Title Notes_Exercise_13.Fluid Mechanics.pdf ✅

13.1 SOLUTIONS TO CONCEPTS CHAPTER 13 1. p = h  g It is necessary to specify that the tap is closed. Otherwise pressure will gradually decrease, as h decrease, because, of the tap is open, the pressure at the tap is atmospheric. 2. a) Pressure at the bottom of the tube should be same when considered for both limbs. From the figure are shown, pg + Hg × h2 × g = pa + Hg × h1 × g  pg = pa + Hg × g(h1 – h2) b) Pressure of mercury at the bottom of u tube p = pa + Hg h1 × g 3. From the figure shown pa + hg = pa + mg/A  hg = mg/A  h = Ap m 4. a) Force exerted at the bottom. = Force due to cylindrical water colum + atm. Force = A × h × w × g + pa × A = A(h w g + pa) b) To find out the resultant force exerted by the sides of the glass, from the freebody, diagram of water inside the glass pa × A + mg = A × h × w × g + Fs + pa × A  mg = A × h × w × g + Fs This force is provided by the sides of the glass. 5. If the glass will be covered by a jar and the air is pumped out, the atmospheric pressure has no effect. So, a) Force exerted on the bottom. = (h w g) × A b) mg = h × w × g × A × Fs. c) It glass of different shape is used provided the volume, height and area remain same, no change in answer will occur. 6. Standard atmospheric pressure is always pressure exerted by 76 cm Hg column = (76 × 13.6 × g) Dyne/cm2 . If water is used in the barometer. Let h  height of water column.  h × w × g 7. a) F = P × A = (h w × g) A b) The force does not depend on the orientation of the rock as long as the surface area remains same. 8. a) F = A h  g. b) The force exerted by water on the strip of width x as shown, dF = p × A = (xg) × A c) Inside the liquid force act in every direction due to adhesion. di = F × r d) The total force by the water on that side is given by F =  1 0 20000 xx  F = 20,000 1 0 2 x[ / ]2 e) The torque by the water on that side will be, Pa Pa Gas Pa h pa 45 kg A =900 cm2 Page 1 UNIT & DIMENSION FLUID MECHANICS
Chapter-13 13.2 i =  1 0 20000 xx (1 – x)  20,000 1 0 2 3 x[ / 2  x / ]3 9. Here, m0 = mAu + mcu = 36 g ...(1) Let V be the volume of the ornament in cm3 So, V × w × g = 2 × g  (Vau + Vcu) × w × g = 2 × g  g m m w au au              = 2 × g  1 9.8 m 19 3. mAu Au         = 2  8.9 mAu + 19.3 mcu = 2 × 19.3 × 8.9 = 343.54 ...(2) From equation (1) and (2), 8.9 mAu + 19.3 mcu = 343.54  m .2 225g (9.8 m m ) 9.8 36 cu Au cu     So, the amount of copper in the ornament is 2.2 g. 10. V g M c w Au Au             = 2 × g (where Vc = volume of cavity) 11. mg = U + R (where U = Upward thrust)  mg – U = R  R = mg – v w g (because, U = vwg) = mg –  m × w × g 12. a) Let Vi  volume of boat inside water = volume of water displace in m3 . Since, weight of the boat is balanced by the buoyant force.  mg = Vi × w × g b) Let, v1  volume of boat filled with water before water starts coming in from the sides. mg + v1 w × g = V × w × g. 13. Let x  minimum edge of the ice block in cm. So, mg + Wice = U. (where U = Upward thrust)  0.5 × g + x3 × ice × g = x3 × w × g 14. Vice = Vk + Vw Vice × ice × g = Vk × k × g + Vw × w × g  (Vk + Vw) × ice = Vk × k + Vw × w  1 V V k w  . 15. Viig = V w g 16. (mw + mpb)g = (Vw + Vpb)  × g  (mw + mpb) =             pb pb w mw m 17. Mg = w  (mw + mpb)g = Vw ×  × g 18. Given, x = 12 cm Length of the edge of the block Hg = 13.6 gm/cc Given that, initially 1/5 of block is inside mercuty. Let b  density of block in gm/cc.  (x)3 × b × g = (x)2 × (x/5) × Hg × g  123 × b = 122 × 12/5 × 13.6  b = 5 13 6. gm/cc Page 2 UNIT & DIMENSION FLUID MECHANICS