Title 2.Mechanical Properties of Fluids-Part1.pdf ✅

1 SANJIVANI JUNIOR COLLEGE,KOPARGAON CHAPTER NO 2 MECHANICAL PROPERTIES OF FLUID Q.1 What is atmospheric pressure? Ans: The force exerted on unit surface area by the weight of air above it is called atmospheric pressure. Q.2 Write a note on fluids. Ans: i. Any substance that can flow is a fluid. ii. Fluid is a phase of matter that includes liquids, gases and plasma. iii. A fluid is a substance that deforms continually under the action of an external force. iv. The shear modulus of a fluid is zero i.e., fluids are substances which cannot resist any shear force applied to them. Air, water, flour dough, toothpaste, etc., are some common examples of fluids. Molten lava is also a fluid. v. A fluid flows under the action of a force or a pressure gradient. Q.3What is an. incompressible fluid? Ans: An incompressible fluid is a fluid whose density does not change with change in pressure i.e., the density remains constant. Q.4 State the properties of an ideal fluid. Ans: An ideal fluid has the following properties: i. It is incompressible i.e., its density is constant. ii. Its flow is irrotational i.e., its flow is smooth with no turbulences in the flow. iii. It is non-viscous i.e., there is no internal friction in the flow and hence the fluid has no viscosity. iv. Its flow is steady i.e., its velocity at each point is constant in time. Q.5 State the properties of a fluid. Ans: A fluid has the following properties: i. It does not oppose deformation i.e., it gets permanently deformed. ii. It has the ability to flow. iii. It has the ability to take the shape of the container. HYDRODYNAMICS Q.6 Define hydrostatics. Ans: The branch of physics which deals with properties of fluids at rest is called hydrostatics. Q.7 Define pressure of a fluid. State its units and dimensions. Ans : i. The normal force (F) exerted by a fluid at rest per unit surface area (A) of contact is called the pressure (P) of the fluid. P= A F ii. Thus, smaller the surface area A on which the given force F acts, greater is the pressure. iii. Pressure is a scalar quantity. iv. Unit: Nm-2 or Pascal in S.I system Pressure is also measured in following units v. 1 bar= 105 Nm-2 1 vi. hectapascal (hPa) = 100 Pa vii. Dimensions :[L-1M 1 T -2 ] Reading between the lines Some other units of pressure are 1 dyne/cm2 (C.G.S. system) = 0.1 Pa 1 torr = 133.32Pa 760 1bar  Q.8 Derive an expression for pressure exerted by liquid column. On which factors does the pressure due to liquid column depend? Ans: Expression for pressure at a point due to liquid column: i. Consider a vessel filled with liquid of density ‘’Suppose a cylinder having area of cross- section ‘A’ and height ‘h’ is imagined to be inside the vessel as shown in figure.
2 SANJIVANI JUNIOR COLLEGE,KOPARGAON ii. Weight of liquid column exerts a downward force on the bottom of the cylinder, given as, F=mg iii. The pressure exerted by the liquid column on the bottom of the cylinder, p= Area Force P= A mg ...... (2) iv. But m = V where m= mass of liquid, V= volume of the cylinder = density of liquid  M=(Ah) .....(since, V = Ah)  .....(3) v. On comparing equation (2) and (3), we have,  P  hg vi. Equation (4) represents the value of pressure P due to a liquid of density ‘’ and at a depth ‘h’ below the free surface. vii. Thus, pressure due to liquid column depends upon: a. height of liquid column b. density of the liquid and c. acceleration due to gravity. Reading between the lines The pressure exerted due to liquid column is independent of the area of the imaginary cylinder. Shape and size of the container P= hpg is true for liquid as well as gases. Q.9Write a note on atmospheric pressure. Ans: i. The atmospheric pressure at any point is equal to the weight of a vertical column of air of unit cross-sectional area starting from that point and extending to the top of the earth’s atmosphere. ii. For a point on the surface of the Earth, the height of the air column is maximum. Thus, the atmospheric pressure is maximum at the surface of the Earth, i.e., at the sea level, and goes on decreasing as we go above the Earth’s surface. iii. The atmospheric pressure at sea level is called normal atmospheric pressure. iv. Vacuum is the region where air pressure is less than the atmospheric pressure. Perfect or absolute vacuum is when no matter i.e., no atoms molecules are present. Q.10 Derive an expression for absolute pressure P at a depth d below the surface of a liquid and at a height h above the liquid surface. Ans: Expression for absolute pressure P at a depth h below the surface of the liquid: i. Consider a tank filled with liquid (water) of density ‘’. Suppose a cylinder having horizontal base area of cross-section ‘A’ and height ‘h’ is imagined to be inside the tank. Here, h=h1-h2 where h1 and h2 are the heights measured from a reference point and h1> h2. ii. Let P1 and P2 be the pressures of liquid at the point’s h1 and h2 respectively. The liquid cylinder is under the action of following vertical forces: iii. Force due to the weight of the water column above the cylinder, acting vertically downwards on the top surface of the cylinder. F1 =P1A .....(1) iv. Force due to the water below the cylinder, acting vertically upwards on the lower surface of the cylinder. F2= P2A ......(2)
3 SANJIVANI JUNIOR COLLEGE,KOPARGAON v. Weight of the liquid cylinder acting vertically downwards. W=mg ...... (3) But, m =V Where, m= mass of liquid, V = volume of the cylinder, = density of liquid m=A (h1-h2) ...... (  V= Ah) .... (4) vi. As the water is in static equilibrium, the forces on the cylinder are balanced. Therefore, the net force on the cylinder is zero. F1+mg-F2=0 F2=F1+mg ........(5) vii. Substituting equations(1),(2),(3) and (4) in equation (5), P2A=P1A+Ag (h1-h2) P2=P1+g (h1-h2) ...... (6) viii. If the top of the cylinder is shifted to the surface of the liquid which is exposed to the atmosphere, then P1- P0 i.e., atmospheric pressure at the surface h1 =0, h2 = - h = -d (depth below the surface and P2= P) Substituting these values in equation (6), P=P0+dpg Equation (7) represents expression for absolute pressure P at a depth d below the surface of the liquid. ix. Similarly, if the cylinder is above the liquid surface by a distance h, then, P2 = P0 i.e., atmospheric pressure at the surface P1=P, h1=h, h2=O and p= Pair Substituting these values in equation (6), P = P0 - hair . . . (8) Equation (8) represents expression for absolute pressure P at a height h above the surface of the liquid. Q.11 Explain the term gauge pressure. Ans: i. The total pressure P at depth d below the surface of liquid at rest, which is open to atmosphere, is given as, P= Pa + dg ...... (1) Where Pa = atmospheric pressure ii. This shows that the total pressure P is greater than atmospheric pressure by an amount equal to dg. iii. This excess of pressure at depth d in liquid is called gauge pressure. P-Pa=dg .... (2) iv. Thus, gauge pressure at a point in a liquid is the difference between the absolute pressure and the atmospheric pressure. Q.12 Explain the term hydrostatic paradox. i. Consider the inter-connected vessels A, B, C and D as shown in the figure (a). ii. iii. When a liquid is poured in any one of the vessels, it is noticed that the level of liquids in all the vessels is the same. This observation is perplexing because the vessels are of different shapes and hold different amounts of liquid. This is known as ‘hydrostatic paradox’.
4 SANJIVANI JUNIOR COLLEGE,KOPARGAON iv. Due to the shape of vessel C, it can be assumed that the pressure at the base of the vessel C would be more than that at the base of vessel B and the liquid from vessel C would rise into the vessel. However, this assumption is never observed. v. The height of the liquid column is the same for all the vessels. Therefore, the pressure of liquid column in each vessel is the same and the system is in equilibrium. vi. Consider the forces exerted by the walls of the vessel on the liquid. These forces are perpendicular to the walls of the vessel at each point as shown in figure (b). vii. Each of these forces can be resolved into their vertical and horizontal components. The vertical component which acts in the upward direction supports the weight of the liquid above the wall. Thus, only the weight in cylindrical column of the liquid in section B is not balanced and contributes towards the pressure at the base. This fact resolves the hydrostatic paradox. Q.13 State Pascal’s law of fluid pressure. Describe the experimental proof for the same. Ans: i. Statement: The pressure applied at any point of an enclosed fluid at rest is transmitted equally and undiminished to every point of the fluid and also on the walls of the container, provided the effect of gravity is neglected. Explanation: ii. Consider a vessel with four arms A, B, C and D having different cross-sectional areas a1, 2a1, 3a1 and a1/2 respectively filled with incompressible fluid and fitted with frictionless, water tight pistons as shown in figure. iii. Initially, the encleosed water is at rest. Now suppose that the piston A is pushed down with a force F1 so that the pressure exerted by it on the fluid is, PA = 1 1 a F iv. It is found that the pistons B,C and D can be prevented from moving backwards only if forces 2F1, 3F1 and F1/2 are exerted on them respectively. v. The pressure on the pistons B, C and D are, piston B: PB = 1 1 1 1 2a 2F a F  piston B: PC = 1 1 1 1 3a 3F a F  piston D: PD = 1 1 1 1 a / 2 F / 2 a F  i.e., PA = PB=PC =PD vi. This shows that the applied pressure on A is transmitted undiminished on pistons B, C and D as required by Pascal’s law. Q.14 Explain the working of a hydraulic lift. Ans: Hydraulic lift: i. A hydraulic lift which is used to lift or support heavy objects such as cars, trucks etc. works on the basis of Pascal’s law. ii. A tank containing a fluid is fitted with two pistons S1 and S2. S1 has a smaller area of cross section A1 while S2 has a much larger area of cross section A2 (i.e., A2>> A1). iii. Piston S1 is used to exert a force F1 directly on the liquid. The pressure P1 = 1 1 A F is transmitted