Title 11. Fluid Mecha Hard Ans.pdf ✅

1. (c) Upthrust on ball = weight of displaced water m V g g      = =     40 1 50 0.8 =   = g g Dyne or 50 gm As the ball is sunk into the water with a pin by applying downward force equal to upthrust on it. So reading of weighing pan = weight of water + downward force against upthrust = 600 + 50 = 650 gm. 2. (b) For streamline flow Reynold's number R r N    should be less. For less value of NR , radius and density should be small and viscosity should be high. 3. (c) If the liquid is incompressible then mass of liquid entering through left end, should be equal to mass of liquid coming out from the right end. ∴ M = m1 + m2  Av1 = Av2 + 1.5A. v  A × 3 = A × 1.5 + 1.5A. v  v = 1 m/s 4.(d)This happens in accordance with equation of continuity and this equation was derived on the principle of conservation of mass and it is true in every case, either tube remain horizontal or vertical. 5. (d) 1 1 2 2 a v a v =  2 2 4.20 5.18 7.60 2.86 /  =   = v v m s 6. (a) Velocity head 2 2 (5) 1.25 2 2 10 v h m g = = =  7. (d) As cross-section areas of both the tubes A and C are same and tube is horizontal. Hence according to equation of continuity A C v v = and therefore according to Bernoulli's theorem P P A C = i.e. height of liquid is same in both the tubes A and C. 8. (b) From Bernoulli's theorem, 1 1 2 2 2 2 P dv dgh P dv dgh A A A B B B + + = + + Here, A B h h = 1 1 2 2 2 2  + = + P dv P dv A A B B  1 2 2 [ ] 2 P P d v v A B B A − = − Now, 0, A B v v r = =  and P P hdg A B − =
1 2 2 2  = hdg dr  or 2 2 2 r h g  = 9. (b) Bernoulli's theorem for unit mass of liquid 1 2 2 P v  + = Constant As the liquid starts flowing, it pressure energy decreases 2 1 2 1 2 P P v  − = 5 5 5 2 2 3 3 1 3.5 10 3 10 2 0.5 10 2 10 10 v v  −     =  = 2  =  = v v m s 100 10 / 10. (c) Time taken for the level to fall from H to H ' 0 2 ' A t H H A g = −     According to problem- the time taken for the level to fall from h to 2 h 1 0 2 2 A h t h A g   = −     and similarly time taken for the level to fall from 2 h to zero 2 0 2 0 2 A h t A g   = −     1 2 1 1 2 2 1. 1 0 2 t t −  = = − − 11. (b) v gh m s = =   = 2 2 10 20 20 / 12. (c) Time required to emptied the tank 0 A H2 t A g =  2 2 1 1 4 2 t H h t H h = = = 2  = t t 2 13. (c) Let A = The area of cross section of the hole, v = Initial velocity of efflux, d = Density of water, Initial volume of water flowing out per second = Av Initial mass of water flowing out per second = Avd A r  B h
Rate of change of momentum = Adv2  Initial downward force on the out flowing water = Adv2 So equal amount of reaction acts upwards on the cylinder.  Initial upward reaction = 2 Adv [As v gh = 2 ]  Initial decrease in weight = Ad gh (2 ) = 2Adgh 1 2 1 980 25 12.5 4   =     =     gm-wt. 14. (a) The height of water in the tank becomes maximum when the volume of water flowing into the tank per second becomes equal to the volume flowing out per second. Volume of water flowing out per second = Av = A gh 2 and volume of water flowing in per second 3 = 70 / sec. cm  A gh gh h 2 70 1 2 70 1 2 980 70 =   =     =  4900 2.5 . 1960 h cm = = 15. (d) 2 2 A m = = (0.1) 0.01 ,  = = 0.01 0.001 Poise decapoise (M.K.S. unit), dv = 0.1 m/s and F = 0.002 N dv F A dx = 0.001 0.01 0.1 0.0005 0.002 Adv dx m F     = = = . 16. (d) When we move from the centre to the circumference the velocity of liquid goes on decreasing and finally becomes zero 17. (b) F rv = 6 18. (b) In the first 100 m body starts from rest and its velocity goes on increasing and after 100 m it acquire maximum velocity (terminal velocity). Further, air friction i.e. viscous force which is proportional to velocity is low in the beginning and maximum at T v v = . Hence work done against air friction in the first 100 m is less than the work done in next 100 m. 19. (c) If two drops of same radius r coalesce then radius of new drop is given by R 4 4 4 3 3 3 3 3 3    R r r = +  3 3 1/3 R r R r =  = 2 2 If drop of radius r is falling in viscous medium then it acquire a critical velocity v and 2 v r 
2 2 1/3 2 1 v R r 2 v r r     = =          2/3 2/3 1/3 2 1 v v m s =  =  =  2 2 (5) 5 (4) / 20. (c) Velocity of ball when it strikes the water surface v gh = 2 ..........(i) Terminal velocity of ball inside the water ( ) 2 2 1 9 v r g   − = ..........(ii) Equating (i) and (ii) we get 2 2 2 ( 1) 9 r g gh   = −  2 2 1 4 81 h r g     − =     21. (b) Rate of flow of liquid P V R = where liquid resistance 4 8 l R r   = For another tube liquid resistance 4 4 8 8 ' .16 16 2 l l R R r r     = = =       For the series combination ' 16 17 New P P P V R R R R R = = = + + 17 V = . 22. (d) From 4 8 P r V l   =  4 V l 8 P r   =  4 2 2 2 1 1 1 1 2 P V l r P V l r   =      4 1 1 2 2 2 4   =   =      1 2 4 4 P P P = = . 23. (c) For parallel combination 1 2 1 1 1 R R R eff = +