46. Water flows through a pipe of diameter 0.30 m. What would be the velocity V for the conditions shown in the figure below?
    

1. 1. 79.2 m/s

   
   
   

   2. 11.6 m/s

   
   
   

   3. 27.6 m/s

   
   
   

   4. 1.084 m/s

   
   
   

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1. View Hint View Answer Discuss in Forum

   h = x		1 −	ρmano
   			ρpipe

   
   

   = 0.3		1 −	800
   			1000

   
   = 0.06 m of water
   V₁ = √2gh = √2 × 9.8 × 0.6
   = 1.084 m/s

   Correct Option: D

   h = x		1 −	ρmano
   			ρpipe

   
   

   = 0.3		1 −	800
   			1000

   
   = 0.06 m of water
   V₁ = √2gh = √2 × 9.8 × 0.6
   = 1.084 m/s

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47. In a hand operated liquid sprayer (figure shown below) the liquid from the container rises to the top of the tube because of
    

1. 1. Capillary effect

   
   
   

   2. Suction produced by the air jet at the top end of tube

   
   
   

   3. Suction produced by the piston during the backward stroke

   
   
   

   4. Pumping of the air into the container

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: C

   NA

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48. An incompressible fluid (kinematic viscosity, 7.4 × 10^–7 m²/s, specific gravity, 0.88) is held between two parallel plates. If the top plate is moved with a velocity of 0.5 m/s while the bottom one is held stationary, the fluid attains a linear velocity profile in the gap of 0.5 mm between these plates: the shear stress in Pascals on the surface of top plate is

1. 1. 0.651 × 10^–3

   
   
   

   2. 0.651

   
   
   

   3. 6.51

   
   
   

   4. 0.651 ×10³

   
   
   

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1. View Hint View Answer Discuss in Forum

   Given:
   υ = 7.4 × 10^–7 m² /s
   ρ = 0.88 × 1000 = 880
   v = 0.5 m/s
   
   

   Shear stress,   τ = μ⋅	du	= (υ × ρ) ×	du
   	dy		dy

   
   

   = 7.4 × 10–7 × 800 ×	0.5	= 0.6512.
   	(0.0005)

   Correct Option: B

   Given:
   υ = 7.4 × 10^–7 m² /s
   ρ = 0.88 × 1000 = 880
   v = 0.5 m/s
   
   

   Shear stress,   τ = μ⋅	du	= (υ × ρ) ×	du
   	dy		dy

   
   

   = 7.4 × 10–7 × 800 ×	0.5	= 0.6512.
   	(0.0005)

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49. The velocity profile in fully developed laminar flow in a pipe of diameter D is given by u = u₀ (1– 4r²/D²), where r is the radial distance from the center. If the viscosity of the fluid is ρ, the pressure drop across a length L of the pipe is

1. 1. μ u0L

      D²

   
   
   

   2. 4μ u0L

      D²

   
   
   

   3. 8μ u0L

      D²

   
   
   

   4. 16μ u0L

      D²

   
   
   

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1. View Hint View Answer Discuss in Forum

   By Hagen – Poiseuille law, for steady laminar flow in circular pipes
   

   τ = –μ	∂u
   	∂r

   
   

   τ =	–∂p	.	r
   	∂x		2

   
   

   μ	∂u	=	∂P	⋅	r
   	∂r		∂x		2

   
   

   μu0		−8r		=	P	.	r
   		D²			L		2

   
   

   ....		∵  u = u0		1 −	4r²
   					D²

   
   

   =	–16μLu0
   	D²

   
   [(–) sign is due to drop]

   Correct Option: D

   By Hagen – Poiseuille law, for steady laminar flow in circular pipes
   

   τ = –μ	∂u
   	∂r

   
   

   τ =	–∂p	.	r
   	∂x		2

   
   

   μ	∂u	=	∂P	⋅	r
   	∂r		∂x		2

   
   

   μu0		−8r		=	P	.	r
   		D²			L		2

   
   

   ....		∵  u = u0		1 −	4r²
   					D²

   
   

   =	–16μLu0
   	D²

   
   [(–) sign is due to drop]

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50. Consider steady laminar incompressible axisymmetric fully developed viscous flow through a straight circular pipe of constant crosssectional area at a Reynolds number of 5. The ratio of inertia force to viscous force on a fluid particle is

1. 1. 5

   
   
   

   2. 1/5

   
   
   

   3. 0

   
   
   

   4. ∞

   
   
   

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1. View Hint View Answer Discuss in Forum

   Reynold’s number =	Inertia force	= 5
   	Viscous force

   Correct Option: A

   Reynold’s number =	Inertia force	= 5
   	Viscous force

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