ICSE Solutions for Chapter 4 Fluids and Atmospheric Pressure Class 9 Physics Selina Publisher
Exerciser 4(A)
1. Define the term thrust. State its S.I. unit.Thrust is the force acting normally on a surface.
Its S.I. unit is ‘newton’.
Pressure is the thrust per unit area of the surface.
Its S.I. unit is ‘newton per meter2‘ or ‘pascal’.
(b) How is the unit bar related to the S.I. unit pascal?
Answer(a) Pressure is measured in ‘bar’.
(b) 1 bar = 105 pascal.
One pascal is the pressure exerted on a surface of area 1 m2 by a force of 1N acting normally on it.
Thrust is a vector quantity.
Pressure is a scalar quantity.
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.
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.
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.
(a) It is easier to cut with a sharp knife than with a blunt one.
(b) Sleepers are laid below the rails.
Answer(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.
A substance which can flow is called a fluid.
Due to its weight, a fluid exerts pressure in all directions; the pressure exerted by the fluid is called fluid pressure.
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.
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
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.
P = Po + ⍴hg
Here, P = Pressure exerted at a point in the liquid
Po = Atmospheric pressure
h = Depth of the point below the free surface
⍴ = Density of the liquid
g = Acceleration due to gravity
Consider a vessel containing a liquid of density. 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 × density × g
= (Area of base PQ × height) × density × g
= (A ×h) ⍴ ×g
= Ah⍴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 = A ⍴hg/A = ⍴hg
Thus, Pressure = depth × density of liquid × acceleration due to gravity
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.
(a) How are P1 and P2 related?
(b)Which is more P1 or P2?
(a) P2 = P 1 + ⍴hg ,
(b) P2 > P 1
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.
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.
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.
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.
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.
(i) the diver moves to the greater depth, and
(ii) The diver moves horizontally?
Answer(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.
Pascal’s law states that the pressure exerted anywhere in a confined liquid is transmitted equally and undiminished in all directions throughout the liquid.
Two applications of Pascal’s law:
(i) Hydraulic press
(ii) Hydraulic jack
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.
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.
(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.
Answer(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.
(v) Valve B will open upwards when the lever arm is moved up.
(vi) valve B will open upwards because the pressure in the cylinder, P decreases.
(vii) Use to squeeze out oil from cotton seeds and linseeds.
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.
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 B1 and 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.
(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.
Answer(a) h⍴ g
(b) same
(c) the same
(d) directly proportional
(e) directly proportional.
Multiple Choice Questions- 4(A)
1. The S.I. unit of pressure is :(a) N cm-2
(b) Pa
(c) N
(d) N m2
Answer(b) Pa
(a) h g
(b) image
(c) image
(d) h
Answer(a) h ρ g
(a) P1 > P2
(b) P1 = P2
(c) P1 < P2
(d) P1 - P2 = atmospheric pressure
Answer(c) P1 < P2
4. 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
Answer(d) P2 - P1 = h - g
Numerical 4(A)
1. 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.
Calculate :
(a) Thrust and
(b) Pressure exerted on the table top
Take 1 kgf = 10 N.
Answer1 kgf = 10 N
∴ g = 10 m s-2
(a) Thrust is
F = mg = 7.5 kg ×10 = 75 N
(b) Pressure is force per area. 
3. 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.
AnswerGiven height, h = 1.5 m
Density of water, ρ = 103 kgm-3
Acceleration due to gravity, g = 9.8 m/s2
Area of base of the vessel, a = 100 cm2 = 100 ×10-4 m2
(a) Pressure, P = hρg
or, P = 1.5 × 103 × 9.8
or P = 1.47 × 104 Nm-2
(b) Thrust = Pressure × area
or, thrust = 1.47 × 104 × 100 × 10-4 N
or, thrust = 147 N
Water (density= 1000 kg m-3) is poured into it up to a depth of 6 cm. Calculate:
Given Area of base of vessel, a = 300 cm2 = 300 × 10-4 m2
Density of water, ρ = 1000 kg m-3
Depth, h = 6 cm = 0.06 m
Acceleration due to gravity, g = 10 ms-2
(a) Pressure = hρg = 0.06 × 1000 ×10 = 600 pascal
(b) Thrust, T = pressure × area = 600 ×300 ×10-4 = 18 N
5. (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?
Answer(a) Given, density of mercury ρ = 13.6 gcm-3
Height of mercury column, h' = 70 cm
Acceleration due to gravity, g = 9.8 ms-2
Let h be the height of the water column.
Density of water ρ = 1 gcm-3
Given, pressure exerted by mercury column = pressure exerted by water column or, h'ρ'g = hρg 
(b) No, the height of the water column shall not change.
6. 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).
Pressure of water on ground floor = 40, 000 pascal
Pressure of water on first floor = 10, 000 pascal
Density of water , ρ = 1000 kg m-3
Acceleration due to gravity, g = 10 ms-2
Let h be the height of the first floor.
Difference in water pressure between ground and first floor = hρg
or, (40000 - 10000) = h (1000)(10)
or h = 3 m
Given : density of mercury = 13.6 x 103 kg m-3 and density of water = 103 kg m-3.
AnswerHeight to which water is poured in one arm, h = 13.6cm
Let h' be the rise in the mercury level in the other arm.
Given, density of mercury = 13.6 × 103 kgm-3
Density of water = 103 kgm-3
Pressure due to water on one arm = Pressure on mercury column in the other arm
or, 13.6 × 103 × g = h' × 13.6 × 103 × g
or, h' = 1 cm
Let the ratio of area of cross - section of the master cylinder and wheel cylinder be A1 : A2
Force on pedal, F1 = 0.5N
Force on break show, F2 = 15N
By the principle of hydraulic machine,
Pressure on narrow piston = pressure on wider piston
10. 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.
Area of small piston A1 = 5 cm
Area of wider piston, A2 = 625 cm
Force on small piston be F1
Force on wider piston or load, F2 = 1250 N
By the principle of hydraulic machine,
Pressure on narrow piston = pressure on wider piston
11. (a) 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?
(a) Diameter of the neck of the bottle, d1 = 2 cm
Diameter of the bottom of the bottle, d2 = 10 cm
Force on the cork in the neck, F1 = 1.2 kgf
Force on the bottom be F2
By the principle of hydraulic machine,
Pressure on neck = pressure on bottom
12. 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.
Ratio of diameter of smaller piston to bigger piston = 5 : 25
∴ Ratio of area of smaller piston to bigger piston = 25 : 625
Force applied on smaller piston, F1 = 50 kgf
Let F2 be the force on the bigger piston.
By the principle of hydraulic machine,
Pressure on narrow piston = pressure on wider piston
13. 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:
Given: Force applied on the narrow piston= 4kg
Area of cross section of A = 8cm2
Area of cross section of B = 320cm2
(i) Pressure on piston A = Thrust/area = 4/8 = 0.5 kg cm-2
(ii) Pressure on piston A = pressure on piston B (as per Pascal's law)
Pressure on piston B is 0.5 kg cm-2
(iii) Thrust on piston B is acting in the upward direction, which is given by,
Pressure/area
⇒ 4kg × (320/8)
⇒ 160kg
In a hydraulic machine
Pressure on narrow piston = Pressure on wider piston
Exercise 4 (B)
1. What do you understand by atmospheric pressure?The thrust exerted per unit area of the earth surface due to column of air, is called the atmospheric pressure on the earth surface.
1.013×105 pascal.
Atmospheric pressure is measured in ‘torr’.
1 torr = 1 mm of Hg.
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 = 1.013×105 pascal
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.
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.
(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.
Answer(i) Why balloon collapses when air is removed from it
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.
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) why Two holes are made in a sealed tin can to take out oil from it
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.
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.
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.
(a) decrease
(b) decrease
A barometer is used to measure atmospheric pressure.
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.
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.
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.
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 = 1.013×105 pascal
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.
(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
Answer(a) The barometric height remains unaffected.
(b) The barometric height remains unaffected.
(c) The barometric height decreases.
Two uses of barometer:
- To measure the atmospheric pressure.
- For weather forecasting
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.
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.
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.
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.
A barometer calibrated to read directly the atmospheric pressure is called an aneroid barometer. It has no liquid, it is light and portable.
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.
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.
Android barometer has no liquid and it is portable. It is calibrated to read directly the atmospheric pressure.
- In a mine, reading of a barometer increases.
- On hills, reading of barometer decreases.
The atmospheric pressure decreases with an increase in the altitude.
Factors that affect the atmospheric pressure are:
- Height of air column
- Density of air
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.
On mountains, the atmospheric pressure is quite low. As such, nose bleeding may occur due to excess pressure of blood over the atmospheric pressure.
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.
(a) gradual fall in the mercury level,
(b) sudden fall in the mercury level,
(c) Gradual rise in the mercury level?
Answer(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.
Multiple Choice Questions 4(B)
(a) 1 torr =l cm of Hg
(b) 1 torr = 0.76 m of Hg
(c) 1 torr = 1 mm of Hg
(d) 1 torr = 1mof Hg
Answer(c) 1 torr = 1 mm of Hg
(a) 76 m of Hg
(b) 76 cm of Hg
(c) 76 Pa
(d) 76 N m-2
Answer(b) 76 cm of Hg
(a) Pl = P2
(b) Pl > P2
(c) Pl < P2
(d) P2 = 0
Answer(c) P l < P2
Numerical 4(B)
1. Convert 1 mm of Hg into pascal. Take density of Hg = 13.6 × 103 kg m-3 and g = 9.8 m s-2 . AnswerDensity of Hg = 13.6 ×103 kgm-3
Acceleration due to gravity, g = 9.8 ms-2
Height of mercury column = 1mm = 0.001 m
∴ Pressure, P = hρg
or, P = (0.001) (13.6 × 103)(9.8) pascal
= 133.28 Pa
Relative Density of Hg = 1.36 = 13.6 ×103 kgm-3
Acceleration due to gravity, g = 9.8 ms-2
Height of mercury column = 0.70 m
∴ Pressure, p = hρg
or, P = (0.7)(1.36 × 103)(9.8) pascal
or, P = 93.3 × 103 Pa
Let h be the height of water column
Then , p = hg
or, 93.3 ×103 = h ×103 ×9.8
or, h = 9.52 m
3. 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.
Atmospheric pressure, P = 76 cm Hg
Rate at which pressure falls = 10 mm of Hg per 120 m of ascent = 1 cm of Hg per 120 m of ascent
Let h be the height of the hill.
Pressure at hill, P' = 70 cm Hg
Total fall in pressure = P - P' = (76 -70)cm Hg = 6 cm Hg
Now, fall in pressure is 1 cm Hg for every 120 m increase in height.
Thus, if the fall in pressure is 6 cm Hg, increase in height shall be (6×120) m = 720 m
∴ Height of the hill = 720 m
Assumption: Atmospheric pressure falls linearly with ascent.
4. At sea level, the atmospheric pressure is 1.04 ×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.
Answer
Atmospheric pressure, P = 1.04 ×105 Pa
Acceleration due to gravity, g = 10 ms-2
Density, ρ = 1.3 kgm-3
Let h be the height of the atmosphere.
P = hρg
5. 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.
