91. The instantaneous stream wise velocity of a turbulent flow is given as follows: u(x, y, z, t) = u(x, y, z) + u'(x, y, z, t) The time average of the fluctuating velocity u (x, y, z, t) is

1. 1. 	u'
      	2

   
   
   

   2. -	u
      	2

   
   
   

   3. zero

   
   
   

   4. 	u
      	2

   
   
   

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1. View Hint View Answer Discuss in Forum

   Time average of fluctuating velocity is zero.

   Correct Option: C

   Time average of fluctuating velocity is zero.

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92. A frictionless circular piston of area 10^-2m² and mass 100 kg sinks into a cylindrical container of the same area filled with water of density 1000 kg/m³ as shown in the figure. The container has a hole of area 10–3m2 at the bottom that is open to the atmosphere. Assuming there is no leakage from the edges of the piston and considering water to be incompressible, the magnitude of the piston velocity (in m/s) at the instant shown is _________ (correct to three decimal places).
    

1. 1. 1.456 m/s

   
   
   

   2. 2.456 m/s

   
   
   

   3. 3.456 m/s

   
   
   

   4. 4.456 m/s

   
   
   

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1. View Hint View Answer Discuss in Forum

   A₁ V₁ = A₂ V₂

   Correct Option: A

   A₁ V₁ = A₂ V₂

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93. A tank open at the top with a water level of 1 m, as shown in the figure, has a hole at a height of 0.5 m. A free jet leaves horizontally from the smooth hole. The distance X(in m) where the jet strikes the floor is
    

1. 1. 0.5

   
   
   

   2. 1.0

   
   
   

   3. 2.0

   
   
   

   4. 4.0

   
   
   

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1. View Hint View Answer Discuss in Forum

   u_x = √2gh
   
   = √2 × 9.8 × 0.5
   u_x = 3.13 m/s
   Now for distance x
   we require u_x so–
   

   x = ux . t +	1	× a × b
   	2

   
   x = 3.13 t
   (∵ acceleration (a) = 0 in horizontal direction)
   

   Now for t ⇒ y = uyt +	1	gt²
   	2

   
   

   ⇒ 0.5 = 0 +	1	× 9.81 × t² [∵ uy = 0]
   	2

   
   t = 0.319 sec.
   so distance, x = 3.13 × 319 = 1m

   Correct Option: B

   u_x = √2gh
   
   = √2 × 9.8 × 0.5
   u_x = 3.13 m/s
   Now for distance x
   we require u_x so–
   

   x = ux . t +	1	× a × b
   	2

   
   x = 3.13 t
   (∵ acceleration (a) = 0 in horizontal direction)
   

   Now for t ⇒ y = uyt +	1	gt²
   	2

   
   

   ⇒ 0.5 = 0 +	1	× 9.81 × t² [∵ uy = 0]
   	2

   
   t = 0.319 sec.
   so distance, x = 3.13 × 319 = 1m

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94. The water jet exiting from a stationary tank through a circular opening of diameter 300 mm impinges on a rigid wall as shown in the figure. Neglect all minor losses and assume the water level in the tank to remain constant. The net horizontal force experienced by the wall is ____ kN. Density of water is 1000 kg/m³. Acceleration due to gravity g = 10 m/s².
    

1. 1. 8.76 N

   
   
   

   2. 7.76 N

   
   
   

   3. 9.76 N

   
   
   

   4. 5.76 N

   
   
   

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1. View Hint View Answer Discuss in Forum

   Force exerted by a jet of water striking fixed wall = ρav²
   

   = 1000 ×	π	× 0.3² × V²
   	4

   
   

   = 1000 ×	π	× 0.3² × (2 × 10 × 6²) [∵ V = √2gh]
   	4

   
   = 8.76 N
   

   Correct Option: A

   Force exerted by a jet of water striking fixed wall = ρav²
   

   = 1000 ×	π	× 0.3² × V²
   	4

   
   

   = 1000 ×	π	× 0.3² × (2 × 10 × 6²) [∵ V = √2gh]
   	4

   
   = 8.76 N
   

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95. Water is coming out from a tap and a falls vertically downwards. At the tap opening, the stream diameter is 20 mm with uniform velocity of 2 m/s. Acceleration due to gravity is 9.81 m/ s². Assuming steady, inviscid flow, constant atmospheric pressure everywhere and neglecting curvature and surface tension effects, the diameter in mm of the stream 0.5 m below the tap is approximately

1. 1. 10

   
   
   

   2. 15

   
   
   

   3. 20

   
   
   

   4. 25

   
   
   

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1. View Hint View Answer Discuss in Forum

   Q = A₁ V₁ =A₂ V₂
   

   ⇒		π	(20)² × 2 =	π	× d² × √2 x 9.8 x 0.5
   		4		4

   
   d ≈ 15mm

   Correct Option: B

   Q = A₁ V₁ =A₂ V₂
   

   ⇒		π	(20)² × 2 =	π	× d² × √2 x 9.8 x 0.5
   		4		4

   
   d ≈ 15mm

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