Question

The density of water is 1.00 x 10³ kg/m³, and the density of mercury is 1.36 x 104 kg

Calculate the absolute pressure p, at the bottom of a 0.250-m
tall graduated cylinder that is three-fourths full of mercury
and one-fourth full of water.

Answer 1

Okay, here are the steps:

1) Density of water: 1.00 x 103 kg/m3

Density of mercury: 1.36 x 104 kg/m3

2) Volume of cylindrical tube: 0.250 m tall

Area of base = pi * r^2 (where r = 0.05 m)

So volume (V) = pi * r^2 * h = 0.005pi * 0.25 = 0.05pi m^3

3) Volume occupied by mercury = 0.05pi * (0.75) = 0.0375pi m^3

Volume occupied by water = 0.05pi * (0.25) = 0.0125pi m^3

4) Mass of mercury = 0.0375pi * 1.36 x 104 kg/m3 = 5 x 10^3 kg

Mass of water = 0.0125pi * 1.00 x 103 kg/m3 = 1.25 x 10^2 kg

5) Total mass (m) = 5 x 10^3 + 1.25 x 10^2 = 7.125 x 10^3 kg

6) Absolute pressure (p) = mg/A = (7.125 x 10^3 kg) * (9.8 m/s^2) / (0.005pi m^2)

= 283750 Pa

So the absolute pressure (p) at the bottom of the graduated cylinder is 283750 Pa.

Let me know if you have any other questions!