Tiêu đề Med-RM_Phy_SP-2_Ch-10-Mechanical Properties of Fluids.pdf ✅

Chapter Contents Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph. 011-47623456 Introduction Pressure Archimedes’ Principle Streamline Flow Bernoulli’s Principle Viscosity Surface Tension Chapter 10 Mechanical Properties of Fluids PRESSURE Fluid Thrust The total normal force exerted by a fluid at rest on a given surface in contact with it is called thrust of the fluid on that surface. Fluids in Equilibrium When a fluid is in equilibrium, forces acting on its surface are always normal to its surface. Pressure Exerted by a Fluid Column : It is defined as normal force per unit area of a surface. F P A  unit : N/m2 or pascal (Pa) Relative Density or Specific Gravity: The relative density or specific gravity of a substance is defined as the ratio of the density of the substance to the density of water at 4oC. The density of water at 4oC is 1.0 × 103 kg m–3. Density of substance Relative density = Density of water at 4°C It has no unit. Pascal’s Law : If external pressure is applied h P h h on any part of an enclosed incompressible fluid, then it will transmit undiminished and equally, to every point of the fluid and the walls of the containing vessel. Introduction Fluid means ‘to flow’. Both liquids and gases can flow, and together, they are called as fluids. Although both can flow, but liquids can be distinguished from gases, as they are approximately incompressible and have a definite volume.
62 Mechanical Properties of Fluids NEET Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph. 011-47623456 Three machines works on principle of Pascal’s law (i) Hydraulic lift (ii) Hydraulic brake (iii) Hydraulic press Variation of Pressure with Depth P P gh 2 1 –   ...(i) If we shift the point 1 to the top of the liquid surface P2 P1 h 1 2 which is open to the atmosphere, P1 becomes the atmospheric pressure Pa. Taking P2 as P then, equation (i) becomes P – Pa = gh P P gh   a ...(ii) The excess pressure (P – Pa), at depth h is called a gauge pressure at that point and P is called absolute pressure at that point. Hydrostatic Paradox The liquid pressure is same at all points at the same horizontal level (same depth) in same liquid although they hold different amounts of liquid. This result is known as hydrostatic paradox. Container is in motion (i) When liquid is at rest or move with constant velocity or in vertical acceleration pressure at same level, with in the same liquid is same PB – PA = hg A C B L h Stationary or moving with constant velocity PC – PB = 0 (ii) If liquid is accelerated horizontally with acceleration a. PB – PA = hg A C B L h a PC – PB = La If container is open at top then angle of surface with horizontal, 1 tan a g          Hydraulic Lift : For equililbrium of the weight W, pressure at M should be equal to pressure at N. W F h g A a  
NEET Mechanical Properties of Fluids 63 Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph. 011-47623456 Area A Area = a h F W N M Equilibrium of different liquids in U-tube: When liquid is at rest or move with constant velocity or in vertical acceleration pressure at same level, with in the same liquid is same In equilibrium P1 = P2 h2 h1 1 2 2 1 2 Pa + h11g = Pa + h22g h11g = h22g Absolute pressure at point P2 P1 h1 P2 Pressure at P2 = Pressure at P1 = Pa + h1g When the U-tube accelerates horizontally, difference of levels of liquid in two arms. P2 – P1 = La h a L  2 1 P2 = P1 + La (Pa + hg) – Pa = La h a L g  tan a g   Example 1 : Find the pressure exerted below a column of water, open to the atmosphere, at depth (i) 5 m (ii) 20 m (Given, density of water = 1 × 103 kg m–3, g = 10 m s–2) Solution : (i) Pressure at a depth of 5 m P = Pa + gh
64 Mechanical Properties of Fluids NEET Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph. 011-47623456 = 1.013 × 105 Pa + (1 × 103 kg m–3)(10 m s–2)(5 m) = 1.013 × 105 Pa + 0.5 × 105 Pa = 1.513 × 105 Pa (ii) Pressure at a depth of 20 m P = Pa + gh = 1.013 × 105 Pa + (1 × 103 kg m–3)(10 m s–2)(20 m) = 1.013 × 105 Pa + 2 × 105 Pa = 3.013 × 105 Pa  3 atm Example 2 : The U-tube in figure contains two different liquids in static equilibrium, water in the right arm and oil of unknown density x in the left. If l = 135 mm and d = 15 mm, what is the density of the oil? Water Oil l d Solution : If the pressure at the oil-water interface in the left-arm is P, then the pressure in the right arm at the level of the interface will also be P. In the left arm, P = P0 + xg(l + d) ...(i) In the right arm, P = P0 + wgl ...(ii) Equating (i) and (ii) 3 kg/m1000 mm 15 mm135 mm135     x  w dl l = 900 kg/m3 Example 3 : Prove that the average pressure of a liquid (density ) on the walls of the container filled upto height h with the liquid is 2 1 hg. Solution : Let us consider a cylindrical container of radius r, filled with a liquid of density  upto height h. Take a small area of the wall of the container between the depths x and (x + dx). This area is dA = 2rdx. The pressure on this area due to the liquid is P = gx  The force on this area is