Title FLUID MECHANICS.pdf ✅

NEET : Physics [ 228 ] www.allendigital.in  Digital Fluid Fluids are the substances that can flow or deforms. Therefore, liquids and gases both are fluids. Study of fluids at rest is called fluid statics or hydrostatics and the study of fluid in motion is called fluid dynamics or hydrodynamics. Fluid statics and fluid dynamics collectively known as fluid mechanics. The intermolecular forces in liquids are comparatively weaker than in solids. Therefore, their shapes can be changed easily. When external force (shear stress) is present, liquid can flow until it conforms to the boundaries of its container. Most liquids resist compression. Unlike a gas, a liquid does not disperse to fill every space of a container and it forms a free surface. The intermolecular forces are weakest in gases, so their shapes and sizes can be changed much easily. Gases are highly compressible and occupy the entire space of the container quite rapidly. Unlike liquid, gases can't form free surface. (i) Density Mass Density Volume =  M V = Units - (i) 3 kg m (SI system) (ii) 3 gm cm (CGS system) Note: (i) water = 1000 kg/m3 = 1 gm/cc (ii) Hg = 13600 kg/m3 = 13.6 gm/cc Density of mixture If two immiscible liquids of mass m1 and m2 and volume V1 and V2 are mixed together then density of mixture is given by : mix Total mass of mixture Total volume of mixture  = 1 2 1 2 1 2 1 1 1 2 m m m m v v m m + + = = + +   Case (1) : If liquid with same masses are mixed i.e. m1 =m2 =m then - 1 2 mix 1 2 2   =  +  (Harmonic mean of individual densities) Case (2) : If liquid with same volumes are mixed i.e. V1 = V2 = V then - 1 2 mix 2  +   = (Arithmetic mean of individual densities) Illustration 15: Two immiscible liquids having density 2 gm/cc and 4 gm/cc are mixed then find density of mixture if - (1) Same volumes are taken (2) Same masses are taken Solution: (1) 1 2 mix 2 4 3gm/cc 2 2  +  +  = = = (2) 1 2 mix 1 2 2 2(2)(4) 8 gm/cc 2 4 3    = = =  +  + Mechanical Properties of Fluids Fluid Mechanics Fluid Statics Fluid dynamics
Properties of Matter and Fluid Mechanics  Digital www.allendigital.in [ 229 ] Illustration 16: Density of mixture is 4 gm/cc when equal volumes are taken and 3 gm/cc when equal masses are taken, then find density of individual liquid. Solution: Let the density of liquids are 1 and 2 If equal volumes are taken, 1 2 4 8 1 2 2  +  =   +  = ...(i) If equal masses are taken, 1 2 1 2 1 2 2 3 12   =    =  +  ...(ii) By solving both equations, 1 = 2 'or' 6 gm/cc ; 2 = 6 'or' 2 gm/cc (ii) Relative density: Density of substance R.D. Density of pure water at 4° C = R.D. is an unitless and dimensionless quantity (iii) Specific weight: Weight of substance mg Specific weight Volume of substance V = = = g unit → N/m3 (iv) Specific Gravity: Specific weight of substance S.G. Specific weight of pure water at 4°C = w w g R.D. g   = = =   Note: The numerical value of specific gravity and relative density are same. Pressure and Pressure due to Liquid Column Pressure P is defined as the magnitude of the normal force acting per unit surface area. F P , A ⊥ = here F⊥ = normal force on a surface of area A SI UNIT : Pascal (Pa) ; 1 Pa = 1 N/m2 Dimensions : [ML–1T–2] Practical units : atmospheric pressure (atm), bar and torr . 1 atm = 1.01325 × 105 Pa = 1.01325 bar = 760 torr = 760 mm of Hg = 10.33 m of water 1 bar = 105 Pa ; 1 torr = 1 mm of mercury column = 133 Pa. Pressure is a scalar quantity. Illustration 17: A 50 kg girl wearing heel shoes balances on a single heel. The heel is circular with a diameter 1 cm. The pressure exerted by the heel on the horizontal floor is (Take g = 10 m s–2) (1) 6.4 × 104 Pa (2) 6.4 × 105 Pa (3) 6.4 × 106 Pa (4) 6.4 × 107 Pa Solution: (1) Here, m = 50 kg, D = 1 cm = 10–2 m, g = 10 ms–1  Pressure exerted by the heel on the horizontal floor is 2 2 F mg 4mg p A (D/2) D = = =   2 2 2 4 50kg 10ms 3.14 (10 m) − −   =  = 6.4 × 106Pa A F
NEET : Physics [ 230 ] www.allendigital.in  Digital Types of Pressures Pressure is of three types : (i) Atmospheric pressure (Po/Patm) (ii) Gauge pressure (Pgauge) (iii) Absolute pressure (Pabs.) (1) Atmospheric Pressure Force exerted by atmospheric column on unit cross-sectional area at mean sea level is called atmospheric pressure (P0). Po = 101.3 kN/m2  Po = 1.013×105 N/m2 Note: Barometer is used to measure the atmospheric pressure. (2) Gauge Pressure Excess Pressure over the atmospheric pressure (P–Patm) measured with the help of pressure measuring instruments is called gauge pressure. Note: Gauge pressure is always measured with the help of a "manometer". (3) Absolute Pressure :- Sum of the atmospheric and gauge pressure is called absolute pressure. Pabs = Patm + Pgauge Pabs = Po + hg Pressure due to liquid at depth 'h' m = mass of liquid element A = Area of cross - section of liquid element Since fluid is at rest  P1A + mg = P2A  P1A + (Ah)g = P2A  P2 = P1 + gh Note: Pressure exerted by a same liquid at any point does not depend on shape and size of the container (it means quantity of liquid). It depends only on the height of liquid column. Pressure due to liquid, PA = PB = PC = PD Pressure due to liquid on a vertical wall of container Pressure due to liquid on a vertical wall is different at different depths, so average fluid pressure on side wall of the container is equal to mean pressure = gh 2  (h = height of wall) A B C D h  h P2A P1A mg
Properties of Matter and Fluid Mechanics  Digital www.allendigital.in [ 231 ] Problems based on Pressure due to Liquid Column and Barometer Illustration 18: A cuboid (a × a × 2a) is filled with a liquid of density '' as shown in figure. Neglecting atmospheric pressure, find out (a) Force on base wall of the cuboid (b) Force on side wall of the cuboid Solution: (a) Force on base wall = Pbase wall × Abase wall = g(2a) × a2 = 2ga3 (b) Force on side wall = Pside wall × Aside wall = 0 g(2a) 2 2a 2   +       = 2ga3 Illustration 19: Find the force acting on the piston of 3 cm2 at point 2 due to the water column of height 10 m. Solution: F = PA = ghA = 103 × 10 × 10 × 3 × 10–4 = 30 N Illustration 20: A tube 1 cm2 in cross-section is attached to the top of a vessel 1 cm high and of cross-section 100 cm2. Water is poured into the system filling it to a depth of 100 cm above the bottom of the vessel as shown in Fig. Take g = 10 ms–2. Solution: P h g 99 1 1 100 =  = +   ( ) = 105 dyne cm–2 F = 105 × 100 = 107 dyne = 100 N V = (1×99) + (1×100) = 199 cm3 W = mg = V.g = (1×199×1000) dyne = 1.99 N Illustration 21: A tank 5 m high is half filled with water and then is filled to the top with oil density 0.85 g-wt/cm3. The pressure at the bottom of the tank, due to these liquids is. Solution: Pressure at the bottom, 1 1 2 2 2 g wt P (h d h d ) cm − = + = [250 × 1 + 250 × 0.85] 2 2 g wt g wt 250[1.85] 462.5 cm cm − − = = 10 m 1 2 a a 2a  1 cm2 100 cm2 99 cm 1 cm Cross sectional area