ICSE Class 9 Physics Chapter 04 Pressure in Fluids and Atmospheric Pressure

Pressure In Fluids And Atmospheric Pressure

Syllabus: Change of pressure with depth (including the formula \(P = h\rho g\)); Transmission of pressure in liquids; Atmospheric pressure. Scope: Thrust and pressure and their units; pressure exerted by a liquid column \(P = h\rho g\); simple daily life examples: (i) broadness of the base of a dam, (ii) Diver's suit etc., some consequences of \(P = h\rho g\); transmission of pressure in liquids; Pascal's law; examples; Atmospheric pressure; common manifestation and consequences. Variation of pressure with altitude, (qualitative only); applications such as weather forecasting and altimeter. (Simple numerical problems).

Pressure In Fluids And Its Transmission

Thrust And Pressure

Thrust: A force can be applied on a surface in any direction. If a force is applied in a direction normal (or perpendicular) to the surface, it is called the thrust. Thus, thrust is the force acting normally on a surface.

The thrust exerted by a body placed on a surface is equal to its weight. The thrust is same in whatsoever position the body is placed on the surface. Thus, thrust exerted by a body on a surface = Weight of the body.

Thrust is a vector quantity.

Unit of thrust: It is measured in the units of force. The S.I. unit of thrust is newton (N) and C.G.S. unit of thrust is dyne, where 1 N = 10⁵ dyne. The gravitational unit of thrust in M.K.S. system is kgf and in C.G.S. system is gf. They are related as: 1 kgf = 9.8 N and 1 gf = 980 dyne

Pressure: The effect of thrust depends on the area of the surface on which it acts. The effect of a thrust is less on a large area, while it is more on a small area.

Example: If you stand on loose sand, your feet sink into the sand, but if you lie on that sand, your body does not sink into the sand. In both the cases, the thrust exerted on the sand is same (equal to your weight). But when you lie on sand, the thrust acts on a larger area and when you stand, the same thrust acts on a smaller area.

The effect of thrust is expressed in terms of thrust per unit area. This quantity is called pressure. Thus, pressure is the thrust per unit area of surface.

If a thrust \(F\) acts on an area \(A\), then \(P = \frac{F}{A}\)

Pressure is a scalar quantity.

Units Of Pressure

From the relation, pressure = thrust / area

Unit of pressure = Unit of thrust / Unit of area

S.I. unit: The S.I. unit of thrust is newton and that of area is metre², so the S.I. unit of pressure is newton per metre² which is abbreviated as N m^-2. This unit is named pascal (symbol Pa) after the french scientist Blaise Pascal. i.e., 1 Pascal = \(\frac{1 \text{ Newton}}{1 \text{ metre}^2}\) or 1 Pa = 1 N m^-2

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

However, if thrust is measured in kgf and area in m², the unit of pressure is kgf m^-2.

C.G.S. unit: The C.G.S. unit of pressure is dyne cm^-2 where 1 dyne cm^-2 = 0.1 N m^-2 or 1 N m^-2 = 10 dyne cm^-2.

Other units: Other units of pressure are bar and millibar, where 1 bar = 10⁵ N m^-2 and 1 millibar = 10^-3 bar = 10² N m^-2. The atmospheric pressure is generally expressed in terms of the height of mercury column in the barometer. At normal temperature and pressure, the barometric height is 0.76 m of Hg (or 760 mm of Hg) at sea level which is taken as one atmosphere. Thus atmospheric pressure is also expressed in a unit atmosphere (symbol atm) where 1 atmosphere (atm) = 0.76 m of Hg = 1.013 x 10⁵ Pa

Sometimes we use torr as unit of atmospheric pressure after the name of the scientist Torricelli where 1 torr = 1 mm of Hg and 1 atm = 760 torr.

Factors Affecting The Pressure

Factors affecting the pressure: The pressure exerted on a surface depends on two factors: (i) the area on which the thrust is applied, and (ii) the thrust.

Examples: (1) A brick of weight 4 kgf having dimensions 20 cm x 10 cm x 5 cm exerts maximum pressure on ground when it is placed with its longest side (20 cm) vertical (Fig. 4.1(a)), while it exerts minimum pressure on ground when it is placed with its shortest side (5 cm) vertical (Fig. 4.1(b)), even though the thrust is same in each case.

In Fig. 4.1(a), Thrust = 4 kgf Area of base = 5 cm x 10 cm = 50 cm²Pressure on base (or ground) = \(\frac{4 \text{ kgf}}{50 \text{ cm}^2}\) = 0.08 kgf cm^-2

In Fig. 4.1(b), Thrust = 4 kgf Area of base = 10 cm x 20 cm = 200 cm²Pressure on base (or ground) = \(\frac{4 \text{ kgf}}{200 \text{ cm}^2}\) = 0.02 kgf cm^-2

Thus pressure on ground in Fig. 4.1(b) is one-fourth of pressure in Fig. 4.1(a).

Obviously, larger the area on which a given thrust acts, lesser is the pressure exerted by it.

(2) In Fig. 4.1(b), if another identical brick is placed over the first brick, the thrust gets doubled (= 8 kgf) and since it acts on same area of base (200 cm²), so the pressure on ground becomes = 8 kgf / 200 cm² = 0.04 kgf cm^-2 (i.e., it gets doubled).

Thus larger the thrust acting on a given area, greater is the pressure exerted on it.

Way of increasing pressure: For the given thrust, the pressure on a surface is increased by reducing the area of surface.

Examples: (a) The ends of nails (or pins) are made pointed so that large pressure is exerted at the pointed ends and they can be driven into, with a less effort. (b) The cutting tools also have either sharp (or pointed) edges so that even a small thrust may cause a great pressure at the edges and cutting can be done with a less effort.

Way of decreasing pressure: For the given thrust, the pressure on a surface is reduced by increasing the area of surface.

Examples: (a) Wide wooden sleepers are placed below the railway tracks so that the pressure exerted by the iron rails on the ground becomes less. (b) The foundations of buildings are made wider than the walls so that the pressure exerted by the building on the ground becomes less.

Pressure In Fluids

A substance which can flow is called a fluid. All liquids and gases are, thus, fluids.

A solid exerts pressure on a surface due to its weight. Similarly, a fluid also exerts pressure due to its weight. A solid exerts pressure only on the surface on which it is placed i.e., at its bottom, but a fluid exerts pressure on the bottom as well as on the walls of the container due to its tendency to flow. A fluid, therefore, exerts pressure in all directions. Thus, a fluid contained in a vessel exerts pressure at all points and in all directions.

Experimental demonstration: Take a vessel fitted with a liquid (say, water). Place it on a horizontal surface. Make several small holes in the wall of the vessel anywhere below the free surface of liquid. It is observed that: (1) The liquid spurts out through each hole. This shows that the liquid exerts pressure at each point on the wall of the vessel. (2) If we put our finger on any of the hole, finger feels a thrust due to liquid. This demonstrates that the liquid contained in the vessel exerts thrust at all points below its free surface. Thrust on unit area at a point gives the pressure due to liquid at that point. (3) If we note the distance from the bottom of the vessel to the point where the liquid from a hole strikes on the horizontal surface, it is noticed that as the depth of the hole below the free surface of liquid increases, the thrust of liquid also increases i.e., the horizontal distance reaches to a greater distance on the horizontal surface. This shows that liquid pressure at a point increases with the increase of depth of point from its free surface (Fig. 4.2).

Pressure Exerted By A Liquid Column (P = Hρg)

The pressure exerted by a liquid of density \(\rho\) at a depth \(h\) is \(P = h\rho g\) where \(g\) is the acceleration due to gravity i.e.,

Pressure \(P = h \rho g\) = depth × density of liquid × acceleration due to gravity

Proof: Consider a vessel containing a liquid of density \(\rho\). Let the liquid be stationary. In order to calculate pressure at a depth \(h\), consider a horizontal circular surface PQ of area \(A\) at depth \(h\) below the free surface XY of the liquid (Fig. 4.3).

The pressure on surface PQ will be due to the weight of the liquid column above the surface PQ (i.e., the liquid contained in cylinder PQRS of height \(h\) with PQ as its base and top face RS lying on the free surface XY of the liquid).

The 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 = \((A \times h) \times \rho \times g = Ah\rho g\)

This thrust is exerted on the surface PQ of area \(A\). Therefore, pressure

\(P = \frac{\text{Thrust on surface}}{\text{Area of surface}} = \frac{Ah\rho g}{A} = h\rho g\)

Note: Since there is atmospheric pressure above the free surface of liquid, so to find the total pressure at a depth inside a liquid, it must also be taken into consideration. If the atmospheric pressure acting on the free surface of liquid is \(P_0\) then,

Total pressure in a liquid at a depth \(h\) = Atmospheric pressure + pressure due to liquid column = \(P_0 + h\rho g\)

Factors affecting the pressure at a point in a liquid: From the equation \(P = h\rho g\), it is clear that the pressure at a point inside the liquid depends directly on the following three factors: (i) depth of the point below the free surface (\(h\)), (ii) density of liquid (\(\rho\)), and (iii) acceleration due to gravity (\(g\)).

At a particular place on the earth surface, the acceleration due to gravity \(g\) is constant, therefore, the pressure at a point in a liquid is (i) directly proportional to the depth \(h\) of the point below the free surface of the liquid, and (ii) directly proportional to the density \(\rho\) of the liquid.

However, the pressure inside a liquid does not depend on (i) the shape and size of the vessel in which the liquid is contained, and (ii) the area of surface on which it acts.

Laws Of Liquid Pressure

Following are the five laws of liquid pressure: (i) Inside the liquid, pressure increases with the increase in depth from its free surface. (ii) In a stationary liquid, pressure is same at all points on a horizontal plane. (iii) Pressure is same in all directions about a point in liquid. (iv) Pressure at same depth is different in different liquids. It increases with the increase in density of liquid. (v) A liquid seeks its own level.

Some Consequences Of Liquid Pressure \(P = h\rho g\)

(i) In sea water, the pressure at a certain depth in sea water is more than that at the same depth in river water: The reason is that the density of sea water is more than the density of river water. (ii) The wall of a dam is broader at the bottom: Fig. 4.4 shows the side view of a dam. The thickness of its wall increases from top towards the bottom. The reason is that the pressure exerted by a liquid increases with its depth. Thus as depth increases, more and more pressure is exerted by water on the wall of dam. A thicker wall is required to withstand a greater pressure, therefore, the wall of the dam is made with thickness increasing towards the base. In Fig. 4.4, the increasing length of arrows in water shows the increasing pressure on the wall of dam towards the bottom. (iii) Water supply tank is placed at high: To supply water in a town (or colony), the tank to store water for supply is made at a sufficient height. The reason is that as greater is the height of tank, more will be the pressure of water in the taps of a house. Thus for a good supply of water, the height of the supply tank must always be a few metre higher than the level at which supply of water is to be made. (iv) Diver's suit: 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, made from glass reinforced plastic or cast aluminium. The pressure inside the suit is maintained at one atmosphere. (v) Size of gas bubble inside the water: It is noticed that as the gas bubble formed at the bottom of a lake, rises, it grows in size. The reason is that when the bubble is at the bottom of lake, total pressure exerted on it is the sum of the atmospheric pressure and the pressure due to water column. As the gas bubble rises, due to decrease in depth, the pressure due to water column decreases, so the total pressure exerted on the bubble decreases. According to Boyle's law, the volume of bubble increases due to the decrease in pressure, i.e., the bubble grows in size. When the bubble reaches the surface of liquid, total pressure exerted on it becomes just equal to the atmospheric pressure only which is minimum and so the size of bubble on surface becomes maximum.

Teacher's Note

When observing a swimming pool or diving, notice how divers feel more pressure as they go deeper. This is because pressure increases with depth due to the weight of water above, directly relating to the formula P = hρg in everyday water activities.

Transmission Of Pressure In Liquids: Pascal's Law

We have read that the pressure due to liquid at a point in a liquid of density \(\rho\) at a depth \(h\) below its free surface is \(P = h\rho g\). Obviously, the pressure difference between any two points \(x\) and \(y\) in a stationary liquid will depend only on the difference in vertical height (\(\Delta h\)) between these points. Now if by some means, the pressure at one point \(x\) is increased, the pressure at other point \(y\) must also increase by the same amount so that the difference in pressure between the two points \(x\) and \(y\) may remain same. Thus pressure exerted at a point \(x\) is equally transmitted to the point \(y\). This is Pascal's law. Thus, Pascal's law states that the pressure exerted anywhere in a confined liquid is transmitted equally and undiminished in all directions throughout the liquid.

This can be demonstrated by the following experiment.

Experiment: Take a glass flask having narrow tubes coming out from its sides and bottom. The flask is provided with an air-tight piston at its mouth as shown in Fig. 4.5. Fill the flask with water. The water in each tube will all be at the same level. The initial level of water in each tube is shown by the dotted black line. Now push the piston down into the flask gently. It is observed that jets of water rise out from each tube, reaching the same height which is shown by the upper dotted coloured line. This shows that the pressure applied to the enclosed liquid is transmitted equally in all directions every where inside the liquid.

Teacher's Note

Hydraulic brakes in cars and lifts in hospitals use Pascal's law to multiply force - a small push on the pedal creates enough pressure to stop a large vehicle by being transmitted equally through the fluid system.

Application Of Pascal's Law

Hydraulic machines such as hydraulic press, hydraulic jack and hydraulic brakes are based on Pascal's law of transmission of pressure in liquids. Principle of a hydraulic machine: The principle of each hydraulic machine is that a small force applied on a smaller piston is transmitted to produce a large force on the bigger piston.

Fig. 4.6 shows two cylindrical vessels P and Q connected by a horizontal tube R. The vessels contain a liquid (or water) and they are provided with water-tight pistons A and B. The vessel P is of smaller diameter as compared to the vessel Q. Let area of cross section of the vessel P be A₁ and that of the vessel Q be A₂. When a weight is placed on the piston A, it exerts a force F₁ on the piston A. Therefore the pressure applied on the piston A is

\(P_1 = \frac{F_1}{A_1}\)

According to Pascal's law, the pressure exerted on piston A is transmitted through the liquid to the piston B. This exerts an upward pressure P₂ on the piston B which is equal to P₁. Thus

\(P_2 = P_1\)

If the upward force exerted on piston B is F₂, Then Pressure on piston B is \(P_2 = \frac{F_2}{A_2}\)

From equations, we get \(\frac{F_1}{A_1} = \frac{F_2}{A_2}\) or \(\frac{F_2}{F_1} = \frac{A_2}{A_1}\)

Since \(A_2 > A_1\), therefore \(F_2 > F_1\). Thus a small force F₁ applied on the smaller piston A can be used to produce a large force F₂ on the bigger piston B. This is the principle of a hydraulic machine which acts as a force multiplier.

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