Question

Mercury is added to a cylindrical container to a depth d and then the rest of the cylinder is filled with water. If the cylinder is 0.4 m tall and the absolute (or total) pressure at the bottom is 1.2 atmospheres, determine the depth of the mercury. (Assume the density of mercury to be 1.36 104 kg/m3, and the ambient atmospheric pressure to be 1.013e5 Pa)

Answer 1

Answer:

0.10839 m

Explanation:

$P_1$ = Atmospheric pressure = 1 atm = 101325 Pa

$P$ = Total pressure at bottom of mecury = 1.2 atm

g = Acceleration due to gravity = 9.81 m/s²

h = d = Depth of mercury

$\rho_m$ = Density of mercury = $1.36\times 10^4\ kg/m^3$

$\rho_w$ = Density of water = $1000\ kg/m^3$

Pressure at the bottom is of the cylinder is given by

$P_2=P_1+\rho_wgh\\\Rightarrow P_2=101325+1000\times 9.81(0.7-d)$

Pressure at the bottom of mercury is

$P=P_2+\rho_mgh\\\Rightarrow 1.2\times 101325=101325+1000\times 9.81(0.7-d)+1.36\times 10^4\times 9.81\times d\\\Rightarrow 1.2\times 101325=123606d+108192\\\Rightarrow d=\dfrac{1.2\times 101325-108192}{123606}\\\Rightarrow d=0.10839\ m$

The depth of the mercury is 0.10839 m