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Physics

Physics. Session. Fluid Mechanics - 1 . Session Objectives. Session Objective. Pressure depth relationship Pressure depth graph plot Pascal's Law Archimedes's Principle Floating Objects Submerged Objects Pressure difference and buoyant force in accelerating fluids.

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Physics

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  1. Physics

  2. Session Fluid Mechanics - 1

  3. Session Objectives

  4. Session Objective • Pressure depth relationship • Pressure depth graph plot • Pascal's Law • Archimedes's Principle • Floating Objects • Submerged Objects • Pressure difference and buoyant force in accelerating fluids

  5. Pressure Depth Relationship P0 h  P Absolute pressure: P=P0+gh Gauge Pressure: gh Atmospheric Pressure: P0 P=P0+gh

  6. P0 P h P=P0+gh P0  P h Depth from surface Pressure Depth Graph Plot P=P0+gh

  7. F1 P0 A2 2 1 A1 F2 3 4 Pascal’s Law P1= P2 =P3= P4=………… From Pascal’s Law P1=P2

  8. Questions

  9. A piston of cross-sectional area 200 cm2 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 103 cm2 (b) 8 x103 cm2 (c) 800 cm2 (d) None of the above Illustrative Example

  10. Solution

  11. 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

  12. Questions

  13. 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 Illustrative Problem

  14. Solution

  15. B W Floating Objects Vs< V

  16. l h1 h2 a0 A B Pressure difference and buoyant force in accelerating fluids

  17. Class Test

  18. Class Exercise - 1 Three identical vessels A, B and C contain equal mass of three liquids. The vessels contain liquids of respective densities PA, PB and PC such that PA < PB < PC. 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

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

  20. 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 m2)

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

  22. 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 Class Exercise - 3

  23. 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. Solution Hence answer is (a).

  24. 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

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

  26. 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

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

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

  29. 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 Class Exercise - 6

  30. Upward force F2 – (F1 + Weight) = ma = \ Pressure difference (F2 – F1) = Solution Using the concept of a liquid in equilibrium Giving pressure difference DP = hrg But if acceleration is upward: Hence answer is (b).

  31. 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.

  32. R.D. of body w.r.t. water R.D. of body w.r.t. water Solution WA = Weight in air = 20 N Ww= Weight in water = 15 N Wl= Weight in liquid = 12.5 N Hence answer is (d).

  33. Consider a cube of sidefilled 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. Class Exercise - 8

  34. Consider an element of width and height dx. Solution (a) Side face Pressure on this element = xrg Thrust on this element dF = (xrg)(ldx)

  35. Solution Cont. = (Pressure at the centroid of area)(Area of the side face) Þ 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 hrg. \ Thrust on bottom face = hrgL2 Hence answer is (a).

  36. 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?

  37. Solution As acceleration a is to the right, any horizontal section will have an excess force to right (F2 – F1) = Mass of liquid element A (So side away from acceleration is higher.)

  38. 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?

  39. Actual volume of gold = Solution Volume of the ornament

  40. Thank you

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