1. A two-dimensional flow field has velocities along the x and y directions given by u = x²t and v = –2xyt respectively, where t is time. The equation of streamline is

1. 1. x²y = constant

   
   
   

   2. xy² = constant

   
   
   

   3. xy = constant

   
   
   

   4. not possible to determine

   
   
   

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1. View Hint View Answer Discuss in Forum

   Given, u = x²t and v = – 2xyt
   

   = v = -2xyt..........(i)

   We know,	δψ
   	δy

   
   = -u = x²t..........(ii)

   δψ
   δy

   
   Integrating equation (i), we get
   ψ = –x²yt + f(y) ...(iii)
   Differentiating equation (iii) with respect to y, we get
   = -x²t + f(y)........(v)

   δψ
   δy

   
   Equating the value of δψ/δy from equations (ii) and (iv), we get
   –x²t =–x²t + f'(y)
   Since, f'(y) = 0, thus f(y) = C
   (where ‘C’ is constant of integration)
   ψ = –x²yt + C
   C is a numerical constant so it can be taken as zero
   ψ = –x²yt
   For equation of stream lines,
   ψ = constant
   –x²yt = constant
   For a particular instance,
   x²y = constant

   Correct Option: A

   Given, u = x²t and v = – 2xyt
   

   = v = -2xyt..........(i)

   We know,	δψ
   	δy

   
   = -u = x²t..........(ii)

   δψ
   δy

   
   Integrating equation (i), we get
   ψ = –x²yt + f(y) ...(iii)
   Differentiating equation (iii) with respect to y, we get
   = -x²t + f(y)........(v)

   δψ
   δy

   
   Equating the value of δψ/δy from equations (ii) and (iv), we get
   –x²t =–x²t + f'(y)
   Since, f'(y) = 0, thus f(y) = C
   (where ‘C’ is constant of integration)
   ψ = –x²yt + C
   C is a numerical constant so it can be taken as zero
   ψ = –x²yt
   For equation of stream lines,
   ψ = constant
   –x²yt = constant
   For a particular instance,
   x²y = constant

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2. In a flow field the stream lines and equipotential lines

1. 1. are parallel

   
   
   

   2. Cut at any angle

   
   
   

   3. Are orthogonal every where in the field

   
   
   

   4. Cut orthogonal except at the stagnation points

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: D

   NA

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3. A fluid flow is represented by the velocity field = ax_î + ay_ĵ, where a is a constant. The equation of stream line passing through a point (1, 2) is

1. 1. x – 2y = 0

   
   
   

   2. 2x + y = 0

   
   
   

   3. 2x – y = 0

   
   
   

   4. x + 2y = 0

   
   
   

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1. View Hint View Answer Discuss in Forum

   Given: u_x = ax and u_y = ay
   Equation of steam line is,
   

   du	=	dy
   ux		uy

   
   

   ⇒	du	=	dy
   	ux		uy

   
   Integrating both sides, we have
   log(ax) = log(ay) + log c
   or ax = c'ay
   or x = cy
   Since the steam line is passed through point (1, 2), therefore
   1 = 2c
   

   ⇒ c =	1
   	2

   
   

   ∴ x =	y
   	2

   
   Hence equation of steam line is
   2x – y = 0.

   Correct Option: C

   Given: u_x = ax and u_y = ay
   Equation of steam line is,
   

   du	=	dy
   ux		uy

   
   

   ⇒	du	=	dy
   	ux		uy

   
   Integrating both sides, we have
   log(ax) = log(ay) + log c
   or ax = c'ay
   or x = cy
   Since the steam line is passed through point (1, 2), therefore
   1 = 2c
   

   ⇒ c =	1
   	2

   
   

   ∴ x =	y
   	2

   
   Hence equation of steam line is
   2x – y = 0.

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4. A streamlined body is defined as a body about which

1. 1. The flow is laminar

   
   
   

   2. The flow is along the streamlines

   
   
   

   3. The flow separation is suppressed

   
   
   

   4. The drag is zero

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: C

   NA

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5. Figure shows the schematic for the measurement of velocity of air (density = 1.2 kg/m³) through a constant-area duct using a pitot tube and a water tube manometer. The differential head of water (density = 1000 kg/ m³) in the two columns of the manometer is 10 mm. Take acceleration due to gravity as 9.8 m/s². The velocity of air in m/s is 10 m m
   

1. 1. 6.4

   
   
   

   2. 9.0

   
   
   

   3. 12.8

   
   
   

   4. 25.6

   
   
   

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1. View Hint View Answer Discuss in Forum

   From Bernoulli's equation
   

   v²1 - v²2	=	p2 - p1
   2g		2g

   
   

   ⇒ v1 = √	(p2 - p1)
   	pa

   
   But p₂ - p₁ = (ρgh)_water
   = 9810 × 10 × 10^{– 3}
   = 98.1 N/m²
   

   ⇒ v1 = √	2 × 98.1	= 12.8 m/s
   	1.2

   Correct Option: C

   From Bernoulli's equation
   

   v²1 - v²2	=	p2 - p1
   2g		2g

   
   

   ⇒ v1 = √	(p2 - p1)
   	pa

   
   But p₂ - p₁ = (ρgh)_water
   = 9810 × 10 × 10^{– 3}
   = 98.1 N/m²
   

   ⇒ v1 = √	2 × 98.1	= 12.8 m/s
   	1.2

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