11. A horizontal jet strikes a frictionless vertical plate (the plan view is shown in the figure). It is then divided into two parts, as shown in the figure. If the impact loss can be neglected, what is the value of θ ?
    
    

1. 1. 15°
      

   
   
   

   2. 30°
      

   
   
   

   3. 45°
      

   
   
   

   4. 60°

   
   
   

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1. View Hint View Answer Discuss in Forum

   Q2	=	1 + cosα
   Q1		1 - cosα

   
   
   α-angle between plate and jet
   

   ⇒	0.75	=	1 + cosα
   	0.25		1 - cosα

   
   

   ⇒ 3 =	1 + cosα
   	1 - cosα

   
   3 - 3cosα = 1 + cosα
   2 = 4cosα
   

   ⇒ cosα =	1
   	2

   
   ∴ α = 60°
   φ = 90 - α = 30°

   Correct Option: B

   Q2	=	1 + cosα
   Q1		1 - cosα

   
   
   α-angle between plate and jet
   

   ⇒	0.75	=	1 + cosα
   	0.25		1 - cosα

   
   

   ⇒ 3 =	1 + cosα
   	1 - cosα

   
   3 - 3cosα = 1 + cosα
   2 = 4cosα
   

   ⇒ cosα =	1
   	2

   
   ∴ α = 60°
   φ = 90 - α = 30°

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12. A laboratory model of a river is built to a geometric scale of 1 : 00. The fluid used in the model is oil of mass density 900 kg/m³. The heighest flood in the river is 10,000 m³ / s. The corresponding discharge in the model shall be
    

1. 1. 0.95 m³ /s

   
   
   

   2. 0.100 m³ / s

   
   
   

   3. 0.105 m³ / s

   
   
   

   4. 10.5 m³ / s

   
   
   

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1. View Hint View Answer Discuss in Forum

   Apply Froude’s law.
   Density of fluid does not affect model study.
   

   Qr =	Qm	= Lr5 / 2
   	Qp

   
   Q_p = 10000 m³ / s
   

   Lr =	1
   	100

   
   

   ⇒ Qm = 10000 ×		1		5 / 2
   		100

   
   

   = 10000 ×		1		5 = 0.1 m3 / s
   		10

   Correct Option: B

   Apply Froude’s law.
   Density of fluid does not affect model study.
   

   Qr =	Qm	= Lr5 / 2
   	Qp

   
   Q_p = 10000 m³ / s
   

   Lr =	1
   	100

   
   

   ⇒ Qm = 10000 ×		1		5 / 2
   		100

   
   

   = 10000 ×		1		5 = 0.1 m3 / s
   		10

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13. For a two-dimensional irrotational flow, the velocity potential is defined as f = log_e (x² + y²). Which of the following is a possible stream function, Ψ , for this flow?
    

1. 1. 1/ 2 tan^–1 (y – x)

   
   
   

   2. tan^–1 (y / x)

   
   
   

   3. 2 tan^–1 (y / x)

   
   
   

   4. 2 tan^–1 (x / y)

   
   
   

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1. View Hint View Answer Discuss in Forum

   We know,

   ∂Ψ	=	∂φ
   ∂y		∂x

   
   

   =	∂	[ loge (x2 + y2) ]
   	∂x

   
   

   1		d	(x2 + y2)
   x2 + y2		dx

   
   

   =	1	× 2x
   	x2 + y2

   
   Integrating both sides,
   

   ∫∂Ψ = ∫	2x	∂y
   	x2 + y2

   
   

   ⇒Ψ = 2x.	1	tan-1		y		+ c
   	x			x

   
   

   ⇒Ψ = 2tan-1		y		+ c
   		x

   Correct Option: C

   We know,

   ∂Ψ	=	∂φ
   ∂y		∂x

   
   

   =	∂	[ loge (x2 + y2) ]
   	∂x

   
   

   1		d	(x2 + y2)
   x2 + y2		dx

   
   

   =	1	× 2x
   	x2 + y2

   
   Integrating both sides,
   

   ∫∂Ψ = ∫	2x	∂y
   	x2 + y2

   
   

   ⇒Ψ = 2x.	1	tan-1		y		+ c
   	x			x

   
   

   ⇒Ψ = 2tan-1		y		+ c
   		x

   We know,

   ∂Ψ	=	∂φ
   ∂y		∂x

   
   

   =	∂	[ loge (x2 + y2) ]
   	∂x

   
   

   1		d	(x2 + y2)
   x2 + y2		dx

   
   

   =	1	× 2x
   	x2 + y2

   
   Integrating both sides,
   

   ∫∂Ψ = ∫	2x	∂y
   	x2 + y2

   
   

   ⇒Ψ = 2x.	1	tan-1		y		+ c
   	x			x

   
   

   ⇒Ψ = 2tan-1		y		+ c
   		x

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14. A pump can lift water at a discharge of 0.15 m³ /s to a head of 25 m. The critical cavitation number (σ_c) for the pump is found to be 0.144. The pump is to be installed at a location where the barometric pressure is 9.8 m of water and the vapour pressure of water is 0.30 of water. The intake pipe friction loss is 0.40 m. Using the minimum value of NPSH (Net Positive Suction Head), the maximum allowable elevation above the sump water surface at which the pump can be located is
    

1. 1. 9.80 m

   
   
   

   2. 6.20 m

   
   
   

   3. 5.50 m

   
   
   

   4. none of the above

   
   
   

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1. View Hint View Answer Discuss in Forum

   σc =	NPSH
   	H

   
   

   ⇒ 0.144 =	NPSH
   	25

   
   ∴ NPSH = 3.6 m
   NPSH = H_atm - h_v - h_s - h₂
   P_atm = H_atm = 9.8 m of water
   P_v = H_v = 0.3 m of water
   h₂ = 0.4
   ⇒ 3.6 = 9.8 – 0.3 – h_s – 0.4
   ⇒ h_s = 5.5 m

   Correct Option: C

   σc =	NPSH
   	H

   
   

   ⇒ 0.144 =	NPSH
   	25

   
   ∴ NPSH = 3.6 m
   NPSH = H_atm - h_v - h_s - h₂
   P_atm = H_atm = 9.8 m of water
   P_v = H_v = 0.3 m of water
   h₂ = 0.4
   ⇒ 3.6 = 9.8 – 0.3 – h_s – 0.4
   ⇒ h_s = 5.5 m

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15. Velocity distribution in a boundary layer flow over a plate is given by (u/u_m) = 1.5η where, η = y / δ ; y is the distance measured normal to the plate; δ is the is the boundary layer thickness; and u_m is the maximum velocity at y = δ. If the shear stress t, acting on the plate is given by
    τ = K( μu_m )/ δ
    where, μ is the dynamic viscosity of the fluid, K takes the value of
    

1. 1. 0

   
   
   

   2. 1

   
   
   

   3. 1.5

   
   
   

   4. none of the above

   
   
   

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1. View Hint View Answer Discuss in Forum

   u	= 1.5 η = 1.5 ×	y
   um		δ

   
   

   ∴ u = 1.5 ×		y		× um
   		δ

   
   

   du	=	1.5 × um
   dy		δ

   
   

   τ = μ	(1.5 × μm)
   	δ

   
   

   = 1.5	(μ × μm)
   	δ

   
   

   τ = k	(μ.μm)
   	δ

   
   ∴ k = 1.5

   Correct Option: C

   u	= 1.5 η = 1.5 ×	y
   um		δ

   
   

   ∴ u = 1.5 ×		y		× um
   		δ

   
   

   du	=	1.5 × um
   dy		δ

   
   

   τ = μ	(1.5 × μm)
   	δ

   
   

   = 1.5	(μ × μm)
   	δ

   
   

   τ = k	(μ.μm)
   	δ

   
   ∴ k = 1.5

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