Chapter 10.4 - The Mercury Barometer

In the previous section, we saw the details about hydrostatic paradox and some of it's applications. We also saw U-tube manometer and a solved example related to it. In this section, we will see a few more solved examples. Later in this section, we will see the details about mercury barometer

Solved example 10.8

A U-tube contains water and methylated spirit separated by mercury. The mercury columns in the two arms are in level with 10.0 cm of water in one arm and 12.5 cm of spirit in the other. What is the specific gravity of spirit ?

Solution:

1. The arrangement is shown in fig.10.19(a) below:

Fig.10.19

• The green dotted line indicates that mercury is at the same level in both the arms

2. At this stage, 10 cm of water and 12.5 cm of spirit are present

• It is clear that:

    ♦ Pressure due to 10 cm of water

    ♦ is balanced by

    ♦ Pressure due to 12.5 cm of spirit

3. Let us write the actual pressures:

    ♦ Pressure due to 10 cm of water = 𝛒_wgh_w = 𝛒_wg × 0.10

    ♦ Pressure due to 12.5 cm of spirit = 𝛒_sgh_s = 𝛒_sg × 0.125

4. Equating the two pressures, we get: 𝛒_wg × 0.10 = 𝛒_sg × 0.125

⇒ 𝛒_w × 0.10 = 𝛒_s × 0.125

5. 'Specific gravity' is another name for 'Relative density'

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

• Thus from (4), we get:

Relative density of spirit = $\mathbf\small{\rm{\frac{\rho_s}{\rho_w}=\frac{0.10}{0.125}}}$ = 0.8

Solved example 10.9

In the previous problem, if 15.0 cm of water and spirit each are further poured into the respective arms of the tube, what is the difference in the levels of mercury in the two arms ? (Specific gravity of mercury = 13.6)

Solution:

1. In this situation, there will be:

    ♦ 25 cm of water in the left arm

    ♦ 27.5 cm of spirit in the right arm

2. Let us calculate the pressures:

(i) Pressure due to 25 cm of water = 𝛒_wgh_w = 1 × 10³ × g × 0.25 = 250g

(ii) Pressure due to 27.5 cm of spirit = 𝛒_sgh_s = 0.8 × 10³ × g × 0.275 = 220g

3. So the pressure due to 27.5 cm of spirit is low

• That means, the spirit side will require some 'contribution from mercury' in order to balance the pressure from the water side

• This situation is shown in fig.10.19(b) above

• We see that: an additional x cm of mercury has risen up into the right arm

4. In this situation, we can write:

    ♦ Pressure due to 25 cm of water

    ♦ is equal to

    ♦ The sum of the pressures due to '27.5 cm of spirit' and 'x cm of mercury'

5. The pressure due to x cm of mercury = 𝛒_mgh_m = 13.6 × 10³ × g × x = 13600xg

• Equating the pressures according to (4), we get:

250g = 220g + 13600xg

⇒ x = 0.002205 m = 0.221 cm

---

• We have seen the basics about atmospheric pressure. We discussed it based on fig. 10.2 at the beginning of this chapter

• Now we will learn about an apparatus which is used to measure atmospheric pressure. It is the mercury barometer. It can be explained in 11 steps:

1. A long glass tube closed at one end is filled with mercury

2. It is then inverted into a trough containing mercury

• This is shown in fig.10.20 below

[Note that, mercury vapours, if inhaled, is dangerous. This experiment should be done only in an advanced lab under the supervision of the lab authorities]

Fig.10.20

3. We see that a column of mercury will remain inside the tube

• The reason for the formation of this column can be written in 2 steps:

(i) The atmospheric pressure P_a presses down on the mercury surface in the trough

• The P_a tries to push more and more mercury into the tube

(ii) But P_a has a certain maximum value

• It cannot push more mercury

    ♦ If the P_a is at the 'maximum possible value' (which occurs at sea level), the height of mercury column will be the maximum

    ♦ If the P_a is at a lower value, the height of mercury column will be less

4. The horizontal green dotted line in fig.b is the datum line

    ♦ Two points A and B are marked in this line

    ♦ The pressure at A must be equal to the pressure at B

5. Let us calculate those pressures

    ♦ The pressure at A = 𝛒_mgh 

    ♦ Pressure at B is the atmospheric pressure Pa

• We can equate the two pressures:

    ♦ Pa = 𝛒_mgh  

6. This equation gives us an easy method to obtain atmospheric pressure

• All we need is to calculate the quantity on the left side

    ♦ For that, we simply measure the height h

    ♦ Then multiply it with 𝛒_m and g

7. Possible values of h are:

• If the apparatus is placed at sea level on a normal day, h will be equal to 76 cm

• If h is 75 cm or less at sea level, it is a sign of approaching storm

• At top of mountains, h will be far less than 76 cm

8. So on a normal day, at sea level, we get '76 cm' by actual reading on the apparatus

• This 76 cm can be obtained theoretically also. It can be explained in steps:

(i) We have: P_a = 𝛒_mgh

(ii) We know that P_a at sea level = 1.013 × 10⁵ N m^-2

(iii) Substituting the known values in (i), we get:

1.013 × 10⁵ = 13.6 × 10³ × 9.81 × h

⇒ h = 0.759 m = 76 cm     

9. This apparatus was invented by the Italian scientist Evangelista Torricelli

• In his honour, a unit called torr is also used to express pressure

• The details about this unit can be written in 4 steps:

(i) We have seen that 76 cm (760 mm) of mercury corresponds to 1.013 Pa

(ii) In a similar way, we can calculate the pressure corresponding to 1 mm of mercury

    ♦ It will be equal to 𝛒_mgh = 13.6 × 10³ × 9.81 × 0.001 = 133.416 Pa 

(iii) A pressure of 133 Pa is called one torr

(iv) But while using torr, we need not mention about the corresponding value in Pa

• We can directly write the ‘torr value’ after measuring the height

• For example, on a normal day, the pressure at sea level can be written as: 760 torr

10. Another unit for measuring pressure is bar

• One bar is equal to 10⁵ Pa

• So on a normal day, the pressure at sea level can be written as: 1.013 bar

11. An important point has to be noted about the mercury column of the barometer

• It can be written in steps:

(i) In fig.10.20, we see that:

A vacant portion is present between the top end of the tube and the top surface of the mercury column

(ii) This vacant portion must be perfect vacuum

• If air is present, the weight of that air will affect the reading

(iii) Normally, since the filled tube is inverted, we do get a vacuum

• Even then some mercury vapours will be present

• This is due to a the mercury molecules separating away from the main liquid mercury

(iv) But the properties of mercury is such that, only a very few molecules separate away from the liquid body

• So the weight due to gaseous mercury will be negligible

Solved example 10.10

Toricelli’s barometer used mercury. Pascal duplicated it using French wine of density 984 kg m^-3. Determine the height of the wine column for normal atmospheric pressure.

Solution:

1. Imagine that, in fig.10.20 above, the liquid is wine instead of mercury

• We can write: P_a = 𝛒_winegh

2. We know that P_a at sea level = 1.013 × 10⁵ N m^-2

• Substituting the known values in (1), we get:

1.013 × 10⁵ = 948 × 9.81 × h

⇒ h = 10.89 m

Solved example 10.11

A manometer reads the pressure of a gas in an enclosure as shown in Fig. 10.21 (a). When a pump removes some of the gas, the manometer reads as in Fig. 10.21 (b). The liquid used in the manometers is mercury and the atmospheric pressure is 76 cm of mercury.

(a) Give the absolute and gauge pressure of the gas in the enclosure for cases (a) and (b), in units of cm of mercury.

(b) How would the levels change in case (b) if 13.6 cm of water (immiscible with mercury) are poured into the right limb of the manometer ? (Ignore the small change in the volume of the gas).

Fig.10.21

Solution:

Part (a), Fig.a:

• The gas is shown in yellow color. Initially, it is completely confined inside the enclosure (red circle)

• When the manometer is attached, some gas comes out of the red circle and pushes the mercury down

    ♦ That means, the gas expands and fill a small portion of the left arm of the manometer

    ♦ That means, there will be some gas present in the the left arm

• If it was a liquid, we must include the 'height of the liquid column in the left arm' in the calculations

• But since it is a gas, we are told to ignore the volume change due to expansion

• That means,

    ♦ the pressure at the bottom end of the gas column

    ♦ can be considered to be the same as

    ♦ the pressure at A

• Now we can write the steps:

1. The two pressures at datum (green dotted line):

• On the left side, we have the pressure at A

• On the right side we have the sum of two pressures:

    ♦ Pressure due to 20 cm of mercury

    ♦ Pressure due to atmosphere (76 cm of mercury)

2. Balancing the two sides, we get:

• Pressure at A =  (20 cm of mercury + 76 cm of mercury) = 96 cm of mercury

• This is the absolute pressure at A

3. When we subtract the atmospheric pressure from absolute pressure, we get gauge pressure

• So the gauge pressure at A = (96 -76) = 20 cm of mercury

Part (a), Fig.b:

1. Consider the situation in which the pump gradually removes gas from the enclosure

• As the gas is being removed, the pressure at A in fig.a gradually falls

• This is because, lesser gas can create only lesser pressure

• So mercury level in fig.a gradually falls

2. When the mercury levels are same in both arms, we can say:

The pressure at A is equal to the atmospheric pressure

3. But in fig.b, we see that:

The mercury level in right arm is below the level in left arm by 18 cm

• That means, the pressure at A is less than the atmospheric pressure

• The action of the pump can be written as two stages:

Stage 1:

The pump removed some gas in such a way that, the pressure at A became equal to atmospheric pressure (76 cm of mercury)

Stage 2:

The pump removed some more gas in such a way that, the pressure at A became less than atmospheric pressure

    ♦ The 'fall in pressure' in stage 2 is equal to '18 cm of mercury'

■ So the absolute pressure at A is (76-18) = 58 cm of mercury

4. As usual, gauge pressure is obtained by subtracting atmospheric pressure from absolute pressure

• So we get:

Gauge pressure = (58-76) = -18 cm of mercury

• The negative sign indicates that, the pressure is below atmospheric pressure

• Note:

We had learnt that, gauge pressure corresponds to the reading in the manometer. It has become true in the case of negative pressure also

Part (b):

1. Imagine that, 13.6 cm of water is poured into the right limb

Now the pressures at the datum in fig.b are:

• On the left side we have the sum of two pressures:

    ♦ Pressure at A

    ♦ Pressure due to 18 cm of mercury

• On the right side also, we have the sum of two pressures:

    ♦ Atmospheric pressure (76 cm of mercury)

    ♦ Pressure due to 13.6 cm of water

2. But the pressure due to 13.6 cm of water is an 'excess pressure'

• That 'excess pressure' will disturb the equilibrium shown in fig.b

• The 'excess pressure is on the right side

• In order to balance that excess pressure on the right side, some mercury will rise up in the left side

• When such a rise of mercury occurs, a new datum will be attained

• This is shown in fig.c

3. Let the mercury rise by x cm in the left arm

The pressures at the datum in fig.c are:

• On the left side we have the sum of three pressures:

    ♦ Pressure at A

    ♦ Pressure due to 18 cm of mercury

    ♦ Pressure due to x cm of mercury

• On the right side, we have the sum of two pressures:

    ♦ Atmospheric pressure (76 cm of mercury)

    ♦ Pressure due to 13.6 cm of water

4. Equating the two sides, we get:

58 cm + 18 cm + x cm = 76 cm + pressure (in cm of mercury) due to 13.6 cm of water

5. Let us calculate the pressure (in cm of mercury) due to 13.6 cm of water. it can be done in 4 steps:

(i) In a vessel, water is taken upto a height 13.6 cm

• The pressure at the bottom of the vessel will be: 𝛒_watergh_water = 1 × 10³ × 9.81 × 0.136 Pa

(ii) In another vessel, mercury is taken upto a height of h cm

• The pressure at the bottom of the vessel will be: 𝛒_mercurygh_mercury = 13.6 × 10³ × 9.81 × h × 0.01 Pa

(iii) We want the pressures in (i) and (ii) to be equal. So we equate them:

1 × 10³ × 9.81 × 0.136 = 13.6 × 10³ × 9.81 × h × 0.01

⇒ h = 1 cm

(iv) So we can write:

13.6 cm of water will produce the same pressure as: 1 cm of mercury

6. So the equation in (4) becomes:

58 cm + 18 cm + x cm = 76 cm + 1 cm

⇒ x = 1 cm

7. So we can write:

The mercury will rise by 1 cm in the left arm and thus the reading will become 19 cm

---

In the next section, we will see Pascal's law

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Chapter 10.3 - The Hydrostatic Paradox

In the previous section, we saw the pressures in the horizontal direction. In this section, we will see hydrostatic paradox. Later in this section we will see gauge pressure and absolute pressure

Hydrostatic paradox can be described in 5 steps:

1. Consider the four vessels A, B, C and D in fig.10.12 below:

Fig.10.12

• They have different shapes

• All four are connected by a horizontal pipe at the bottom

2. Water (or any other liquid) poured in any one of the vessels, will reach all the other three vessels

• This is due to the interconnection through the horizontal pipe at the bottom

3. When the poured water settle down, we will see that:

• The level of water is same in all the vessels

• This is known a hydrostatic paradox

4. If some more water is poured, a new common level will be attained

5. The word ‘paradox’ is used because, different vessels hold different quantities of water. Even then, the levels are the same

---

A basic explanation can be written in 3 steps

1. We have Eq.10.1: P_a + 𝛒gz

• This equation gives the pressure at any depth z

2. We derived this equation using the three cylinders in fig.10.7 in a previous section

• The red, yellow and green cylinders have the same base area A

• But this ‘A’ is not coming in the final result

• What ever be the area, we will get the same Eq.10.1

• That means, the size or shape have no role in determining the pressure

3. Let us apply this equation to the vessels in fig.10.12 above

• We want the pressure at the bottom of each vessel

    ♦ The height of water is the same 'h' in all vessels

    ♦ So for all the vessels, z will be equal to h

    ♦ 𝛒 will be the same because, we are using the same liquid in all the vessels

• So at the bottom, pressure will be the same for all the vessels

---

In the above fig.10.12, different vessels hold different quantities of liquid. Yet, the pressure is same at the bottom. Upon seeing this situation, some questions arise in our minds. Those questions and their answers are given in the form of the solved example 10.6 below:

Solved example 10.6

Two vessels have the same base area but different shapes. The first vessel takes twice the volume of water that the second vessel requires to fill up to a particular common height. Is the force exerted by the water on the base of the vessel the same in the two cases ? If so, why do the vessels filled with water to that same height give different readings on a weighing scale ?

Solution:

• The solution has two parts: The analysis part and the actual solution part

• The analysis part helps to get a good understanding about the question. It can be written in 4 steps:

1. There are two vessels. Those two vessels satisfy three conditions:

(i) Both have the same base area

(ii) Both are filled water to the same height h

(iii) Volume of water in the first vessel = 2 × Volume of water in the second vessel

2. Let us try to make two vessels satisfying the above three conditions. It can be written in 6 steps:

(i) Consider fig.10.13(a) below:

Fig.10.13

• A vessel is filled with water upto a height h

    ♦ The vessel has a rectangular shape

    ♦ So the water also has a rectangular shape

          ✰ The rectangle representing water is shown in blue color

(ii) This blue rectangle is divided into two equal parts. This is shown in fig.b

• The division is done diagonally. So we get two equal triangles

    ♦ They are shown in yellow and magenta colors

(iii) The yellow triangle is flipped both vertically and horizontally

• Then it is attached to the left side of the blue rectangle

(iv) The magenta triangle is flipped horizontally

• Then it is attached to the right side of the blue triangle

(v) So the blue rectangle now has to triangles attached to it's sides

• The final shape is shown in fig.c

(vi) Now take the two vessels:

• The vessel in fig.a and the vessel in fig.c

■ When taken together, they will satisfy all the three conditions written in (1)

    ♦ They have the same water height h

    ♦ They have the same base area A

    ♦ Volume in one is twice than the other  

3. We are asked this question:

Is the force exerted by the water on the base of the vessel the same in the two cases ?

• Analysis of this question can be written in 4 steps:

(i) Which is the force and where is it applied?

    ♦ The force is exerted by water

    ♦ The force is exerted at the base

(ii) Calculation of the force in first vessel:

• We have: Force = Pressure × Area

    ♦ So Force = 𝛒gz × A = 𝛒ghA

(iii) Calculation of the force in second vessel:

• We have: Force = Pressure × Area

    ♦ So Force = 𝛒gz × A = 𝛒ghA

(iv) So we conclude that, force is same in both the cases 

4. We are asked this question:

If so, why do the vessels filled with water to that same height give different readings on a weighing scale ?

• Analysis of this question can be written in 5 steps:

(i) We are asked to compare the weights. So we put them on a weighing scale

This is shown in fig.10.14 below:

Fig.10.14

(ii) It is clear that, reading in fig.a will be greater

(iii) The weighing apparatus records the force exerted on it by the vessels

• The force is exerted through the base of the vessel

(iv). So a confusion arises:

    ♦ In (3), we see that: Forces at the base are the same

    ♦ But now we see that, weights are different

(v) We are asked to give an explanation for such a difference

---

■ The analysis part is complete. Now we write the solution part. It can be written in 5 steps:

1. Consider fig.10.15(a) below:

Fig.10.15

• The rectangular portion is in the middle

    ♦ It's weight is denoted as F_R

• The triangular portions are on the sides

    ♦ Each of them has a weight denoted as F_T

2. F_R acts directly on the base of the vessel

• But F_T acts on the sides of the vessel

    ♦ F_T does not act directly on the base

3. We know that, the fluid pressure '𝛒gh' actually arises due to the weight of the fluid

• But here, the 'weight F_T due to the triangular portions' does not act on the base

4. F_T is first transmitted to the sides

• From the sides, they are transmitted to the base

• So the weighing apparatus will experience the total force: (F_T + F_R + F_T)

5. In the case of the ordinary rectangular vessel in fig.10.13(a):

    ♦ the reading in the weighing apparatus

    ♦ will be same as

    ♦ the force due to hydrostatic pressure

---

We have solved the above problem. At this stage, it is better to see the basics of another problem also, which is closely related. This can be written in steps:

1. Consider the ordinary rectangular vessel that we saw earlier in fig.10.13(a) 

• In the above solved example, two triangles were added to it’s sides

2. Now we remove two triangles

• This is shown in fig.10.15 (b) above

• The result is a conical vessel with same base area

• This is shown in fig.c

3. As before, we will be mislead by the resulting forces

    ♦ The force at the base of the rectangular vessel = 𝛒ghA 

    ♦ The force at the base of the conical vessel = 𝛒ghA

• We get the same force because, base area and heights are same

4. The weighing apparatus will of course show different readings

• An explanation can be written in 3 steps:

(i) Consider the conical vessel in fig.10.15(c)

• The water will exert pressure in a direction perpendicular to the walls

• The walls are sloping inwards

• So the pressure will not be horizontal. It will be sloping. This is denoted by the two green arrows

(ii) So there will be a net upward force on the walls

• Due to the upward force, the weighing apparatus will not experience the full force (Accurate mathematical steps can be written using calculus. We will see them in higher classes) 

(iii) We can write:

• For ordinary rectangular vessel:

    ♦ Reading in the weighing apparatus = 𝛒ghA 

• For vessel with sides sloping outwards:

    ♦ Reading in the weighing apparatus = 𝛒ghA + weights of the two triangular portions 

• For vessel with sides sloping inwards:

    ♦ Reading in the weighing apparatus = 𝛒ghA - weights of the two triangular portions

---

Let us write a summary of what we discussed so far in this section. It can be written in two steps:

1. In fig.10.12, we saw that, inter connected vessels will always have the same height of liquid

2. Using figs.10.13, 10.14 and 10.15, we gave a satisfactory explanation of how ‘same height’ is maintained even when shapes are different

---

Interconnected vessels have many practical applications. Here we will see one such application. It can be written in 20 steps:

1. Consider the water tank in fig.10.16(a) below

• The tank is shown in yellow color

• It is kept at an elevated position

Fig.10.16

2. Water is taken out from the bottom of that tank

• For taking the water out, a pipe of medium diameter is used

• This pipe is shown in red color 

3. This red pipe goes down to below ground level

• A pipe of small diameter is taken out from the red pipe

• This small diameter pipe is shown in magenta color

4. If proper support is available, this magenta pipe can be taken vertically upwards to a large height

• Water will rise in the magenta pipe, up to the same height as in the tank

    ♦ This 'same height' is indicated by the white dotted line 

    ♦ All houses which are below the white dotted line will get water supply

    ♦ But houses above the white dotted line will not get water supply

5. Now consider fig.b

• A tap is taken out from an intermediate point along the magenta pipe

    ♦ The tap is at a vertical distance of h from the water surface in the tank

• Now we recall a common phenomenon that we experience at our homes

    ♦ If the tap is in the first floor of the building, h will be less

          ✰ The water from the tap will be flowing at a low speed

          ✰ This indicates low pressure

    ♦ If the tap is in the ground floor of the building, h will be high

          ✰ The water from the tap will be flowing at a high speed

          ✰ This indicates high pressure

6. We want to know the exact pressure available at the tap

• We have Eq.10.1: P_a + 𝛒gz

• In our present case, z = h

• So the pressure available at the tap is equal to P_a + 𝛒gh

7. So we want the value of h

    ♦ The water tank may be far away from the tap

    ♦ So it may not be easy to measure h directly

■ In such situations, we use a device known as: The open tube manometer or U-tube manometer

• Let us see how it works. The following steps from (8) to (20) explains the working of a manometer

8. The manometer consists of a U-tube

• The left limb of the U-tube is attached to the tap. Then the tap is opened. This is shown in fig.10.17(a) below:

Fig.10.17

• The U-tube is shown in white color

9. The right limb of the U-tube do not have much height

• So when the tap is opened, water will flow out through the right limb

• In order to avoid such a ‘flow out’, we fill a high density fluid (such as mercury) on the right limb

10. This is shown in fig,b

    ♦ Fig.b shows the 'portion of the tap and the manometer' separately

• We see that, the column of mercury can effectively suppress the outward pressure from the tap

11. Now we mark the important points:

• Point A indicates the bottom tip of the water column in the left limb of the manometer

• We draw a green dotted line through A

• This dotted line intersects the mercury column in the right limb at B

• This dotted line is the datum

12. Next we measure the heights:

• Height of water (above datum) in the left limb = h₁

• Height of mercury (above datum) in the right limb = h₂

13. Next we write the pressures:

• The total pressure acting at the datum on the left side

= Atmospheric pressure + Pressure due to water of height h + Pressure due to water of height h₁

= P_a + 𝛒_wgh + 𝛒_wgh₁ 

• The total pressure acting at the datum on the right side

= Atmospheric pressure + Pressure due to mercury of height h₂

= P_a + 𝛒_mgh₂

• Where:

    ♦ 𝛒_w is the density of water

    ♦ 𝛒_m is the density of mercury

14. At equilibrium, the two pressures will be equal. So we get:

P_a + 𝛒_wgh + 𝛒_wgh₁ = P_a + 𝛒_mgh₂

• P_a is to be included in the left side because:

    ♦ The atmosphere is pushing down on the water in the tank

    ♦ The effect of that push will be felt at the tap

• P_a is included in the right side because:

    ♦ The atmosphere is pushing down at the top of the mercury column

    ♦ That push also contributes to suppress the water in the left limb

• Since P_a is present on both sides, they cancel out

Thus we get: 𝛒_wgh + 𝛒_wgh₁ = 𝛒_mgh₂

⇒ 𝛒_wgh = 𝛒_mgh₂ - 𝛒_wgh₁

⇒ 𝛒_wh = 𝛒_mh₂ - 𝛒_wh₁

15. In the above equation, h is the only unknown. So it can be easily calculated

• Note that, h₁ and h₂ are the only readings that we have to take

• An example can be written in 4 steps:

(i) Let h1 = 0.3 m and h2 = 0.65 m

(ii) Substituting the values, we get:

1 × 10³ × h = 13.6 × 10³ × 0.65 - 1 × 10³ × 0.3

⇒ h = 8.54 m

(iii) So the pressure available at the tap = P_a + 𝛒_wgh = 1.01 × 10⁵ + 1 × 10³ × 9.8 × 8.54 = 184692 Pa 

(iv) The height of a two storeyed building is approximately equal to 8.54 m

• So we can write:

The tap is at the ground floor and the tank is placed on top of the first floor

16. Using the equation in (14), we get h

• Once we get h, we can find 𝛒gh

• That means:

    ♦ Using just the two readings h₁ and h₂ from the manometer, we can calculate h

    ♦ And using that h, we can calculate 𝛒gh

17. Now consider Eq.10.1: P_a + 𝛒gh

• We see that 𝛒gh is one of the two components in Eq.10.1 (the other component being P_a)

• The component '𝛒gh' is easily obtained from the readings in the manometer

• The 'manometer' is a type of 'pressure gauge'

• So the 'manometer pressure gauge' readings are used to calculate 𝛒gh

■ So '𝛒gh' is called gauge pressure

• In Eq.10.1, we are actually adding this gauge pressure to the atmospheric pressure Pa

■ The sum thus obtained (Pa + 𝛒gh) is called absolute pressure

18. In the above discussion, we calculated the pressure in a system

• It is called a 'system' because, it consists of various components:

A water tank, proper supports to keep the tank at an elevated position, pipes of various diameters, tap etc.,

• These components work together for the proper functioning of the system

• Boilers, Distillation plants etc., are some other examples of system

    ♦ Many components work together harmoniously in such systems

• For simple problems involving manometers, we do not have to show the details of the system

• This can be explained in 3 steps:

(i) Consider the red circle in fig.10.18(a) below:

Fig.10.18

• This 'red circle and the blue liquid inside it' together denotes the system

(ii) Point A is the center of the circle

• We will be asked to find the pressure at A

(iii) Thus we represent a large system by a simple circle filled with fluid

19. Pressure at A can be calculated in 4 simple steps:

(i) Let P be the pressure at A

(ii) This pressure pushes the mercury down in the left limb

• So mercury rises in the right limb

(iii) The mercury column in the right limb, together with the atmospheric pressure, balances the pressures on the left side

(iv) From the fig., it is clear that:

P + 𝛒gh₁ = P_a + 𝛒_mgh₂

⇒ P - P_a = 𝛒_mgh₂- 𝛒gh₁

Where:

    ♦ 𝛒 is the density of the liquid in the system

    ♦ 𝛒_m is the density of mercury

20. Consider the result in (19)

• On the left side, P_a is subtracted from the total pressure P

    ♦ We know that, when P_a is subtracted, the remaining pressure is the gauge pressure

          ✰ We saw this in step (17)

• For computing the right side, h₁ and h₂ are the only information we need

    ♦ We know that, h₁ and h₂ are simple manometer readings

• That means:

    ♦ The left side gives the gauge pressure

    ♦ The right side is calculated using manometer readings

■ Thus it is proved again that:

The manometer readings give gauge pressure

---

Now we will see a solved example:

Solved example 10.7

The right limb of a U-tube manometer containing mercury is open to the atmosphere. The left limb is connected to a system filled with a liquid of relative density 0.9. The level of mercury in the right limb is 12 cm above the center of the system. The difference in levels of mercury in the two limbs is 20 cm. Calculate the pressure at the center of the system

Solution:

1. The arrangement is shown in fig.10.18(b) above

• It is clear that h₁ = 8 cm and h₂ = 20 cm

2. The absolute pressure at the center point A will be given by: P = P_a + 𝛒_mgh₂- 𝛒gh₁ 

• Substituting the values, we get:

P = 1.01 × 10⁵ + 13.6 × 10³ × 9.8 × 0.20 - 0.9 × 10³ × 9.8 × 0.08 = 126950.4 Pa

---

In the next section, we will see some solved examples. We will also see barometer

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Chapter 10.2 - Pressure in Horizontal Direction

In the previous section, we saw the pressures in the vertical direction. In this section, we will see pressures in the horizontal directions

• Eqs.10.1 and 10.2 that we saw in the previous section, are related to 'pressures in the vertical direction'

• We can relate those equations to the 'pressures in the horizontal directions' also. This can be written in 11 steps:

1. The red cylinder that we saw in fig.10.7 of the previous section, is shown again in fig.10.10(a) below

• We already know how to calculate the pressure at it’s top and bottom points

Fig.10.10

2. Let us now consider intermediate points along that cylinder

• For that, a small slice is marked in that cylinder

    ♦ This slice is in the form of a disc

    ♦ This disc has a very small thickness

    ♦ It is situated at a depth of z₁ from the free surface

3. We have already seen that:

    ♦ A downward pressure will be acting at the top of the red cylinder

    ♦ A upward pressure will be acting at the bottom of the red cylinder

4. In a similar way:

    ♦ A downward pressure will be acting at the top of the disc

    ♦ A upward pressure will be acting at the bottom of the disc

5. But there is a major difference between (3) and (4). This can be explained in 2 steps:

(i) The height of the disc is very small

(ii) So the two pressures mentioned in (4) will be the same

(iii) The two pressures mentioned in (3) will be different because, there is a large height difference between their points of applications

■ We can write: 

Both the pressures mentioned in (4) will be equal to: P_a + 𝛒gz₁

6. At the beginning of the previous section, we saw that, if the element is very small, the pressure experienced by it in all directions will be the same

• We proved this using figs.10.5 and 10.6

7. So it is clear that, the disc will be experiencing the pressure of ‘P_a + 𝛒gz₁’ in the horizontal directions also

• So in fig.10.10(a) above, two horizontal arrows are drawn on either sides of this disc

    ♦ Note that, there will be horizontal arrows all around the disc. We have drawn only two of them

    ♦ Those arrows will try to compress the disc

8. In a similar way we can draw horizontal arrows for another disc at a depth z₂

    ♦ The horizontal arrows at z₂ will be larger than those at z₁

    ♦ This is because, as depth increases, pressure also increases

9. Assuming the red cylinder to be made up of a large number of discs, we can draw a correspondingly large number of horizontal arrows

    ♦ This is shown in fig.10.10(b)

    ♦ We can see that, as depth increases, the size of arrows increases

10. We can find the magnitude of any arrow that we see in fig.b

• All we need are:

    ♦ the depth (z) of that arrow from the free surface

    ♦ the density of the fluid

■ Then the magnitude of that arrow will be equal to: P_a + 𝛒gz

11. The walls of the container will also be experiencing the same horizontal forces

• That means, the fluid will be trying to push the walls outwards

    ♦ The upper portions of the walls will be experiencing lower pressures

    ♦ The bottom portions of the walls will be experiencing higher pressures

---

Now we will see some solved examples

Solved example 10.4

A submarine is at a depth of 1000 m below the surface of the ocean. The interior of the submarine is maintained at sea level atmospheric pressure. What is the force acting on a window of the submarine, if the area of the window is 400 sq.cm? The relative density of sea water is 1.03. Take sea level atmospheric pressure to be 1.01 × 10⁵ Pa and g to be 10 m s^-2. 

Solution:

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

2. This P_z acts equally in all directions

    ♦ So the ocean will exert an inward pressure of P_z on the window

    ♦ So the inward force experienced by the window = (P_z × area) = 0.04P_z

3. But there is a pressure of P_a inside the submarine

    ♦ The air inside, will exert an outward pressure of P_a on the window

    ♦ So the outward force experienced by the window = (P_a × area) = 0.04P_a

4. So net force = (0.04P_z - 0.04P_a)

= 0.04[P_z-P_a] = 0.04[(P_a + 𝛒gz) - P_a] = 0.04𝛒gz = 0.04 × 1.03 × 10³ × 10 × 1000 = 412000 N

Solved example 10.5

A tank with a square base of area 1.0 m² is divided by a vertical partition in the middle. The bottom of the partition has a small-hinged door of area 20 cm² . The tank is filled with water in one compartment, and an acid (of relative density 1.7) in the other, both to a height of 4.0 m. compute the force necessary to keep the door close.

Solution:

1. The arrangement is shown in fig.10.11 below:

2. The pressure at the bottom of the right side compartment = P_{right(4 m)} = P_a + 𝛒_acid × g × 4

• So the force experienced by the door from the right side = (P_{right(4 m)} × area) = 0.002 × P_{right(4 m)}

3. The pressure at the bottom of the left side compartment = P_{left(4 m)} = P_a + 𝛒_water × g × 4

• So the force experienced by the door from the left side = (P_{left(4 m)} × area) = 0.002 × P_{left(4 m)}

3. So net force = 0.002[P_{right(4 m)} - P_{left(4 m)}]

= 0.002[(Pa + 𝛒_acid × g × 4) - (P_a + 𝛒_water × g × 4)]

= 0.002 × g × 4[𝛒_acid - 𝛒_water]

= 0.002 × g × 4 × 10³[1.7 - 1] = 54.88 N

4. This net force will be acting from right to left

So this force will push the door towards the left

We will have to provide a force of 54.88 N from right to left so that, the door remains closed

5. Note that, the base area of the tank is not coming into the calculations

• That means:

    ♦ We can provide the partition in such a way that, the acid compartment is small

    ♦ Even then, if the depth is 4 m in both compartments, we will have to provide the same force of 54.88 N

---

In the next section, we will see hydrostatic paradox

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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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