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

---

• 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

---

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

---

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

---

• 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

---

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

---

• 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

---

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 = $\mathbf\small{\rm{\frac{13.6 \times 10^3\;(kg\;m^{-3})}{1.0 \times 10^3\;(kg\;m^{-3})}=13.6}}$

---

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

PREVIOUS           CONTENTS          NEXT

Copyright©2020 Higher Secondary Physics. blogspot.in - All Rights Reserved

 

Chapter 9.3 - Shear modulus and Bulk modulus

In the previous section, we saw the basic details about Young's modulus. In this section we will see Shear modulus and Bulk modulus

Shear modulus

Basic details about shear modulus can be written in 4 steps:

1. We have already seen the basic details about shear stress and shear strain

(See figs.9.7 and 9.8 of section 9)

• We saw the expression for shearing stress: $\mathbf\small{\rm{Shearing \;stress=\frac{F}{A}}}$

• We saw two expressions for shear strain: $\mathbf\small{\rm{Shearing \;strain =\frac{\Delta x}{L}}}$ OR $\mathbf\small{\rm{Shearing \;strain =\theta}}$

• Thus we can easily find the shearing stress and corresponding shearing strain

2. Now consider the ratio: $\mathbf\small{\rm{\frac{Shearing \; stress}{Shearing \;strain}}}$

■ If the applied shearing stresses are less than or equal to the proportional limit, this ratio will be a constant

■ This constant is called shear modulus of the material

• So we can write: $\mathbf\small{\rm{Shear \;modulus =\frac{Shearing \; stress}{Shearing \;strain}}}$

• The symbol for shear modulus is: G

• Another name for shear modulus is: Modulus of rigidity

3. Now we can write an expression for shear modulus:

$\mathbf\small{\rm{G=\frac{Shearing \; stress}{Shearing \; strain}=\frac{\frac{F}{A}}{\frac{\Delta x}{L}}}}$

• Thus we get: $\mathbf\small{\rm{G=\frac{F \times L}{A \times \Delta x}}}$

• We can use the other expression for strain also:

$\mathbf\small{\rm{G=\frac{Shearing \; stress}{Shearing \; strain}=\frac{\frac{F}{A}}{\theta}}}$

• Thus we get: $\mathbf\small{\rm{G=\frac{F}{A \theta}}}$

4. We know that, strain has no unit

• So the unit of G is Nm^-2

 

We will see some solved examples after completing the discussion on bulk modulus

 

Bulk modulus

1. We have already seen the basic details about hydraulic stress and volume strain

(See figs.9.9 and 9.10 of section 9)

• We saw that, hydraulic stress is the force per unit area

• 'unit area' in this case is:

One unit of the total surface area of the body

• So 'force per unit area' can be replaced by pressure (p)

• We saw the expression for volume strain: $\mathbf\small{\rm{Volume \;strain =\frac{\Delta V}{V}}}$

• Thus we can easily find the hydraulic stress and corresponding volume strain

2. Now consider the ratio: $\mathbf\small{\rm{\frac{Hydraulic \; stress}{Volume \;strain}}}$

■ If the applied hydraulic stresses are less than or equal to the proportional limit, this ratio will be a constant

■ This constant is called bulk modulus of the material

• So we can write: $\mathbf\small{\rm{Bulk \;modulus =\frac{hydraulic \; stress}{Volume \;strain}}}$

• The symbol for bulk modulus is: B

3. Now we can write an expression for bulk modulus:

$\mathbf\small{\rm{B=\frac{Hydraulic \; stress}{Shearing \; strain}=\frac{p}{\frac{-\Delta V}{V}}}}$

• Thus we get: $\mathbf\small{\rm{B=-\frac{p \times V}{\Delta V}}}$

■ The -ve sign is provided because, when pressure increases, volume decreases. This can be elaborated in 3 steps:

(i) When a pressure p is applied, the volume decreases from V₁ to V₂

(ii) V₂ will be smaller than V₁

(iii) So the change in volume (ΔV = V₂-V₁) will be negative

4. We know that, strain has no unit

• So the unit of B is Nm^-2

---

Compressibility

Basic details of compressibility can be understood using an example. It can be written in 7 steps: 

1. Let us take two spheres of the same volume V

    ♦ One is made of iron and the other is made of glass

2. We want to decrease the volume of the steel sphere by ΔV

• We want the same decrease (ΔV) to happen to the glass sphere also

3. Let us calculate the required pressures:

    ♦ Let P_I be the pressure required to reduce the volume of the iron sphere by ΔV

    ♦ Let B_I be the bulk modulus of iron

• Then we can write: $\mathbf\small{\rm{B_I=-\frac{p_I \times V}{\Delta V}}}$

    ♦ Let P_G be the pressure required to reduce the volume of the glass sphere by ΔV

    ♦ Let B_G be the bulk modulus of glass

• Then we can write: $\mathbf\small{\rm{B_G=-\frac{p_G \times V}{\Delta V}}}$

4. Taking ratios, we get: $\mathbf\small{\rm{\frac{B_I}{B_G}=-\frac{p_I \times V}{\Delta V}\times -\frac{\Delta V}{p_G \times V}=\frac{P_I}{P_G}}}$

• From the data book, we have the B of iron and glass:

    ♦ B_I is 100 × 10⁹ N m^-2

    ♦ B_G is 37 × 10⁹ N m^-2

• Substituting the values of B, we get:

$\mathbf\small{\rm{\frac{P_I}{P_G}=\frac{B_I}{B_G}=\frac{100 \times 10^9}{37 \times 10^9}=2.7}}$

    ♦ That means, the pressure required in the case of iron is greater

    ♦ To be precise, the pressure required by iron is 2.7 times that required by glass

5. We can think about the 'reverse of the result in (4)':

• Pressure required by glass is lesser than that required by iron

    ♦ That means, it is easier to compress glass than steel

    ♦ That means, glass has greater compressibility

6. Compressibility is the reciprocal of bulk modulus

• The symbol of compressibility is k

• So we can write: $\mathbf\small{\rm{k=\frac{1}{B}}}$

• But $\mathbf\small{\rm{B=-\frac{p}{\Delta V/V}}}$

■ So we get: $\mathbf\small{\rm{k=\frac{1}{B}=-\frac{\Delta V/V}{p}}}$

7. A comparison can be written as follows:

    ♦ If a material has a high value for B, that material is difficult to compress

    ♦ If a material has high value for k, that material is easy to compress

• Solids have high values for B

    ♦ So they cannot be compressed easily

• Gases have high values of k

    ♦ So they can be compressed easily

---

Now we will see some solved examples on shear and bulk moduli:

 

Solved example 9.17

A square lead slab of side 50 cm and thickness 10 cm is subject to a shearing force (on its narrow face) of 9.0 × 10⁴ N. The lower edge is riveted to the floor. How much will the upper edge be displaced?

Solution:

• Fig.9.20(a) below shows a 3D view

    ♦ The bottom face is riveted to the platform

    ♦ It is clear that, the shearing force F will be acting on an area of (0.5× 0.1) = 0.05 m²

• The deformed shape is shown in fig.9.20(b)

Fig.9.20

• Fig.9.21 below shows the 2D view

    ♦ The displacement Δx can be clearly seen in fig.9.21(b)

Fig.9.21

Data given is:

• F = 9.0 × 10⁴ N

• L = 50 cm = 0.50 m

• A = (0.5 × 0.1) m² = 0.05 m²

• G = 5.6 × 10⁹ N m^-2 

1. We have: $\mathbf\small{\rm{G=\frac{F \times L}{A \times \Delta x}}}$

$\mathbf\small{\rm{\Rightarrow \Delta x=\frac{F \times L}{A \times G}}}$

2. Substituting the known values, we get:

Δx = 1.6 × 10^-4 m

 

Solved example 9.18

The edge of an aluminium cube is 10 cm long. One face of the cube is firmly fixed to a vertical wall. A mass of 100 kg is then attached to the opposite face of the cube. The shear modulus of aluminium is 25 GPa. What is the vertical deflection of this face?

Solution:

• Fig.9.22(a) below shows a 3D view

    ♦ The aluminium cube is firmly fixed on the vertical side of a wall

Fig.9.22

• In fig.9.22(b), a plate (shown in red color) is glued firmly on to the free face of the cube

    ♦ A wt of 100 kg (shown in cyan color) is attached to the plate through a rod. The rod is shown in yellow color

    ♦ Due to the 100 kg wt, the free face will deflect in the vertical direction

• Fig.9.23 below shows the 2D view
    ♦ The vertical deflection Δx can be clearly seen in fig.9.23(b)

Fig.9.23

Data given is:

• F = 100g N

• L = 10 cm = 0.10 m

• A = (0.1 × 0.1) m² =  0.01 m²

• G = 25 × 10⁹ N m^-2 

• g = 10 ms^-2 

1. We have: $\mathbf\small{\rm{G=\frac{F \times L}{A \times \Delta x}}}$

$\mathbf\small{\rm{\Rightarrow \Delta x=\frac{F \times L}{A \times G}}}$

2. Substituting the known values, we get:

Δx = 4 × 10^-7 m

 

Solved example 9.19

The average depth of Indian Ocean is about 3000 m. Calculate the fractional compression, ΔV/V, of water at the bottom of the ocean, given that the bulk modulus of water is 2.2 × 10⁹ N m^-2. (Take g = 10 m s^-2 )

Solution:

Data given is:

• Depth h = 3000 m

    ♦ So pressure p = 𝝆gh = 1000 × 10 × 3000 =  3 × 10⁷ N

    ♦ 𝝆 is the density of water in kg m^-3

• B = 2.2 × 10⁹ N m^-2 

1. We have: $\mathbf\small{\rm{B=-\frac{p}{\Delta V/V}}}$

$\mathbf\small{\rm{\Rightarrow \Delta V/V=-\frac{p}{B}}}$

2. Substituting the known values, we get:

ΔV/V = 1.36 × 10^-2

 

Solved example 9.20

Compute the bulk modulus of water from the following data: Initial volume = 100.0 litre, Pressure increase = 100.0 atm (1 atm = 1.013 × 10⁵ Pa), Final volume = 100.5 litre. Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large.

Solution:

Data given is:

• Initial volume V₁ = 100 litre

• Final volume V₂ = 100.5 litre

    ♦ So ΔV = V₂-V₁ = 0.5 litre

          ✰ 1 litre = 10^-3 m3

          ✰ So V₁ = 100 litre = (100 × 10^-3) = 0.1 m³

          ✰ So ΔV = 0.5 litre = (0.5 × 10^-3) = 5 × 10^-4 m³

• Pressure increase = Δp = 100 atm = 1.013 × 10⁷ Pa = 1.013 × 10⁷ N m^-2

1. We have: $\mathbf\small{\rm{B=-\frac{\Delta p \times V_1}{\Delta V}}}$

2. Substituting the values, we get: B of water = 2.026 × 10⁹ N m^-2 

3. We have: B of air = 1 × 10⁵ N m^-2 

• Taking ratios, we get: $\mathbf\small{\rm{\frac{B\;of\;water}{B\;of\;air}}}$ = 2.026 × 10⁴

    ♦ That means, 'B of water' is very large when compared to 'B of air'

    ♦ 'B of water is 2.026 × 10⁴ times the 'B of air'

• This is because, when compared to air, it is very difficult to compress water

    ♦ When compared to air, the ‘space available between the molecules’ is very small in water

 

Solved example 9.21

What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is 1.03 × 10³ kg m^-3  ?

Solution:

Data given is:

• Initial pressure p₁ = 1 atm (∵ the pressure at the surface is 1 atm)    

• Final pressure p₂ = 80.0 atm

    ♦ So Δp = p₂-p₁ = 79 atm = 79 × 1.013 × 10⁷ N m^-2

• Density at the surface = 1.03 × 10³ kg m^-3

• B of water = 2.026 × 10⁹ N m^-2 

 1. Given that, density at the surface = 1.03 × 10³ kg m^-3

    ♦ So 1 m³ of water at the surface will have a mass of 1.03 × 10³ kg

    ♦ So V₁ = 1 m³

2. We take this 'same mass' and 'same volume' to a depth, where the pressure is 80 atm

    ♦ Due to the high pressure at that depth, the volume will be reduced to a new value (V₂) which will be less than 1 m³

    ♦ But mass will remain the same

■ We have to find V₂

3. We have: $\mathbf\small{\rm{B=-\frac{\Delta p \times V_1}{\Delta V}}}$

$\mathbf\small{\rm{\Rightarrow \Delta V=-\frac{\Delta p \times V_1}{B}}}$

4. Substituting the values, we get: ΔV = -0.00395 m³

• So (V₂ - V₁) = -0.00395

⇒ (V₂ - 1) = -0.00395

⇒ V₂ = (1 - 0.00395) = 0.99605 m³

5. So new density = $\mathbf\small{\rm{\frac{Mass}{New\;volume}=\frac{1.03 \times 10^3}{0.99605}}}$ = 1.034 × 10³ kg m^-3 

 

Solved example 9.22

Compute the fractional change in volume of a glass slab, when subjected to a hydraulic pressure of 10 atm.

Solution:

Data given is:

• Pressure p = 10 atm = 10 × 1.013 × 10⁵ N m^-2

• B of glass = 37 × 10⁹ N m^-2 

1. We have: $\mathbf\small{\rm{B=-\frac{p}{\Delta V/V}}}$

$\mathbf\small{\rm{\Rightarrow \Delta V/V=-\frac{p}{B}}}$

2. Substituting the values, we get: ΔV/V = 2.74 × 10^-5

 

Solved example 9.23

Determine the volume contraction of a solid copper cube, 10 cm on an edge, when subjected to a hydraulic pressure of 7.0 × 10⁶ Pa.

Solution:

Data given is:

• Pressure p = 7 × 10⁶ N m^-2

• B of copper = 140 × 10⁹ N m^-2 

• Initial volume V = (0.1)³ = 0.001 m³ 

1. We have: $\mathbf\small{\rm{B=-\frac{p}{\Delta V/V}}}$

$\mathbf\small{\rm{\Rightarrow \Delta V/V=-\frac{p}{B}}}$

2. Substituting the values, we get: ΔV/V = 5.0 × 10^-5

So ΔV = V × 5.0 × 10^-5 = 5.0 × 10^-8 m³

 

Solved example 9.24

How much should the pressure on a litre of water be changed to compress it by 0.10%?

Solution: 

Data given is:

• B of water = 2.2 × 10⁹ N m^-2 

• ΔV/V = 0.10 % =$\mathbf\small{\rm{\frac{0.1}{100}}}$ =0.001

1. We have: $\mathbf\small{\rm{B=-\frac{\Delta p}{\Delta V/V}}}$

$\mathbf\small{\rm{\Rightarrow \Delta p=B \times \Delta V/V}}$

2. Substituting the values, we get: Δp = 2.2 × 10⁶ N m^-2

 

Solved example 9.25

Two strips of metal are riveted together at their ends by four rivets, each of diameter 6.0 mm. What is the maximum tension that can be exerted by the riveted strip if the shearing stress on the rivet is not to exceed 6.9 × 10⁷ Pa? Assume that each rivet is to carry one quarter of the load

Solution:

• Fig.9.24 below shows the two strips

    ♦ One is shown in cyan color. The other is shown in blue color

Fig.9.24

• The four hemispherical parts are the rivets

    ♦ There are four more hemispheres at the bottom. But they are not visible in this view

• The strips are pulled by F on either ends

    ♦ This will produce shearing stresses in the four rivets 

Data given is:

• Diameter of each rivet = 6 mm

    ♦ So cross sectional area of each rivet = $\mathbf\small{\rm{\frac{\pi \times 6^2}{4}}}$ = 28.26 mm² = 28.26 × 10^-6 m²  

• Maximum shearing stress allowable = 6.9 × 10⁷ N m^-2 

1. We have: Shearing stress = $\mathbf\small{\rm{Shearing \; stress=\frac{Force}{Area}}}$

• So maximum force that a rivet can take = (Allowable Shearing stress × Area) 

= (6.9 × 10⁷ × 28.26 × 10^-6) = 1949.94 N

2. So the four rivets together can take (4 × 1949.94) = 7800 N

3. That means, the maximum force (F) with which, we can pull from one end of the riveted strips is 7800 N

Solved example 9.26

The Marina trench is located in the Pacific Ocean, and at one place it is nearly eleven km beneath the surface of water. The water pressure at the bottom of the trench is about 1.1 × 10⁸ Pa. A steel ball of initial volume 0.32 m³ is dropped into the ocean and falls to the bottom of the trench. What is the change in the volume of the ball when it reaches to the bottom?

Solution:

Data given is:

• Pressure p = 1.1 × 10⁸ N m^-2

• B of steel = 160 × 10⁹ N m^-2 

• Initial volume V = 0.32 m³ 

1. We have: $\mathbf\small{\rm{B=-\frac{p}{\Delta V/V}}}$

$\mathbf\small{\rm{\Rightarrow \Delta V/V=-\frac{p}{B}}}$

2. Substituting the values, we get: ΔV/V = 6.875 × 10^-4

3. So ΔV = V × 6.875 × 10^-4 = 2.148 × 10^-3 m³

---

In the next chapter, we will see mechanical properties of fluids

PREVIOUS           CONTENTS          NEXT

Copyright©2020 Higher Secondary Physics. blogspot.in - All Rights Reserved