Question

1.2 A high cylinder with a cross-sectional area of ​​12.0 cm ^ 2 was partially filled with mercury whose density is 13.6 x 10^3 kg / m ^ 3; The surface of the mercury is at a height of 5.00 cm above the base of the cylinder. Water (1.00 x 10 ^ 3 kg / m ^ 3) is slowly poured over the mercury and it is observed that these two liquids do not mix. What volume of water should be added to double the gauge pressure at the base of the cylinder?

Answer 1

Consider the situation before pouring the water :

= Density of mercury = 13600 kg/m³

h = height of the surface of mercury = 5 cm = 0.05 m

Gauge pressure on the base is given as

P = gh

P = (13600) (9.8) (0.05)

P = 6664 Pa

After water is poured :

P' = Gauge pressure at the base = 2 P

= Density of water = 1000 kg/m³

h' = height of water above the surface of mercury

Gauge pressure at the base can now be given as

P' = g h' + gh

2P = g h' + P

P = g h'

6664 = (1000) (9.8) h'

h' = 0.8 m

A = Area of cross-section = 12 cm² = 12 x 10^-4 m²

Volume of water added is given as

V = A h'

V = (12 x 10^-4 ) (0.8)

V = 9.6 x 10^-4 m³