Question

The heights of mercury surfaces in the two arms of the manometer shown in figure (13-E1) are 2 cm and 8 cm. Atmospheric pressure = $1.01\times {10}^{5}N{m}^{-}2$. 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

Given $h_1$ = 8 cm
$h_2$ =2 cm
$\rho$ (Hg) = 13.6 gm/cc= $13.6\times {10}^{3}kg/m^3$
$\left(P_0\right)=1.01\times {10}^{5}N/m^2$
a.The pressure of the gas in the cylinder

$\Rightarrow P_0+\rho \left(\delta H\right)g$
= $1.01\times {10}^5+13.6\times {10}^3\times 9.8\times (8-2)\times {10}^{-2}N/m^2$

=$1.01\times {10}^5+0.079968\times {10}^5$
= $1.09\times {10}^{5}N/m^2$ (approximate value )

(b)The pressure of mercury at the bottom of the U tube.

= $\Rightarrow P_0+\rho Hg$
=$1.01\times {10}^5+13.6\times {10}^3\times 9.8\times 8\times {10}^{-2}N/m^2$

$=1.01\times {10}^5+0.106624\times {10}^5$

$=1.1167\times {10}^{5}N/m^2$ (approximate value )