Chapter 10.1 - Pressure Exerted by a Fluid

In the previous section, we saw that, pressure always acts perpendicular to the surface. In this section, we will see a few more details about pressure

We know that:

• If a body is kept in the interior of a fluid, it will be compressed from all directions

• That means, the body will experience pressure from all directions

• We want to know the relation between those pressures

• That is., we want the answers to these questions:

    ♦ Are all those pressures equal in magnitude?

    ♦ Do some of those pressures have greater magnitudes than others?

• The following analysis will help us to find the answers. It is written in 14 steps

1. Consider the right angled wedge shown in fig.10.5(a) below

    ♦ This wedge is situated at the interior of a fluid

    ♦ The fluid is at rest

Fig.10.5

2. The forces acting on the wedge are:

(i) Yellow forces (F_Y) on the vertical sides

(ii) Blue force (F_B) on the rear side

(iii) Green force (F_G) on the bottom side

(iv) Red force (F_R) on the sloping side

3. In fig.a, some sides of the wedge are hidden from view

• So we give a bit of transparency to the sides. This is shown in fig.b

• Now the sides can be named:

    ♦ The two vertical sides are: ABC and DEF

          ✰ Note that, DEF is not visible in fig.a

    ♦ The rear side is: ABED

          ✰ Note that, this rear side is not visible in fig.a

    ♦ The bottom side is BCFE

          ✰ Note that, this bottom side is not visible in fig.a

    ♦ The sloping side is ACFD

4. There is a total of four forces. They were mentioned in (2)

• For our present discussion we consider only three forces:

    ♦ F_R which acts on ACFD

    ♦ F_B which acts on ABED

    ♦ F_G which acts on BCFE

5. Let us give specific names to the areas:

    ♦ The red force acts on ACFD

          ✰ So we will denote the 'area of ACFD' as: A_R

    ♦ The blue force acts on ABED

          ✰ So we will denote the 'area of ABED' as: A_B

    ♦ The green force acts on BCFE

          ✰ So we will denote the 'area of BCFE' as: A_G

6. Resolving the forces into components:

• We see that F_B is horizontal and F_G is vertical

    ♦ So these two forces need not be resolved into their components

• We will resolve F_R into it's 'horizontal and vertical components'

    ♦ For that, we need the 2D view of the wedge. This is shown in fig.10.6 below:

Fig.10.6

• The angle of the wedge is 𝛳

    ♦ So the horizontal component of F_R will be F_R cos 𝛳

    ♦ Also, the vertical component of F_R will be F_R sin 𝛳

7. Next, we apply the conditions of equilibrium:

(i) Since the wedge is in equilibrium, the horizontal components must cancel each other

So we get: F_B = F_R cos 𝛳

(ii) Since the wedge is in equilibrium, the vertical components must cancel each other

So we get: F_G = F_R sin 𝛳

8. Now we apply the principles of trigonometry to the areas of the wedge. We get:

(i) A_B = A_R cos 𝛳

(ii) A_G = A_R sin 𝛳

9. Let us divide both sides of 7(i) by A_B

    ♦ We get: $\mathbf\small{\rm{\frac{F_B}{A_B}=\frac{F_R \cos \theta}{A_B}}}$ 

• But from 8(i), we have: A_B = A_R cos 𝛳

• So we can replace the A_B on the right side denominator. We get:

$\mathbf\small{\rm{\frac{F_B}{A_B}=\frac{F_R \cos \theta}{A_R \cos \theta}}}$ 

$\mathbf\small{\rm{\Rightarrow \frac{F_B}{A_B}=\frac{F_R}{A_R}}}$

10. Let us divide both sides of 7(ii) by A_G

    ♦ We get: $\mathbf\small{\rm{\frac{F_G}{A_G}=\frac{F_R \sin \theta}{A_G}}}$ 

• But from 8(ii), we have: A_G = A_R sin 𝛳

• So we can replace the A_G on the right side denominator. We get:

$\mathbf\small{\rm{\frac{F_G}{A_G}=\frac{F_R \sin \theta}{A_R \sin \theta}}}$ 

$\mathbf\small{\rm{\Rightarrow \frac{F_G}{A_G}=\frac{F_R}{A_R}}}$

11.From (9) and (10), we get:

$\mathbf\small{\rm{\frac{F_B}{A_B}=\frac{F_R}{A_R}=\frac{F_G}{A_G}}}$

⇒ P_B = P_R = P_G

■ That means, pressure experienced by the three sides are the same

12. We chose a right angled wedge (fig.10.5) for our analysis

• By choosing another suitable shape, we can include the yellow forces also in our calculations

• But then, the red force will have to be resolved into three components. It is a 3D problem

• However, from such a 3D problem also, we will get the same result:

■ The pressure experienced by all sides of the wedge is the same

13. In the above step (12), we concluded that:

■ The pressure experienced by all sides of the wedge is the same

• We arrived at this result by analyzing the wedge in fig.10.5 above

• But we need to recognize some important points about this wedge

• They can be written in 4 steps

(i) The wedge in fig.5.10 is very very small

    ♦ We were doing all the analysis on an enlarged shape

(ii) The edges are very close to each other:

    ♦ The edge AD is very close to edge BE

    ♦ Also, edge DE is very close to edge AB

• In effect, the side ABED is very small

    ♦ That is., the area A_B is very small

(iii) It is compulsory to do the analysis on such a small area because:

• If the area A_B is large, F_B will not be uniform

    ♦ F_B will be having smaller magnitudes near the top edge AD

    ♦ F_B will be having larger magnitudes near the bottom edge BE

(We will see the reason for this variation, later in this section)

• If F_B varies in such a manner, the ratio ($\mathbf\small{\rm{P_B = \frac{F_B}{A_B}}}$) is meaning less

• If the area is very small, F_B can be assumed to be uniform

    ♦ Then the ratio ($\mathbf\small{\rm{P_B = \frac{F_B}{A_B}}}$) will be meaningful

          ✰ It will give the actual pressure on the area A_B

(iv) In a similar way, all sides of the wedge must be very small

• This can be achieved easily if the 'wedge as a whole' is very small

14. Now we can answer the questions asked at the beginning of this section. The questions were:

    ♦ Are all those pressures equal in magnitude?

    ♦ Do some of those pressures have greater magnitudes than others?

• The answer is that:

    ♦ If the object is very very small, all pressures will be of the same magnitude

    ♦ If the object is large, some of the pressures will be larger in magnitude than the others

---

• In the above discussion, we concluded that:

    ♦ If the object is large, some of the pressures will be larger in magnitude than the others

• That means, a large object will be experiencing different pressures

    ♦ Some portions of that body will experience greater pressures

    ♦ Some other portions will experience lesser pressures

• We need to find the reason for such a variation

• Also, we need to find a method to calculate pressure at the various portions of a body

• The following analysis will help us to achieve those goals. It is written in 7 steps:

1. In fig.10.7(a) below, a fluid is taken in a container

    ♦ A point ‘1’ is marked in the interior of the fluid

    ♦ Another point ‘2’ is marked vertically below ‘1’

    ♦ The distance between the two points is h

• We want to find the pressures at points 1 and 2

Fig.10.7

2. Consider the red cylinder in fig.b

• This cylinder has 3 properties:

(i) The point 1 is at the exact center of the top surface

(ii) The point 2 is at the exact center of the bottom surface

    ♦ So height of the cylinder will be h

(iii) The area of cross section is A

3. Consider the yellow cylinder in fig.b

• This cylinder has 4 properties:

(i) The top surface coincides with the free surface of the liquid

(ii) The bottom surface rests on top of the red cylinder

(iii) Height of the cylinder is z

(iv) The area of cross section is the same A as that of the red cylinder

4. Consider the green cylinder in fig.b

• This cylinder has 3 properties

(i) The top surface is shown to be irregular 

    ♦ This is to indicate that, the height extends up to the top of the atmosphere

    ♦ That is., height of the green cylinder is same as the ‘height of atmosphere’

(ii) The bottom surface rests on top of the yellow cylinder

(iii) The area of cross section is the same A as that of the red and yellow cylinders

5. Next, we calculate the force and pressure at the top of the red cylinder. It can be done in 8 steps:

(i) The top surface of the red cylinder is a circular area A

(ii) This area supports the weight of two cylinders: green and yellow

(iii) Let the weight of the green cylinder be W_a

    ♦ The subscript ‘a’ indicates that, it is the weight due to atmospheric air

(iv) Weight of the yellow cylinder can be calculated as follows:

Weight = mass × g

⇒ Weight = volume × density × g

⇒ Weight = base area × height × density × g

⇒ Weight = A × z × 𝛒 × g = A𝛒gz

    ♦ Where 𝛒 is the density of the fluid 

(v) So the total force on the top of the red cylinder = W_a + A𝛒gz

(vi) So pressure on the top of the red cylinder = $\mathbf\small{\rm{\frac{Force}{Area}=\frac{W_a+A \rho g z}{A}=\frac{W_a}{A}+\rho gz}}$ 

(vii) But $\mathbf\small{\rm{\frac{W_a}{A}}}$ is the atmospheric pressure P_a

• Also, 'pressure on the top of the red cylinder' is same as 'pressure at point 1'

• So we get:

Pressure at point 1 = P_a + 𝛒gz

(viii) We can use this equation to find the pressure at any point which is at a depth of 'z' from the surface

■ If P_z is the pressure at a depth 'z' from the free surface, we can write:

Eq.10.1: P_z = P_a + 𝛒gz

6. Next, we calculate the force and pressure at the bottom of the red cylinder. It can be done in 8 steps:

(i) The bottom surface of the red cylinder is a circular area A

(ii) This area supports the weight of three cylinders: green, yellow and red

(iii) We know that weight of the green cylinder is W_a

(iv) We calculated the weight of the yellow cylinder as: A𝛒gz

(v) In a similar way, the weight of the red cylinder will be: A𝛒gh

(vi) So the total force at the bottom of the red cylinder = W_a + A𝛒gz + A𝛒gh

(vii) So pressure at the bottom of the red cylinder = $\mathbf\small{\rm{\frac{Force}{Area}=\frac{W_a+A \rho g z+A \rho g h}{A}=\frac{W_a}{A}+\rho gz+\rho gh}}$ 

• But $\mathbf\small{\rm{\frac{W_a}{A}}}$ is the atmospheric pressure P_a

• Also, 'pressure at the bottom of the red cylinder' is same as 'pressure at point 2'

• So we get:

Pressure at point 2 = P_a + 𝛒gz + 𝛒gh = P_a + 𝛒g(z + h)

(viii) Note that, the equation in (vii) above is a mere extension of Eq.10.1

• This can be explained in 4 steps:

    ♦ Eq.10.1 gives the pressure at a depth z

    ♦ Point 2 is at a depth of (z+h)

    ♦ So, to find the pressure at point 2, we simply replace z by (z+h)

    ♦ This will give the equation in (vii)

7. Now we know the method to find the pressure at any given depth

• Next, let us calculate the ‘pressure difference’ between point 1 and point 2

• This can be done in 3 steps:

(i) We have: pressure at point 1 = Pa + 𝛒gz

(ii) We have: pressure at point 2 = Pa + 𝛒g(z + h)

(iii) So the difference in pressure = [Pa + 𝛒g(z + h)]-[Pa + 𝛒gz] = 𝛒gh

■ Thus we can write:

Eq.10.2: Pressure difference between two points at a vertical distance of h apart = 𝛒gh

---

The above Eq.10.2 can be derived by another method also. It can be written in 5 steps:

1. Consider the red cylinder that we saw in fig.10.7(b) above

• It is shown again in fig.10.8 below:

Fig.10.8

2. This cylinder will be experiencing pressure from all directions

• First we will consider the vertical direction

    ♦ Let the pressure at the top surface be P₁

    ♦ Let the pressure at the bottom surface be P₂

• Then we get:

    ♦ Force acting on the top surface = F₁ = P₁ × A

          ✰ This force acts vertically downwards 

    ♦ Force acting on the bottom surface = F₂ = P₂ × A

          ✰ This force acts vertically upwards

3. We know that weight of the cylinder will be acting vertically downwards

• If m is the mass of the cylinder, the weight will be mg

4. So we have three forces in the vertical direction: F₁, F₂ and mg

Considering equilibrium in the vertical direction, we get:

F₁ + mg = F₂

⇒ F₁ - F₂ = mg

⇒ P₁ × A - P₂ × A = mg

⇒ (P₁-P₂)A = mg

5. If 𝝆 is the density of the fluid, we can calculate m in just two steps:

(i) Volume of the cylinder = Ah

(ii) Mass = volume × density = Ah𝝆

• So the equation in (4) becomes: (P₁-P₂)A = Ah𝝆g

• Thus we get: P₁ – P₂ = 𝝆gh

• This is same as Eq.10.2 that we derived earlier

---

Now we will see two solved examples

Solved example 10.2

What is the pressure on a swimmer 10 m below the surface of a lake? Take atmospheric pressure to be 1.01 × 10⁵ Pa and g = 10 Nm^-2

Solution:

1. The pressure at a depth z is given by Eq.10.1: P_z = P_a + 𝛒gz

• In our present case, z = 10 m

2. Substituting the known values, we get:

P₁ = 1.01 × 10⁵ + 1000 × 10 × 10 = 2.01 × 10⁵ Pa

Solved example 10.3

An open vessel contains water and oil in layers. The bottom water layer has a height of 20 cm. The top oil layer has a height of 10 cm. Find the pressure at (i) the interface of the two liquids (b) at the bottom of the vessel. The relative density of oil is 0.9. Take atmospheric pressure to be 1.01 × 10⁵ Pa and g = 9.8 Nm^-2 

Solution:

• The arrangement is shown in fig.10.9(a) below. We have to find the pressures at points 1 and 2

Fig.10.9

• The relative density of oil is given to be 0.9

    ♦ We have: $\mathbf\small{\rm{Relative \;Density\;=\;\frac{Density\;of\;substance}{Density\;of\;water}}}$

    ♦ Substituting the values, we get: $\mathbf\small{\rm{0.9\;=\;\frac{Density\;of\;oil}{1000\;(kg\;m^{-3})}}}$

    ♦ Thus we get: Density of oil (𝛒) = 900 kg m^-3

Part (a)

1. The pressure at a depth z is given by Eq.10.1: P_z = P_a + 𝛒gz

• In our present case, z = 10 cm = 0.10 m

2. Substituting the known values, we get:

P₁ = 1.01 × 10⁵ + 900 × 9.8 × 0.10 = 101882 Pa

Part (b)

1. We have to find the pressure at the bottom of the red cylinder in fig.10.9(b)

2.So we need to consider the weights of the red, yellow and green cylinders

• Thus there will be three terms

3. The pressure at point 2 will be given by: P_z = P_a + 𝛒_ogz₁ + 𝛒_wgz₂ 

• In our present case:

    ♦ 𝛒_o = 900 kg m^-3 and z₁ = 0.10 m

    ♦ 𝛒_w = 1000 kg m^-3 and z₂ = 0.20 m

4. Substituting the known values, we get:

P₂ = 1.01 × 10⁵ + 900 × 9.8 × 0.10 + 1000 × 9.8 × 0.20 = 103842 Pa

---

• Eq.10.1 and 10.2 that we saw above, were derived based on pressures in the vertical direction

• We want to know how they are related to the pressures in the horizontal directions

• We will see it in the next section

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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

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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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