Hydrostatic.

tutorial

1. Find the pressure due to the fluid at a depth of 76 cm in still
Water (ρw = 1.00g/cm^3)
Mercury (ρ = 13.6g/cm^3)
Solution :
P = ρgh
= (1000 kg/m^3) (9.8m/s^2) (0.76)
= 7.448kPa @ 7.5kPa
P = ρgh
= (13600 kg/m^3) (9.8m/s^2) (0.86)
= 1.01 x 10^5Nm^2 @ 101.3 x 10^3N/m^2

2. A weight piston confines a find density ρ in a closed continue. The combined weight of piston and weight iss 200N, the total cross-sectional area of the piston is A = 8.0cm^2. Find the total pressure at point B if the fluid is mercury and h = 25cm. What would and ordinary pressure gauge read at B?

Solution :
P = (F/A) + ρgh
= (2000/8 x 10^ -4) + (13 600) (9.81m/s^2) (25 x 10^ -2)
= 250 000 + 33 354
= 2.8 x 10^5 Pa
P m = (13.6g/cm^3) (9.81m/s^2) (76cm)
= (0.136kg/m) (9.81m/s^2) (0.76cm)
= 1.01 x 10^5N/m
P = (2.8 x 10^5Pa) + Pm
= 92.8 x 10^5Pa) + (1.01 x 10^3N/m)
= 3.8 x 10^5Pa

3. A vertical test tube has 2.0cm of oil (ρ = 0.8g/cm^2) floating on 8.0cm of water. What is the pressure at the bottom of the tube due to the fluid in it?

Solution :
P1 = ρgh
= (800kg/m) (9.81m/s^2) (2 x10^ -2)
= 156.96
P2 = (1000) (9.81m/s^2) (8 x 10^ -2)
= 784.8
P1+P2 = 156.96 + 784.8
= 0.94kPa

4. The U-tube device connected to the tank. What is the pressure in the tank if atmosphera ρ is 76cm of mercury? The density of mercury is 13.6kg/m^3?

Solution :
P = (76cm) – (13.6g/cm^3) (9.81m/s^2) (5 x 10^ -2)
= (76cm) – (13 600kg/cm^3) (9.81m/s^2) (5 x 10^ -2)
= 94.65 x 10^3
= 95kPa


5. The mass of a block of a aluminium is 25.0g.
What is its volume?
What will be tthe tension in a string that suspends the block when the block is totally submerged in water? The density of aluminium is 2700kg/m^3.
Solution :
V = m/p
= (0.025) / (2700)
= 9.26cm^3
ft = v,( l-Pal )
= (9.26 x 10^ -6) (9.81) (1000-2700)
= 0.154N


6. A solid aluminium cylinder with ρ=2700kg/m^3 has a measured mass of 67g in air and 45g when immersed in turpentine. Determine the density of turpentine?

Solution :
Pt = RB / VTg
FB+FR = mg
FB = mg – FR
= (0.067 x 9.81) – (0.045 x 9.81)
= 0.21582
Vr = m / p
= (0.067) / (2700)
= 2.48 x 10^ -5
Pt = 0.21582
= (2481 x 10^ -5) (9.81)
= 886.7
= 8.9 x 10^2 kg/m^3