Question

A 7.59-m-tall container is filled to the brim, partway with mercury and the rest of the way with water. The container is open to the atmosphere. What must be the depth of the mercury so that the absolute pressure on the bottom of the container is twice the atmospheric pressure?

Answer 1

Answer:

1.0335 m

Explanation:

Atmospheric pressure, Po = 1.01 x 10^5 Pa

Let the depth of mercury is h

the depth of water is 7.59 - h

density of mercury = 13.6 x 10^3 kg/m^3

density of water = 1000 kg/m^3

Pressure due to mercury + pressure due to water = 2 x Po

height of mercury x density of mercury x g + height of water x density x g

                                        = 2Po

h x 13.6 x 1000 x g + (7.59 - h) x 1000 x g = 2 x 1.01 x 10^5

h x 13.6 x 9.8 + (7.59 - h) x 9.8 = 2 x 1.01 x 100

133.28 h + 74.382 - 9.8 h = 202

123.48 h = 127.618

h = 1.0335 m

Thus, the height of mercury column is 1.0335 m.

Answer 2

Answer:

0.22 m

Explanation:

$P_{o}$ = Atmospheric pressure = 101325 Pa

$P$ = Pressure at the bottom = tex]2 P_{o}[/tex] = 2 (101325) = 202650 Pa

$h$ = height of the container = 7.59 m

$x$ = depth of the mercury

Pressure at the bottom = Atmospheric pressure + Pressure due to mercury + Pressure due to water

$P = P_{o} + \rho _{w} g (h - x) + \rho _{hg} g x\\2 P_{o} = P_{o} + \rho _{w} g (h - x) + \rho _{hg} g x\\P_{o} = \rho _{w} g (h - x) + \rho _{hg} g x\\101325 = (1000) (9.8) (7.59 - x) + (13600) (9.8) x\\x = 0.22 m$