76. Three rigid buckets, shown as in the figures (1), (2) and (3), are of identical heights and base areas. Further, assume that each of these buckets have negligible mass and are full of water. The weights of water in these buckets are dented as W₁, W₂ and W₃ respectively. Also, let the force of water on the base of the bucket be denoted as F₁, F₂ and F₃ respectively. The option giving an accurate description of the system physics is
    

1. 1. W₂ = W₁ = W₃ and F₂ > F₁ > F₃

   
   
   

   2. W₂ > W₁ > W₃ and F₂ > F₁ > F₃

   
   
   

   3. W₂ = W₁ = W₃ and F₁ = F₂ = F₃

   
   
   

   4. W₂ > W₁ > W₃ and F₁ = F₂ = F₃

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   The buckets have same height and base area.
   From the figure it is clear that,
   W₂ > W₁ > W₃
   Force at base = γ_w⋅h × A
   Since, γ_w,h and A is same for all buckets,
   F₁ = F₂ = F₃

   Correct Option: D

   
   The buckets have same height and base area.
   From the figure it is clear that,
   W₂ > W₁ > W₃
   Force at base = γ_w⋅h × A
   Since, γ_w,h and A is same for all buckets,
   F₁ = F₂ = F₃

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77. A plane flow has velocity components u =	x	; v =	y
    	T1		T2

    
    w = 0 along x, y and z directions respectively, where T₁ (#0) and T₂ (#0) are constants having the dimension of time. The given flow is incompressible if

1. 1. T₁ = −T₂

   
   
   

   2. T1 = −	T2
      	2

   
   
   

   3. T1 =	T2
      	2

   
   
   

   4. T₁ = T₂
      

   
   
   

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1. View Hint View Answer Discuss in Forum

   For incompressible flow,
   

   ∂u	+	∂v	= 0
   ∂x		∂y

   
   

   	u =	x	; v =	y
   		T1		T2

   
   

   1	−	1	= 0
   T1		T2

   
   ∴ T₁ = T₂.

   Correct Option: D

   For incompressible flow,
   

   ∂u	+	∂v	= 0
   ∂x		∂y

   
   

   	u =	x	; v =	y
   		T1		T2

   
   

   1	−	1	= 0
   T1		T2

   
   ∴ T₁ = T₂.

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78. A venturimeter having a throat diameter of 0.1 m is used to estimate the flow rate of a horizontal pipe having a diameter of 0.2 m. For an observed pressure difference of 2 m of water head and coefficient of discharge equal to unity, assuming that the energy losses are negligible, the flow rate (in m³ /s) through the pipe is approximately equal to

1. 1. 0.500

   
   
   

   2. 0.150

   
   
   

   3. 0.050

   
   
   

   4. 0.015
      

   
   
   

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1. View Hint View Answer Discuss in Forum

   A1 =	π	× (0.2)2 = 0.01 π
   	4

   
   

   A2 =	π	× (0.1)2 = 0.0025 π
   	4

   
   

   Q =	CDA1A2√2gh	= 0.5 m3/s
   	√A1² − A2²

   Correct Option: C

   A1 =	π	× (0.2)2 = 0.01 π
   	4

   
   

   A2 =	π	× (0.1)2 = 0.0025 π
   	4

   
   

   Q =	CDA1A2√2gh	= 0.5 m3/s
   	√A1² − A2²

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79. A circular pipe has a diameter of 1m, bed slope of 1 in 1000, and Manning’s roughness coefficient equal to 0.01. It may be treated as an open channel flow when it is flowing just full, i.e., the water level just touches the crest. The discharge in this condition is denoted by Q_full. Similarly, the discharge when the pipe is flowing half-full, i.e., with a flow depth of 0.5m, is denoted by Q_half.
    The ratio Q_full/Q_half is:

1. 1. 1

   
   
   

   2. √2

   
   
   

   3. 2

   
   
   

   4. 4

   
   
   

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1. View Hint View Answer Discuss in Forum

   Discharge, Q =	1	A R2/3S1/2
   	n

   
   

   Qfull =	1		π	D²			D		2/3	S1/2
   	n		4				4

   
   

   Qhalf =	1	.		π	D²			D		2/3	S1/2
   	n			8				4

   
   

   Qfull	= 2
   Qhalf

   Correct Option: C

   Discharge, Q =	1	A R2/3S1/2
   	n

   
   

   Qfull =	1		π	D²			D		2/3	S1/2
   	n		4				4

   
   

   Qhalf =	1	.		π	D²			D		2/3	S1/2
   	n			8				4

   
   

   Qfull	= 2
   Qhalf

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80. In a two-dimensional steady flow field, in a certain region of the x-y plane, the velocity component in the x-direction is given by

    vx = x² and the density varies as ρ =	1	.
    	x

    
    Which of the following is a valid expression for the velocity component in the y-direction, v_y?

1. 1. v_y = – x/y

   
   
   

   2. v_y = x/y

   
   
   

   3. v_y = –xy

   
   
   

   4. v_y = xy

   
   
   

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1. View Hint View Answer Discuss in Forum

   Continuity equation
   

   ∂	(ρvx)	∂	(ρvy) = 0
   ∂x		∂y

   
   

   ⇒	∂	(x) +	∂		1	.vy		= 0
   	∂x		∂y		x

   
   

   ⇒ 1 +	∂		vy		= 0
   	∂y		x

   
   

   ⇒	∂		vy		= −1
   	∂y		x

   
   ⇒ v_y = – xy

   Correct Option: C

   Continuity equation
   

   ∂	(ρvx)	∂	(ρvy) = 0
   ∂x		∂y

   
   

   ⇒	∂	(x) +	∂		1	.vy		= 0
   	∂x		∂y		x

   
   

   ⇒ 1 +	∂		vy		= 0
   	∂y		x

   
   

   ⇒	∂		vy		= −1
   	∂y		x

   
   ⇒ v_y = – xy

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