Physics

Fluid Mechanics - 1 Session

Session Objectives

Session Objective 1.Pressure depth relationship 2.Pressure depth graph plot 3.Pascal's Law 4.Archimedes's Principle 5.Floating Objects 6.Submerged Objects 7.Pressure difference and buoyant force in accelerating fluids

Pressure Depth Relationship P=P 0 +gh h P0P0 P  Gauge Pressure: gh Absolute pressure: P=P 0 +gh Atmospheric Pressure: P 0

Pressure Depth Graph Plot P=P 0 +gh h P0P0 P  P0P0 h Depth from surface P

Pascal’s Law P 1 = P 2 =P 3 = P 4 =………… F1F1 A1A1 A2A2 F2F2 From Pascal’s Law P1=P2P1=P2 P0P

Questions

Illustrative Example A piston of cross-sectional area 200 cm 2 is used in a hydraulic press to exert a force of 100N on the water. The cross- sectional area of the other piston which supports an object having a mass 400kg is: (a) 4 x 10 3 cm 2 (b) 8 x10 3 cm 2 (c) 800 cm 2 (d) None of the above

Solution

Archimedes’ Principle (Submerged Objects) B W For any floating body: Weight of the body=Buoyant force W=B B= weight of the fluid displaced by the body =Vg Where V = volume of liquid displaced  = density of fluid

Questions

Illustrative Problem Two solids A and B float in same liquid. A floats with half its volume immersed and B floats with 2/3 of its volume immersed. Then densities of A and B are in the ratio: (a) 4:3(b) 3:2 (c) 3:1(d) 3:4

Solution

Floating Objects B W V s < V

Pressure difference and buoyant force in accelerating fluids l AB a0a0 h1h1 h2h2

Class Test

Class Exercise - 1 Three identical vessels A, B and C contain equal mass of three liquids. The vessels contain liquids of respective densities P A, P B and P C such that P A < P B < P C. How will the forces on the bases be related to? (a) Maximum on base of C (b) Maximum on base of B (c) Maximum on base of A (d) Equal on bases of all vessels

Solution Hence answer is (b). As all the masses are same, the force due to each of them will be mg, which are equal. (Remember: Pressure = h  g and with same area F = Ah g = V g = mg)

Class Exercise - 2 Two vessels, as shown, are open vessels. Connected by a siphon arrangement containing water the vertical separation between water levels in the vessels (= XY) is 1 m. What is the pressure difference between X and Y? (Area of the vessel is 0.4 m 2 )

Solution Both points X and Y are open to the atmosphere and so have atmospheric pressure. So the pressure difference is zero.

Class Exercise - 3 Consider the two equations: (a) The first equation is correct but the second is not (b) The second equation is correct but the first is not (c) Both equations are correct (d) Both equations are incorrect

Solution Hence answer is (a). The first is the definition of pressure. In the second, the acceleration is the total acceleration of the elevation. So, where a is the acceleration of the elevation.

Class Exercise - 4 A rectangular block is 5 cm × 5 cm × 10 cm. The block is floating in water with 5 cm side vertical. If it floats with 10 cm side vertical, what change will occur in the level of water? (a) No change (b) It will rise (c) It will fall (d) It may rise or fall depending on the density of the block

Solution Hence answer is (a). The same volume of water is displaced in both cases. So level will not change.

Class Exercise - 5 A body is floating in a liquid with some portion outside the liquid. If the body is slightly pushed down and released, it will (a) start oscillating (b) sink to the bottom (c) remain at the depressed position (d) just comes back to the same position

Solution At equilibrium condition: When pressed down by a small distance x:

Solution Cont. Hence answer is (a). F up (upward force) = Upthrust – Weight Substituting (i) in (ii), Displacement is downward.  Oscillation (SHM) will occur.

Class Exercise - 6 A beaker containing liquid of density moves up with acceleration a. The pressure due to liquid at a depth h below the surface of the liquid is

Solution Hence answer is (b). Using the concept of a liquid in equilibrium Giving pressure difference  P = h  g But if acceleration is upward: Upward force F 2 – (F 1 + Weight) = ma =  Pressure difference (F 2 – F 1 ) =

Class Exercise - 7 A heavy box weighs 20 N in air, 15 N in water and 12.5 N when immersed in a liquid. Calculate the density of the liquid, density of water being 1 g/cc.

Solution Hence answer is (d). W A = Weight in air = 20 N W  = Weight in water = 15 N W = Weight in liquid = 12.5 N R.D. of body w.r.t. water

Class Exercise - 8 Consider a cube of side filled with a liquid of density to a height h. The side faces are vertical. Find the thrust on (a) side face, (b) bottom face of the cube.

Solution (a) Side face Consider an element of width and height dx. Pressure on this element = xg Thrust on this element dF = (x  g)( dx)

Solution Cont. Hence answer is (a). = (Pressure at the centroid of area)(Area of the side face)  Thrust on the vertical face = (Pressure at the centroid of area) × (Area of the vertical face) (b)Bottom face is uniform at all points and pressure is equal to equal h  g.  Thrust on bottom face = h  gL 2

Class Exercise - 9 A liquid is kept in a container, which accelerates horizontally with an acceleration a. What angle will the liquid surface subtend with the horizontal?

Solution As acceleration a is to the right, any horizontal section will have an excess force to right (F 2 – F 1 ) = Mass of liquid element A (So side away from acceleration is higher.)

Class Exercise - 10 A gold ornament weighs 50 g in air, but weighs 45 g in water. The specific gravity of gold is 20. Some cavities are known to be present in the ornament. What is the total volume of the cavities?

Solution Actual volume of gold = Volume of the ornament

Thank you