressure = 1.01 × 10⁵ N⁻². Find (a) the pressure of the gas in the cylinder and (b) the pressure of mercury at the bottom of the U tube.

Answer 1

(a) Given:
Height of the first arm, h₁ = 8 cm
Height of the second arm, h₂ = 2 cm
Density of mercury, ρ_Hg = 13.6 gm/cc
Atmospheric pressure, p_a = 1.01 × 10⁵ N/m² = 1.01 × 10⁶ dyn/cm²
Now,
Let p_g be the pressure of the gas.
If we consider both limbs, then the pressure at the bottom of the tube will be the same.



According to the figure, we have:
$$\begin{gathered} p_g+{\rho }_{\mathrm{H}g}\times h_2\times g=p_a+{\mathrm{\rho }}_{\mathrm{Hg}}\times h_1\times g \\ \Rightarrow p_g=p_a+{\mathrm{\rho }}_{\mathrm{H}g}\times g(h_1-h_2) \\ =1.01\times {10}^6+13.6\times 980\times (8-2)\mathrm{dyn}/{\mathrm{cm}}^2 \\ =(1.01\times {10}^6+13.6\times 980\times 6)\mathrm{dyn}/{\mathrm{cm}}^2 \\ =1.09\times {10}^5\mathrm{N}/{\mathrm{m}}^2 \end{gathered}$$

(b) Pressure of the mercury at the bottom of the U-tube:
p_Hg = pa + ρ_Hg× h₁ × g
$$\begin{gathered} =1.01\times {10}^6+13.6\times 8\times 980 \\ =1.12\times {10}^5\mathrm{N}/{\mathrm{m}}^2 \end{gathered}$$