Selina ICSE Solutions for Class 9 Physics Chapter 4 Pressure in Fluids and Atmospheric Pressure

Selina ICSE Solutions for Class 9 Physics Chapter 4 Pressure in Fluids and Atmospheric Pressure

Exercise 4(A)

1S.Define the term thrust. State its S.I. unit.

Solution 1S.

Thrust is the force acting normally on a surface.

Its S.I. unit is ‘newton’.

2S.What is meant by pressure? State its S.I. unit.

Solution 2S.

Pressure is the thrust per unit area of the surface.

Its S.I. unit is ‘newton per metre2‘ or ‘pascal’.

3S.(a)What physical quantity is measured in bar?
(b)How is the unit bar related to the S.I. unit pascal?

Solution 3S.

(a) Pressure is measured in ‘bar’.

(b) 1 bar = 105 pascal.

4S.Define one pascal (Pa), the S.I. unit of pressure.

Solution 4S.

One pascal is the pressure exerted on a surface of area 1 m2 by a force of 1N acting normally on it.

5S.State whether thrust is a scalar or vector?

Solution 5S.

Thrust is a vector quantity.

6S.State whether pressure is a scalar or vector?

Solution 6S.

Pressure is a scalar quantity.

7S.Differentiate between thrust and pressure.

Solution 7S.

Thrust is the force applied on a surface in a perpendicular direction and it is a vector quantity. The effect of thrust per unit area is pressure, and it is a scalar quantity.

8S.How does the pressure exerted by thrust depend on the area of surface on which it acts? Explain with a suitable example.

Solution 8S.

Pressure exerted by thrust is inversely proportional to area of surface on which it acts. Thus, larger the area on which the thrust acts, lesser is the pressure exerted by it.

Example: If we stand on loose sand, our feet sink into the sand, but if we lie on that sand, our body does not sink into the sand. In both the cases, the thrust exerted on the sand is equal (equal to the weight of the body). However, when we lie on sand, the thrust acts on a large area and when we stand, the same thrust acts on a small area.

9S.Why is the tip of an allpin made sharp?

Solution 9S.

The tip of an allpin is made sharp so that large pressure is exerted at the sharp end and it can be driven into with less effort.

10S.Explain the following statements:

(a) It is easier to cut with a sharp knife than with a blunt one.

(b) Sleepers are laid below the rails.

Solution 10S.

(a) It is easier to cut with a sharp knife because even a small thrust causes great pressure at the edges and cutting can be done with less effort.

(b) Wide wooden sleepers are placed below the railway tracks so that the pressure exerted by the rails on the ground becomes less.

11S.What is a fluid?

Solution 11S.

A substance which can flow is called a fluid.

12S.What do you mean by the term fluid pressure?

Solution 12S.

Due to its weight, a fluid exerts pressure in all directions; the pressure exerted by the fluid is called fluid pressure.

13S.How does the pressure exerted by a solid and fluid differ?

Solution 13S.

A solid exerts pressure only on the surface on which it is placed, i.e. at its bottom, but a fluid exerts pressure at all points in all directions.

14S.Describe a simple experiment to demonstrate that a liquid enclosed in a vessel exerts pressure in all directions.

Solution 14S.

Take a can or large plastic bottle filled with water. Place it on a horizontal surface. Make a series of holes in the wall of the vessel anywhere below the free surface of the liquid. The water spurts out through each hole. This shows that the liquid exerts pressure at each point on the wall of the bottle.

Liquid exerts pressure at all points in all directions

15S.State three factors on which the pressure at a point in a liquid depends.

Solution 15S.

Pressure at a point in a liquid depends upon the following three factors:

Depth of the point below the free surface.

Density of liquid.

Acceleration due to gravity.

16S.Write an expression for the pressure at a point inside a liquid. Explain the meaning of the symbols used.

Solution 16S.

P = Po + hρg

Here, P = Pressure exerted at a point in the liquid

Po = Atmospheric pressure

h = Depth of the point below the free surface

ρ = Density of the liquid

g = Acceleration due to gravity

17S.Deduce an expression for the pressure at depth inside a liquid.

Solution 17S.

Consider a vessel containing a liquid of density ρ. Let the liquid be stationary. In order to calculate pressure at a depth, consider a horizontal circular surface PQ of area A at a depth h below the free surface XY of the liquid. The pressure on the surface PQ will be due to the thrust of the liquid contained in cylinder PQRS of height h with PQ as its base and top face RS lying on the frees surface XY of the liquid.

Total thrust exerted on the surface PQ

= Weight of the liquid column PQRS

= Volume of liquid column PQRS x density x g

= (Area of base PQ x height) x density x g

= (A x h) x ρ x g

This thrust is exerted on the surface PQ of area A. Therefore, pressure is given as shown below.

P = Thrust on surface / Area of surface

P = Ah ρg / A = hρg

Thus, Pressure = depth x density of liquid x acceleration due to gravity

18S.How does the pressure at a certain depth in sea water differ from that at the same depth in river water? Explain your answer.

Solution 18S.

Due to dissolved salts, density of sea water is more than the density of river water, so pressure at a certain depth in sea water is more than that at the same depth in river water.

19S.Pressure at the free surface of a water lake is P₁, while at a point at depth h below its free surface is P₂. (a) How are P₁ and P₂ related? (b)Which is more  P₁ or P₂?

Solution 19S.

(a) P2 = P1 + h ρ g,

(b) P2 > P1

20S.Explain why a gas bubble released at the bottom of a lake grows in size as it rises to the surface of the lake.

Solution 20S.

The reason is that when the bubble is at the bottom of the lake, total pressure exerted on it is the atmospheric pressure plus the pressure due to water column. As the gas bubble rises, due to decrease in depth the pressure due to water column decreases. By Boyle’s law, PV = constant, so the volume of bubble increases due to decrease in pressure, i.e., the bubble grows in size.

21S.A dam has broader walls at the bottom than at the top. Explain.

Solution 21S.

The pressure exerted by a liquid increases with its depth. Thus as depth increases, more and more pressure is exerted by water on wall of the dam. A thicker wall is required to withstand greater pressure, therefore, the thickness of the wall of dam increases towards the bottom.

22S.Why do sea divers need special protective suit?

Solution 22S.

The sea divers need special protective suit to wear because in deep sea, the total pressure exerted on the diver’s body is much more than his blood pressure. To withstand it, he needs to wear a special protective suit.

23S.State the laws of liquid pressure.

Solution 23S.

Laws of liquid pressure:

Pressure at a point inside the liquid increases with the depth from its free surface.

In a stationary liquid, pressure is same at all points on a horizontal plane.

Pressure is same in all directions about a point in the liquid.

Pressure at same depth is different in different liquids. It increases with the increase in the density of liquid.

A liquid seeks its own level.

24S.A tall vertical cylinder filled with water is kept on a horizontal table top. Two small holes A and B are made on the wall of the cylinder, one near the bottom and other just below the free surface of water. State and explain your observation.

Solution 24S.

The liquid from hole B reaches a greater distance on the horizontal surface than that from hole A.

This explains that liquid pressure at a point increases with the depth of point from the free surface.

25S.How does the liquid pressure on a diver change if:

(i) the diver moves to the greater depth, and

(ii) The diver moves horizontally?

Solution 25S.

(i) As the diver moves to a greater depth, pressure exerted by sea water on him also increases.

(ii) When the diver moves horizontally, his depth from the free surface remains constant and hence the pressure on him remains unchanged.

26S.State Pascal's law of transmission of pressure.

Solution 26S.

Pascal’s law states that the pressure exerted anywhere in a confined liquid is transmitted equally and undiminished in all directions throughout the liquid.

27S.Name two applications of Pascal's law.

Solution 27S.

Two applications of Pascal’s law:

Hydraulic press

Hydraulic jack

28S.Explain the principle of a hydraulic machine. Name two devices which work on this principle.

Solution 28S.

The principle of a hydraulic machine is that a small force applied on a smaller piston is transmitted to produce a large force on the bigger piston.

Hydraulic press and hydraulic brakes work on this principle.

29S.Name and state the principle on which a hydraulic press works. Write one use of hydraulic press.

Solution 29S.

Hydraulic press works on principle of hydraulic machine.

It states that a small force applied on a smaller piston is transmitted to produce a large force on the bigger piston.

Use: It is used for squeezing oil out of linseed and cotton seeds.

30S.The diagram below in Fig. shows a device which makes the use of the principle of transmission of pressure.

 

(i) Name the parts labelled by the letters X and Y.

(ii) Describe what happens to the valves A and B and to the quantity of water in the two cylinders when the lever arm is moved down

(iii) Give reasons for what happens to the valves

A and B

(iv) What happens when the release valve is opened?

(v) What happens to the valve B in cylinder P when the lever arm is moved up?

(vi) Give a reason for your answer in part (v).

(vii) State one use of the above device.

Solution 30S.

(i) X : Press Plunger; Y: Pump Plunger

(ii) When the lever is moved down, valve B closes and valve A opens, so the water from cylinder P is forced into the cylinder Q.

(iii) Valve B closes due to an increase in pressure in cylinder P. This pressure is transmitted to the connecting pipe and when the pressure in connecting pipe becomes greater than the pressure in the cylinder Q, valve A opens up.

(iv) When the release valve is opened, the ram (or press) plunger Q gets lowered and water of the cylinder Q runs out in the reservoir.

31S.Draw a simple diagram of a hydraulic jack and explain its working.

Solution 31S.

Working: When handle H of the lever is pressed down by applying an effort, the valve V opens because of increase in pressure in cylinder P. The liquid runs out from the cylinder P to the cylinder Q. As a result, the piston B rises up and it raises the car placed on the platform. When the car reaches the desired height, the handle H of the lever is no longer pressed. The valve V gets closed (since the pressure on the either side of the valve becomes same) so that the liquid may not run back from the cylinder Q to cylinder P.

32S.Explain the working of a hydraulic brake with a simple labelled diagram.

Solution 32S.

Working: To apply brakes, the foot pedal is pressed due to which pressure is exerted on the liquid in the master cylinder P, so liquid runs out from the master cylinder P to the wheel cylinder Q. As a result, the pressure is transmitted equally and undiminished through the liquid to the pistons B1and B2 of the wheel cylinder. Therefore, the pistons B1 and B2 get pushed outwards and brake shoes get pressed against the rim of the wheel due to which the motion of the vehicle retards. Due to transmission of pressure through the liquid, equal pressure is exerted on all the wheels of the vehicle connected to the pipe line R.

On releasing the pressure on the pedal, the liquid runs back from the wheel cylinder Q to the master cylinder P and the spring pulls the break shoes to their original position and forces the pistons B1 and B2 to return back into the wheel cylinder Q. Thus, the brakes get released.

33S.Complete the following sentences :

(a) Pressure at a depth h in a liquid of density p is......................

(b) Pressure is.................... in all directions about a point in a liquid.

(c) Pressure at all points at the same depth is....................

(d) Pressure at a point inside the liquid is

...................... To its depth.

(e) Pressure of a liquid at a given depth is

.................. To the density of the liquid.

Solution 33S.

(a) h ρ g (b) same (c) the same (d) directly proportional (e) directly proportional.

1M.The S.I. unit of pressure is :

1. N cm-2

2. Pa

3. N

4. N m2

Solution 1M.

Pa

2M.The pressure inside a liquid of density p at a depth h is:

1. h  g

2.  

3.  

4. h 

Solution 2M.

h ρ g

3M.The pressure P1 at a certain depth in river water and P2 at the same depth in sea water are related as:

1. P1 > P2

2. P1 = P2

3. P1 < P2

4. P1 - P2 = atmospheric pressure

Solution 3M.

P1 < P2

4M.The pressure P1 at the top of a dam and P2 at a depth h form the inside water (density p) are related as :

1. P₁  P₂

2. P₁ = P₂

3. P₁  P₂ = h g

4. P₂  P₁ = h g

Solution 4M.

P2 – P1 = h ρ g

1N.A hammer exerts a force of 1.5 N on each of the two nails A and B. The area of cross section of tip of nail A is 2 mm2 while that of nail B is 6 mm2. Calculate pressure on each nail in pascal.

Solution 1N.

2N.A block of iron of mass 7.5 kg and of dimensions 12 cm × 8 cm × 10 cm is kept on a table top on its base of side 12 cm × 8 cm.

Calculate :

a. Thrust and

b. Pressure exerted on the table top

Take 1 kgf = 10 N.

Solution 2N.

3N.A vessel contains water up to a height of 1.5 m.

Taking the density of water 103 kg m-3,

acceleration due to gravity 9.8 m s-2 and area of base of vessel 100 cm2, calculate: (a) the pressure and (b) the thrust at the base of vessel.

Solution 3N.

4N.The area of base of a cylindrical vessel is 300 cm2.

Water (density= 1000 kg m-3) is poured into it up to a depth of 6 cm. Calculate: (a) the pressure and (b) the thrust of water on the base. (g = 10m s-2.

Solution 4N.

5N.(a) Calculate the height of a water column which

 will exert on its base the same pressure as the

 70 cm column of mercury. Density of mercury is

 13-6 g cm-3.

(b) Will the height of the water column in part (a) change if the cross section of the water column is made wider?

Solution 5N.

6N.The pressure of water on the ground floor is

40,000 Pa and on the first floor is 10,000 Pa. Find the height of the first floor.

(Take : density of water = 1000 kg m-3, g = 10 m s-2)

Solution 6N.

7N.A simple U tube contains mercury to the same level in both of its arms. If water is poured to a height of 13.6 cm in one arm, how much will be the rise in mercury level in the other arm?

Given : density of mercury = 13.6 x 103 kg m-3

and density of water = 103 kg m-3.

Solution 7N.

8N.In a hydraulic machine, a force of 2 N is applied on the piston of area of cross section 10 cm2. What force is obtained on its piston of area of cross section 100 cm2 ?

Solution 8N.

9N.What should be the ratio of area of cross section of the master cylinder and wheel cylinder of a hydraulic brake so that a force of 15 N can be obtained at each of its brake shoe by exerting a force of 0.5 N on the pedal ?

Solution 9N.

10N.The areas of pistons in a hydraulic machine are

5 cm2 and 625 cm2. What force on the smaller piston will support a load of 1250 N on the larger piston? State any assumption which you make in your calculation.

Solution 10N.

11N.(i) The diameter of neck and bottom of a bottle are 2 cm and 10 cm, respectively. The bottle is completely filled with oil. If the cork in the neck is pressed in with a force of

1.2 kgf, what force is exerted on the bottom of the bottle?

(ii) Name the law/principle you have used to find the force in part (i).

Solution 11N.

12N.A force of 50 kgf is applied to the smaller piston of a hydraulic machine. Neglecting friction, find the force exerted on the large piston, if the diameters of the pistons are 5 cm and 25 cm respectively.

Solution 12N.

13N.Two cylindrical vessels fitted with pistons A and B of area of cross section 8 cm2 and 320 cm2 respectively, are joined at their bottom by a tube and they are completely filled with water. When a mass of 4 kg is placed on piston A, Find : (i) the pressure on piston A, (ii) the pressure on piston B, and (iii) the thrust on piston B.

Solution 13N.

Data is incomplete

14N.What force is applied on a piston of area of cross section 2 cm2 to obtain a force 150 N on the piston of area of cross section 12 cm2 in a hydraulic machine?

Solution 14N.

Exercise 4(B)

1S.What do you understand by atmospheric pressure?

Solution 1S.

The thrust exerted per unit area of the earth surface due to column of air, is called the atmospheric pressure on the earth surface.

2S.Write the numerical value of the atmospheric pressure on the surface of earth in pascal.

Solution 2S.

1.013 x 10 5 pascal

3S.What physical quantity is measured in torr? How is it related to the S.I. unit of the quantity?

Solution 3S.

Atmospheric pressure is measured in ‘torr’.

1 torr = 1 mm of Hg.

4S.Name the physical quantity which is expressed in the unit 'atm'. State its value in pascal.

Solution 4S.

At normal temperature and pressure, the barometric height is 0.76 m of Hg at sea level which is taken as one atmosphere.

1 atmosphere = 0.76 m of Hg = 1.013 x 105 pascal

5S.We do not feel uneasy even under enormous pressure of the atmosphere above as well as around us. Give a reason.

Solution 5S.

We do not feel uneasy under enormous pressure of the atmosphere above as well as around us because of the pressure of our blood, known as blood pressure, is slightly more than the atmospheric pressure. Thus, our blood pressure balances the atmospheric pressure.

6S.Describe an experiment to demonstrate that air exerts pressure.

Solution 6S.

Experiment to demonstrate that air exerts pressure:

Take a thin can fitted with an airtight stopper. The stopper is removed and a small quantity of water is boiled in the can. Gradually the steam occupies the entire space of can by expelling the air from it [Fig (a)]. Then stopper is then tightly replaced and simultaneously the flame beneath the can is removed. Cold water is then poured over the can.

It is observed that the can collapses inwards as shown in fig (b).

The reason is that the pressure due to steam inside the can is same as the air pressure outside the can [Fig (a)]. However, on pouring cold water over the can, fitted with a stopper [fig (b)], the steam inside the can condenses producing water and water vapour at very low pressure. Thus, the air pressure outside the can becomes more than the vapour pressure inside the closed can.

Consequently, the excess atmospheric pressure outside the can causes it to collapse inwards.

7S.Explain the following :

(i) A balloon collapses when air is removed from it.

(ii) Water does not run out of a dropper unless its rubber bulb is pressed.

(iii) Two holes are made in a sealed tin can to take out oil from it.

Solution 7S.

(i) When air is removed from the balloon, the pressure inside the balloon (which was due to air in it) is much less than the atmospheric pressure outside and hence the balloon collapses.

(ii) Water is held inside the dropper against the atmospheric pressure because the pressure due to height column of liquid inside the dropper is less than the atmospheric pressure. By pressing the dropper we increase the pressure inside the dropper and when it becomes greater than the atmospheric pressure the liquid comes out of the dropper.

(iii) There is no air inside a completely filled and sealed can. When a single hole is made to drain out the oil from the can, some of the oil will come out and due to that the volume of air above the oil will increase and hence the pressure of air will decrease. But if two holes are made on the top cover of the can, air outside the can will enter it through one hole and exert atmospheric pressure on the oil from

inside along with the pressure due to oil column, and it will come out of the can from the other hole.

8S.Why does the liquid rise in a syringe when its piston is pulled up?

Solution 8S.

When syringe is kept with its opening just inside a liquid and its plunger is pulled up in the barrel, the pressure of air inside the barrel below the plunger becomes much less than the atmospheric pressure acting on the liquid. As a result, the atmospheric pressure forces the liquid to rise up in the syringe.

9S.How is water drawn up from a well by a water pump?

Solution 9S.

In a water pump, on pulling the piston up, the pressure of air inside the siphon decreases and the atmospheric pressure on the water outside increases. As a result, the atmospheric pressure pushes the water up in pump.

10S.A partially inflated ballon is placed inside a bell jar connected to a vaccum pump. On creating vaccum inside the bell jar, ballon gets more inflated. How does the pressure change : increase, decrease or remain same, inside the (a) bell and (b) ballon?

Solution 10S.

(a) Pressure increases inside the bell jar.

(b) Pressure decreases inside the balloon.

11S.What is the purpose of a barometer?

Solution 11S.

A barometer is used to measure atmospheric pressure.

12S.What is a barometer? How will you construct a simple barometer?

Solution 12S.

A barometer is an instrument which is used to measure the atmospheric pressure.

Construction of a simple barometer:

A simple mercury barometer can be made with a clear, dry, thick-walled glass tube about 1 metre ling. The glass tube is sealed at one end and is filled with mercury completely. While filling the tube with mercury care has to be taken so that there are no air bubbles present in the mercury column. Close the open end with thumb and turn the tube upside down carefully over a trough containing mercury. Dip the open end under the mercury level in the trough and remove the thumb.

The mercury level in the tube falls until it is about 76 cm (h =760 mm) vertically above the mercury level. It is the atmospheric pressure acting on the surface of the mercury in the trough that supports the vertical mercury column. The empty space above the mercury column is called the ‘Torricellian vacuum’.

13S.Explain how is the height of mercury column in tube of a simple barometer is a measure of the atmospheric pressure.

Solution 13S.

In given figure, at all points such as C on the surface of mercury in trough, only the atmospheric pressure acts. When the mercury level in the tube becomes stationary, the pressure inside tube at the point A, which is at the level of point C, must be same as that at the point C. The pressure at point A is due to the weight (or thrust) of the mercury column AB above it. Thus, the vertical height of the mercury column from the mercury surface in trough to the level in tube is a measure of the atmospheric pressure.

The vertical of the mercury column in it (i.e., AB = h) is called the barometric height.

Had the pressure at points A and C be not equal, the level of mercury in the tube would not have been stationary.

14S.Illustrate with the help of a labelled diagram of a simple barometer that the atmospheric pressure at a place is 76 cm of Hg.

Solution 14S.

15S.Why is the barometric height used as unit to express the atmospheric pressure?

Solution 15S.

It is the atmospheric pressure acting on the surface of the mercury in the trough that supports the vertical mercury column. Hence, barometric height is used as unit to express the atmospheric pressure.

16S.What is meant by the statement 'the atmospheric pressure at a place is 76 cm of Hg'? State its value in Pa.

Solution 16S.

The atmospheric pressure at a place is 76 cm of Hg means at normal temperature and pressure, the height of the mercury column supported by the atmospheric pressure is 76 cm.

76 cm of Hg = 1.013 x 10⁵ pascal

17S.How will you show that there is vacuum above the surface of mercury in a barometer? What name is given to this vacuum?

Solution 17S.

The space above mercury is a vacuum. This empty space is called ‘Torricellian vacuum’.

This can be shown by tilting the tube till the mercury fills the tube completely. Again when the mercury column becomes stationary, the empty space is created above the mercury column. If somehow air enters into the empty space or a drop of water gets into the tube, it will immediately vaporize and the air will exert pressure on mercury column due to which the barometric height will decrease.

18S.How is the barometric height of a simple barometer affected if

(a) Its tube is pushed down into the trough of mercury?

(b) Its tube is slightly tilted from vertical?

(c) A drop of liquid is inserted inside the tube?

Solution 18S.

(a) The barometric height remains unaffected.

(b) The barometric height remains unaffected.

(c) The barometric height decreases.

19S.State two uses of a barometer.

Solution 19S.

Two uses of barometer:

To measure the atmospheric pressure.

For weather forecasting

20S.Give two reasons for use of mercury as a barometric liquid.

Solution 20S.

Two advantages of using mercury as barometric liquid:

The density of mercury is greater than that of all the liquids, so only 0.76m height of mercury column is needed to balance the normal atmospheric pressure.

The mercury neither wets nor sticks to the glass tube therefore it gives the correct reading.

21S.Give two reasons why water is not a suitable barometric liquid.

Solution 21S.

Water is not a suitable barometric liquid because:

The vapour pressure of water is high, so its vapours in the vacuum space will make the reading inaccurate.

Water sticks with the glass tube and wets it, so the reading becomes inaccurate.

22S.Mention two defects of a simple barometer and state how they are removed in a Fortin barometer.

Solution 22S.

In a simple barometer, there is no protection for the glass tube but in Fortin’s barometer, this defect has been removed by enclosing the glass tube in a brass case.

In a simple barometer, a scale cannot be fixed with the tube (or it cannot be marked on the tube) to measure the atmospheric pressure but Fortin’s barometer is provided with a vernier calipers to measure the accurate reading.

23S.Draw a simple labelled diagram of a Fortin barometer and state how it is used to measure the atmospheric pressure.

Solution 23S.

To measure the atmospheric pressure, first the leather cup is raised up or lowered down with the help of the screw S so that the ivory pointer I just touches the mercury level in the glass vessel. The position of the mercury level in the barometer tube is noted with the help of main scale and the vernier scale. The sum of the vernier scale reading to the main scale reading gives the barometric height.

24S.What is an aneroid barometer? Draw a neat and labelled diagram to explain its construction and working.

Solution 24S.

A barometer calibrated to read directly the atmospheric pressure is called an aneroid barometer. It has no liquid, it is light and portable.

Construction: Figure above shows the main parts of an aneroid barometer. It consists of a metallic box B which is partially evacuated. The top D of the box is springy and corrugated in form of a diaphragm as shown. At the middle of diaphragm, there is a thin rod L toothed at its upper end. The teeth of rod fit well into the teeth of a wheel S attached with a pointer P which can slide over a circular scale. The circular scale is initially calibrated with a standard barometer so as to read the atmospheric pressure directly in terms of the barometric height.

Working: When atmospheric pressure increases, it presses the diaphragm D and the rod L gets depressed. The wheel S rotates clockwise and pointer P moves to the right on the circular scale. When atmospheric pressure decreases, the diaphragm D bulges out due to which the rod L moves up and the wheel S rotates anti-clockwise. Consequently, the pointer moves to the left.

25S.State two advantages of an aneroid barometer over a simple barometer.

Solution 25S.

Aneroid barometer has no liquid and it is portable. It is calibrated to read directly the atmospheric pressure.

26S.How is the reading of a barometer affected when it is taken to (i) a mine, and (ii) a hill?

Solution 26S.

(i) In a mine, reading of a barometer increases.

(ii) On hills, reading of barometer decreases.

27S.How does the atmospheric pressure change with altitude? Draw an approximate graph to show this variation.

Solution 27S.

28S.State two factors which affect the atmospheric pressure as we go up.

Solution 28S.

Factors that affect the atmospheric pressure are:

Height of air column

Density of air

29S.Why does a fountain pen leak at high altitude?

Solution 29S.

A fountain pen filled with ink contains some air at a pressure equal to atmospheric pressure on earth’s surface. When pen is taken to an altitude, atmospheric pressure is low so the excess pressure inside the rubber tube forces the ink to leak out.

30S.Why does nose start bleeding on high mountains?

Solution 30S.

On mountains, the atmospheric pressure is quite low. As such, nose bleeding may occur due to excess pressure of blood over the atmospheric pressure.

31S.What is an altimeter? State its principle. How is its scale calibrated?

Solution 31S.

An altimeter is a device used in aircraft to measure its altitude.

Principle: Atmospheric pressure decreases with the increase in height above the sea level; therefore, a barometer measuring the atmospheric pressure can be used to determine the altitude of a place above the sea level.

The scale of altimeter is graduated with height increasing towards left because the atmospheric pressure decreases with increase of height above the sea level.

32S.What do the following indicate in a barometer regarding weather :

(a) gradual fall in the mercury level,

(b) sudden fall in the mercury level,

(c) Gradual rise in the mercury level?

Solution 32S.

(a) It indicates that the moisture is increasing i.e., there is a possibility of rain.

(b) It indicates the coming of a storm or cyclone.

(c) It indicates that the moisture is decreasing i.e., it indicates dry weather.

1M.The unit torr is related to the barometric height as:

1. 1 torr = l cm of Hg 

2. 1 torr = 0.76 m of Hg

3. 1 torr = 1 mm of Hg

4. 1 torr = 1m of Hg

Solution 1M.

1 torr = 1 mm of Hg

2M.The normal atmospheric pressure is :

1. 76 m of Hg

2. 76 cm of Hg

3. 76 Pa

4. 76 N m-2

Solution 2M.

76 cm of Hg

3M.The atmospheric pressure at earth's surface is P1 and inside mine is P2. They are related as :

1. P _l =P₂

2. P _l P₂

3. P _l  P₂

4. P₂ = 0

Solution 3M.

P _l  P₂

1N.Convert 1 mm of Hg into pascal. Take density of

Hg = 13.6 x 103 kg m-3 and g = 9.8 m s-2 .

Solution 1N.

2N.At a given place, a mercury barometer records a pressure of 0.70 m of Hg. What would be the height of water column if mercury in barometer is replaced by water? Take R.D. of mercury = 13.6.

Solution 2N.

3N.At sea level, the atmospheric pressure is 76 cm of Hg. If air pressure falls by 10 mm of Hg per 120m of ascent, what is the height of a hill where the barometer reads 70 cm Hg. State the assumption made by you.

Solution 3N.

Assumption: Atmospheric pressure falls linearly with ascent.

4N.At sea level, the atmospheric pressure is

1.04 x 105 Pa. Assuming g = 10 m s-2 and density of air to be uniform and equal to 1.3 kg m-3, find the height of the atmosphere.

Solution 4N.

5N.Assuming the density of air to be 1.295 kg m-3, find the fall in barometric height in mm of Hg at a height of 107 m above the sea level. Take density of mercury = 13.6 × 103 kg m-3.

Solution 5N.

Selina ICSE Solutions for Class 9 Physics Chapter 4 Pressure in Fluids and Atmospheric Pressure