Chapter 10 - Mechanical Properties of Fluids

In the previous section, we completed a discussion on the mechanical properties of solids. In this chapter we will see mechanical properties of fluids

■ When we study the mechanical properties of a body, we need to examine how pressure affects that body

• In the previous chapter, we have seen this:

    ♦ When external pressure is applied on a solid, it’s volume decreases

• If solids can be subjected to such a decrease in volume, then surely, we can do it with liquids and gases also

• But there are some differences:

    ♦ In the case of solids and liquids:

          ✰ Even a very high pressure can bring about only very small decrease in volume

    ♦ In the case of gases

          ✰ Even a small pressure can bring about a very large decrease in volume

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• Let us see some basic details about pressure

• It can be written in 4 steps:

1. Consider the platform in fig.10.1(a) below:

Fig.10.1

• Some forces are acting on the top surface of the platform

    ♦ All those forces are of the same magnitude

    ♦ All those forces are perpendicular to the top surface of the platform

2. In such a situation, we draw a square on the platform

    ♦ We can draw the square at any convenient place on the top surface

    ♦ The side of the square must be 1 m. This is shown in blue color in fig.b

• The total force experienced in that 1 m square area is called pressure

3. Obviously, we need not actually draw the square

• We can obtain the result in (2) just by:

    ♦ Dividing the total force acting on the entire top surface

    ♦ By the area of the top surface

• This is because:

    ♦ The no. of '1 m squares' that can be drawn on the top surface

    ♦ is equal to

    ♦ The area (measured in m²) of the top surface

■ So we can write: $\mathbf\small{\rm{Pressure\;(N\;m^{-2})=\frac{Force\;(N)}{Area\;(m^2)}}}$

4. We see that, area is in the denominator

■ So, if force remains the same and area is increased, the pressure will decrease

• We see a practical application of this principle in the two rear wheels of a tractor

    ♦ Those wheels are made broader

    ♦ So the contact surface area increases

    ♦ This reduces the pressure on the ground

    ♦ Since the pressure is reduced, the wheels will not sink into the muddy or sandy soil

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Units of pressure

• We see that, the unit of pressure is N m^-2

• Another name for this ‘N m^-2’ is: pascal

    ♦ 1 N m^-2 = 1 pascal

• So we can express pressure in either Nm^-2 or pascal

    ♦ The symbol for pascal is pa

• The name pascal is given n honour of the French scientist Blaise Pascal

    ♦ It was Blaise Pascal who carried out pioneering studies in fluid pressure

■ Another unit for pressure is atmosphere

    ♦ It’s symbol is atm

• The basis of this unit can be written in 11 steps:

1. Consider a platform. We know that atmospheric air will be present all around that platform

2. Air has weight. That means, air is exerting weight on the top surface of the platform

3. Now consider a column made up of air. It is shown in yellow color in fig.10.2 below:

Fig.10.2

• The column is in the form of a square prism. That is., the top and bottom bases are squares

    ♦ The side of these top and bottom squares should be exactly 1 m

4. We know that:

• To completely define a square prism, we need two items:

    ♦ The side of the square base of the prism

    ♦ The height of the prism

(Some basic details about prisms can be seen here)

5. We already know the side of the square base. It is 1 m

• What is the height?

• Answer can be written in 2 steps:

(i) The height is equal to the height of the atmosphere

(ii) The atmosphere extends to many kilo meters above the surface of the earth

    ♦ We need to take that total height of the atmosphere

6. Now our prism is completely defined

• So we can calculate it’s volume

7. Next, we multiply that volume by ‘density of air’

• The product will be equal to the mass of air in that air column

8. Next, we multiply that mass by the acceleration due to gravity g

• The product will be equal to the weight of that air column

• Scientists have calculated this weight as: 1.013 × 10⁵ N

9. The base of our prism is a 'square of side 1 m'

• So, every 1 m² area of the platform is subjected to a weight of 1.013 × 10⁵ N

■ That means, the pressure exerted by the atmosphere on the platform is 1.013 × 10⁵ N m^-2

10. But we need to consider the height factor

• This can be explained in 3 steps:

(i) The weight of prism we saw in (8), is obtained when the platform is placed at sea level

(ii) If the platform is at the top of a mountain, the height of the prism will be less

(iii) Consequently, the weight of the prism will be less

11. So we can write:

■ When height from the sea level increases, the pressure decreases

■ The pressure experienced by the platform at sea level is called: 1 atmosphere pressure

■ So we can write: 1 atm = 1.013 × 10⁵ N m^-2

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Let us see a solved example:

Solved example 10.1

Density of the atmosphere at sea level is 1.29 kg m^-3. Assume that, this density does not change with altitude. Then how high would the atmosphere extend?

Solution:

1. From the data book, we have:

Pressure at sea level = 1 atm = 1.013 × 10⁵ N m^-2 

2. So the weight of a square prism having:

    ♦ side of the base 1 m

    ♦ height same as 'height of atmosphere'

• is equal to 1.013 × 10⁵ N

• So mass of the square prism = $\mathbf\small{\rm{\frac{1.013 \times 10^5}{9.8}}}$ kg  

3. Mass of the square prism = Volume of that square prism × density of air 

⇒ Mass of the square prism = (Base area × Height of atmosphere) × density of air

• Substituting the known values, we get:

$\mathbf\small{\rm{\frac{1.013 \times 10^5}{9.8}=(1 \times Height\;of\;atmosphere)\times1.29}}$ 

• Thus we get:

Height of atmosphere = 8013 m 

4. So we get a height of approximately 8 km

• But the actual height is more than 100 km

• We get a very different result due to two wrong assumptions:

    ♦ We assumed the density to be uniform. In reality, as we go upwards, the density decreases

    ♦ We assumed g to be uniform. In reality, as we go upwards, g decreases

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• We know that, solid, liquid and gas are the three states of matter

• Out of the three, liquids and gases are together called fluids

    ♦ This is because, liquids and gases have the ability to flow

• Solids cannot be called fluids because, solids cannot flow

■ Now we will see the 'force exerted by fluids' on submerged bodies

• We will first see the direction of this force

• It can be written in 7 steps:

1. Consider the wedge in fig.10.3(a) below:

Fig.10.3

• It is submerged in a fluid

• The fluid is stationary. That is., it is not flowing

2. A force is exerted by the fluid on the sloping surface of the wedge

• This force is shown to be perpendicular to the sloping surface

3. Let us see what happens if the force is not perpendicular

• In fig.10.3(b) the force is not perpendicular

• Since it is not perpendicular, there will be two components for that force

    ♦ The green component which is parallel to the sloping surface

    ♦ The blue component which is perpendicular to the sloping surface

4. Once a force is resolved into it's components, we can ignore the original force

• So in fig.c, we ignore the red force

    ♦ We say this:

          ✰ The fluid exerts the green force parallel to the sloping surface

          ✰ The fluid exerts the blue force perpendicular to the sloping surface

• The 'combined effect of the green and blue forces' will be same as the 'effect of the red force' 

(Recall that, this type of resolution is not possible in fig.a because, the red force is perpendicular to the surface) 

5. Now consider the green force

• The presence of this green force indicates that, the fluid is exerting a force parallel to the sloping surface

• If the fluid exert this parallel force, the wedge will exert an equal and opposite green force on the fluid

    ♦ As a result, the fluid will move (flow)

• But in (1), we said that, the fluid is stationary

6. So it is clear that, if the fluid is stationary,

    ♦ the force exerted by the fluid

    ♦ on a submerged body

    ♦ will be perpendicular to the surface (as shown in fig.10.3.a)

7. In fig.10.3(a), only one force is shown. It is the force on the sloping surface

• But in fact, the fluid exerts forces on all the surfaces of the wedge

• All those forces will be perpendicular to the surfaces on which they are acting

---

• Now we know the 'direction of the force'

    ♦ We saw that: the direction is always perpendicular

• So it is easy to calculate the pressure

    ♦ By definition, we can consider only 'perpendicular forces' for pressure calculation

• The forces exerted by fluids are indeed perpendicular

• So we can confidently divide those forces by corresponding areas

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Next we will see the basics of a ‘device used to measure pressure’

It can be written in 7 steps:

1. In fig.10.4(a) below, a cylinder is shown in silver color

Fig.10.4

• A blue disc moves inside the cylinder

• The portion below the disc is vacuum

2. The disc rests on a spring

• A needle is attached to the top end of the spring

• When there is no force acting on the disc, the spring will be at it’s normal position

• In this position, the number zero is marked near the tip of the needle

3. Let the area of the disc be 5 cm²

• Apply a force of 0.05 N on the disc

• Then the pressure experienced by the disc will be: $\mathbf\small{\rm{\frac{0.05\;(N)}{5(cm^2)}=0.01\;N\;cm^{-2}=100\;N\;m^{-2}}}$

• Due to the pressure of 100 N m^-2, the spring will get compressed

    ♦ The needle will lower to a new position

    ♦ In this position, '100' is marked near the tip of the needle

    ♦ This is shown in fig.b

4. Apply a force of 0.1 N on the disc

• Then the pressure experienced by the disc will be: $\mathbf\small{\rm{\frac{0.1\;(N)}{5(cm^2)}=0.02\;N\;cm^{-2}=200\;N\;m^{-2}}}$

• Due to the pressure of 200 N m^-2, the spring will get compressed

    ♦ The needle will lower to a new position

    ♦ In this position, '200' is marked near the tip of the needle

    ♦ This is shown in fig.c

5. Apply a force of 0.15 N on the disc

• Then the pressure experienced by the disc will be: $\mathbf\small{\rm{\frac{0.15\;(N)}{5(cm^2)}=0.03\;N\;cm^{-2}=300\;N\;m^{-2}}}$

• Due to the pressure of 300 N m^-2, the spring will get compressed

    ♦ The needle will lower to a new position

    ♦ In this position, '300' is marked near the tip of the needle

    ♦ This is shown in fig.d

6. In this way, the green rectangle can be filled with appropriate values

• When all the values are marked, the device is ready to measure pressure values 

• Calibration of a measuring device involves two steps:

(i) Finding the values to be filled inside the green rectangle

(ii) Marking those values accurately in the green rectangle

7. To measure the pressure at a particular point 'A', inside a fluid:

• We place the device in such a way that, the top surface of the disc coincides with A

• The pressure will push down the disc and we will get a reading

• Note that, the fig.10.4 shows only a schematic arrangement. We will see the actual arrangement in higher classes

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• Next, we will learn about density

• Density is an important property related to fluids

• We know that, density is obtained when we divide mass by volume

• The symbol for density is: $\mathbf\small{\rm{\rho}}$ (Greek letter 'rho')

    ♦ If m is the mass and V the volume, we can write: $\mathbf\small{\rm{\rho=\frac{m}{V}}}$ 

• The SI unit of $\mathbf\small{\rm{\rho}}$ is: kg m^-3

■ In general, liquids are incompressible. That is, we cannot decrease their volume easily by applying pressure

    ♦ So for liquids, volume remains constant

    ♦ Since the volume remains constant, the density also remains constant

■ But gases are highly compressible

    ♦ So the density of a sample of a gas can be varied easily

• This can be explained in 3 steps:

1. Consider a sample of a gas

    ♦ It will have a definite mass m and a definite volume V₁

    ♦ So the density of that gas sample will be: $\mathbf\small{\rm{\rho_1=\frac{m}{V_1}}}$

2. Apply a pressure and thus decrease the volume of the sample to V₂

    ♦ Even though the volume is decreased, the mass m remains the same

    ♦ The new density of the sample will be: $\mathbf\small{\rm{\rho_2=\frac{m}{V_2}}}$

3. Compare the results in (1) and (2)

    ♦ The numerator remains unchanged. But the denominators are different

    ♦ So $\mathbf\small{\rm{\rho_1}}$ will be different from $\mathbf\small{\rm{\rho_2}}$

    ♦ Thus we can write: The density of a gas sample can be changed by applying pressure

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Next, we will learn about relative density

It can be written in steps:

1. Relative density is defined as

    ♦ The ratio of

    ♦ The density of a substance

    ♦ to The density of water at 4 ^oC

2. The temperature of 4 ^oC is specified because, the density of water tends to change slightly with temperature

• At 4 ^oC, the density of water is 1.00 × 10³ kg m^-3

3. Let us find the relative density of some common substances:

(i) The density of aluminium is 2.7 × 10³ kg m^-3

So the relative density of aluminium = $\mathbf\small{\rm{\frac{2.7 \times 10^3\;(kg\;m^{-3})}{1.0 \times 10^3\;(kg\;m^{-3})}=2.7}}$

(ii) The density of mercury is 13.6 × 10³ kg m^-3

So the relative density of mercury = $\mathbf\small{\rm{\frac{13.6 \times 10^3\;(kg\;m^{-3})}{1.0 \times 10^3\;(kg\;m^{-3})}=13.6}}$

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In the next section, we will see more details about pressure

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Chapter 2.10 - Significant figures - Solved examples

In the previous section, we saw the 'rules for arithmetic operations with significant figures'. We saw some solved examples also. In this section, we will see a few more solved examples.

Solved example 2.15
Fill in the blanks
(Note: In stating numerical answers, take care of significant figures)
(a) The volume of a cube of side 1 cm is equal to .....m³
(b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to  ...(mm)²
(c) A vehicle moving with a speed of 18 km h^-1 covers ....m in 1 s
(d) The relative density of lead is 11.3. Its density is ....g cm^-3 or ....kg m^-3
Solution:
Part (a):
1. We have: Volume  = s × s × s
• Where 's' is the length of side
2. The side is given in cm. But we want the volume in m³
• 1 cm = 0.01 m
    ♦ 1 cm have 1 significant figure
    ♦ 0.01 m also have 1 significant figure
3. So volume in m³ = (0.01)³ = 10^-6 m³  
Part (b):
1. We have: Total surface area of a solid cylinder = 2𝞹r² + 2𝞹rh = 2𝞹r(r + h)
See details here
2. First we evaluate the quantity within the brackets
• r = 2.0 cm
• h = 10.0 cm
(r + h) = 2.0 + 10.0 = 12.0 cm
(Since both values have the same number of digits after the decimal point, we do not need to draw the magenta line that we saw in fig.2.18 above)
3. So total surface area = 2𝞹r × 12.0 = (2 × 𝞹 × 2.0 × 12.0) cm²
• 2 is a factor
It has infinite number of significant figures
• 2.0 has 2 significant figures
• 10.0 has 3 significant figures
• So the final result must have only 2 significant figures
• We will take the value of 𝞹 in such a way that, it has 3 (one more than what is required in the final result) significant figures
• So 𝞹 = 3.14
4. Thus total surface area = (2 × 3.14 × 2.0 × 12.0) cm² = 150.72 cm²
5. one cm = 10 mm
• So 1 cm² = 100 mm² 
• Here 100 is a factor. It has infinite number of significant figures
6. We can now convert the 'area in cm²' to 'area in m²'
• Total surface area in m² = 150.72 × 100
• In scientific notation, this value is 1.5072 × 10⁴ m²
7. But in (2) we saw that, the final result must have only 2 significant digits. This is valid even if units change
• So total surface area = 1.5 × 10⁴ m².
Part (c):
1. Distance traveled in 1 h = 18 km
• '18' has 2 significant figures
• 1 h = (60 × 60) = 3600 s
• So distance traveled in 1 s = (¹⁸⁄₃₆₀₀) km
2. 3600 is a factor. It has infinite number of significant figures
• So the result of (¹⁸⁄₃₆₀₀) must have only 2 significant figures
3. One km = 1000 m
• So (¹⁸⁄₃₆₀₀) km = [(¹⁸⁄₃₆₀₀) × 1000] m
4. 1000 is a factor. It has infinite number of significant figures
• So the result of [(¹⁸⁄₃₆₀₀) × 1000] must have only 2 significant figures 
5. We have: [(¹⁸⁄₃₆₀₀) × 1000] = 5 
• If '5' is to have two significant figures, we must write it as 5.0
• So we get: 
A vehicle moving with a speed of 18 km h^-1 covers 5.0 m in 1 s
Part (d):
• Relative density of a material is it's density expressed in relation to the 'density of another material' (usually water)
• Relative density of a material A is given by: ^{Density of A}⁄_{Density of water}
■ Numerator and denominator must be in the same units: g cm^-3  OR  kg m^-3
• So relative density does not have a unit. It is just a ratio
• Now we can write the steps:
1. When the unit is g cm^-3: 
• From data book, density of water is 1.000 g cm^-3
• 1.000 has 4 significant figures
2. We can write: ^{Density of lead}⁄_{Density of water} = 11.3
⇒ ^{Density of lead}⁄_1.000 = 11.3
⇒ Density of lead = (11.3 × 1.000) g cm^-3
• 11.3 has 3 significant figures
• 1.000 has 4 significant figures
• So the result (11.3 × 1.000) must have only 3 significant figures
3. In scientific notation, we can write (11.3 × 1.000) as 1.13 × 10¹
• (1.13 × 10¹)  = 11.3
It has only 3 significant figures
• So we get: Density of lead = 11.3 g cm^-3  
4. When the unit is kg m^-3: 
• From the data book, density of water is 1000 kg m^-3
• 1000 has 4 significant figures
6. We can write: ^{Density of lead}⁄_{Density of water} = 11.3
⇒ ^{Density of lead}⁄₁₀₀₀ = 11.3
⇒ Density of lead = (11.3 × 1000) kg m^-3
• 11.3 has 3 significant figures
• 1000 has 4 significant figures
• So the result (11.3 × 1000) must have only 3 significant figures
7. In scientific notation, we can write (11.3 × 1000) as 1.13 × 10⁴
• It has only 3 significant figures
• So we get: Density of lead = 1.13 × 10⁴ kg m^-3

Solved example 2.16
The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m, and 2.01 cm respectively. Give the area and volume of the sheet to correct significant figures.
Solution:
1. Area of top and bottom faces = length × breadth = 4.234 × 1.005 = 4.25517 m².
• 4.234 has 4 significant figures
• 1.005 has 4 significant figures
• So the result 4.25517 should have only 4 significant figures
• Rounding off, we get: 4.255
• So area of the top and bottom faces = 4.255 m².
2. Area of front and rear faces = length × thickness
• Thickness is given to us in cm. We have to convert it into m
• 1 cm = 0.01 m
• So we have to multiply each cm by 10^-2 
• So 2.01 cm = 2.01 × 10^-2 m
3. Now we can find the area:
• length × thickness = 4.234 × (2.01×10^-2) = 8.51034 ×10^-2 m³.
• 4.234 has 4 significant figures
• (2.01×10^-2) has 3 significant figures
• So the result (8.51034 ×10^-2) should have only 3 significant figures
• Rounding off, we get: (8.51 ×10^-2)
• So area of the front and rear faces = (8.51 ×10^-2) m².
4. Area of left and right faces = breadth × thickness
= 1.005 × (2.01×10^-2) = 2.02005 ×10^-2 m³.
• 1.005 has 4 significant figures
• (2.01×10^-2) has 3 significant figures
• So the result (2.02005 ×10^-2) should have only 3 significant figures
• Rounding off, we get: (2.02 ×10^-2)
• So area of the left and right faces = (2.02 ×10^-2) m²    
5. So total area = 2 [4.255 + (8.51 ×10^-2) + (2.02 ×10^-2)]
= 2 [4.255 + 0.0851 + 0.0202]
6. The addition inside the square brackets is shown in fig.2.19 below:

Fig.2.19

• The digits on the right side of the magenta line should be rounded off
• After rounding off, we get: 4.360
7. Thus the total area becomes: 2 × [4.360] = 8.72
• 2 is a factor. It has infinite number of significant figures
• 4.360 has 4 significant figures
• So the result 8.72 must also have 4 significant figures
• Thus the total area becomes: 8.720 m².
• We can write:
Total surface area of the metal sheet = 8.720 m².
8. Calculation of volume:
• We have: Volume = area of top face × thickness
= 4.255 × (2.01×10^-2) = 8.55255 ×10^-2
• 4.255 has 4 significant figures
• (2.01×10^-2) has 3 significant figures
• So the result (8.55255 ×10^-2) should have only 3 significant figures
• Rounding off, we get: (8.55 ×10^-2)
• So volume = (8.55 ×10^-2) m³.

Solved example 2.17
The mass of a box measured by a grocer’s balance is 2.3 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box. What is (a) the total mass of the box, (b) the difference in the masses of the pieces to correct significant figures?
Solution:
Part (a):
1. Mass of the box = 2.3 kg
• Mass of gold piece 1 = 20.15 g
• Mass of gold piece 2 = 20.17 g
2. We will convert g into kg
• In this way, we will get measurements with decimal places
• If we convert kg into g, the decimal point will vanish
(2.3 kg = 2300 g)
• For problems involving addition/subtraction, we need to have numbers with decimal places. This will enable us to apply the rule more easily
3. 20.15 g = (20.15 ×10^-3) kg = 0.02015 kg
    ♦ There is no change in the number of significant figures
20.17 g = (20.17 ×10^-3) kg = 0.02017 kg
    ♦ There is no change in the number of significant figures
4. Sum of all the weights is shown in fig.2.20 (a) below:

Fig.2.20

• The digits on the right side of the magenta line should be rounded off
• After rounding off, we get: 2.3 kg
■ This result does not show the effects of adding the gold pieces to the box. This is because, the box is weighed with a device of low precision
Part (b):
• The difference in weights is shown in fig.2.20(b) above
• There are no digits on the right side of the magenta line. So there is no rounding off to be done
• We get: The difference in the masses of the pieces = 0.02 g

Solved example 2.18
The radius of a circle is 2.12 m. What is it's area according to rules of significant rules?
Solution:
1. We have: Area of circle = 𝞹r² 
2. So in our present case, area = (𝞹 × 2.12 × 2.12) m²
• 2.12 has 3 significant figures
• So the final result must have 3 significant figures
• We will take the value of 𝞹 in such a way that, it has 4 (one more than what is required in the final result) significant figures
• So 𝞹 = 3.141
3. Substituting the value, we get:
• Area = (3.141 × 2.12 × 2.12) = 14.1169104 m²
• Rounding off to 3 significant figures, we get:
Area = 14.1 m².

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In the next section, we will learn about errors

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