Answer:
0.05m
Explanation:
Density of water = ρ(w) = 1000 kg/ m³ ;
Density of Mercury = ρ(m) = 13628.95 kg/ m³
Total pressure at bottom of cylinder=1.1atm
Therefore, pressure due to water and mercury =1.1-1 =atm
0.1atm=10130pa
The pressure at the bottom is given by,
ρ(w) x g[0.4 - d] + ρ(m) x g x d = 10130
1000 x 9.8[0.4 - d] + 13628.95 x 9.8 d = 10130
3924 - 9810d + 133416d= 10130
123606d= 6206
d= 6202/123606
d= 0.05m
Depth of mercury alone = d = 0.05m
We have that for the Question " determine the depth of the mercury."
It can be said that
From the question we are told
the cylinder is 0.4 m tall and the absolute (or total) pressure at the bottom is 1.1 atmospheres, determine the depth of the mercury. (Assume the density of mercury to be 1.36 104 kg/m^3, and the ambient atmospheric pressure to be 1.013e5 Pa)
Therefore,
Absolute pressure at the bottom of the container =
[tex]P = 1.1 atm = 1.1 * (1.013*105) Pa\\\\= 1.1143 * 10^5 Pa[/tex]
Where,
Height of the cylinder = H = [tex]0.4 m[/tex]
Height of the water in the cylinder = [tex]H_1[/tex]
Height of the mercury in the cylinder = [tex]H_2[/tex]
Therefore,[tex]H = H_1 + H_2\\\\H_1 = H - H_2\\\\P = P_{atm} + \rho_1gH_1 + \rho_2gH_2\\\\P = P_{atm} + \rho_1g(H - H_2) + \rho_2gH_2\\\\1.1143*10^5 = 1.013*10^5 + (1000)(9.81)(0.4 - H_2) + (1.36*10^4)(9.81)H_2\\\\1.013*10^4 = 3924 - 9810H_2 + 133416H_2\\\\143226H_2 = 6206\\\\H_2 = 4.333*10^{-2} m[/tex]
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