aContents:Definition of a fluidProperties of a fluidHow is a fluid different from a solid?What is the basic difference between liquid and gas?Density & Specific gravityPressure Pascal’s Law and its applicationsPressure Exerted by a Liquid ColumnAtmospheric Pressure10.1 Introduction:Liquids and gases are called fluids. Existence of fluid is very important for all of us. Fluid is all around the earth as it has the gaseous envelope called atmosphere and also two third of the earth is covered by water. The prime the content of our body is water which is a fluid. The physics of fluid is the basis of the hydraulic engineering. The physics of fluid is studied two different branches:(i) Fluid statics or hydrostatics [The branch of Physics which deals with the study of various properties of fluid when a fluid is at rest] and(ii) Fluid dynamics or hydrodynamics [The branch of physics which deals with the study of fluids in motion].There are applications of mechanics of fluid in the field of basic science, engineering and medical science etc.10.2 Definition of a fluid:A fluid is a substance that can flow. It assumes the form of the vessel containing it. Examples: Liquids and gases10.3 Properties of a fluid:(i) Atoms and molecules of a fluid move randomly.(ii) Fluid cannot withstand hearing stress or the force applied tangentially to its surface and this is the reason that a fluid starts moving once such force is applied. (iii) Fluid does not have any definite shape of its own. It takes the shape of the container and it is possible for a fluid because it has no rigidity modulus.(iv) Fluid exerts force in a direction perpendicular its surface.10.4 How a solid is different from a fluid?A solid cannot flow like a fluid. It has a definite shape and size and it cannot be deformed easily as it has definite modulus of rigidity. Atoms and molecules in a crystalline solid are arranged a particular fashion. Atoms and molecules in a solid are strongly bound to each other and cannot move in a random manner independent of each other.10.5 What is the basic difference between Liquid and gas?Liquid is incompressible in nature and hence it has a definite volume and boundary surface of its own whereas gas is compressible i.e., gas can be compressed by applying high pressure and put into a container of a very low volume and the same gas can occupy a very large volume when the pressure is released.10.6 Density and Specific gravityDensity:Mass per unit volume of any substance is called the density of that substance.If the density of any substance does not vary from one place to another (i.e., density is uniform), then density of that substance can be mathematically defined asP=$P=$M

V$\frac{M}{V}$where, M be the mass of the substance and the V be the volume occupied by the substance. Density is a scalar quantity and has a positive value always. Density of a liquid remains constant at any pressure as it is incompressible in nature. Since gas is compressible in nature, its volume changes with pressure and hence the density of the gas varies with pressure.Units: SI unit of density iskgm-3$kg{m}^{-3}$CGS unit of density gcm-3$gc{m}^{-3}$Dimension of density is [ML-3]$\left[M{L}^{-3}\right]$Specific gravity: Specific gravity is a relative density. Specific gravity of a substance is defined as the ratio of the density the substance to the density of water at 4°$}^{\xb0$. aSpecific gravity =density of the substance

density of water at 4°C$\begin{array}{l}\phantom{\rule{0.44em}{0ex}}\\ \phantom{\rule{0.22em}{0ex}}Specific\phantom{\rule{0.22em}{0ex}}gravity\phantom{\rule{0.22em}{0ex}}=\frac{density\phantom{\rule{0.22em}{0ex}}of\phantom{\rule{0.22em}{0ex}}the\phantom{\rule{0.22em}{0ex}}substance}{density\phantom{\rule{0.22em}{0ex}}of\phantom{\rule{0.22em}{0ex}}water\phantom{\rule{0.22em}{0ex}}at\phantom{\rule{0.22em}{0ex}}{4}^{\xb0}C}\phantom{\rule{0.22em}{0ex}}\\ \phantom{\rule{0.66em}{0ex}}\end{array}$Density of water at 4°C is 103 kg/m3$\phantom{\rule{0.22em}{0ex}}{4}^{\xb0}C\phantom{\rule{0.22em}{0ex}}is\phantom{\rule{0.22em}{0ex}}{10}^{3}\phantom{\rule{0.22em}{0ex}}kg/{m}^{3}$10.7 PressureThrust:There is a natural tendency of any fluid to exert force on the surfaces of the vessel containing it. The force exerted by a liquid or gas on any surface in contact is called the thrust (F) and thrust exerted by a liquid is perpendicular to the surface of the container.Pressure: Pressure at any point on the surface is defined as the thrust per unit area around the point. Pressure also acts normally to the surface of the container.If a uniform thrust, F acts on a surface of area A then the pressure is uniform and mathematically pressure can be expressed asP=$P=$F

A$\frac{F}{A}$If the thrust is not uniform then pressure also will not be uniform. In that case the pressure has to be measured at each point on the surface having infinitesimal area 𝛥$\mathit{\Delta}$A. If the thrust at each point on the surface is 𝛥$\mathit{\Delta}$F then pressure at a particular point can be defined asP=lim𝛥A→0𝛥F

𝛥A=dF

dA$P=li{m}_{\mathit{\Delta}A\to 0}\phantom{\rule{0.22em}{0ex}}\frac{\mathit{\Delta}F}{\mathit{\Delta}A}=\frac{dF}{dA}$Pressure is a scalar quantity, as pressure at a point has equal magnitude in all directions.Units and dimension of Pressure:SI unit of pressure is Nm-2$SI\phantom{\rule{0.22em}{0ex}}unit\phantom{\rule{0.22em}{0ex}}of\phantom{\rule{0.22em}{0ex}}pressure\phantom{\rule{0.22em}{0ex}}is\phantom{\rule{0.22em}{0ex}}N{m}^{-2}$also called Pascal [Pa]).Dimension of pressure is [ML-1T-2]$\left[M{L}^{-1}{T}^{-2}\right]$10.8 Pascal’s LawPascal’s law gives an idea how pressure is transmitted from one point to the in an incompressible fluid.Pascal’s law states that ‘a change in pressure applied to an enclosed incompressible fluid is transmitted in undiminished to every point of the fluid and to the wall of the container’.Experimental Evidences of the Pascal’s Law:(a) Let us take a rubber ball and fill that with water completely such that there is no air inside the ball. Now if a numbers of holes are pierced identically on the surface of the ball and apply pressure by hand as shown in the figure-10.1(a), water will come out through all the holes with equal force. This happens because the pressure applied on the confined water has transmitted to each point on the wall without any change.

Figure-10.1(a)

(b) Let us consider a spherical vessel with four cylindrical openings A, B, C and D. These openings A, B, C, and D are having cross-sectional area A, A/2, 2A and 3A respectively and they are fitted with frictionless moveable and air tight as well as water tight pistons. Suppose, the spherical container is filled with water such that no air is there between the water and pistons. If an additional force, Fis applied to the piston fitted with the opening A. It is observed that to keep the pistons B, C and D at their original positions, respectively F/2, 2F and 3F forces are to be applied externally. This is another experimental evidence Pascal’s law.

Figure-10.1(b)

Applications of Pascal’s Law:Hydraulic Lift and Hydraulic brakes of automobiles work on Pascal’s law.(a) Hydraulic Lift:Hydraulic lift is used to lift heavy objects. It acts as a force multiplier. A hydraulic lift consists of two cylindrical arms C1$}_{1$ and C2$}_{2$ which are connected to each other by a pipe. Area of cross-section of the arm C1$}_{1$ is A1$}_{1$ and that of the arm C2$}_{2$ is A2$}_{2$ and A2>A1$A}_{2}>{A}_{1$. Both the arms are fitted with frictionless moveable water and air tight pistons. Both arms and the connecting pipe are filled with liquid. Suppose a force F1$}_{1$ is applied at the smaller piston fitted to the arm C1$}_{1$.Then the pressure exerted on the liquid is P1=F1

A1$P}_{1}=\frac{{F}_{1}}{{A}_{1}$According to the Pascal’s law, the same pressure,P1$P}_{1$ will be transmitted to the larger piston of cross-sectional area A2$A}_{2$ through the incompressible liquid. Therefore the force on the larger piston will beF2=(P1×A2)=a

F1

A1×A2

. Since A2>A1, F2>F1.${F}_{2}=({P}_{1}\times {A}_{2})=\left(\begin{array}{c}\frac{{F}_{1}}{{A}_{1}}\times {A}_{2}\\ \end{array}\right).\phantom{\rule{0.22em}{0ex}}Since\phantom{\rule{0.22em}{0ex}}{A}_{2}>{A}_{1},\phantom{\rule{0.22em}{0ex}}{F}_{2}>{F}_{1}.$Therefore, making the factor,A2

A1$\frac{{A}_{2}}{{A}_{1}}$ very large, a large force can be obtained even by applying considerably less amount of force, which enablesus to lift very heavy load like lifting a car or truck or very big boxes etc.

Figure-10.2(a): Hydraulic lift lifting a car

Is energy law of conservation violated in case of hydraulic lift?No, the law of conservation of energy is not violated in hydraulic lift. Since the liquid is incompressible in nature, the volume of the liquid that is displaced by smaller piston is the same as that of the volume of the liquid displaced by the larger piston. Therefore there will be no gain of work i.e., the work done the force F1$F}_{1$ will be equal to the work done by the force F2$F}_{2$. Let us prove that the work done by both the pistons is the same.Suppose the smaller piston or the input piston is displaced downwards by a distance d1$d}_{1$ and the larger piston or the output piston is displaced upwards by a distance d2$d}_{2$ such that the volume of the incompressible liquid displaced in both the cylinders is the same. Therefore, aA1d1=A2d2....................................(10.1)Now work done by force F1 is given by W1=F1d1=a

P1

A1×d2

..................(10.2)Now work done by force F2 is given byW2=F2d2=a

P1

A2×d2

=a

P1×A1d1

d2

×d2=P1×A1d1=W1(Using equation 10.1)$\begin{array}{l}{A}_{1}{d}_{1}={A}_{2}{d}_{2}....................................(10.1)\\ \\ Now\phantom{\rule{0.22em}{0ex}}work\phantom{\rule{0.22em}{0ex}}done\phantom{\rule{0.22em}{0ex}}by\phantom{\rule{0.22em}{0ex}}force\phantom{\rule{0.22em}{0ex}}{F}_{1}\phantom{\rule{0.22em}{0ex}}is\phantom{\rule{0.22em}{0ex}}given\phantom{\rule{0.22em}{0ex}}by\phantom{\rule{0.22em}{0ex}}\\ \\ {W}_{1}={F}_{1}{d}_{1}=\left(\begin{array}{c}\frac{{P}_{1}}{{A}_{1}}\times {d}_{2}\\ \end{array}\right)..................(10.2)\\ \\ Now\phantom{\rule{0.22em}{0ex}}work\phantom{\rule{0.22em}{0ex}}done\phantom{\rule{0.22em}{0ex}}by\phantom{\rule{0.22em}{0ex}}force\phantom{\rule{0.22em}{0ex}}{F}_{2}\phantom{\rule{0.22em}{0ex}}is\phantom{\rule{0.22em}{0ex}}given\phantom{\rule{0.22em}{0ex}}by\\ \\ {W}_{2}={F}_{2}{d}_{2}=\left(\begin{array}{c}\frac{{P}_{1}}{{A}_{2}}\times {d}_{2}\\ \end{array}\right)=\left(\begin{array}{c}{\frac{{P}_{1}\times {A}_{1}{d}_{1}}{{d}_{2}}}_{}\\ \end{array}\right)\times {d}_{2}={P}_{1}\times {A}_{1}{d}_{1}={W}_{1}\phantom{\rule{0.22em}{0ex}}(Using\phantom{\rule{0.22em}{0ex}}equation\phantom{\rule{0.22em}{0ex}}10.1)\end{array}$Thus it is proved that the work done by input piston is equal to the work done by the output piston and hence law of conservation of energy is satisfied.(b) Hydraulic Brakes:Hydraulic brakes used in the automobiles are based on the Pascal’s law of transmission of pressure through liquid. Construction: A hydraulic brake consists of a tube filled with brake oil. At one end of the tube a master piston P is connected to the brake paddle through lever system and the other end of the tube is connected to the wheel cylinder having two pistons B1$}_{1$ and B2$}_{2$. The pistons B1$}_{1$ and B2$}_{2$ are connected to the brake shoes S1$}_{1$ and S2$}_{2$. The area of cross-section of the wheel cylinder is greater than that of the master cylinder. The construction of a hydraulic brake is shown in the Figure-10.2(b).

Figure-10.2(a): Schematic diagram of Hydraulic brake of an automobile.

Working Principle:When the pedal is pressed, the piston P is pushed into the master cylinder by the lever action of the lever system connected to it. According to Pascal’s law, the pressure is transmitted to wheel cylinder through the brake oil and pushed out the pistons B1$B}_{1$ and B2$B}_{2$. The brake shoes get pressed against the inner rim of the wheel by these pistons and the motion of the wheel is retarded. Since the area of the pistons at the wheel cylinder has larger area of cross-section, even a small force applied to the brake paddle produces a larger force which decelerates the automobile. When the pressure on paddle is released, the brake shoes are pulled off the rim of the wheel by a spring connected to the brake shoes and the pistons in both the cylinders moves to their normal positions.10.9 Pressure exerted by a Liquid columnLet us consider a vessel of height ‘h’ and area of cross-section ‘A’ which is filled with a liquid of density 𝜌$\mathit{\rho}$. The liquid column exerts a pressure at the bottom of the vessel. This pressure can be calculated from the downward thrust that the weight of the liquid exerts on bottom face of the vessel. The vessel is shown in the figure-10.3.

Figure-10.3: Pressure exerted by a liquid column

Calculation of the pressure: The weight of the liquid column in the vessel isW=V×𝜌×g=Ah𝜌g$W=V\times \mathit{\rho}\times g=Ah\mathit{\rho}g$Therefore, the pressure exerted by the liquid column on the bottom of the vessel is given byP=W

A=h𝜌g.$P=\frac{W}{{A}_{}}=h\mathit{\rho}g.$10.10 Atmospheric PressureEnvelop of gases that surrounds the earth or any other planet is called atmosphere. Here we will discuss the earth atmosphere.The pressure exerted by the atmosphere is called atmospheric pressure. The atmospheric pressure at the sea level is1.013×105 Nm-2 orPa.$1.013\times {10}^{5}\phantom{\rule{0.22em}{0ex}}N{m}^{-2}\phantom{\rule{0.22em}{0ex}}orPa.$Height of the atmosphere:To calculate the height of the atmosphere, we need to make following two assumptions.(a) Air has a uniform density of 𝜌=1.3kgm-3$\mathit{\rho}=1.3kg{m}^{-3}$ , (Actually density decrease as height increases)(b) Temperature remains uniform throughout(c) The value of g$g$ does not change appreciably up to certain height.Now if the height of the air column is h$h$ , thenhpg=1.013Nm-2$hpg=1.013N{m}^{-2}$⇒ h=1.013Nm-2

𝜌g=1.013Nm-2

1.3×9.8=7951m ≈8km$\Rightarrow \phantom{\rule{0.22em}{0ex}}h=\frac{1.013N{m}^{-2}}{\mathit{\rho}g}=\frac{1.013N{m}^{-2}}{1.3\times 9.8}=7951m\phantom{\rule{0.22em}{0ex}}\approx 8km$Therefore the height of the atmosphere is approximately 8 km.

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