CSEC Physics Syllabus - Effective for examinations from May - June 2015
Section A - Mechanics
Hydrostatics
Specific Objective 6.1
define pressure and apply definition;
Specific Objective 6.2
relate the pressure at a point in a fluid to its depth and the density;
Specific Objective 6.3
apply Archimedes' principle to predict whether a body would float or sink in a given fluid.
Pixabay
Example 1
a) Explain what is meant by pressure.
b) A rectangle block 0.01 m by 0.02 m by 0.04 m has a mass of 0.064 kg. Calculate:
1. the density of the material of the block
2. the weight of the block
3. the pressure the block would exert when resting on its smallest side.
Acceleration of free fall is 10 m/s2
Solution
a) Explain what is meant by pressure.
Pressure is defined as the force per unit area acting on an object or surface.
Pressure = Force ÷ Area
b) 1. The Density of the material is given by:
Density = Mass ÷ Volume
The volume of the rectangular block is given by:
Volume = Area of Base × Height
Volume = Length × Width × Height
Volume = 0.01 m × 0.02 m × 0.04 m
⸫ Volume = 0.000008 m3
Therefore,
Density = Mass ÷ Volume
Density = 0.064 kg ÷ 0.000008 m3
⸫ Density = 8 000 kg/m3
2. The Weight of the block is given by:
Weight = Mass × gravity
Weight = 0.064 kg × 10 m/s2
⸫ Weight = 0.64 N
3. The Pressure exerted by the block on its smallest side is given by:
Area of the block on its smallest side = Length × Width
Area of the block on its smallest side = 0.01 m × 0.02 m
⸫ Area of the block on its smallest side = 0.0002 m2
Therefore,
Pressure = Force ÷ Area
Pressure = 0.64 N ÷ 0.0002 m2
Pressure = 3 200 Pa
Example 2
A block of metal of density 3 000 kg/m3 is 2 m high and stands on a square base of side 0.5 m, as shown in the diagram below.
a) What is the base area of the block?
b) What is the volume of the block?
c) What is the mass of the block?
d) What is the weight of the block?
e) What is the pressure exerted by the weight of the block on the surface on which it stands?
Solution
a) The base area is given by:
Base Area = Length × Width
Base Area = 0.5 m × 0.5 m
Base Area = 0.25 m2
b) The volume of the block is given by:
Volume = Area of Base × Height
Volume = 0.25 m2 × 2 m
⸫ Volume = 0.5 m3
c) The mass of the block is given by:
Density = Mass ÷ Volume
Mass = Density × Volume
Mass = 3 000 kg/m3 × 0.5 m3
⸫ Mass = 1 500 kg
d) The weight of the block is given by:
Weight = Mass × gravity
Weight = 1 500 kg × 10 m/s2
⸫ Weight = 15 000 N
e) The pressure exerted by the weight of the block on the surface it stands on:
Pressure = Force ÷ Area
Pressure = 15 000 N ÷ 0.25 m2
⸫ Pressure = 60 000 Pa
Example 3
The pressure at a point below the surface of the sea is caused by both the height of water above that point and the air pressure on the surface. If a diver reaches a depth of 20 m below the surface of the sea, calculate:
a) the additional pressure on him due to the height of seawater alone
b) the total pressure on him.
The density of seawater = 1150 kg/m3
Air pressure = 1.0 × 105 Pa
g = 9.8 m/s2
Solution
a) The additional pressure due to the height of seawater alone is given by:
Pressure = Density of seawater × gravity × Height
Pressure = 1150 kg/m3 × 9.8 m/s2 × 20 m
⸫ Pressure = 225 400 Pa
b) The total pressure acting on him is given by:
Total Pressure = Air Pressure + Pressure
Total Pressure = 1.0 × 105 Pa + 225 400 Pa
⸫ Total Pressure = 325 400 Pa
Example 4
A diver reaches a depth in seawater where the pressure on him is five times greater than at the surface. If the air pressure at the surface of the sea is 1.0 × 105 Pa, what depth has the diver reached in the sea? (density of seawater is 1030 kg/m3, g = 9.8 N/kg)
Solution
The pressure on the diver at the depth is given by:
Pressure = 5 × air pressure at the surface of the sea
Pressure = 5 × 1.0 × 105 Pa
⸫ Pressure = 5.0 × 105 Pa
Therefore, the depth the diver reached is given by:
Pressure = Density of seawater × gravity × Depth
5.0 × 105 Pa = 1030 kg/m3 × 9.8 N/kg × Depth
1030 kg/m3 × 9.8 N/kg × Depth = 5.0 × 105 Pa
Depth = 5.0 × 105 Pa ÷ (1030 kg/m3 × 9.8 N/kg)
⸫ Depth = 48.5 m
Example 5
a) State Archimedes' principle.
b) A solid cylinder of wood has a cross-sectional area of 20 cm2 and a height of 90 cm. It floats upright in water. If two-thirds of the length of the cylinder is submerged, calculate:
1. the weight of water displayed
2. the weight of the wooden cylinder
3. the density of the wood.
Acceleration due to gravity, g = 10 m/s2
Density of water = 1 000 kg/m3
Solution
a) State Archimedes' Principle
The upthrust force on an object wholly or partially immersed in a fluid is equal and opposite to the weight of the fluid displaced by the object.
b) 1. Using Archimedes' principle, the weight of the water displaced is equal to 2/3 of the weight of the wooden cylinder, which is equal to the upthrust on the submerged cylinder.
The weight of the water displaced is given by:
Weight = Mass × gravity
We, need to find the mass of the water displaced by using the volume of water displaced and the density of water:
The volume of water displaced is given by:
The volume of water displaced = 2/3 of the volume of the cyclinder
The volume of water displaced = 2/3 × Area of Base × Height
The volume of water displaced = 2/3 × Area of circle × Height
The volume of water displaced = 2/3 × 20 cm2 × 90 cm
⸫ The volume of water displaced = 1 200 cm3
Converting the volume of water displaced into its SI unit:
100 cm = 1 m
100 cm × 100 cm × 100 cm = 1 m × 1 m × 1 m
1 000 000 cm3 = 1 m3
1 cm3 = [1 ÷ 1 000 000] m3
1 cm3 = 0.000001 m3
1 200 cm3 = [0.000001 × 1 200] m3
⸫ 1 200 cm3 = 0.0012 m3
Recall,
Density = Mass ÷ Volume
Mass = Density × Volume
Mass = 1 000 kg/m3 × 0.0012 m3
Mass = 1.2 kg
Therefore,
Weight = Mass × gravity
Weight = 1.2 kg × 10 m/s2
⸫ Weight = 12 N
2. The weight of the wooden cylinder is given by:
The weight of the water displaced = 2/3 × the weight of the wooden cylinder
The weight of the wooden cylinder = 3/2 × the weight of the water displaced
The weight of the wooden cylinder = 3/2 × 12 N
⸫ The weight of the wooden cylinder = 18 N
3. The density of the wood is given by:
Density = Mass ÷ Volume
We need to find the mass and volume of the wooden cylinder in its S.I. units:
The mass is given by:
Weight = Mass × gravity
Mass = Weight ÷ gravity
Mass = 18 N ÷ 10 m/s2
⸫ Mass = 1.8 kg
The volume is given by:
Volume = Area of Base × Height
Volume = 20 cm2 × 90 cm
Volume = 1 800 cm3
Converting the volume in its S.I. unit:
1 cm3 = 0.000001 m3
1 800 cm3 = [0.000001 × 1 800] m3
⸫ 1 800 cm3 = 0.0018 m3
Therefore,
Density = Mass ÷ Volume
Density = 1.8 kg ÷ 0.0018 m3
⸫ Density = 1 000 kg/m3
Example 6
a) A rectangular block of wood measures 4 cm × 4 cm × 10 cm and has a mass of 128 g.
1. Calculate its volume in SI units.
2. Express this volume in standard form.
3. Express the mass in SI units and standard form.
4. Calculate the density of the wood.
b) Using Archimedes' principle, explain why this block of wood will float in water of density 1 000 kg/m3.
c) Do you think the same block of wood would float in a liquid of relative density 0.85? Explain.
Solution
a) 1. The volume of the rectangular block is given by:
Volume = Area of Base × Height
Volume = Length × Width × Height
Volume = 4 cm × 4cm × 10 cm
Converting the measurements to their S.I. unit:
Volume = 0.04 m × 0.04 m × 0.1 m
⸫ Volume = 0.00016 m3
2. The volume in standard form is expressed as:
Volume = 0.00016 m3
⸫ Volume = 1.6 × 10-4 m3
3. The mass of the rectangular block in its S.I. unit and standard form is expressed as:
Mass = 128 g
Converting the measurements to their S.I. unit:
1 000 g = 1 kg
1 g = [1 ÷ 1 000] kg
1 g = 0.001 kg
128 g = [0.001 × 128] kg
128 g = 0.128 kg
⸫ 128 g = 1.28 × 10-1 kg
4. The density of the wood is given by:
Density = Mass ÷ Volume
Density = 1.28 × 10-1 kg ÷ 1.6 × 10-4 m3
⸫ Density = 800 kg/m3
b) Archimedes' principle states that the upthrust force on an object wholly or partially immersed in a fluid is equal and opposite to the weight of the fluid displaced by the object. Therefore, this block of wood will float in water of density 1 000 kg/m3 because the weight of the water displaced, which is equal to the upthrust force on the block of wood, is greater than the weight of the block of wood.
c) This same block of wood would float in a liquid of relative density 0.85
We shall find the density of the liquid:
Relative Density = Density of Liquid ÷ Density of water
0.85 = Density of Liquid ÷ 1 000 kg/m3
Density of Liquid = 0.85 × 1 000 kg/m3
⸫ Density of Liquid = 850 kg/m3
The Relative Density of the block of wood, when placed in this liquid, is given as:
Relative Density = Density of the block of wood ÷ Density of Liquid
Relative Density = 800 kg/m3 ÷ 850 kg/m3
⸫ Relative Density = 0.94
Recall,
If Relative Density ≥ 1, then the object will sink and
If 0 ＜ Relative Density ＜ 1, then the object will float.
Therefore, the block of wood would float in a liquid of relative density of 0.85 because the relative density of the block of wood, and this liquid is less than 1.
0 Comments