**Selina Concise Class 9 Physics Solution Chapter No. 4- ‘Pressure in Fluids and Atmospheric Pressure’ For ICSE Board Students.**

**1.) Solution:**

Ans:

Thrust is nothing but the force which is acting normally on surface.

Thus, thrust exerted by a body on surface= weight of the body

Thrust is the vector quantity.

The SI unit of thrust is newton or N.

**2.) Solution:**

Ans:

Pressure is nothing but the thrust per unit area of the surface.

Hence, pressure is given by

Pressure= thrust/area

P= F/A

Pressure is the scalar quantity.

The SI unit of pressure is N/m2.

**3.) Solution:**

Ans:

a) pressure is measured in bar.

b) 1bar= 105 pascal

**4.) Solution:**

Ans:

The SI unit of pressure is pascal.

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

**5.) Solution:**

Ans:

Thrust is the force which is acting normally on surface.

Hence, thrust is the vector quantity.

**6.) Solution:**

Ans:

Pressure is the thrust per unit area of the surface.

Pressure is the scalar quantity.

**7.) Solution:**

Ans:

Thrust:

Thrust is nothing but the force which is acting normally on the surface.

It is the vector quantity.

The SI unit of thrust is newton.

Pressure:

Pressure is nothing but the thrust per unit area of the surface.

It is the scalar quantity.

The SI unit of pressure is N/m2 or pascal.

**8.) Solution:**

Ans:

Pressure is nothing but the thrust per unit area of surface.

Hence, the pressure exerted by the thrust is inversely proportional to the area of surface on which it is acting.

Thus, the effect of thrust is more on less area while it is less effective on large area.

For example:

When we stand on the loose sand, our feet directly sink into the sand.

While when we lie on the same sand we observed that our body does not sink into the sand.

Here, in both the cases thrust exerted on the sand is same. Only the difference is that thrust is exerted on different areas of sand.

When we stand on the sand, there is less area on which thrust is acting more effectively and hence we sink into the sand.

But, when we lie on the stand the total area of our body exerts thrust on sand due to which we can’t sink into it.

**9.) Solution:**

Ans:

We know that, the pressure exerted by the thrust is inversely proportional to the area of surface on which it is acting.

Thus, smaller the area the pressure exerted by the thrust will be more.

In case of allpin, tip is made sharp due to which surface area becomes small and hence more pressure will be exerted by the thrust on this small area in order to driven with less effort.

**10.) Solution:**

Ans:

a) we know that,

The pressure exerted by the thrust is inversely proportional to the area.

If less is the area more pressure will be exerted.

In case of sharp, by making them sharp the area of contact is decreased and so that when we cut any vegetables with sharp knife more pressure is exerted on small area and hence the vegetables get easily cuted.

While in case of blunt knife the area of contact is more so that the pressure exerted will be less and hence by using it we can’t any object or material easily.

b) sleepers of wide wooden block are laid below the railway tracks in order to decrease the pressure exerted by the train on the track. Due to which the track doesn’t go inside the ground. Since, the pressure exerted by the thrust is inversely proportional to the area of contact.

**11.) Solution:**

Ans:

The substance which may flow is called as the fluid. Hence, liquid and gases are the best examples of fluids.

**12.) Solution:**

Ans:

We know that solid exerts pressure on the surface on which it is placed i.e. at its bottom. Due to the weight of solid. But, fluids have the tendancy to flow, hence when they are poured in a container they exerts the pressure on the bottom and also on the walls of the container. Hence fluid can exerts pressure in all directions.

**13.) Solution:**

Ans:

We know that solid exerts pressure on the surface on which it is placed i.e. at its bottom. Due to the weight of solid. But, fluids have the tendancy to flow, hence when they are poured in a container they exerts the pressure on the bottom and also on the walls of the container. Hence fluid can exerts pressure in all directions.

**14.) Solution:**

Ans:

Following is the experimental demonstration which explains the how liquid exerts pressure when it is poured into the container.

We have taken a vessel in which liquid is poured and it is placed on the horizontal surface as shown in figure. And we have made the small holes at various places on the wall of container below its liquid surface.

We observed that, liquid is comes out through this holes and which is only due to the pressure exerted by the liquid on each point of the wall of container.

And also when we put our finger on the hole we feel the thrust and which is exerted by the liquid. Thus, we can say that liquid exerts thrust at each point on the wall of container. Hence, the thrust acting per unit area at a point is the pressure exerted by the liquid at that point.

In this way, we feel more pressure as we go in higher depth i.e. more below the liquid surface level.

**15.) Solution: **

Ans:

The following are the three factors which affects the pressure at a point in liquid.

a) depth of the point below the free surface

b) density of the liquid

c) acceleration due to gravity

**16.) Solution:**

Ans:

The expression for the pressure exerted by the liquid at a point inside the liquid is given by,

P = h*q*g

That means,

Pressure= depth*density of liquid*acceleration due to gravity

Thus, h- is the depth

q – is the density of liquid

g – is the acceleration due to gravity

P – is the pressure exerted by liquid

**17.) Solution:**

Ans:

Following is the derivation for the pressure at a depth inside a liquid.

Consider the liquid of density q which is poured in a container. To find the pressure exerted let us the liquid as stationary. To find pressure inside the liquid at depth h we consider the horizontal surface AB at depth h below the free surface PQ of the liquid as shown in figure.

The pressure on the horizontal surface AB is due the weight of the liquid column above the horizontal surface AB.

And it is in the form of cylinder ABCD whose height is h and AB is the base. While the surface CD lies on the free surface PQ.

Thus, we can find the thrust exerted on the surface AB which is given by,

Thrust on surface AB= weight of the liquid column ABCD

= Volume of the liquid column ABCD* density*g

= (Area of base AB*height)*density*g

= (A*h)*q*g

= Ahqg

This is the thrust exerted on surface AB of area A.

Thus, pressure exerted is given by,

P= Thrust on surface/Area of surface

P= Ahqg/A= hqg

Thus, this is expression for pressure at a depth h inside the liquid.

From this expression we conclude that, the pressure exerted by liquid depends on the depth, density of the liquid and acceleration due to gravity g.

**18.) Solution:**

Ans:

We know that, the pressure exerted by the liquid at depth h is given by,

P=h*q*g

Where h-is the depth

q-is the density of the liquid

g- is the acceleration due to gravity

Hence, we can say that the pressure exerted is directly proportional to the depth, density and acceleration due to gravity.

And in case of sea water due to more salts present in it, it has high density than the river water.

Thus, the pressure exerted at the same depth in sea water is more than that of the river water.

**19.) Solution:**

a) how are P1 and P2 related?

b) which is more P1 or P2?

Ans:

Since, the pressure exerted depends on the depth h. If the depth is more pressure exerted will be more.

Hence, the pressure at a free surface of water lake P1 is less than the pressure at a point at depth h below its free surface P2.

Hence,

a) P1 and P2 are related as,

Change in pressure =∆P = P2 – P1

= hqg

Hence, P2= P1 + hqg

b) and P2 is more than that of P1.

Thus, P1<P2

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

Ans:

We know that, the pressure exerted by the liquid is directly proportional to the depth of the liquid surface.

And when the gas bubble is at the bottom of lake, there will be two pressure acting on it. Atmospheric pressure and also the pressure exerted by the liquid.

We know that, as depth decreases the pressure exerted by the liquid also decreases. Hence, when gas bubble rises to the surface of lake the depth of the liquid decreases due to which pressure exerted by liquid also decreases.

But, we know that, according to Boyles law

PV= constant

As pressure decreases and in order to make product of P and V constant the volume of gas bubble increases indirectly.

Because of this reason, the gas bubble released at bottom of lake grows in size as it rises to the surface of lake.

**21.) Solution:**

Ans:

We know that, the pressure exerted by the liquid increases with depth of the of the point at which we have to find pressure below the free liquid surface.

At the bottom of the dam the pressure exerted by the water is more due to large depth and it is same in all directions at a particular point. Thus, the resultant pressure at the bottom will be more as compared to the top of dam. Hence, the dam has broader walls at the bottom than at the top.

**22.) Solution:**

Ans:

As we know that, the pressure exerted by the liquid is directly proportional to the depth. As depth increases the pressure exerted also increase.

The sea divers needs a special protective suit because they in the deep sea water there will be more pressure exerted on the body of divers than its his blood pressure. And to maintain that pressure the divers need special protective suit.

**23.) Solution:**

Ans:

There are total 5 laws of liquid pressure which are explained as follows.

1) the pressure inside the liquid increases with increase in the depth from its surface.

2) in case of stationary liquid, the pressure will be same at all the points on the horizontal surface.

3) inside the liquid, pressure is same in all directions at a particular point.

4) as pressure exerted by liquid also depends on the density of liquid, it’s value is different in different liquids at same depth also. Since the liquids are having different density.

4) the liquid has tendancy to seeks its own level.

**24.) Solution:**

Ans:

As in given, the hole A is near in the middle and B just below the free surface of water, the water from hole A goes to larger distance as pressure exerted on it is more than the water through hole B.

Thus, this observation conclude that, the pressure exerted by the liquid increases with the depth.

**25.) Solution:**

Ans:

a) we know that, the pressure exerted by the liquid increases with increasing depth from the free surface of liquid.

The diver which moves to greater depth should have increase in pressure.

b) and we also know that,

The pressure exerted by the liquid at a particular point is same in all directions i.e. the horizontal plane containing that point has same pressure.

Thus, the diver which is moving horizontally has pressure which remains as it is.

**26.) Solution:**

Ans:

According to Pascal’s law, the pressure exerted at anywhere in a case of confined liquid is always transmitted equally and undiminished in all directions through out that liquid.

**27.) Solution:**

Ans:

The principal of hydraulic machines such hydraulic press, hydraulic jack and hydraulic brakes are all based on the Pascal’s law of transmission of pressure in liquid.

**28.) Solution:**

Ans:

According to principle of hydraulic machine, a small force applied on a smaller piston will be transmitted in order to produce a large force on the bigger piston.

Hydraulic press and hydraulic brakes works on this principle.

**29.) Solution:**

Ans:

Hydraulic press works on the principle of hydraulic machine.

According to it, a small force applied on a smaller piston will be transmitted in order to produce large force on the bigger piston.

Hydraulic press is used to extract the juice from sugarcane, sugar beet etc.

**30)**

The given diagram is the diagram of hydraulic press and it is working on the principle of Pascal’s law.

a)

X – is the press plunger or RAM

Y- is the pump plunger

b) valve B closes and at the same time valve A opens, when we move down the lever and hence due to which water is forced from the cylinder P into the cylinder Q directly.

c) when the pressure in the cylinder P is increased valve B starts to close and this pressure will be transmitted through connecting pipe.

When this pressure become more through connecting pipe than pressure in cylinder Q then valve A starts to open.

d) when the release valve is opened the water from the cylinder Q get out from through the reservoir, due to the lowering of press plunger.

g) the uses of hydraulic press are as follows:

It is used for pressing cotton bales and goods such as quilts, books etc.

It is also used for extracting the juice from sugarcane, sugar beet etc.

It can be used for squeezing oil from linseed and cotton seeds.

It may used for engraving monograms on goods.

**31.) Solution:**

Ans:

The following figure shows the neat labelled diagram of hydraulic jack.

It is used mainly for lifting vehicles like cars, trucks etc. in service station for their repairing. Hydraulic press works on the principle of Pascal’s law.

The working of hydraulic press is as given below.

When we press down the handle of H of lever then valve V opens as pressure in the cylinder P is increased. And hence the liquid starts to run from cylinder P to cylinder Q. And piston B will be moved up and with it the car also rises on the platform. After reaching the car at a required height we slows down the applying pressure on the handle H. And finally valve V gets closed in order to stop the liquid flowing from cylinder Q to the cylinder P.

**32.) Solution:**

Ans:

The hydraulic brakes are used in cars etc. are also based on Pascal’s law of pressure.

The following figure shows the neat labelled diagram of hydraulic brakes.

The working of hydraulic brake is as given below.In order to apply the brakes the the foot pedals are pressed due to which the pressure is exerted on the liquid present in the master cylinder P. And hence, the liquid start to run through the cylinder P and reach the wheel cylinder Q resulting into the transmitting pressure equally and undiminished through the liquid to the pistons B1 and B2 of wheel cylinder Q.

In this way, a small force produced on the foot pedals produces a greater force on the pistons B1 and B2 of cylinder Q. This greater force produced regards the motion of vehicle.

In this way, from this working we conclude that, due to the transmission of pressure through liquid will exerts the same pressure on the all wheel of vehicle connecting with pipeline R.

**Multiple choice type:**

1.) The SI unit of pressure is

Ans: b) Pa

2.) The pressure inside a liquid of density q at a depth h is

Ans: a) hqg

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

Ans: c) P1<P2

4.) The pressure P1 at the top of dam and P2 at depth h from the top inside water are related as

Ans: d) P2 – P1 = hqg

**Numerical:**

**1.) Solution:**

Ans:

Given that, force on each of the two nails A and B is 1.5 N.

Area of cross section of tip of nail A= 2mm2= 2*10-6 m2

Area of cross section of tip of nail B= 6mm2 = 6*10-6 m2

The pressure exerted on tip of nail A is given by,

P= F/A= 1.5/2*10-6= 7.5*105 pascal

Similarly, the pressure exerted on tip of nail B is given by,

P= F/A= 1.5/6*10-6 = 2.5*105 pascal

** 2.) Solution:**

Ans:

Given that,

Mass of iron block m= 7.5kg

g=10m/s2

1kgf= 10N

a)

The thrust is nothing but the force acting

Hence, thrust= m*g= 7.5*10= 75N

b) the pressure exerted on table top is given by,

It is the pressure acting on the base area.

Thus, pressure= F/Area of base= 75/(12*8*10-4)= 7812.5Pa

**3.) Solution:**

Ans:

Given that,

Height of vessel h= 1.5m

Density of water q= 103kg/m3

Acceleration due to gravity g= 9.8m/s2

Area of base of vessel = 100cm2 = 100*10-4m2

a)

We know that, pressure is given by,

P=qhg= 103*1.5*9.8

= 14.7*103 = 1.47*104 N/m2

b)

We know that, pressure is nothing but the thrust per unit area,

Thus, pressure=thrust/area

Hence, thrust at the base of vessel is given by,

Thrust at base= pressure*area of base

= 1.47*104*100*10-4

=147N

Thus, the thrust acting at the base of vessel will be 147N.

**4.) Solution:**

Ans:

Given that,

Area o the base of cylinderical vessel = 300cm2= 300*10-4m2

Depth h= 6cm= 0.06m

Density of water q= 103 kg/m3

a)

We know that,

The pressure exerted is given by,

Pressure= hqg= 0.06*103*10= 600Pa

b) and the thrust on the base of the cylindrical vessel is given by,

Thrust= pressure*area of base

= 600*300*10-4= 18N

Thus , the thrust acting on the base of cylinder will be 18N.

** 5.) Solution:**

Ans:

a)

Given that,

Density of mercury= 13.6gm/cm3 = 13*103Kg/m3

Acceleration due to gravity g= 9.8m/s2

height h= 70cm= 0.7m

We know that,

The pressure exerted by column of mercury will be given by,

P=hqg= 0.7*13.6*103*9.8= 93.296*103= 9.32*104 N/m2

Let, h be the height of water column.

Density of water=103kg/m3

Then pressure exerted by the water column is given by,

P=hqg= h*103*9.8

But,given that pressure exerted by mercury column and water column is the same.

Hence, h*103*9.8= 9.32*104

Thus, h= 93.2/9.8= 9.5m

Thus, the height of water column will be 9.5m.

b) if the cross section of the water column is made wider, the height of water column in part a) will not change.

**6.) Solution:**

Ans:

Given that,

Pressure of water on the ground floor is = 40,000 Pa

And on the first floor the pressure is = 10,000Pa

Density of water=103kg/m3

Acceleration due to gravity g= 10m/s2

We know that,

The pressure difference between the pressure exerted by water on the ground floor and first floor is given by,

∆P= hqg

40,000 – 10,000= h*1000*10

Thus, h= 30,000/10,000= 3m

Thus, the height of the first floor will be found as 3m.

**7.) Solution: **

Ans:

Given that,

Height of water h= 13.6cm = 0.136m

Density of water=1000kg/m3

Density of mercury= 13.6*103kg/m3

Acceleration due to gravity g=10m/s2

From the given data, we conclude that the pressure exerted by mercury and the water will be same. Hence, we can write as

P=hqg

Thus, 0.136*1000*10= h*13.6*103*10

Thus,h= 0.136/13.6

Hence, h= 0.01m = 1cm

Thus, the rise in mercury level in the other arm will be 1cm.

** 8.) Solution:**

Ans:

Given that,

Force applied on piston first = 2N

Area of first piston= 10cm2= 10-3m2

Area of piston 2= 100cm2= 10-2m2

But, according to the principle of hydraulic machine, both the pressure acting on piston are equal.

Hence, we write as

2/10-3= F/10-2

Thus, F= 20N

Thus, the force acting on the piston of cross section 100cm2 will be 20N

**9.) Solution:**

Ans:

Given that, force on each brake shoe = 15N

Force acting on pedal = 0.5N

We know that, according principle of hydraulic machine the pressure on brake shoe and the pedal will be the same.

Hence,

Pressure on brake shoe= pressure on pedal

Force/area= force/area

15/area 1= 0.5/area 2

Thus, area 2/ area 1= 0.5/15= 5/150= 1/30

Thus, their ratio will be 1:30.

**10.) Solution:**

Ans:

Given that,

Area of small piston = 5cm2 = 5*10-4m2

Area of large piston = 625cm2 = 625*10-4m2

Force on small piston = 1250N

We know that, according to the principle of hydraulic machine, the pressure on smaller piston will be equal to the pressure on larger piston.

Hence,

1250/625*10-4= force on smaller piston/5*10-4

Force on smaller piston= 1250*5/625 = 10N

Thus, the force acting on the smaller piston will be 10N.

And the assumption we have taken here is there is no friction or leakage of liquid occurs.

**11.) Solution:**

Ans:

Given that,

Diameter of the neck= 2cm= 0.02m

Diameter of bottom of bottle= 10cm = 0.1m

Force on the neck is = 1.2kgf

We know that, according to the principle of hydraulic machine, the pressure on neck is equal to pressure on bottom.

And here area is πr2.

Thus,

1.2/3.14*(0.02/2)2= force on bottom/3.14*(0.1/2)2

1.2/3.14*10-4= force on bottom/3.14*25*10-4

Force on bottom = 1.2*25= 30kgf

Thus, the force exerted on the bottom of bottle is 30kgf.

b) here we have used Pascal’s of pressure to find the force exerted on the bottom of the bottle.

**12.) Solution:**

Ans:

Given that,

Force applied on the smaller piston= 15kgf

Diameter of smaller piston = 5cm = 0.05cm

Diameter of larger piston= 25cm = 0.25cm

And we know that, according to principle of hydraulic machine the pressure on smaller piston will be equal to the pressure on larger piston.

Hence, we can write as

50/25= force on larger piston/625

Force on larger piston= 50*625/25= 50*25= 1250kgf

Thus, the force acting on the larger piston will be 1250kgf.

**13.) Solution:**

Ans:

Given that,

Area of cross section of piston A= 8*10-4m2

Area of cross section of piston B= 320*10-4m2

Mass placed on piston A= 4kg

- a)

The pressure on piston A is given by,

Pressure= thrust/area= 4kg/8cm2

Pressure= 0.5kg/cm2

Thus, the pressure acting on piston A is 0.5kg/cm2.

b) and according to Pascal’s law of pressure the pressure on piston A is equal to the pressure exerted on piston B.

Hence, we can say that the pressure exerted on the piston B is 0.5kg/cm2.

c) the thrust acting on piston is given by,

Thrust= pressure on B *area of B

Thrust= 0.5*320/8= 160kgf

Thus, the thrust acting on the piston B is 160kgf.

**14.) Solution:**

Ans:

Given that,

Area of smaller piston= 2*10-4m2

Area of larger piston= 12*10-4m2

Force on larger piston= 150N

We know that,

According to the principle of hydraulic machine the pressure on smaller piston is equal to the pressure on larger piston.

Hence, we can write as

150/12*10-4= force on smaller piston/2*10-4

Thus force on smaller piston= 300/12= 25N.

In this way, the force applied on the smaller piston will be 25N.

**Exercise-B Solution**

**1.) Solution:**

Ans:

The column of air exerts thrust on the surface of earth and that thrust exerted on unit area of surface of earth is the atmospheric pressure on the surface of the earth.

**2.) Solution:**

Ans:

The numerical value of the atmospheric pressure on the earth is 1.013*105 pascal

**3.) Solution:**

Ans:

Atmospheric pressure is the physical quantity measured in torr.

And 1 torr = 133.28pa

** 4.) Solution:**

Ans:

At NTP, the barometer height is 0.76 m of Hg at the sea level and which is called as the one atmosphere i.e. atm.

And, 1 atm = 0.76m of Hg= 1.013*105 pascal

**5.) Solution:**

Ans:

We do not feel uneasy even under the enormous pressure of atmosphere above as well as around us because of the pressure of our blood called as blood pressure. The blood pressure is slightly more than the atmospheric pressure and which balances the atmospheric pressure.

** 6.) Solution:**

Ans:

The following is the experiment which demonstrates the existence of atmospheric pressure. The experiment is called as the collapsing tin can experiment.

- We have taken a thin tin can which we fit by airtight stopper. When we remove the stopper small amount of water will be boiled in that can.
- In this way, gradually the steam covers the entire space within the can and air will be expelled from it.
- Then we tightly replaced the stopper and simultaneously the flame beneath and can is removed. And when we pour cold water on the can it collapses inwards as shown in figure. Because, initially the pressure inside the can which is due to heated steam will be same as the pressure outside the can.
- But, when we pour cold water on that can then it inwards and the steam inside the can starts to condense and produces water and water vapour at very low pressure.
- Now, at that time the pressure inside the can is very low than outside the can and hence the increased outside pressure results in inwarding it.
- From this experiment we conclude that, the atmosphere outside the can will exerts the pressure on the can which is called as atmospheric pressure.

**7.) Solution:**

Ans:

a)

The balloon rises in air which is due the pressure of air filled inside it which is more than the outside atmospheric pressure. But when we remove the air from the balloon its inner pressure becomes more decreased than outside atmospheric pressure which causes the collapsing in balloon.

b)

The water remains in dropper when the inner pressure of water and outer atmospheric pressure both balances each other. But, when we press dropper the pressure on the water inside the dropper increases than the outside atmospheric pressure and hence water will be removed from the dropper.

c)

If we have made only one hole on the tin can then after coming some oil through it the volume of oil inside the can decreases and volume of air above the oil inside the tin can increases. Due to which the pressure inside the tin can decreases and oil doesn’t comes out. So to maintain the pressure inside the tin can we have made two holes from one hole air enters which also exerts pressure inside and pressure of oil column is also there and hence the oil easily comes out.

** 8.) Solution:**

Ans:

When we pulled up the piston, the liquid volume increases inside the syringe and hence air pressure decreases. At the same time the outer atmospheric pressure is more than the pressure inside the syringe and hence the liquid will be easily rises in syringe when its piston is pulled up.

**9.) Solution:**

Ans:

When water is pumped the volume of air inside it increases and hence pressure decreases. But, in the well the region in which water is present is at high pressure and we know that any object moves from high to low pressure. Thus, when start pumping the water moves from high pressure region to low pressure region and comes out from the well.

**10.) Solution:**

Ans:

a) the pressure inside the bell jar increases.

b) the pressure inside the balloon decreases.

**11.) Solution:**

Ans:

Barometer is used to measure the atmospheric pressure.

**12.) Solution:**

Ans:

- The simple barometer is the instrument used for measuring the atmospheric pressure.
- In 1643, Torricelli first constructed the simple barometer using mercury as the barometric liquid.
- The construction of simple barometer is as given below:
- The following figure shows the construction of simple barometer.

- It consist of a hard glass tube which is about 1m in length and closed at the one end.
- This glass tube is completely filled with the mercury in order to maintain there will be no air bubble inside it. We close the one open end of the glass tube with thumb and making upside down many times we remove the air bubble from inside the tube.
- Now, the completely filled tube whose one open end we have closed by pressing thumb is then inserted into the trough of mercury so that the open end of the tube will be inserted well inside the trough of mercury and tube will be standing as shown in figure.
- And we have removed the thumb in a such way that there will be no air bubble entering inside the tube.
- Finally we observe that, mercury level in the glass tube falls till it’s height above the mercury level in trough becomes h= 76cm as indicated in the figure.

**13.) Solution:**

Ans:

To understand how the height of mercury column in the tube of simple barometer is a measure of the atmospheric pressure see the following figure and the explanation given.

- When the atmospheric pressure increases the pressure at point C also increases due to which the mercury from the trough flows inside the tube in order to increase the height of mercury column in the glass tube and thereby maintain the equal pressure at points A and C.
- But, when atmospheric pressure decreases the vertical height of the mercury column inside the tube also decreases to balance the pressure.
- In this way, the vertical height of mercury column from the mercury surface in the trough to the level in the glass tube is the measure of atmospheric pressure.

**14) Solution:**

**15.) Solution:**

Ans:

- The density of mercury is greater than all other liquids and also 0.76m height of mercury is required to balance the normal atmospheric pressure.
- The vapour pressure of mercury is negligible and it does not affect the height of barometer.
- And hence, the total pressure acting on the mercury is which is in the tube is the atmospheric pressure, hence barometric height is used as a unit to express atmospheric pressure.

**16.) Solution:**

Ans:

The atmospheric pressure at a place is 76cm of Hg, this statement gives the height of mercury column in the barometer indicating barometric height at normal temperature and pressure at sea level is 0.76m or 76cm or 760mm of mercury.

And 76cm or 0.76m or 760mm of Hg = 1.013*105 pascal

**17.) Solution:**

Ans:

- To show there is a vacuum above the surface of mercury in a barometer we tilt the tube till the glass tube is completely filled with mercury.
- While when the mercury column becomes stationary, there will be a empty space above the mercury column. If air is entered or water drop is entered into the tube then it will be vaporised and the hence the air exerts more pressure on the column of mercury and hence the height of the barometer will be decreased. Thus, this shows that there is a vacuum above the surface of mercury in a barometer.
- This vacuum is called as Torricellian vacuum.

**18.) Solution:**

Ans:

a)

If the tube is pushed down into the trough of mercury the barometric height of the simple barometer is remains unaffected.

b)

When the tube is slightly tilted from vertical the barometric height of the simple barometer remains unaffected.

c)

When the drop of liquid is inserted inside the tube the barometric height of the simple barometer is decreases.

**19.) Solution:**

Ans:

- Barometer is used mainly for measuring atmospheric pressure at a place.
- It is also used for whether forecasting.
- It is also used as an altimeter to measure the height.

**20.) Solution:**

Ans:

The following are the two reasons for the use of mercury as a barometric liquid.

- The density of mercury is 6*103kg/m3 and it is greater than that of density of any liquid. So the 0.76m height of mercury will be required to balance the normal atmospheric pressure. And hence if we used other liquids than mercury then the tube required will be very long.
- Also the mercury never wets and never sticks to the glass tube and hence the reading observed will be 100% accurate.

**21.) Solution:**

Ans:

Since, the vapour pressure of water is high due to which the vapours present in the vacuum space will make the readings inaccurate.

Also, water sticks with the glass tube and also wet it, due to which the readings taken will be inaccurate.

**22.) Solution:**

Ans:

There are many demerits of simple barometer which are given as follows.

- In simple barometer there is no protection for the glass tube.
- And the measuring scale is also not fixed with the tube in simple barometer.
- All these demerits are overcomes in Fortin barometer as follows.
- In Fortin barometer the glass tube is protected with brass covering.
- And the measuring scale is fixed to the glass tube so that we can get accurate readings of it.

**23.) Solution:**

Ans:

The following is the neat labelled diagram of Fortin barometer.

- To measure the atmospheric pressure, initially the level of mercury in the leather cap is raised up or sometime lowered down with the help of screw S due to which mercury level in the glass tube touches the ivory point I.
- And here the position of the mercury level in the glass tube is measured with the help of main scale and Vernier scale.
- By summing the Vernier scale reading and the mains scale reading we get the barometric height.

**24.) Solution:**

Ans:

__Aneroid barometer:__

- Aneroid barometer does not contain liquid and it is light in weight so that it can be easily handled from one place to another also. It gives directly the atmospheric pressure readings.

__Construction:__

- The following figure shows the neat labelled diagram of Aneroid barometer.
- In fig. a there is a main part of aneroid barometer consisting of metallic part B which is partially evacuated.
- The top D of the box is springy and it is in the form of diaphragm as shown in the figure.
- In figure b) at the middle there is a thin rod L which is toothed at the upper end.
- This teeth of the rod L are fixed well into the teeth of the wheel S which is attached to the pointer P moving freely on a circular scale.
- This scale is already graduated and calibrated with the standard barometer so as to take direct accurate readings.

__Working:__

- As the atmospheric pressure increases the diaphragm D will be pressed and due to which the rod L totally get depressed.
- While wheel S moves clockwise and at the same time the pointer P moves on the circular scale.
- But, as the atmospheric pressure decreases, the diaphragm D bulged out and hence the rod L moves in upward direction and due to which wheel S start rotating in anticlockwise direction.
- And in similar way, the pointer P moves to left and the corresponding readings of pressure are measured from the calibrated scale.

**25.) Solution:**

Ans:

The Aneroid barometer has no liquid and also it is light in weight due to which it can be easily carried from one place to another place.

**26.) Solution:**

Ans:

a)

When the barometer is taken to a mine the readings of barometer will be increases.

b)

When the barometer is taken to a hill the readings of barometer will be decreases.

**27.) Solution:**

Ans:

The following graph shows the variation of atmospheric pressure with altitude.

__From the above graph we conclude the following things.__

- As we go to higher altitude the atmospheric pressure decreases and breathing becomes more difficult as the blood pressure is increased which is more than the atmospheric pressure.
- Therefore, high blood pressure patients are advised not to go hills and at higher altitude.
- Thus, we can say that, with increase in altitude the atmospheric pressure decreases as shown in graph.

**28.) Solution:**

Ans:

There are two main factors on which atmospheric pressure depends as we go up.

__Altitude: __

Because as we go up i.e. at higher altitude the atmospheric pressure decreases.

__Temperature__:

And as we go higher altitude the atmospheric pressure decreases, due to low atmospheric pressure the temperature also decreases and hence we feel more cool at places which are at higher altitude.

**29.) Solution:**

Ans:

At higher altitude the fountain pen leaks because the ink filled in the pen also has the air column which is at the pressure equal to the atmospheric pressure of earth’s surface.

But, when we go to higher altitude the atmospheric pressure decreases and hence the pressure inside the column in ink will be more than outside atmospheric pressure and hence it forces to leak the ink from the pen.

**30.) Solution:**

Ans:

As we know that, when we go to higher altitude i.e. on high mountains the atmospheric pressure decreases and our blood pressure becomes more than that atmospheric pressure.

When the blood pressure becomes more over the atmospheric pressure our nose starts bleeding. Because of this reason the blood pressure patients becomes advised not to go at higher altitude.

**31.) Solution:**

Ans:

__Altimeter:__

An altimeter is the Aneroid barometer which is used in aircraft for measuring it’s altitude.

__The principle of altimeter is explained as follows.__

- As atmospheric pressure decreases as we go to higher altitude from sea level hence the barometer which measure the atmospheric pressure at higher altitude aneroid barometer is used.
- The scale of altimeter is calibrated in the form of height of ascent with height increasing towards left because the atmospheric pressure decreases as the height from the sea level increases.
- As we go to higher altitude the atmospheric pressure will be very less where the barometer like aneroid barometer are used for pressure measurement.

**32.) Solution:**

Ans:

__a) gradual fall in the mercury level:__

The gradual fall in the mercury level in the barometer indicates there is increase in moisture which shows the possibility of rain.

Since, as we go to higher altitude the atmospheric pressure decreases i.e. height of mercury column decreases and we feel cold as temperature deceased.

__b) sudden fall in the mercury level:__

The sudden fall in the mercury level indicates there is huge increase in moisture which shows the possibility of storm or cyclone.

__c) gradual rise in the mercury level:__

The gradual rise in the mercury level indicates decrease in the moisture which shows the possibility of dry whether.

Since, as the atmospheric pressure increases height of mercury level in the barometer increases which shows the increase in temperature like in mines.

**Multiple choice type:**

1.) The unit torr is related to barometric height as

Ans: c) 1torr= 1mm of Hg

2.) The normal atmospheric pressure is

Ans: b) 76cm of Hg

3.) The atmospheric pressure at earth surface is P1 and inside a mine is P2, they are related as

Ans: c) P1< P2

**Numerical:**

**1.) Solution:**

Ans:

Given that, height of column h= 1mm= 0.001m

Density of Hg= 13.6*103kg/m3

Acceleration due to gravity g= 9.8m/s2

The pressure exerted is given by,

P=hqg= 0.001*13.6*103*9.8

P= 133.28pa

** 2.) Solution:**

Ans:

Height of mercury column h= 0.7m

Acceleration due to gravity g= 9.8m/s2

Density of mercury= 13.6*103kg/m3

The pressure exerted by mercury column is given by,

P=hqg= 0.7*13.6*103*9.8= 93.29*103pa

If we consider the height of water column as h then, pressure exerted by water column is given by,

P=hqg

93.29*103= h*1000*9.8

Thus, h= 93.29/9.8= 9.5m

Thus, the height of water column is 9.5m.

**3.) Solution:**

Ans:

Given that,

Atmospheric pressure = 76cm of Hg

Pressure falls at the rate = 10mm of Hg per 120m of ascent= 1cm of Hg per 120m of ascent

Given that, pressure at hill = 70cm of Hg

Thus, the total decrease in pressure is given by,

∆P= atmospheric pressure- pressure at hill

∆P= 76-70= 6cm of Hg

Given that, the fall in pressure is 1cm of Hg when height increases by 120m.

Hence, here the fall in pressure is 6cm of Hg.

Then the height increased will be = 6*120= 720m

Here, the assumption we have made is that atmospheric pressure is falling linearly with ascent.

**4.) Solution:**

Ans:

Given that,

Atmospheric pressure is = 1.04*105Pa

Acceleration due to gravity g= 10m/s2

Density = 1.3kg/m3

We know that, the pressure is given by,

P=hqg

Thus, h= P/qg= 1.04*105/1.3*10= 0.08*105= 8000m

**5.) Solution:**

Ans:

Given that,

Density of air = 1.295kg/m3

The change in pressure is given by,

Pressure at a height h – pressure at sea level= density of air*g*h

∆h= density of air*h/ density of mercury= 1.295*107/13.6*103= 0.01 m of Hg

= 1mm of Hg

Thus, the fall in barometric height will be 1mm of Hg.