Question

A cylindrical container 1.3 m tall contains mercury to a certain depth, d. The rest of the cylinder is filled with water.

Part A If the pressure at the bottom of the cylinder is two atmospheres, what is the depth d?

Answer 1

Density of water = ρ(w) = 1000 kg/ m³ ;
Density of Mercury = ρ(m) = 13628.95 kg/ m³
The pressure at the bottom = ρ(w).g.[1.3- d] + ρ(m).g.d + 1 atm = 2 atm.
(The pressure at the bottom is taken as Absolute pressure and
, 1 Atm = 1.01*10⁵ Pa.)
plugging the values:
=»1000*9.81[1.3 - d] + 13628.95*9.81*d + 1 Atm = 2 Atm
=»1000{ 12.753 - 9.81d + 133.7d + 101 } = 2*101*1000
=»12.753 - 123.89d = 101

⇒ d = 88.247 / 123.89 = 0.71 m.
Depth of mercury alone ,

d= 0.71 m.