Question

A tank contains a pool of mercury 0.30 m deep, covered with a layer of water that is 1.2 m deep. Determine the pressure exerted by the double layer of liquids on the bottom of the tank.

Answer 1

Density of mercury, $\rho_m$ = 13560 kg/m³

Density of water $\rho_w$ = 1000 kg/m³

Water layer thickness, H_w = 1.2m

Mercury Layer thickness, H_m = 0.30 m

Gravity, g = 9.8 m/s²

Thus, the pressure P is given as,

= P=PwgHw+ PmgHm P = (1000kg/m)(9.8m/s)(1.20m) + (13560kg/m?)(9.8m/s)(0.30m) P=51626kg/m.s2 = 51626Pa 51.6kPa

Thus the Pressure exerted by the two layers of liquid will be 51.6 kPa.

The total pressure, if we add atmospheric pressure of 101.3 kPa as well will be 101.3kPa+51.6 kPa= 152.9 kPa