Handbook p7 pressure and density

Agastya International Foundation

Pressure and Density Handbook P7

“Measure what is measurable, and make measureable what is not so.�

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Handbook P7 Pressure & Density OVERVIEW OF HANDBOOK ABL

CONCEPT

NO. OF ACTIVITIES

TIME (MIN)

ABL 1 ABL 2 ABL 3

Force and Pressure Atmospheric Pressure Interesting demonstrations depicting fluid Pressures.

4 3 6

65 65 65

ABL 4

Fluids in motion, Pascal’s law and Bernoulli’s principle with applications Density, Buoyancy and Archimedes’s principle

4

50

6

70

ABL 5

ABLs WITH REFERENCE TO STANDARD S.No. 1 2 3 4 5

STANDARD 6 6 8 8 9

PREFERRED ABL ABL 1 ABL 2 ABL 3 ABL 4 ABL 5

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PAGE NO.

3

LIST OF FIGURES, CHARTS AND WORKSHEETS

S. No

Name

Page No

Fig 1

Force exerted by bricks with different areas of contact

7

Fig 2

Pressure depends on depth

10

Fig 3

Manometer

16

Fig 4

Working of manometer

16

Fig 5

Pressure exerted by liquids at different position but same depth

17

Fig 6

Pressure exerted by liquid column and atmospheric pressure

18

Fig 7

Atmosphere exerts upward pressure

21

Fig 8

Atmospheric pressure acting on palm

25

Fig 9

Atmospheric pressure acts equally on all sides

25

Fig 10

Pressure inside lungs

25

Fig 11

Atmospheric pressure deforms the can

27

Fig 12

Set up for ABL 3.1

29

Fig 13

Set up for ABL 3.2

33

Fig 14

Set up for ABL 3.3

36

Fig 15

Water fountain in round bottom flask

39

Fig 16

Set up for ABL 2.4

40

Fig 17

Liquid pressure depends on height of the column

44

Fig 18

Set up for ABL 3.6

46

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Fig 19

Simple siphon system

50

Fig 20

Pascal’s law model

53

Fig 21

Hydraulic pump

54

Fig 22

Boy demonstrating Bernoulli’s principle

58

Fig 23

Set up for ABL 4.4

61

Fig 24

Submarine model

73

Fig 25

Set up for ABL 5.5

76

Chart 1

Earth’s atmosphere

23

Chart 2

Magdeberg’s hemispheres

24

Worksheet 1

Observations for calculating density

65

Worksheet 2

Observations for calculating density for irregularly shaped objects

66

Worksheet 3

Observations for ABL 5.5

77

Worksheet 4

Observations for calculating relative density

80

Note to Instructor: All the figures in this handbook are for the Instructor’s reference only. The Charts need to be printed and shown to the learners during the course of the activity. Worksheets need to be printed out in advance for the learners. The number of worksheets required is mentioned in the Material List.

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ABL 1: Force and Pressure Activity

1.1

Learning objective

To understand the concept of pressure and study its relation to the force causing it.

Key messages



Time (min)

The ratio of normal force (F) 15 acting on a surface and the area of the surface on which force acts is defined as the pressure over the surface.

Pressure, P = Force/Area = N/m2

or Nm-2

Hence pressure can be understood as the effect of force on a surface.

1.2

Understanding the pressure exerted by solids on the surface of contact.



Pressure is the continuous normal force exerted on unit area of the surface in contact.

10

1.3

Understanding of the pressure exerted due to liquids To demonstrate the variation of pressure exerted by liquids at different depths.



Like solids liquids also exert pressure.

20



The liquid pressure depends on the height of the water column.



The pressure exerted by a liquid column at certain depth is, P = ρgh



Liquid molecules like air bombard against each other and also against surface of contact and exert pressure. Pressure exerted by liquid is larger at larger

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depths. (P



h)

To 1.4study the To study some of the features of the pressure exerted by a liquid using a manometer

i)

same at all points in a given horizontal plane and

1.At different positions in a horizontal plane and

ii)

is same in all directions.

i)

Pressure in a liquid varies with different depths.

2.In different directions at a given point.

The liquid pressure is

20

3.At different depths

Total Time

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65 mins

7

ABL 1.1

Time: 10 min Time: 15 min

Learning Objective: To understand the concept of pressure and study its relation to the force causing it; Understanding the pressure exerted by solids on the surface of contact.

ADVANCE PREPARATION Material List s.no 1 2

Material Rubber sheet Brick

Quantity 1 1

Things to do Not Applicable Safety Precautions Not Applicable

SESSION 1.1 a Procedure: This is a Group Discussion. Imagine situations like,    

Driving a nail into a wall by hitting with a hammer.

Providing larger area of the foundation to raise the height of the building.

Increasing the diameter of a pillar or column to support larger roof weights and A person standing on foot rest which has a number of sharp needles fixed on it and remaining unhurt.

In all these cases when considered individually it is seen that

i) A nail can be driven easily when hit on its head and its tip pressed against wall. ii) A tall building has certain weights that acts on the ground through the foundation area and possibly sink, on increasing the foundation area the sinking of the building can be minimized.

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iii) A roof resting on a pillar has a huge weight. It may cause penetration of pillar into the earth leading to sinking of the structure. This is minimized either by making the pillars of lager diameter or increasing the number of pillars. iv) A person with his foot on a sharp nail tip gets hurt because of the tip easily trenchant the skin. But he remains unhurt when the foot rests on a number of sharp pins in the same foot rest.

What we notice in all these situations is that the effect of force on a surface is more when the area of contact is less and the effect of force is less when the area of contact is more. In other words there is a relationship between force acting, area of contact and the effect on the surface. When a force acts on smaller area, the effect is more pronounced and we say the pressure exerted is large. When the same force acts on a large area or multiple smaller areas, the effect is less pronounced and we say the pressure exerted is small. Surprisingly, the force in both the cases is same. Pressure seen as the effect of force is varied.

SESSION 1.1 b Procedure 1.

2.

3.

4.

Take a thin rubber sheet and ask two students to stretch horizontally and hold it.

Place a brick on the rubber sheet with broad side in contact with rubber sheet.(fig1) Observe the depression caused on the sheet.

Next, place the brick on the rubber sheet as shown in fig2 and fig3 separately and observe the depression each time.

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Fig 1- Force exerted by bricks with different areas of contact

UNDERSTANDING THE ACTIVITY Leading questions 1.

2.

3.

4.

What happens when a brick is placed on the rubber sheet?

Is the depression same when brick is placed in different positions?

For which position do you think the depression is maximum?

How do you explain the behaviour in question 2?

Discussion and explanation 1. A brick because of its weight exerts a force F = mg, on the rubber sheet. This force causes stretching of the rubber which yields. There is a depression in the region.

2. The depression is not the same in the three positions though the weight of the brick is same.

3. It is maximum in position3 and minimum in position1.

4. The pressure on the surface of the rubber sheet is P = F/a ---------------(1) Here F is weight of the brick and ‘a’ is area in contact with the rubber sheet.

The area of contact is a1 in position 1, a2 in position 2 and a3 in position 3.

Also a1>a2 >a3.

From equation 1 it can be seen P1 = F/a1 ,P2 = F/a2 andP3 = F/a3. F is same and ‘a’ is decreasing P1< P2 < P3 or P3 > P2 >P1.

KEY MESSAGES 

The ratio of normal force (F) acting on a surface and the area of the surface on which force acts is defined as the pressure over the surface.

Pressure, P = Force/Area = N/m2 or Nm-2

Hence pressure can be understood as the effect of force on a surface.



Pressure is the continuous normal force exerted on unit area of the surface in contact.

LEARNING CHECK Consider a situation where weight of the brick is 0.4 Kg and dimensions Agastya International Foundation. For Internal Circulation only.

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25cm x 12cm x 6cm a1

= .25 x .12 = .0300 m2

a2 = .25 x .06 = .015 m2 a3 = .12 x .06 = .0072 m2 P1 = F/a1

= 0.4 x 9.8/0.03

= 130.66 Nm-2 P2 = F/a2

= 0.4 x 9.8/0.015

= 261.32 Nm-2 P3 = F/a3

= 0.4 x 9.8/0.0072

= 544.4 Nm-2 Pressure in position 3 is 4.1 times the pressure in position 1

ABL 1.2 LEARNING OBJECTIVE: Understanding the pressure exerted due to liquids; To demonstrate the variation of pressure exerted by liquids at different depths.

ADVANCE PREPARATION Material List S.No 1

2 3 4 5 6

Material Measuring jar or acrylic transparent tube Balloon Thread or rubber band Water Large beaker Long straw or glass

Quantity 1

3 1 2 buckets 1 1

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tube Things to do: Do this activity before class to check that the experiment works Safety Precautions Not Applicable

SESSION 1.2a Link to known information/previous activity: In this section we will learn how liquids also exert pressure on the surfaces in contact.

Procedure:    

Take an open tube of length 10 to 12 cm.

On lower end of the tube tie a thin stretched rubber membrane.

Now hold the tube in vertical position and pour water into the tube.

Pour more water into the tube and observe what happens to the membrane.

Fig 2: Pressure depends on depth

UNDERSTANDING THE ACTIVITY: Leading questions: 1.

2.

3.

4.

5.

What happens to the balloon membrane when water is poured into the tube?

How was the balloon membrane before pouring the water into the tube?

When you pour water into the tube why did the membrane bulge downwards?

Why did the balloon membrane bulge greater when more water is poured into the tube?

What is the learning from this activity?

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  



When we pour water into the tube the membrane bulges outwards.

Before pouring water the balloon was either flat or slightly bulging downwards.

When we pour water into the tube the balloon bulges out wards, because the water in the tube exerts force on the balloon due to gravity and causes stretching. The balloon bulged greater when more water is poured into the tube because water exerts a greater force on the balloon membrane and causing a larger stretch.

Water exerts stretching force on the balloon membrane due to its weight. This force acting per unit area is called pressure.

Let m be the mass of the water in the tube.

The downward weight due to this water column on the membrane is w = mg, which is the downward force causing the membrane to bulge.

P = F/A P = mg/A

since

P = Vx ρx g /A

m

F = mg = Vxρ

=(Axh)ρ

Where ρ is density of the liquid and h is the height of liquid column.

V is volume of water Ρ is density of water G is acceleration due to gravity.

P= A x h x ρ x g/A = ρgh P = ρgh The above equation suggests that pressure is directly proportional to the height of the water column and density. i.e., P ρ andP

h

Thus when more water is poured into the water a larger weight acts on the same area of the membrane causing a greater bulge. Likewise when a liquid of larger density due to larger weight acting on the membrane is poured into the tube it creates a larger pressure on the balloon.

Note: The pressure exerted by liquid is called hydrostatic pressure.

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SESSION 1.2b Procedure: Take a large beaker and fill it with water. Tie a balloon to the free end of a long straw or glass tube (long enough to be dipped deep into water).

Blow air into the balloon through the open end. Deflate the balloon and immerse it just under water level. Blow air and try to inflate as much as you can.

Deflate the balloon and take it little deeper into water (B). Try inflating the balloon by blowing air.

Try the same way by taking the balloon still deeper inside the water (C).

UNDERSTANDING THE ACTIVITY Leading questions 1. With what effort could you inflate the balloon while in the air?

2. What was your experience while inflating the balloon at different depths inside the liquid?

3. At which position inside the liquid was it easier or most difficult to inflate the balloon?

Discussion and Explanation When air is blown in to the balloon, the air molecules having gained kinetic energy, bombard the walls of the balloon exerting a continuous pressure. The balloon expands creating more space for air. While the balloon expands one can feel a kind of resistance due to outside air trying to oppose the expansion. Once the atmospheric pressure is overcome by forcefully blowing air the balloon begins to expand. On the hole it is much easier to inflate balloon against atmospheric pressure than inside water.

When we try to inflate the balloon by just immersing under water there is a greater resistance from outside liquid. It is more difficult to inflate the balloon in the liquid than in air. It becomes more and more difficult to inflate the balloon when it is taken to deeper positions (B and C).

Liquid molecules which are pulled to the base due to gravity, strike the base and rebound. They in turn collide with other molecules of the liquid and create a condition where the molecules are in incessant motion, colliding against each other and moving in random directions. They thus exert pressure not only on the base but also on the walls as well as on the surface of any object immersed inside.

When we try to inflate the balloon immersed under water, the water molecules bombard the balloon wall from all the directions. The liquid thus exerts a pressure on the balloon wall from all the directions and tries to resist the expansion of the balloon.Since this liquid pressure is greater than the pressure exerted by atmospheric pressure (while in air) it is more difficult to inflate the balloon while it is under water.

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When the balloon is moved deeper and deeper into the liquid it becomes more and much more difficult to inflate the balloon. It only means, the liquid pressure on the balloon increases as it is moved deeper and is greatest when it is near the base.

Hence it is easier to inflate the balloon while it is in air and most difficult while it is inside the water at the base.

KEY MESSAGES:   

 

Like solids liquids also exert pressure.

The liquid pressure depends on the height of the water column.

The pressure exerted by a liquid column at certain depth is,P = ρgh Liquid molecules, like molecules in air, bombard against each other and also against surface of contact and exert pressure from all directions from all directions.

Pressure exerted by liquid is larger at larger depths. (P h)

LEARNING CHECK: Take two similar test tubes, fill one of them with about 10 cm of water and another with 10 cm of Hg. Do think the pressure exerted on the base is same in both the cases? Compare.

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ABL 1.3 LEARNING OBJECTIVE: To understand the Pressure exerted by air on the walls of an enclosure.

ADVANCE PREPARATION Material List:

S.no 1 2 3 4 5

Material Balloon Cycle pump Cycle tube or football Thread Scissors

Quantity 2 per class 1 per class 1 per class 1 per class 1 per class

Things to do Not Applicable Safety Precautions Not Applicable

SESSION Link to known information/previous activity As a prerequisite, students should understand that air occupies space. This concept will Be reviewed throughout this ABL. Procedure Complete the Instructor demo and facilitate the student activity.

1.4a Instructor demo Take one balloon and start blowing air into it. Ask learners to observe carefully what is happening.

After the balloon is big enough, tie it with a thread.

Repeat with the second balloon but this time keep blowing air until the balloon bursts.

1.4b Student activity Call any two students and ask them to pump air into the cycle tube/football with the help of cycle pump.

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Ask the other students to observe carefully.

UNDERSTANDING THE ACTIVITY Leading questions 1.

2.

What happened to the balloon and the tyre when air is pumped in?

What happens if you put too much air in the balloon or tyre? Why did the Second balloon burst?

3. There is air all around us. Does it act on us the same way as it acts on a balloon or a tyre?

Discussion and Explanation 

The balloon and the tyre inflated when we put air into them.

o

Air is a mixture of gases. All the gases are made of tiny particles called molecules. Molecules are so small that we cannot see them!

o

Molecules are in constant motion and move around freely in all directions. As they move about some of them will actually hit the walls of the balloon or the tyre and will push the walls.

o

When a molecule strikes a surface for a small time called “collision time” and bounces off, it transfers some momentum to the surface. The ratio of the momentum transferred to the collision time is the force exerted by the molecule that causes a push.

o

Although they are tiny and only push the walls with a tiny force, there are a very large number of molecules bouncing against the walls. Together, they add up to create a larger force.

o

As we blow air into the balloon or tyre, the number of air molecules increases. More molecules are bouncing against the wall. The rubber walls will stretch, causing the balloon or tyre to inflate.

o

As we pump more air, more molecules are pushed in, which means more of them crashing against the wall, causing the balloon or tyre to expand further.



The force that air molecules exert on the walls of the balloon or tyre from inside creates air pressure.

Air pressure makes the balloon and the tyre inflate.

o

Air pressure is defined as the force exerted by air on unit surface area in contact with it. In this case, the air was exerting pressure on balloon wall or cycle tube wall.

Pumping more and more air means increasing the pressure exerted on the walls inside.

o o

The balloon bursts because there is a limit to which the balloon rubber can stretch. As we blew more airinto the balloon, the pressure increased inside the balloon until it was more than what the rubber could withstand. The rubber gave out and the balloon burst.

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KEY MESSAGES  

Pressure is the normal force exerted on unit area of the surface in contact.

Solids, liquids and gases exert pressure.

LEARNING CHECK Why do tyres sometimes burst in summer? What is the connection between this question and the earlier one about balloon bursting?

(Answer: In the summer, the air can get very warm. When air molecules heat up, they move faster. On a hot day, the molecules inside a tyre will hit the walls of the tyre more often than they would on a cold day, so the air pressure in the tyre is greater. Occasionally, this air pressure gets so great that the tyre rubber cannot withstand the pressure, so the tyre bursts. Thus the tyre bursts for the same reason that a balloon bursts – the air pressure inside was too great!)

INTERESTING INFORMATION When a tyre rolls over stones on a road, it does not burst or get punctured. But when the same tyre rolls over a nail or thorn on the road it gets punctured. Why? When tyre is in contact with the surface of a stone it exerts a force that constitutes action. The stone in turn exerts an equal form of reaction on the tyre. The reactional force acts on a larger area (compared to the tip of the normal). When the same tyre rolls over a nail or thorn the same reaction acts over a small surface of the tip. This creates a very large pressure on the tyre that causes a puncture.

ABL 1.4 LEARNING OBJECTIVE: To study some of the features of the pressure exerted by a liquid using a manometer At different positions in a horizontal plane and ii) In different directions at a given point. iii) At different depths i)

ADVANCE PREPARATION Material List Agastya International Foundation. For Internal Circulation only.

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S.no 1 2

Materials Manometer Bucket of water

Quantity 1

SESSION Link to known information/previous activity In the previous activities, we have seen that pressure is exerted by gases. Let us see if pressure is exerted by liquids as well.

Procedure: Pour some colour liquid into the ‘U’ tube and fill it partially. The level of the liquid on both sides of the ‘U’ tube will be the same. Now connect one end of the ‘U’ tube with a rubber tube and attach a thistle funnel at its open end.

Fix a stretched elastic membrane to the thistle funnel covering its mouth. Press the membrane slightly and observe the liquid levels in the two arms of the manometer and

Fig 3: Manometer immerse the thistle funnel of the manometer into water taken in a large beaker. Measure the depth of the funnel below water level and also the difference in the liquid levels in the two arms.

Take the funnel to different depths inside water and note down the manometer readings each time.

Fig 4: Working of manometer

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Keep the funnel at a particular depth inside the water in the beaker and turn the position of its mouth from up to down and left to right observe the manometer readings in all positions.

Keep the funnel at a particular depth inside water with its mouth either upwards or downwards. Move the funnel to different position in the same horizontal plane inside the liquid. Observe manometer readings for each position.

In all these cases calculate the difference in liquid levels in the two limbs.

Fig 5: Pressure exerted by a liquid at different positions but same depth

UNDERSTANDING THE ACTIVITY: Leading questions: 1. Why were the levels of the liquid in the two limbs of the manometer same when both the ends of the ‘U’ be is open?

2. What happens to the liquid levels in the two limbs when the membrane covering the thistle funnel was pressed? Why did it happen so?

3. What did you observe in the difference in the liquid levels when the thistle funnel was moved to different depths inside the liquid?

4. What did you observe in the difference in the liquid levels when the thistle funnel is moved from right to left and up or down?

5. What did you observe in the difference in the liquid levels when the thistle funnel is moved to different points in the same horizontal plane inside the liquid?

6. Do you find any similarity in the behaviour of gases and liquids while dealing with the pressures exerted?

Discussion and explanation: 



When the thistle funnel is not connected the liquid surfaces in the two limbs of the ‘U’ tube are exposed freely to the atmospheric air. The atmospheric pressure is the same on the liquid surfaces in the two limbs. Hence the liquid levels stand to the same height in both the limbs and difference in the levels is zero. The same is true even when the thistle funnel rubber tubing is fitted to the manometer.

When the membrane covering the thistle funnel is pressed the air inside is pushed into the ‘U’ tube limb. It exerts a larger pressure on the liquid level and causes a depression, consequently the liquid

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

 







level in the other limb raises. The difference in the levels of the liquid (h) is directly proportional to the excess pressure created across the thistle funnel membrane.

When the thistle funnel is immersed under water the liquid exerts pressure on the membrane. It pushes the air in the thistle funnel into the ‘U’ tube limb. Causing again a difference in the levels of the liquid in the two limbs. Now as the funnel is taken to different depths (l) in the liquid a greater pressure is exerted on the membrane. The difference in the levels of the liquid increases as the depth of the funnel below water increases.

Difference in levels is directly proportional to the liquid pressure on the membrane or the depth of funnel inside. (h l) When the thistle funnel at a given position is moved to left or right or up and down there is no change in the difference in the liquid levels. It means the liquid pressure at a point remains the same in all directions.

When the thistle funnel at a given depth is moved to different points in the same horizontal plane inside the liquid there was no change in the difference in the levels of the liquid in the two limbs, this observation suggests that the liquid pressure is the same at all points in a horizontal plane within the liquid.

Gases and liquids called as fluids behave same way. In the case of gases the molecules in random motion constantly bombard a surface in contact and exert a pressure. Likewise the liquid molecules also in random motion bombard the surface in contact and exert pressure. In general the fluids exert pressure on the base as well as sides of the container.

When we consider pressure at a point inside a liquid it is actually the sum of the pressure exerted by the liquid column above the point and the atmospheric pressure acting on the liquid surface.

Fig 6: Pressure exerted by liquid column and the atmospheric pressure

KEY MESSAGES:  The liquid pressure is ii) same at all points in a given horizontal plane and iii) is same in all directions at a given point. iv) Pressure in a liquid varies with different depths.

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LEARNING CHECK ď&#x201A;ˇ The pressure exerted by a liquid at a point inside is 2000 pascals. If the atmospheric pressure is 100 kilo Pascal, what is the total pressure at that point?

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ABL 2 Atmospheric Pressure 2.1

To understand the pressure exerted by air on the walls of an enclosure.





Pressure is the normal force exerted on unit area of the surface in contact.

Solids, liquids and gases exert pressure.

15

To 2.2demon

To demonstrate that the atmospheric air exerts pressure.



Air molecules in the atmosphere bombard a surface in contact constantly and this average force exerted on unit surface area of a body is called the atmospheric pressure.

15

2.3

How large is atmospheric pressure?



Atmospheric air exerts pressure on all objects present on the surface of the Earth from all sides.

Atmospheric pressure decreases as we move higher.

Standard air pressure is 1.03 kg/cm2.

20



15

 

2.4

Understanding through an experiment how the atmospheric air exerts pressure on an object from all sides

Air exerts pressure in all directions.

Total Time

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65 min

23

ABL 2.1

Time: 15 min

LEARNING OBJECTIVE: To demonstrate that the atmospheric air exerts pressure.

ADVANCE PREPARATION Material List S.no 1 2 3

Materials Water tumbler Small square chart paper Bucket of water

Quantity 1 1 1

Things to do

Not Applicable Safety precautions Not Applicable Procedure: Take a glass tumbler filled with water up to its brim and place a chart paper on it. Now press the palm of your one hand on top of the chart paper. Then invert the water filled glass upside down and gently remove your hand from the chart paper to release it. Observe what happens?

UNDERSTANDING THE ACTIVITY: Leading questions: 1.

2.

3.

4.

What do you observe when your hand holding the paper is taken off?

Did the paper fall off from the tumbler when the hand holding it is released?

Why do you think the paper was not pushed down above?

What are the forces acting on the paper in contact with the liquid?

Discussion and explanation:

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  

Fig 7: Atmosphere exerts upward pressure When hand is taken off the paper stays in contact with the tumbler.

The paper did not fall off.

The liquid in the tumbler above the paper has a weight w = mg. This is the gravitation force acting on the paper and tries to push it downward. If A is the area of the surface of the paper W/A is the down ward pressure, at every point on the surface of the paper.

If this is the only force acting on the paper it should start moving down according to Newton’s first law. The fact that the paper remains stationary suggests that the air below the paper is constantly bombarding it and exerting an upward force which is either equal to or greater than the weight of the liquid. This experiment clearly demonstrates the air molecules which constantly bombard the surface of the paper exert a pressure which is called atmospheric pressure. 

Clearly the forces acting on paper are The downward weight of the liquid column The upward thrust exerted by the molecules of the air below the paper.

KEY MESSAGES: 

Air molecules in the atmosphere in incessant random motion bombard a surface in contact constantly from all directions and this average force exerted on unit surface area of a body is called the atmospheric pressure.

LEARNING CHECK Ask learners to list the key things they have learnt. Guide them to the key messages listed and then put up the chart of key messages. If you have time during the class, make up a small game, quiz or match the following as a learning check. This may have to be done as part of advance preparation.

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ABL 2.2

Time: 20 min

LEARNING OBJECTIVE: How large is atmospheric pressure?

Note to Instructor â&#x20AC;&#x201C; The focus of this activity is to demonstrate how immense the atmospheric pressure is. If your audience is from upper classes, you should also bring to their notice that the atmospheric pressure acts on the hemispheres in all directions.

ADVANCE PREPARATION Material List Material

Quantity

1

Magdeburg hemispheres

1 set, consisting of two hemispheres, per class

2

Vacuum bows or darts or hooks

1 set per class

Things to do Not Applicable

Safety Precautions Not Applicable

SESSION Link to known information/previous activity In the previous activity, we learned that the atmosphere exerts pressure on everything on earth. In this activity, we will learn how strong atmospheric pressure really is!

Procedure Take a set of Magdeburg hemispheres; show that when they are simply held together they can be easily separated.

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Press the two hemispheres together and lock them. Call any two students and ask them to separate the spheres by pulling them apart (without sliding).

Let all the students try this in groups of two.

UNDERSTANDING THE ACTIVITY Leading questions 1. Why it is hard to separate the hemispheres? What holds the two hemispheres together?

2. How is atmospheric pressure so strong?

3. If atmospheric pressure can prevent us from pulling the hemispheres apart, why are we not affected by the pressure of the atmosphere on us?

Discussion and Explanation o

The earth is surrounded by air, which we call as the atmosphere. The atmospheric air exerts pressure on every object on earth; this is called atmospheric pressure.

o

Atmospheric air extends to 100 km*from the surface of the earth. That means we live at the bottom of vast ocean of air. This must exert a great pressure on us.

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o

Chart 1: Earthâ&#x20AC;&#x2122;s Atmosphere The air is dense near the surface of the earth and becomes thinner and thinner as one goes higher.

o

For convenience scientists have divided the atmosphere into different layers. These layers are simply for our reference; actually, the atmosphere is continuous. The layer nearest to the earth is called the troposphere.

o

In the above diagram, Mt. Everest, our tallest mountain is drawn so small, just like a hill. This is correct! Mt. Everest is only a few kilometres tall and earthâ&#x20AC;&#x2122;s atmosphere extends far, far beyond that.

When the hemispheres are simply held together, the air pressure inside them is the same as the air pressure outside, and they can be easily pulled apart. When we press the hemispheres together and lock them, the air from inside the sphere is removed, creating a partial vacuum. This means that there is highly reduced air pressure inside the sphere. o The atmospheric pressure from outside is acting on the sphere from all sides (as seen in the figure) and is much greater than the pressure of the air inside. This pressure pushes the sphere from all sides, making it difficult to separate. o To separate the two hemispheres, one would have to apply a pressure greater than atmospheric pressure.

Chart 2: Magdebergâ&#x20AC;&#x2122;s hemispheres o When Otto van Ma Otto von Guericke did this experiment in 1657, he used 4 pairs of horses to pull them apart and could not. o To understand the strength of the atmospheric pressure, imagine you have outstretched your palm.

I put a 130km long tube on your hand, filled with air. We think the air has negligible weight but when we take the weight of such huge quantity (130 km x cross-sectional area of the tube) it does become non-negligible. The weight of the air alone in the tube would be around 1 kg (assuming the tube to have area of 1 sq. cm), ignoring the tube weight. That is one way to measure air pressure: A pressure of Agastya International Foundation. For Internal Circulation only.

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1.03 kg per square centimetre is called the standard pressure. This is also written as 101.325 kPa (kilopascals). ď&#x201A;ˇ Stretch your hand (palm up) and think about the force exerted on your hand. The area of an average personâ&#x20AC;&#x2122;s palm is about 150 square centimetres. Since the area of the palm is 150 sq. cm and each sq. cm has about 1 kg of air above it, there must be 150 kg wt force pushing down on it. o Then why are you able to hold your hand up so easily? The answer is simple: Air pressure does not act in the downward direction alone; it acts in all directions equally. Ifthere is 150 kg of force pushing down on your palm from above, there is also 150 kg of force pushing up on it from below.

There is no net force on the hand so you can move your hand easily.

Fig 8: Atmospheric pressure acting on the palm o This gives rise to next question: Why does your hand not get crushed by all the pressure around it?

There are two reasons. One, our bones are strong enough to withstand the pressure. Two, our body is made of cells, and the fluid in a cell is also at atmospheric pressure, so there is no net force on the cell.

Fig 9: Atmospheric pressure acts equally on all sides o You can extend this reasoning to our lungs too. You know our lungs have many tiny air sacks.

The air in these sacks is at atmospheric pressure, so there is no net force on the lung. In fact, when the lung muscles contract, they create a pressure greater than atmospheric pressure and the gases in the lung come out. This is how we breathe out. And when we want to breathe in, the lung muscles relax, making the lung expand so outside air goes into it.

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f Fig 10: Pressure inside the lungs

KEY MESSAGES   

Atmospheric air exerts pressure on all objects present on the surface of the Earth from all sides.

Atmospheric pressure decreases as we move higher.

Standard air pressure is 1.03 kg/cm2.

LEARNING CHECK 1.

2.

3.

4.

Do you use vacuum wall hooks to hang clothes? How do they work?

How much load do you think you can hang on these hooks?

Why do they come off easily on a rough surface and stick very well on smooth surface?

One gets toys like guns, bows and arrows that shoot arrow or darts which stick to a wall or any other surface they hit. What is the principle behind this? [If such a toy is available in the kit materials, demonstrate and then ask the question.]

ABL 2.3

Time: 20 min

LEARNING OBJECTIVE: Understanding through an experiment how the atmospheric air exerts pressure on an object from all sides ADVANCE PREPARATION Material List Agastya International Foundation. For Internal Circulation only.

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S.no 1 2 3 4 5 6 7

Materials Empty metal can Sprit lamp Tripod stand Wire mesh Match box Water tub Sand/water dustbin

Quantity 1 1 1 1 1 1 1

SESSION Link to known information/ previous activity

Procedure: Take an empty metal can made of thin iron sheet and fill with water up to its one third of the volume.

Heat the can, when water in the can boils, steam is formed which drives out the air particles present in the upper space of the can. In fact, when boiling water converts into steam, its volume increases and so it pushes the air out of the can. After boiling the water for sufficient time close the mouth of the can with an air tight cap and immediately cool the can in cold water bath.

Fig 11: Atmospheric pressure deforms the can

UNDERSTANDING THE ACTIVITY: Leading questions: Agastya International Foundation. For Internal Circulation only.

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1.

2.

3.

4.

What happens to the tin when it is kept inside the cold water immediately after heating it?

Why the can bulges inwards when it is kept under the water and taken out?

Why should the can with water be heated for some time?

What do you learn from this activity?

Discussion and explanation:  



We observe that the can bulges inwards when it is cooled inside the water after heating it.

When the can is heated to boiling, the air inside is driven out along with steam. The same can when cooled in water; the steam condenses leaving a partial vacuum inside the can. Thus the pressure inside is very low, when the can is taken out, the atmospheric air bombarding the can wall from all direction outside, exerts a much larger pressure. The wall bulges inwards or can is crushed.

This experiment demonstrates that air exerts pressure in all directions on the objects in contact.

KEY MESSAGES:  Air exerts pressure in all directions.

LEARNING CHECK Ask learners to list the key things they have learnt. Guide them to the key messages listed and then put up the chart of key messages. If you have time during the class, make up a small game, quiz or match the following as a learning check. This may have to be done as part of advance preparation.

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ABL 3: Interesting demonstrations depicting fluid Pressures.

Demonstrations to understand Fluid Pressures Activity

Learning objective

3.1

What is the cause for air flow?



3.2

In what direction does air exert pressure?



How does air pressure affect the flow of liquids?



3.3

Key messages



3.4

3.5

Relation between pressure difference and flow of liquids.



Demonstration of liquid pressure exerted horizontally.







Does the pressure of a liquid depend on the shape of its container?

3.6

Air flows from areas of high pressure to areas of low pressure.

Air pressure acts in all directions: upwards, downwards, forwards, backwards, and sideways.

Air pressure can cause water to flow from a region of high pressure to a region of low pressure.

A partial vacuum is created when the air is sucked out of a region.

Liquids flow from areas of higher pressure to areas of lower pressure.

The pressure exerted by a liquid is called hydrostatic pressure.

Water exerts pressure horizontally.

The horizontal pressure of the water at a point depends on the height of the water column above that point.  The shape of the container has no bearing on the force driving the water through an out let as long as the hydrostatic pressure (P = ρgh) remains the same.

Total Time

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Time (min) 10

15

10

10

10

10

65 min

33

Time: 10 min

ABL 3.1 LEARNING OBJECTIVE: What is the cause for air flow?

Note to Instructor â&#x20AC;&#x201C; These activities will show students that airflow occurs as a result of an air pressure difference between two regions.

ADVANCE PREPARATION Material List Material

Quantity

1

Conical flask

1 per group

2

Rubber cork with two holes

1 per group

3

Balloon

1 per group

4

Tubes (10 cm long) to fit in the holes of the rubber cork

2 per group

5

Thread

1 per group

6

Scissor

1 per group

Things to do Try this activity before class to check that the tubes fit into the cork with no leakage of air.

Fix two tubes into the holes in the rubber cork as shown in the figure. Tie a balloon to one of the tubes.

Insert the rubber cork into the conical flask such that the balloon is inside the flask.

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Fig 12: Set up for ABL 3.1

Safety Precautions Not Applicable

SESSION Link to known information/previous activity There is a standard value for the atmospheric pressure at sea level. If all air were always at standard atmospheric pressure, then we would never have to worry about air pressure because it would have a constant effect on all objects. But air can be at different pressures. In this activity we will learn about what happens when there is a difference in air pressures between two regions.

Procedure Divide students into groups and facilitate the following group activity.

One student should blow air into the tube with the balloon. Have the other students observe what happens.

Have another student blow air into the balloon. This time, have a third student close the second tube with a finger (or if you are using a rubber tube, with a clip or by folding it). Have the other students observe what happens.

Tell the students that it is possible to blow up the balloon without blowing into the tube with the balloon.

Ask them to think about how to do this, and let students try out their solutions.

Note to Instructor: The solution is to suck air from the other tube.

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UNDERSTANDING THE ACTIVITY Leading questions 1 – Blowing into the tube with the balloon 1. What happens to the balloon when you blow air in to the tube tied with balloon?

2. What is happening to the air inside the flask, when the balloon is getting inflated?

3. Why can’t you inflate the balloon after closing the other tube?

4.

Leading Questions 2 – Sucking air from the other tube 5. When you suck air from the tube, what happens to the air inside the flask?

6. What happens to the balloon when you suck air from the other tube?

Discussion and Explanation Discuss the second discussion item first. It is easier for students to understand high and low pressure in the context of the second discussion item. Then you can explain the first discussion item using the concept of air pressure.

Discussion Item 1 – Blowing into the tube with the balloon



 

The balloon inflates when you blow into the tube tied to the balloon. o You created a higher air pressure inside your mouth (with your lungs), and pushed the highpressure air into the tube and the balloon. The air entered the balloon and the increased air pressure causes the balloon to inflate.

As the balloon gets inflated, the air inside the flask goes out from the other tube. o As the balloon inflated, it occupied the additional space inside the flask by pushing air out of the flask. This air escaped through the other tube.

When we close the second tube and blow, the air inside the flask cannot go out so there is no space for the balloon to inflate. o When we tried to blow air, the balloon started to inflate, but the air in the flask was unable to escape. The pressure inside the flask increased and became equal to the pressure of the air inside our mouth, so we were not able to push any more air into the tube.

Discussion Item 2 – Sucking air from the other tube 



When we suck the tube, the air inside the flask comes out, thus making the flask emptier. There is less air but the flask is the same. Less air occupying the same volume means the air pressure inside the flask is reduced. The air outside the flask is at a higher pressure than the air inside the flask, so it rushes in through the tube attached to the balloon. o The air flowed from a higher pressure to a lower pressure. Air always flows from high pressure to low pressure.

Sucking air out from the second tube causes air from outside to enter into the balloon and inflate.

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KEY MESSAGES 

Air flows from areas of high pressure to areas of low pressure.

LEARNING CHECK Our breathing is a demonstration of the fact that air flows from high-pressure areas to low-pressure areas.

When we breathe out our lungs muscles compress then it creates an area of higher pressure inside the lungs. The air outside is at a lower pressure so the air in the lungs come out through the nose.

Write a similar explanation, using the words ‘high pressure’ and ‘low pressure’, for the process of breathing in. Compare this process with what we did in this activity.

(Answer: We breathe in when our lungs expand, reducing the air pressure inside them. The air pressure of the atmosphere is higher than the pressure in our lungs, so air flows into our lungs from outside. Our lungs when we breathe in are similar to the flask in this activity – we sucked air out of the tube so that the flask was at a lower pressure than the atmosphere.)

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ABL 3.2

Time: 15 min

LEARNING OBJECTIVE: In what direction does air exert pressure?

Note to Instructor â&#x20AC;&#x201C; This demonstration will show students that air exerts pressure in all direction at any point.

ADVANCE PREPARATION Material List Material

Quantity

1

Funnel

1 per class

2

Table tennis ball

1 per class

Things to do Try this activity beforehand to get a feel for the force with which you have to blow into the funnel.

Safety Precautions N. A.

SESSION Link to known information/previous activity In previous activities, we have learned that atmospheric pressure is very strong but we usually do not feel it because it acts with the same force in all directions. In this activity we will learn why.

Procedure Demonstrate the activity to the entire class.

Take a funnel. Hold a table tennis ball in the funnel. Blow air into the stem of the funnel. Release the ball when you start blowing; the ball will remain stuck to the funnel.

You can change the direction of the funnel â&#x20AC;&#x201C; holding it vertical, slanting, horizontal, even inverted, the ball will remain stuck to the funnel as long as you are blowing the air.

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UNDERSTANDING THE ACTIVITY Leading questions Why did the ball not fall out of the funnel?

Discussion and Explanation ď&#x201A;ˇ When we blow air, it moves through the stem of the funnel with a certain speed. When it comes into the conical region, it has more space and it expands, reducing the pressure at the centre of the funnel. That is, a partial low-pressure area is created at the centre of the funnel, causing the surrounding air to move toward the centre and push the ball back into the funnel.

Fig 13: Set up for ABL 3.2 o Since air pressure acts in all directions, this phenomenon is seen even when we are holding the funnel in different directions. When we hold the funnel vertically upright, we see that the air pressure is acting downwards. When we hold the funnel upside down the air pressure is acting from below. When the funnel points sideways the air pressure acts from the side.

KEY MESSAGES ď&#x201A;§

Air pressure acts in all directions: upwards, downwards, forwards, backwards, and sideways.

LEARNING CHECK Would you see the same effect if you replaced the table tennis ball with a steel ball? What if you had a machine that could blow air into the funnel much faster than your teacher can?

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TRY IT YOURSELF 1. Fill a flask with water; close it with a cork with a single hole and a tube attached to it (as before).

Invert the flask; you will see that water does not come out of it. Now instead of a single-hole cork, use a two-hole cork and put a second glass tube in the additional hole. Make sure that the second glass tube is long enough to reach the bottom of the flask. Now if you invert the flask water will come out continuously. Try to explain why.

2. Try activity 1.5 with a balloon instead of plastic pouch. What happens?

INTERESTING INFORMATION Magdeburg Hemispheres

1. This experiment was first performed in 1657 at Magdeburg, Germany by Otto von Guericke (16021686), the Burgomaster (mayor) of Magdeburg.

2. The hemispheres were made of bronze and were about 1.2 feet across. Von Guericke calculated that a force of almost 2700 pounds would be needed to pull the two hemispheres apart. He used four pairs of horses but failed to pull them apart.

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3. Otto von Guericke was also first to show that bells cannot be heard in a vacuum, animals cannot survive in a vacuum, fruit can be preserved if it is placed in a vacuum, and a flame cannot be maintained in a vacuum.

Meteorological impact of air pressure differences The very fact that air flows from regions of high pressure to regions of low pressure has vast effects on our life. The greatest example of this is in monsoon winds. A low-pressure region of air is created by the heating of the land in central India and China. This causes air from a high-pressure region over the Indian Ocean to flow towards India, bringing moisture and clouds along with it.

WEB RESOURCES http://www.srh.noaa.gov/jetstream/atmos/pressure.htm (an introduction to air pressure with information about weather) http://www.albany.edu/faculty/rgk/atm101/structur.htm (the structure of the atmosphere) http://geography.about.com/od/climate/a/highlowpressure.htm (high and low pressure and their impact on weather) http://www.wxdude.com/page15.html (air pressure and wind)

VOCABULARY 1) Pressure – The pressure exerted by air on any surface in contact with it, is defined as force acting on a unit area 2) Atmosphere – The ocean of air that surrounds the earth and extends for 100 km above the surface of the earth 3) Atmospheric pressure – The pressure exerted by the weight of earth’s atmosphere, 1.03 kg/cm 2or 101.325 kPa at sea level 4) Partial vacuum – A state in which the air pressure in a space is far less than atmospheric pressure

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Time: 10 min

ABL 3.3 LEARNING OBJECTIVE: How does air pressure affect the flow of liquids?

Note to Instructor â&#x20AC;&#x201C; This activity shows that water will flow to remedy a difference in air pressure.

ADVANCE PREPARATION Material List S.NO.

Material

Quantity

1

Round bottom flask

1 per group

2

Water tub

1 per group

3

Spirit lamp

1per class

4

Match box

1per class

5

Single-hole rubber cork

1 per group

6

Glass tube

1 per group

Things to do Try this activity before facilitating it to check that the tube fits in the cork with no air leakage.

Take a round bottom flask and fix a single-hole rubber cork in the opening. (Make sure that the flask is airtight.) Fix a glass tube in the hole in the cork.

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Fig 14: Set up for ABL 3.3 We will be asking students to come one after one andblowinto the tube, so keep a separate bowl/glass of water to rinse the tube after each student blows into it. If required, fix a rubber tube to the glass tube so that it is easier for students to use.

Safety Precautions When using the spirit lamp, heat the flask gently and carefully so that it does not break.

SESSION Link to known information/previous activity In the previous activity, we saw that when one area has a higher air pressure than another region, air will flow from the high-pressure region to the low-pressure region until they have uniform pressures. In this activity, we see that a difference in air pressures can also cause water to move into a low-pressure region to balance the air pressures.

Procedure Have students complete the group activity, with several trials at completing their task. For the Instructor demo, have them suggest possibilities to you for how to complete their second assigned task, and demonstrate for the class what happens when those possibilities are attempted.

2.3a Group activity Take the flask with the rubber cork and the tube and place it next to a bowl of water.

Tell the students that their task is to fill the flask with water without removing the rubber cork.

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Give students time to think. Encourage each student to come forward and try out his/her solution.

If a group of students does not find the solution after a few trials, give a hint (e.g.to make water enter the flask you must somehow remove the air from the flask).

Note to Instructor: The solution is to suck air from the flask, close the tube with a finger, insert the tube in the water, and release the finger.

2.3b Instructor demo Empty the flask and fix the rubber cork with the tube in it into the top of the flask again.

Tell the students that the next task is to fill the flask with water, without sucking air from the flask.

Note to Instructor: The solution is to heat the flask. Use a spirit lamp to heat the flask from the bottom. When you see steam coming out, invert the flask into the water bowl.

Before heating, ensure that the flask is completely dry on the outside. Heat it gently so that the flask does not crack.

UNDERSTANDING THE ACTIVITY Leading questions 1. Why was it not possible to fill the flask with water without removing the cork?

2. When you sucked the air out of the flask, why was it necessary to immediately close the opening with a finger?

3. What happened when we sucked the air from the flask and inverted it into the water?

4. What caused the water inside the bowl to rush inside the flask?

5. What happened when the flask was heated?

6. Why does water stop rushing into the flask after some time?

Discussion and Explanation ď&#x201A;ˇ When the flask was inverted directly into water the air inside prevented water from rushing in. ď&#x201A;ˇ When the air was sucked out of the flask, the air pressure inside the flask decreased so that it was lower than the air pressure in the room outside the flask. We learned in the last activity that air flows from regions of high pressure to regions of low pressure, which means that air would tend to flow into the flask from the room. It was necessary to close the flask opening so that air would not flow into it and undo the work we did to suck the air out of the flask.

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 





When the air was sucked from the flask, water flowed into the flask through the tube.

Water flowed into the flask because the air pressure was lower inside the flask than outside it. o Air was removed from the flask but the flask size remained the same, so the pressure inside it was reduced. Outside the flask the air pressure did not change, so it was greater than the pressure inside. o Atmospheric air pressure acts on everything, so the water in the bowl was under atmospheric pressure. The pressure on the water forced it towards the lower pressure area, which is inside the flask. Thus water flowed into the flask.

Fig 15: Water fountain in round bottom flask o We can say that when we sucked air out of the flask we created a partial vacuum.

(It is not possible to create a total vacuum, but do not start a discussion of this with students.)

When the flask was heated, the water and air present inside the flask got heated up. The water was converted into steam and escaped through the tube. At the same time, the heated air rose up, reducing the pressure in the flask area. Hence, as in the previous situation, when you invert the flask, water rushes in to the tube.

The water entered the flask because there was a pressure difference. As the water entered the flask, the pressure difference was reduced; the water continued to enter the flask until the water tub and the flask attained the same pressure. (This will be explored further in the next activity.)

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KEY MESSAGES  

Air pressure can cause water to flow from a region of high pressure to a region of low pressure.

A partial vacuum is created when the air is sucked out of a region.

LEARNING CHECK Why did we ensure that the cork fit very tightly in the flask? What would happen if the cork were loose?

(Answer for Instructor’s Reference: If the cork were loose, then we could not create a region in the flask at a lower pressure than the atmosphere – air would flow back into the flask immediately after it was sucked out of the flask. There would be no pressure differential, so no water would flow into the flask.)

ABL 3.4

Time: 10 min

LEARNING OBJECTIVE: Relation between pressure difference and flow of liquids Note to Instructor – This activity will show students that water exerts pressure the same way that air does.

ADVANCE PREPARATION Material List Material

Quantity

1

Water bottles

2per class (both should be identical)

2

Water

Enough to fill a water bottle

3

Saline tube with stopper

1 per class

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Things to do Keep the model ready. Take two plastic bottles and connect them using a saline tube with a stopper (see the figure below).

Fig 16: Set up for ABL 3.4 Safety Precautions Not Applicable

SESSION Link to known information/previous activity In the previous ABLs we have seen that if a flask has air removed from it and is then immersed in a tub of water, water enters the flask, replacing the missing air. We have also seen, in the activity in which we created a manometer that pushing a balloon deeper into a bowl of water increased the pressure on the balloon. Here we will learn why these things happened.

Procedure Show the model and explain how it was constructed. Close the stopper. Ask one student to pour water in one bottle. Ask students to note their observations.

After the bottle is filled, open the stopper carefully. Ask students to note their observations again.

UNDERSTANDING THE ACTIVITY Leading questions 1. Why did the water flow to the empty bottle?

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2. Why did all the water not go into the other bottle? Why did the flow stop after some time?

3. Was the level of water the same in both the bottles? Why? 4. Did you notice any change in the rate at which the water filled in the second bottle?

Discussion and Explanation  One of the two bottles was filled with water. The pressure at the bottom was equal to the sum of the atmospheric pressure and the pressure due to the water in the bottle. The pressure at the bottom of the other bottle was equal to only the atmospheric pressure.

Thus there is a pressure difference between the two bottles. Once we opened the stopper, the two bottles were connected. Just like air would, the water too flowed from a region of high pressure to a region of lower pressure. As a result, the water flowed to the empty bottle when the stopper was opened. o The pressure exerted by a liquid is called hydrostatic pressure or liquid pressure.  As the water entered the second bottle, the pressure in the second bottle started increasing and at the same time the pressure in the first bottle started decreasing. This continued until both pressures were equal.  As the pressure difference in the two bottles gradually decreased the rate of flow also decreased and finally flow stops.  The water stopped flowing when the height of the water columns in both the bottles became equal. This is because when the heights of the water columns were equal, the water in each bottle exerted the same amount of pressure. Once the pressure in each bottle became the same, the water no longer needed to flow to balance the pressure, so the water stopped flowing.

KEY MESSAGES  

Liquids flow from areas of higher pressure to areas of lower pressure.

The pressure exerted by a liquid is called hydrostatic pressure.

LEARNING CHECK What will happen if I pour more water into one of the bottles after the water levels are balanced?

(Answer: If you pour more water into one of the bottles, the water will flow through the saline tube to the other bottle until the water levels become equal.)

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ABL 3.5

Time: 10 min

LEARNING OBJECTIVE: Demonstration of liquid pressure exerted horizontally?

Note to Instructor â&#x20AC;&#x201C; This activity demonstrates that liquids, like gases, exert pressure in all directions.

ADVANCE PREPARATION Material List Material

Quantity

1

Water bottles

2 identical bottles per class

2

Plastic straw

2 per class

3

Water

2 litres

4

Meter sticks

2 per class

Things to do Assemble the model before starting the class.

Take two identical water bottles. Make a hole on the side of one bottle, in the middle. Make a hole on the side of the second bottle, about 2 cm from the bottom. Fix a straw (about 5 cm long) in each hole. Ensure that both straws are of the same length and project out from the bottles by the same distance.

Safety Precautions Not Applicable

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SESSION Link to known information/previous activity In the previous activities, we have seen that water pressure is related to the height of the water column. How the pressure of a column of water is related to the distance that water will flow if released from the bottle?

Procedure Ask one student to close the openings of both straws.

Pour 1litre of water into each bottle and place the bottles side by side on the ground. The bottles should not be moved hereafter.

Place a meter stick just below each straw. The edges of the measuring sticks should sit beneath the ends of the straws, and when the water is released, it should fall on the measuring sticks so that you are able to measure the distance travelled by the water beams.

Ask the student release his/her fingers from both of the straws simultaneously. A water jet should rush out from the straws.

Measure the distance of the water jet from each bottle. After the water is released, the distance it will flow will begin to decrease, so you have to measure the initial distance quickly.

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Fig 17: Liquid pressure depends on height of the column

UNDERSTANDING THE ACTIVITY Leading questions

1. What happens to the water when fingers are released from the straws?

2. From which bottle did the water form a longer stream?

3. Why was the stream of water shorter from the bottle in which the straw was attached in the middle?

4. What is the relation between the distance of the water jet and the height of the water column?

5. Does water exert pressure on the walls of the bottle?

Discussion and Explanation  When the student removed his fingers from the straws, water came out of the bottles.

This is because the water exerted pressure on the sides of the bottles. This sideways pressure causes the water to flow sideways. The sideways pressure is uniform in all horizontal directions.  The water formed a longer stream from the bottle with the hole at the bottom.  In the previous activity, we observed that water pressure depends on the height of the column of water. In this activity, a straw was placed in the middle of the first bottle and on the bottom of the second bottle. Although the bottles contained equal amounts of water, there was a difference in the heights of the water column above the level of the straw in the two bottles; the second straw had a higher water column above the straw than the first bottle, so the pressure at the straw was greater in the second bottle than in the first bottle.  The second bottle, with the straw at the bottom, had a greater water pressure at the straw. This higher pressure caused the water to eject out faster, causing a longer jet stream. Thus a higher water column yields a greater distance of the water jet.  Water exerts pressure in all directions, so it must exert pressure on the walls of the bottle.

KEY MESSAGES  

Water exerts pressure horizontally.

The horizontal pressure of the water at a point depends on the height of the water column above that point.

LEARNING CHECK 1. Why do dams have a broad base?

2. Why are the outlets of tanks always kept at the bottom?

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ABL 3.6

Time: 10 min

LEARNING OBJECTIVE: Does the pressure of a liquid depend on the shape of its container?

ADVANCE PREPARATION Material List S.no 1

2 3 4 5

Material Bottles of different shapes(at least 3) fitted with glass tube at the base Water Straws Plastic bucket Stoppers

Quantity 1 set

1bucket 10 1 3

Things to do Set up the model (see the figure below).

Fig 18: Set up for ABL 3.6 Take 3 or 4 bottles of different shapes. Make a hole at the base of each bottle. Fix straw of equal lengths at the base of each.

Procedure Take 3 or 4 bottles of different sizes supplied to you. Close the open end of the tubes with a tape.

Pour water into each up to a same height, for all bottles. Keep the bottles open. Assign each Agastya International Foundation. For Internal Circulation only.

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student a job to measure the distance up to which the jet of water stretches in air. Do not spill water, collect in a tank.

UNDERSTANDING THE ACTIVITY Leading questions 1.

2.

3.

4.

What is obviously different when we look at all the bottles?

What is common to all of them?

What happened when the tape was opened?

Did the water jet stretch out to the same distance for all the bottles?

Discussion and explanation:  Obviously the bottles are all of different shapes but contain water raised to same height in all of them.  When the tapes sealed were opened, the water driven by the hydrostatic pressure (P =ρgh) jets through the open end to what extent it stretches is determined by the pressure or height of water column.  The water jet stretches to same distance in air for all the bottles, suggests the shape of the bottle has no bearing on the force driving the water out. It solely depends on hydrostatic pressure or height of water column..

KEY MESSAGES  The shape of the container has no bearing on the force driving the water through an out let as long as the hydrostatic pressure (P = ρgh) remains the same.

LEARNING CHECK Ask learners to list the key things they have learnt. Guide them to the key messages listed and then put up the chart of key messages. If you have time during the class, make up a small game, quiz or match the following as a learning check. This may have to be done as part of advance preparation.

INTERESTING INFORMATION You will notice that the water level gradually falls as water continues to flow out. Decrease in water level (h) means a decrease in pressure driving water. The rate of flow of water naturally slows down till the bottle is emptied. If we can add water to the bottle from the top and maintain the ‘height’ always the same ( like in overflowing jar ), then the water continues to flow out at the same rate throughout.

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ABL 4: Fluids in motion,Pascal’s law and Bernoulli’s principle with applications.

Activity

Learning objective

Key messages



4.1

How can we utilize what we know about liquid flow to create a siphon?

What is Pascal’s law?



What is Bernoulli’s principle?



4.2

4.3

4.4

How does an atomizer work? – Application of Bernoulli’s principle.

Time (min)

A siphon is a device that exploits the 10 principle that water flows from regions of high pressure to regions of low pressure.

15 Pascal’s Law: A change in pressure at any point in an enclosed, incompressible fluid at rest is transmitted undiminished to all points in the fluid.

15 The relationship between the velocity of flow and pressure exerted by a moving fluid is described by the Bernoulli’s principle as the velocity of a fluid increases, the pressure exerted by that fluid decreases. 



An atomizer is a device that converts a stream of liquid into a fine spray.

Atomizers are used for perfumes, medications, spray-drying, and insecticides.

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10

Time: 15 min

ABL 4.1

LEARNING OBJECTIVE: How can we utilize what we know about liquid flow to create a siphon?

Note to Instructor â&#x20AC;&#x201C; This activity uses the tendency of liquids to flow from high pressure to low pressure to demonstrate how a siphon works.

ADVANCE PREPARATION Material List

S.no

Material

Quantity

1

Water

1 litre

2

Water tub

1 per class

3

Large beaker

1 per class

4

Stand

1 per class

5

Flexible tube

1 per class

Things to do

Set up the apparatus as shown in the figure below.

Safety Precautions Not Applicable

SESSION Procedure

Fill large beaker with water and place it on a wooden stand at a large height. Take the rubber tubing and fill it completely with water close the two ends with thumbs. Keep one end of the tube dipped into water in beaker. Open up the other end into a water tub on floor. What do you observe?

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UNDERSTANDING THE ACTIVITY Leading questions

1.

2.

3.

4.

What do you observe on opening the other free end into tub?

How long does water flow out?

What is the cause for the flow?

Why water cannot be siphoned out if the empty tube is tried with instead of filling it with water?

Discussion and Explanation ď&#x201A;ˇ On opening the free end of the tube in the tub the liquid starts flowing into the tub. The flow continues till the water in the beaker is emptied out.

Initially the liquid column in the beaker exerts an extra pressure near the end A than at the end B which is open to atmosphere (pressure at A = atmospheric pressure + pressure due to liquid column and pressure at B = atmospheric pressure). The liquid starts flowing from A to B, the point of higher pressure to point of lower pressure. As the liquid level falls the excess pressure decreases lowering the rate of flow. When the liquid is emptied out the pressures at A and B become equal and there is no further flow. If the tube is empty initially, there is only air column from A to B. It will rise up to level in the tube till it is equal to level in the beaker. When it is filled with water, there is a liquid stream from A to B. The liquid at end A is pushed because of excess pressure due to liquid column in the beaker.

Atmospheric pressure

Fig 19: Simple siphon system o The tube is bent so that the shorter limb is dipped into water in the beaker. If this part is longer, the liquid column in it may exert a larger back pressure and prevent the flow of liquid.

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KEY MESSAGES 

A siphon is a device that exploits the principle that water flows from regions of high pressure to regions of low pressure.

LEARNING CHECK 1. How high can I pull the water up by siphon, can it be any height or there is a limit?

(Answer for Instructor’s Reference: It depends on the atmospheric pressure. At one atmosphere pressure, the typical pressure at sea level, one can support a column of 33 feet of water or 76 centimetres of mercury.)

2. Suppose in the simple siphon experiment, instead of inserting the filled tube just introduce one end of the empty tube into the water and suck from the other end till water start flowing out, would i still see the siphon action?

(Answer for Instructor’s Reference: Yes, as long as their are no air blocks in the liquid column.)

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Time: 15 min

ABL 4.2 LEARNING OBJECTIVE: What is Pascal’s law?

Note to Instructor – This activity demonstrates Pascal’s law by using a piston and cylinder to put pressure on the top of a ball and force water out of holes all over the ball.

ADVANCE PREPARATION Material List Material

Quantity

1

Pascal’s law model

1 per class

2

Water

Enough to fill the ball in the model

Things to do Keep the model ready. See the figure below.

Safety Precautions Not Applicable

SESSION Link to known information/previous activity We have learned that liquids can exert pressure downwards, upwards, and horizontally. What happens if we put pressure on a liquid from one point?

Procedure Remove the piston from the cylinder in the model. Fill the ball and the cylinder with water and replace the piston.

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Fig 20: Pascal’s law model Ask a student to come forward and tell him/her to push down on the piston. Observe what happens.

UNDERSTANDING THE ACTIVITY Leading questions

1. What happened to the water in the ball when the piston was pushed?

2. Did water come out of each hole with equal force? Were the jet streams from each hole of equal length?

3. Can we conclude that the pressure applied at the piston got transmitted equally to every part of the liquid in the ball?

4. Did the pressure transmitted remained same or decrease?

Discussion and Explanation  When the piston was pushed downwards, a pressure is exerted on the water at the top.

Water being in compressible the pressure exerted on it is transmitted down and then equally in all directions; this causes the water to flow out from all the holes.  Water came out of each hole in a jet stream of the same length. This is because the pressure forcing the water through holes is the same everywhere. o This was discovered by the French scientist Blaise Pascal and hence is called Pascal’s law.

Pascal’s law: When a pressure is applied at one point on a liquid at rest, it is transmitted equally in all direction. 

Notice here that we mention pressure and not force.

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 

Note that the law holds only for incompressible fluids. All liquids are regarded as incompressible. The compression if measured is very very small and can be ignored.

One of the most important applications is in hydraulic jacks and lifts (see the figure below). Here one arm of the U tube has a smaller diameter and the other a much bigger diameter. The tubes are filled with liquid. Assume that the cross-sectional area of the wider tube is ‘n’ times bigger than the cross-sectional area of the smaller one. We set weights on pistons on both sides. Since the pressure created by the weight on one side is transmitted equally to all points in the liquid, the weight required to balance the two sides will be n times larger on the larger side than on the smaller side. Thus, with a small weight we can balance a much bigger weight. This is the principle of hydraulic lifts and jacks that is commonly used in garages to lift cars. Note that this creates a multiplier effect in pressure, not in force. The law of conservation of energy is not violated. If we want to move the bigger weight up by k units, the smaller weight must be pushed down by a distance n*k. This is similar to the mechanical advantage we get from levers and pulleys.

Let a1 be the area of cross section of the narrow tube and a2 be the area of cross section of the wider tube. If p is the downward pressure on the narrow tube, P = F1/a1 where F1 is the force causing the pressure P When the same pressure is transmitted and acts upwards in the wider limb, P = F2/a2 F2/a2 = F1/a1 F2 F2

= (a2/a1 )F1 = n x F1

Apply F1 downward on shorter limbs, you get an upward push of F2 = nF1 in the wider limb

Fig 21:Hydraulic pump

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KEY MESSAGES ď&#x201A;§

Pascalâ&#x20AC;&#x2122;s Law: A change in pressure at any point in an enclosed, incompressible fluid at rest is transmitted undiminished to all points in the fluid.

LEARNING CHECK If in the above diagram a1 = 10 sqcm, a2 = 200sqcm then what is the lifting force in the wider limb if a 20kg weight is placed in the narrow limb? ( It can lift a weight of 400 kgs)

The photograph above shows an old hydraulic press. By carefully studying the picture, explain how it functions.

TRY IT YOURSELF As discussed in earlier activities take two identical water bottles and a saline tube. Instead of connecting the bottles at the bottoms, see what will happen if you connect them in the middle.

Then see what will happen if you attach the tube in the middle of one bottle and the bottom of the other bottle.

As a follow-up to this activity, take two identical bottles. Remove the lids and fix a long straw pipe to one of the lids and a short straw pipe to another lid. Make a hole in the bottom of each bottle.

Cover the holes and fill the bottles with an equal volume of water. Fix the caps to the bottles.

Remove the covers on the bottom holes and invert the two bottles at the same time. Observe which bottle empties first. Determine the reason.

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INTERESTING INFORMATION Applications of the siphon

Siphons have been used since ancient times to draw water from a river over a dike. They were used in old wineries in Europe to draw liquid wine from the drums without disturbing the solids settled at the bottom. They were used in other such applications where it is not possible to put a tap at the bottom due to sediment.

Historically, siphons have been used from a very long time ago to pump water from rivers over a dike. Here is an interesting interaction between two scientists about a siphon: On July 27, 1630, Giovanni Battista Baliani wrote a letter to Galileo Galileiexplaining an experiment he had made in which a siphon, led over a hill about twenty-one meters high, failed to work. Galileo responded with an explanation of the phenomenon: he proposed that it was the power of a vacuum that held the water up, and at a certain height the amount of water simply became too much and the force could not hold any more, like a cord that can support only so much weight.

WEB RESOURCES An animated demonstration of hydraulic lift: http://www.youtube.com/watch?v=A3ormYVZMXE

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ABL 4.3 LEARNING OBJECTIVE: What is Bernoulli’s principle?

Note to Instructor – Pressure is not an independent quantity; for liquids it is linked to velocity, volume and temperature. Bernoulli’s principle describes the inverse relationship between the velocity of a fluid and the flow pressure. When a fluid flows through a region of lower pressure, it speeds up, and when it flows through a region of higher pressure, it slows down; this is demonstrated through two activities where one gets a result that is intuitively not expected.

ADVANCE PREPARATION Material List Material

Quantity

1

Table tennis ball

1 per class

2

Air pump

1 per class

3

Bendable straw

1 per group

4

Plastic beads (1cm diameter, hollow)

1 per group

.

Things to do You will need AC power to complete this activity.

Safety Precautions N. A.

SESSION Link to known information/previous activity In the previous activity we saw that a table tennis ball could be held in a funnel by blowing air into the funnel. Here we will do a similar demonstration using an air blower that is more powerful than your teacher breathing.

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Procedure Complete the demonstration with the table tennis ball and air pump. This is a more powerful version of the demonstration in shown in 2.2. Then tell students that they will do another experiment to understand why the ball remained stuck to the air pump. Demonstrate the second activity and then ask students to do it themselves.

3.3a Instructor demo

Connect the blower (air pump). Place a ball at the mouth (or nozzle) of the tube of the air blower.

Initially you may place the ball a little bit away from the mouth.

Switch on the blower and observe the ball. Demonstrate that the blower is strong enough to blow the ball away.

Ask students what might happen if the ball is brought closer to the mouth of the blower.

Switch off the blower; place the ball close to the mouth of the tube. Switch on the blower and observe what happens.

Change the direction of the tube of the air pump and observe what happens to the ball.

3.3b Group activity Take a flexible straw and bend it into â&#x20AC;&#x2DC;Lâ&#x20AC;? shape. Place the longer end in your mouth and place a plastic bead on the other end (see the figure below).

Gently blow air into the tube and observe what happens to the ball.

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Fig 22: Boy demonstrating Bernoulli’s Principle

UNDERSTANDING THE ACTIVITY Leading questions 1. Although the blower was blowing air, the ball did not get blown away. Instead it remained stuck to the tube. Why did this happen?

2. What is the relationship between the second activity with the straws and the one with the blower?

Discussion and Explanation  In the first activity, air was moving with great speed inside the tube. When itflows past the open end it creates a fall in pressure, in accordance with Bernoulli’s theorem.  Statement: For a fluid in streamlined flow the pressure in a region decreases if thevelocity of flow increases.

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 

As we learned in earlier activities, air flows from high-pressure areas to low-pressure areas.

When there was a low-pressure area directly above the tube of the blower, air flowed towards this area, pushing the ball with it.

The same principle is involved in straw pipe, plastic bead activity.

KEY MESSAGES 

The relationship between the velocity of flow and pressure exerted by a moving fluid is described by the Bernoulli’s principle: as the velocity of a fluid increases, the pressure exerted by that fluid decreases.

LEARNING CHECK 1. Explain the second activity using the Bernoulli’s principle.

2. There are a number of devices based on Bernoulli’s principle, including the atomizer or water spray.

Give additional examples of devices where this principle is applied. (One example is a carburettor.)

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ABL 4.4

Time: 15 min

LEARNING OBJECTIVE: How does an atomizer work-application of Bernoulli’s Principle.

Note to Instructor – An atomizer is one application of Bernoulli’s principle. In this activity we make an atomizer spray and have fun playing with it.

ADVANCE PREPARATION Material List Material

Quantity

1

Plastic straw

1 per student

2

Water cup

1 per student

3

Scissors

1 per group

4

Water

About 100 ml per student

5

Water tub

1

Things to do Check that the experiment works beforehand. This will give you an idea of how much air you have to blow to get a good airbrush spray.

Note to Instructor: If you have a sufficient number of straws it is best that each student carries out this activity individually. Otherwise form groups of 5-6 students and work with the groups. Encourage the students to try this activity on their own later.

Safety Precautions Not Applicable

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SESSION Link to known information/previous activity In the previous activity, students learned about Bernoulliâ&#x20AC;&#x2122;s principle. Here they will play with an atomizer to understand how atomizers make use of Bernoulliâ&#x20AC;&#x2122;s principle.

Procedure Take a plastic straw. Cut it at about 1/3 of its length.

Put some water in a glass or a cup and insert the shorter piece of the straw vertically into the cup. If necessary, attach the straw to the edge or the cup so that it remains vertical.

Blow air through one end of the longer piece of the straw, holding it horizontal(see the figure below).The open end of the horizontal straw should be close to (but not covering) the end of the vertical straw.

Observe what happens.

Note to Instructor: If the water does not lift, encourage students to try lowering the vertical straw. Then have them try blowing harder. Finally have them try with a straw with a narrower hole.

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Fig 23: Set up for ABL 4.4

UNDERSTANDING THE ACTIVITY Leading questions 1. What happened when you blew air through the straw?

2. Why did the water lift from the cup to the mouth of the straw?

3. Why did the water come out in the form of a fine spray?

4. What applications can you think of for this device?

Discussion and Explanation  When you blew air through the horizontal straw, you should have seen water rise to the top of the vertical straw and spray into the air.  When the air is blown through the straw, air is forced out at a larger speed and a low-pressure zone is created at the point where it opens (the point where it is bent because of large velocity). In accordance with Bernoulli’s theoremthe water in the cup remains under atmospheric pressure. Because of the low-pressure area at the upper end of the vertical piece of the straw, water gets sucked in to the straw (or the pressure at the end of the straw dipped in liquid is at a pressure slightly higher than the atmospheric pressure and the open end at the top is at a pressure much lower than the atmospheric pressure. As result the liquid is pushed up the tube). This is similar to activity 2.3, in which removing air from a flask forced water to enter it.  Once the water rises to the top of the straw, it hits a jet of air. This jet spreads out through air in all directions, causing the water to spread out as well. Refer to the diagram in the previous activity. Thus the water will form a fine spray.  An atomizer is a device that converts a stream of liquid into a fine spray. Atomizer nozzles are used to spray perfumes, apply pain reliever, perform injections, and spray-dry installations. o Flit pumps use the same principle to spray insecticide. o The device is called an atomizer because the spray it creates is very fine. Although it is called an atomizer, it creates very small droplets of water, each droplet containing thousands of atoms.

KEY MESSAGES:  An atomizer is a device that converts a stream of liquid into a fine spray.

Atomizers are used for perfumes, medications, spray-drying, and insecticides.

LEARNING CHECK Ask learners to list the key things they have learnt. Guide them to the key messages listed and then put up the chart of key messages. If you have time during the class, make up a small

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game, quiz or match the following as a learning check. This may have to be done as part of advance preparation.

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ABL 5: Density, Buoyancy, Archimedes’ principle Activity

Learning objective

5.1

What is density?

Key messages

 

5.2

Why do some objects float on water and some objects sink?





5.3

Do liquids exert pressure upwards?

5.4

How to make objects travel up and down in water?

5.5

How does the weight of an object appear to change when it is immersed in water?

5.6

How do you find the relative density of an object using Archimedes’ principle?

15 The ratio of mass and volume of an object is called its density.

Density is a property of a material and is constant irrespective of the quantity of the material.

In a system with several different objects, the 10 object with the greatest density sinks to the bottom while the object with the least density will float to the top.

Objects with density greater than the density of water will sink in water, and objects with a density less than the density of water will float in water.



Liquids exert an upward thrust on an object that is immersed inside.  This is called upward buoyancy and decides whether the object sinks or floats.  The weight of the immersed body and the buoyant force on it decide if the body sinks or floats.  

Time (min)

25

5

5 When an object is immersed in a fluid it loses some weight due to buoyancy.

The apparent loss in the weight of a body when it is immersed completely or partially in a fluid is equal to the weight of the fluid displaced by it- This is Archimedes’s principle.  The relative density of an object can be found by using the formula Relative density = weight of the object/weight of the water it displaces.

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Time: 15 min

ABL 5.1 LEARNING OBJECTIVE: What is density?

Note to Instructor â&#x20AC;&#x201C; This activity introduces students to the concept of density. It also shows them how to determine the density of regular and irregularly shaped objects.

ADVANCE PREPARATION Material List Material

Quantity

1

Stone

1 per group

2

Measuring jar

1 per group

3

Water tub with water

1 per group

4

Spring balance

1 per group

5

Thread, scissors

1 per group

6

Cuboids iron blocks of different sizes

3 blocks of different sizes for each group

7

Aluminium block

1 per group

8

Wooden block

1 per group

Things to do

Keep the model ready. See the figure below.

Safety Precautions Not Applicable

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SESSION Prerequisite

Students should be familiar with the concept of volume and should be able to calculate the volume of simple regular solids.

Link to known information/previous activity

In this activity we will see a new way to measure the volume of an object in terms of volume of displaced liquid.

Procedure

This activity is split in two parts. The first activity asks students to calculate the density of iron by measuring the mass and volume of three iron blocks. The second activity asks students to find the density of a stone, using the water displacement method to determine its volume.

4.1a Group activity

Take one block of iron. Ask students to measure all sides (length, breadth and height) of the iron block using a slide callipers and find Volume of a cuboid = Length x Breadth x Height.

Ask students to do the measurement very carefully as even a small error may make a difference on our result. Ask them to find the volumes of the other iron blocks and make a table as follows: Material

1

Iron block A

Volume

l=

Mass

Mass/Volume = density

V = lbh

b= h= 2

Iron block B

“

V=

3

Iron block C

“

V=

4

Aluminium block

“

V=

5

Wooden block

“

V=

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Ask them to measure the mass of each block using a spring balance. Use a thread to tie the blocks to the hook.

Calculate the ratio of mass and volume which is the density of the material of the block. Demonstrate that this ratio is same for a given material within limits of experimental errors..

As an additional exercise, ask students to find the density of the aluminium and wooden blocks by the same method.

4.1b Group activity

Take a stone and suspend it from the spring balance in the air. Record the mass of the stone as m.

Take some water in a measuring jar and record its initial volume as V 1. Immerse the stone in the water then find the volume of the water; record this volume as V 2.

Subtract the final volume from the initial volume to get V = V 2 â&#x20AC;&#x201C; V1. This gives the volume of the stone.

Calculate the ratio of mass to volume of the stone (m/V). Record your data in your observation sheet.

Mass of the stone m (g)

Initial volume of water V1 3

(cm )

Final volume of water V2

Volume of the stone V=V2-V1

3

(cm )

Density of the stone D = m/V (g/cm3)

The density of the stone is found to be _____ g/cm3.

Worksheet 2: Observations for calculation of density of irregularly shaped objects

UNDERSTANDING THE ACTIVITY Leading questions

1. Did you see (approximately) the same values of density for each of the three iron blocks?

Why? 2. How does the density of the iron blocks relate to the density of the aluminium and the wood?

How does it relate to the density of the stone?

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3. What does density of a material mean?

4. Why did we use immersion method to find the volume of stone?

Discussion and Explanation



    

Each of the iron blocks has the same density. This is because all iron made up of same kind of atoms has the same density. Density is an intrinsic property of a material so it does not depend on the size of the block.It depends on how the atoms or molecules are packed together. It is different for different materials.

The density of iron is 7.8 g/cm3, the density of wood is 0.4-0.9 g/cm3, the density of aluminium is 2.7 g/cm3, and the density of stone is 2.3-2.8 g/cm3.

The density of an object is its mass per unit volume. It is the mass of the object in kg that occupies a volume of 1 m3. Density can be represented as D= m/V (kg/m3), where D is density, m is mass, and V is volume.

Piece of stone has no regular shape. Its volume can be measured by immersion or displacement method. Make sure the solid is not soluble in liquid. This method can also be used for regular bodies also.

The density of stone is more than that of wood and water, less than that of iron but comparable to that of aluminium.

The density of a material simply means mass of the material that occupies unit volume. It shows how massive each molecule or atom is and also the closeness with which they are packed.

KEY MESSAGES  

The ratio of mass and volume of an object is called its density.

Density is a property of a material and is constant irrespective of the quantity of the material.

LEARNING CHECK  If the density of wrought iron is 7850 kg/m3, what is the volume of 3 kg of iron?

(Answer for Instructor’s Reference: 0.00038 m3)  If the density of gold is 19300 kg/m 3. What would be the mass of a can of one litre capacity filled with gold?

(Answer for Instructor’s Reference:19.3 kg)

ABL 5.2

Time: 10 min

LEARNING OBJECTIVE: Why do some objects float on water and some objects sink?

Note to Instructor – In this activity we will apply Archimedes’ principle to explain why things float.

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ADVANCE PREPARATION Material List

1

Material

Quantity

Density model

1 per class

Things to do

Set up the model. Check that it contains kerosene, water, thick piece of plastic, and stones. You may add any other materials you think are suitable.

Safety Precautions Not Applicable

SESSION Link to known information/previous activity

In the previous activity we learned about the property of density, which is the intrinsic property of a material that describes how much mass it contains per unit volume. In this activity we will learn how objects of different densities behave when they are immersed into liquids.

Procedure Shake the bottle containing the different materials and allow it to settle. Ask the students to note which material is on the top and which is at the bottom.

Now ask any student to try change the order of the materials in the bottle. Students can try shaking it again or may change the orientation of the bottle.

UNDERSTANDING THE ACTIVITY Leading questions

1. Why do the straws/piece of plastic float on water and sink in kerosene?

Why do the stones sink to the bottom of the water and the kerosene?

Why does kerosene float on water?

2. When will an object float on water? Or any liquid for that matter?

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Discussion and Explanation Note to Instructor: The learners are likely to say that some objects float because they are lighter.

During the discussion you have to bring up the point – lighter than what? The straws float in water but sink in kerosene. Ask them to answer the question using the word ‘density’. 

 

In our model, straws/plastic pieces, kerosene, water, and stones each have different density. o Kerosene has the lowest density among all the materials so it floats on the top. o Straws/plastics have a lower density than water but a greater density than kerosene so they will occupy the 2ndposition, below the kerosene and above the water. o The stones have the highest density; they sink to the bottom. o Water has a density less than stones and greater than straws/plastics and kerosene, hence it settles on stones and sinks under kerosene.

If an object has its density greater than the density of water (or any liquid) it floats.

If the density of the material of an object is less than that of water (or liquid) it sinks in water (or in that liquid).

This discussion holds good for homogeneous solid object (without any cavity or hallow shapes).

KEY MESSAGES  

In a system with several different objects, the object with the greatest density sinks to the bottom while the object with the least density will float to the top.

Objects with density greater than the density of water will sink in water, and objects with a density less than the density of water will float in water .

LEARNING CHECK 1. If we stir some salt into water what will be the effect on its density?

2. It is said that it is easier to swim in seawater than in river water. Can you think of a reason for this?

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10 min

ABL 5.3 LEARNING OBJECTIVE: Do liquids exert pressure upwards?

Note to Instructor â&#x20AC;&#x201C; This activity introduces the concept of buoyancy as a force in the upward direction when an object is immersed under water.

ADVANCE PREPARATION Material List Material

Quantity

1

Hollow plastic ball

1 per class

2

Water

2 litres

3

Bucket

1 per class

Things to do Not Applicable

Safety Precautions Not Applicable

SESSION Link to known information/previous activity

In the previous activities we have learned that liquids exert downward pressure as well as equal pressure in all directions at a point.

Here we will use this concept to calculate upward pressure on an object.

Procedure Call one student and ask him/her to place the ball in the water. The ball will float.

Ask him/her to push the ball under the water and then release it. The ball will not stay under the water.

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Ask the student to keep pushing the ball in water again and again. Ask him/her to hold the ball deep in the water for 4 to 5 minutes.

Ask the student to share with the class what he/she felt while holding the ball underwater.

UNDERSTANDING THE ACTIVITY Leading questions

1.

2.

3.

4.

5.

How do you feel when you press the ball into water?

When you release the ball in the water what happens to the ball?

Why it is hard to push or hold the ball under the water?

Why did ball not stay under the water when left free?

Is this behaviour confined to only water or applicable to all fluids in general?

Discussion and Explanation  When the ball is pressed under the water, we feel that it is being pushed upwards.  When the ball is released under the water, it rises to the surface. This is because the water pushes the ball upwards.  When an object is immersed under water the downward pressure on the upper surface (P1) is always less than the upward pressure on the lower surface. As a result there is a net upward pressure (P = P2 – P1) acting on the body. Because of this the body experiences an upward thrust (or force), Fb = PA where A is area of the cross section of base or top. This net upward force experienced by a body when immersed in a liquid is called the force of buoyancy.

The object under water is pulled down by weight, w = mg. Thus, there are two Forces Fb the force of buoyancy and weight (gravitational force W) acting in opposite directions.

The force of buoyancy Fb = (hρg)(A) where h is the height of the object, ρ is the density of the liquid and A is the area of cross-section. Fb depends on the dimensions of object and density of the liquid.

If

 

W > Fb, the body sinks W < Fb the body floats on surface W = Fb the body just floats where ever you leave.

In the case of a ball underwater, the buoyant force upwards is far greater than its weight. It therefore rises quickly to the top and floats. This phenomenon holds good for objects immersed in any liquid.

Sometimes people have the misconception that the buoyancy force exists only in liquids like water. In fact, the buoyant force exists in all fluids, including air. But since the density of air is very low compared to the density of water we do not notice the buoyancy of air. o Gas balloons go up when released; this is a good demonstration of the buoyancy of air.

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o Interestingly, the buoyant force of air is felt very strongly by small insects. Most small insects can fly very easily with small wings.

KEY MESSAGES  

Liquids exert an upward thrust on an object that is immersed inside.

This is called upward buoyancy and decides whether the object sinks or floats.

LEARNING CHECK Why do we use air-tubes or wooden planks while learning to swim?

Is the buouant force the same when a stone is immersed in different liquids? ( hint : Liquid of larger density exerts a larger buoyant force on the object. Like what a swimmer experiences when he goes from river water to sea water)

Time: 5 min

ABL 5.4

LEARNING OBJECTIVE: How to make objects travel up and down in water?

Note to Instructor – In this activity you will assemble a Cartesian diver model and let students play with it. They will learn that by changing the density of an object they can make it sink or float.

ADVANCE PREPARATION Material List Material

Quantity

1

Empty 2-litre bottle

1 per class

2

Water

2 litres plus a cupful

3

Small glass

1 per class

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4

Ink dropper

1 per class

Things to do

Assemble the Cartesian diver model, as follows: Take a 2-litre water/soft drink bottle. Fill it completely with water.

Take an ink dropper. If required, paint it with a dark colour so that students are able to see it from a distance. Press the top and let in some water. Make sure that it is not completely filled with water and an air bubble remains in the dropper. Dip the ink dropper into a glass of water and see if it floats or sinks; if it is floating you may want to add some weight to it so that it floats about midway. Carefully dip this dropper in the filled water bottle. Ensure that the water bottle is completely full. Close the lid tight. Ensure that no air has entered the bottle. Your Cartesian diver model is ready.

Variation: If you are able to twist the open tip of the dropper, the water, while entering or leaving the dropper, will give it a twist so the diver will spin while moving. This will add to the fun.

Safety Precautions Not Applicable

SESSION Link to known information/previous activity In the previous activity we learned that density is the factor that causes some objects to sink in water and some to float. We have also learned that density is an intrinsic property of a substance, so the density of a material cannot be changed. Here we will investigate into the reason why a suitably shaped model or non-homogeneous body or bodies enclosing a hollow region, float or sink in water.

Procedure You may perform the demonstration like a magic show. While performing the demo, hold the bottle in your hand, so that the all the students are able to see what is happening inside the bottle.

Chant a few magic words and gently press the bottle; the ink dropper will shoot down. The students should not know that you are pressing the bottle.

Again chant some magic words and release the pressure; the ink dropper comes back up.

Repeat this a few times.

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Fig 24: Submarine model Ask student to try their hand and see if they can also do it. After some trials, reveal the trick.

UNDERSTANDING THE ACTIVITY Leading questions

1.

2.

3.

4.

5.

How is the Cartesian diver positioned initially?

What happens when the bottle is pressed?

What happens when the pressure is released?

How do you explain the behaviour of the diver in either case?

How do submarines exploit this to travel up and down in the water?

Discussion and Explanation

  

Initially the Cartesian diver just floats in water, wherever it was positioned. When the bottle is pressed the diver begins to sink towards the bottom. When the pressure is released it moves and starts floating near the top.

The ink pillar with some water filled initially forms a system that has some weight. But, when it is immersed in water it experiences a force of buoyancy in the upward direction. Since both are equal, the system just floats wherever you drop in to water.

On pressing the bottle, the pressure of the liquid increases and as a result some water rushes into the pillar and increase the weight of the diver system. The upward buoyant force which depends on volume of the diver is the same at all positions.

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

Since the weight of the body is more than the buoyant force the body sinks. When the pressure is released, the water rushes out of the pillar due to decrease in water pressure outside. The weight of the diver decreases and is less than upward thrust. The diver therefore rises up or floats near the top.  The volume of diver is fixed and does not change with the change in position inside the liquid.

The buoyant force which is equal to the weight of an equal volume of water is same everywhere. It is only the weight of the diver that changes on pressure or releasing the bottle.  In the case of submarines, seawater is pumped into a special compartment of a submarine.

This increases its weight so the submarine sinks into the water. When the submarine is to be brought to the surface, the water is pumped out into the sea; this makes the submarine lighter, resulting in it rising up. o We thus arrive at the true condition for floating or sinking.

If the weight of the body > the buoyant force - it sinks If the weight of the body < the buoyant force - it floats at the top.

If both are equal it just floats wherever it is positioned.

KEY MESSAGES 

The weight of the immersed body and the buoyant force on it decide if the body sinks or floats.

LEARNING CHECK 1. If you do not fill the water bottle up to the brim and some air remains, will the diver still work? Try it out and explain why or why not!

2. If you seal the opening of the dropper, will the diver work? Try it out and explain why or why not!

ABL 5.5

Time: 25 min

LEARNING OBJECTIVE: How does the weight of an object appear to change when it is immersed in water?

ADVANCE PREPARATION Material List Agastya International Foundation. For Internal Circulation only.

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Material

Quantity

1

Spring balance

1

2

Stone

1

3

Overflow vessel

1

4

Measuring cylinder

1

5

Thread

1 reel

Things to do

Try the activity beforehand to understand the pitfalls and possible mistakes one can make while doing the measurements. A correct measurement of the weight is critical to get good results.

Safety Precautions Not Applicable

SESSION Link to known information/previous activity Not Applicable Procedure

Take the stone, suspend it with a thread, and attach it to the hook of the spring balance. Measure the mass of the stone. Let us call this W1 (in grams).

Take an overflow vessel and fill it with water until over flowing water spills out. Measure the mass of this overflow vessel filled with water. Let us call this W3 (in grams).

Place a measuring cylinder under the overflow tube and slowly dip the stone into the overflow vessel containing water. Some of the water will spill into the measuring cylinder.

Note to Instructor: If the measuring cylinder does not fit below the overflow tube, use something to raise the overflow vessel. You must collect the overflow water in the measuring cylinder. Do not collect the water in a smaller beaker and then transfer it to the measuring cylinder. This would introduce too much error.

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Fig 25: Set up for ABL 5.5

Read the weight of the stone while it is in the water from the spring balance. Let us call the weight of the stone in the water W2 (in grams).

Remove the stone and now measure the weight of the overflow vessel with the remaining water. Let us call this W4 (in grams).

Now you should have values for W1, W2, W3, and W4.W1 – W2 gives the decrease in the weight of the stone when it was immersed in water. W3 – W4 gives the weight of water that overflowed or displaced by the object.

Compare W4 - W3and W1 - W2.

Note to Instructor: Ideally, W4 – W3 = W1 – W2, but usually students do not get very good results the first time. If time permits, ask students to repeat the activity, this time taking precautions to measure all the weights more carefully.

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Trial Number

W1 (g) (weight of the stone in air)

W2 (g) (weight of the stone in water)

W1 – W2 (g) (loss of weight of the stone)

W3 (g) (weight of the full overflow bucket)

W4 (g) (weight of the overflow bucket without the overflow water)

W3 – W4 (g) (water lost from the overflow bucket)

1 2 Worksheet 3: Observations for ABL 5.5

UNDERSTANDING THE ACTIVITY Leading questions: 1. What happened to the weight of the stone when it was immersed in water? Why? 2. Compare W4 – W3 with W1 – W2.Do all the objects of same volume displace equal volume of water? And would the loss in weight be same of all?

3. How could we reduce the error inherent in this experimental design?

Discussion and Explanation  The weight of the stone decreased in the water because the water exerts upward pressure on the stone. o When an object is immersed in water, the water exerts a net pressure upward on the object. The force of this upward pressure is called buoyancy. Because of the buoyancy of the water, the stone is pushed upwards. The down ward weight is opposed by the upward buoyant force. The weight of the body therefore appears to decrease by an amount equal to the upward buoyant force. o Weight in liquid = weight in air – force of buoyancy.

It is already discussed in earlier section that the force of buoyancy is equal to the weight of equal volume of water or weight of displaced water.

Weight in air – weight in liquid = loss of weight or force of buoyancy or weight of displaced liquid. o How large is this force of buoyancy? The force of buoyancy is equal to the loss of weight. o The force of buoyancy is why we feel light when we swim in water.

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

The two differences W4 – W3 and W1 – W2 are the same (within a reasonable amount of experimental error).W1 – W2 is the loss of weight of the stone in the water. W3 –W4 is the weight of the displaced water. This means that the loss of weight of the stone is equal to the weight of the water displaced by the stone. o This was discovered by Archimedes and is commonly known as Archimedes’ principle. o Archimedes’ principle: The apparent loss in the weight of a body when it is immersed completely or partially in a fluid is equal to the weight of the fluid displaced by it.



To reinforce this third question, ask the following two follow-up questions. o If, instead of a stone, I had used an object of the same size but of a less dense material, would the loss in weight be the same? o



Instead of a stone, if I had used an object of the same weight, but having more dense material (that is, the object would be smaller but would have the same weight), would the loss in weight be the same?

o Let the students think. After a few moments, give them the following hint: They should look at their observation tables and view the results of the calculations. o The answer is that objects of the same volume do displace equal volumes of water. This means that they displace equal weights of water, because water has a constant density. Thus the loss of weight of the objects is the same. This has nothing to do with the actual weight of the objects, just the amounts by which their weights decrease in water.

In this experiment we found the weight of the displaced water by measuring the weight of the overflow vessel before and after immersing the stone. This method yields a significant proportion of error because we are subtracting two big numbers to get value of a small number. A more accurate method to find the weight of the displaced water is the following: Find the volume of water displaced. We know that the density of water is 1 g/cm 3 and we know that density = mass/volume. In other words, we know that 1000 cm 3 of water weighs 1000 g. So if we know the volume of water we can find its weight!



The loss of weight depends upon i) volume of the object immersed and ii)density of the liquid displaced.

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KEY MESSAGES  

When an object is immersed in a fluid it loses some weight due to buoyancy.

The apparent loss in the weight of a body when it is immersed completely or partially in a fluid is equal to the weight of the fluid displaced by it. – This is Archimedes’s principle.

LEARNING CHECK 1.

A piece of iron sinks in water. But a boat made of same material can float. Why?

2.

It is observed that a wooden block floats with 1/3 of its volume above water. What would be the weight of water displaced by the wooden block when pushed in water completely?

3.

Would the loss in weight be same if object of different materials but of same weight, when immersed in a liquid?

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Time: 5 min

ABL 5.6

LEARNING OBJECTIVE: How do you find the relative density of an object using Archimedesâ&#x20AC;&#x2122; Principle?

ADVANCE PREPARATION Material List Material

Quantity

1

Water

2 litres

2

Spring balance

1 per group

3

Thread

1 piece per group

4

Stone

1 per group

5

Water tub

1 per group

Things to do

Try the experiment yourself beforehand.

Safety Precautions Not Applicable

SESSION Link to known information/previous activity In the previous activity we learned Archimedesâ&#x20AC;&#x2122; Principle, which states that the apparent loss of weight of an object immersed in water is equal to the weight of the water it displaces. In this activity we will use that principle to make it easier for us to calculate the relative density of a stone with respect to water.

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Procedure

Facilitate the group activity with 5to6 learners in each group.

Take some water in a measuring cylinder. Suspend a stone from a spring balance with the help of a thread and measure the weight of the stone. Call this weight (in grams) W 1.

Now immerse the stone in the water and measure its weight. Call this weight (in grams) W 2.

Calculate W1 – W2.Calculate the value of W1/(W1 – W2). This gives the relative density of the stone.

Weight of the object in air, W1 (g)

Weight of the stone in water, W2 (g)

Loss of weight of the stone, W1 – W2 (g)

Relative density, W1/(W1 – W2)

Worksheet 4: Observations for calculating relative density

UNDERSTANDING THE ACTIVITY Leading Questions: 1. What is Relative Density?

Discussion and Explanation Relative density is the ratio of the mass of the object to the mass of an equal volume of water. The mass of the object is found by direct measurement (W1).

From the previous activity, we know that when an object is immersed in water it will displace a volume of water equal to the volume of the object, and we know that the weight of the water so displaced is equal to the apparent loss in the weight of the object. Thus the weight of a volume of water equal to the volume of the object is the same as the loss in the weight of the object, in this case W1 – W2.

The relative density of the stone relative to that of water is thus W1/ (W1 – W2).

Note: Relative density is actually defined as the ratio of the density of a substance to the density of water. But in this activity we did not calculate densities at any time. How did we get relative density?

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Relative density = (mass of solid/volume of solid)/(mass of water/volume of water) Relative density = mo / V0 ÷ mw / Vw But V0 = Vw Relative density = mo /mw = mass of stone /mass of equal volume of water = mass of stone / loss of weight in water = W1/ (W1 – W2) Note: Specific gravity is another term for relative density; it usually means relative density with respect to water. The term relative density is generally preferred in modern scientific usage.

KEY MESSAGES 

The relative density of an object can found by using the formula Relative density = weight of the object/weight of the water it displaces.

LEARNING CHECK Here we calculated the relative density of the stone with respect to water. One can calculate the relative density using substances other than water. If we were to find the relative density of the stone with respect to kerosene, would it be greater, equal, or less than the value you calculated in the above activity?

TRY IT YOURSELF Make a hydrometer yourself. Use your hydrometer to find the relative densities of regular milk and skimmed milk. Alternatively, investigate the densities of different liquids like kerosene, oil, and vinegar.

Web resources: http, http://www.youtube.com/watch?v=H5AN3-Yt_G4

INTERESTING INFORMATION The story of Archimede’s eureka King Hiero had commissioned a new crown to be gifted to the royal goddess. He provided solid gold to the goldsmith for it. When the crown arrived, King Hiero was suspicious that the goldsmith used some Agastya International Foundation. For Internal Circulation only.

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of the gold and kept the rest for himself. He thought the goldsmith had added silver to make the crown the correct weight. Archimedes was asked to determine whether or not the crown was pure gold without harming it in the process. Archimedes was perplexed. His main difficulty was to find the correct volume of the crown. He found an inspiration while taking a bath. He noticed that the full bath overflowed when he lowered himself into it, and suddenly realized that he could measure the crown's volume by the amount of water it displaced. He knew that since he could measure the crown's volume, all he had to do was discover its weight in order to calculate its density and hence its purity.

Archimedes was so exuberant about his discovery that he ran down the streets of Syracuse naked shouting, “Eureka!” which meant “I've found it!” in Greek.

Situations that are explained by Archimedes’ principle: 2.

Hot air balloons: A hot air balloon consists of a large bag (called an envelope) with a wicker basket suspended underneath. The basket contains a burner that heats the air in the envelope through an opening. The hot air in the envelope is less dense than the surrounding cooler air, so the surrounding air has a buoyant force on the balloon. By Archimedes’ principle, the force of buoyancy of the balloon is equal to the weight of the cooler surrounding air displaced by the balloon. The air inside the balloon is less dense than the surrounding air so the buoyant force due to the surrounding air is greater than the weight of the air inside the balloon. To generate lift, the buoyant force must exceed the weight of the air inside the balloon, plus the weight of the envelope, plus the weight of the wicker basket, plus the weight of the passengers and equipment in the basket. To lower the balloon, an operator can stop firing the burner, causing the hot air in the envelope to cool and increasing the weight of the air.

3. Submarines: The buoyancy of a submarine is controlled by the ballast tanks found between the submarine’s inner and outer hulls. When the submarine is resting on the surface, the ballast tanks are mostly full of air and the weight of submarine is lessthan equal volume of waterdisplaced. To submerge the submarine, operators open vents on top of the ballast tanks.

Seawater comes in through the flood ports on the bottom and forces air out through the vents.

Now the ballast tanks are filled with seawater, the submarine is heavier than the buoyant force.

The exact depth can be controlled by adjusting the water-to-air ratio in the ballast tanks.

4. Ships: A piece of steel will sink in water, but ships made of steel float. Based on Archimedes’ principle, the buoyant force of the water must be greater than the mass of the ship – a huge mass of water must be displaced. The hull of the ship is shaped in such a way that as the ship sinks into the water it displaces more and more liquid until a balance is reached between the mass of water displaced and the mass of the ship.

The book How to Weigh Appukuttan [published by Eklavya] is an interesting story based on Archimedes’ Principle.

WEB RESOURCES Agastya International Foundation. For Internal Circulation only.

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The densities of common substances: http://hyperphysics.phy-astr.gsu.edu/hbase/tables/density.html A Cartesian diver: http://www.ccmr.cornell.edu/education/modules/documents/CartesianDiver.pdf This is an excellent pdf about how to make a Cartesian diver [activity 3.3] with some variations to demonstrate the effect of an increase or decrease of pressure.

Making a Cartesian diver (slightly different version)–a demo video: http://www.youtube.com/watch?v=BiI7DhFgoNQ A spinning Cartesian diver: http://www.youtube.com/watch?v=9aMFKL0VXu0 Archimedes’ Principle: http://physicsprinciples.tripod.com/ArchimedesPrinciple/id8.html Applications of Archimedes’ principle: http://www.climatechangematters.net.au/LOTS/Phy/sub/floating/floating.htm Archimedes’ principle on hot air balloons: http://www.real-world-physics-problems.com/hot-air-balloon-physics.html Archimedes’ principle on submarines: http://www.onr.navy.mil/focus/blowballast/sub/work3.htm How to build a hydrometer: http://www.wikihow.com/Build-a-Hydrometer http://www.youtube.com/watch?v=H5AN3-Yt_G4

REFERENCES Haliday and Resnick. Fundamentals of Physics. John Willes & Sons.

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Request to Readers- Kindly mail details of any discrepancies to handbooks.agastya@gmail.com