Fluid Mechanics and Hydraulic Machinery Miscellaneous
- In a flow field the stream lines and equipotential lines
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NA
Correct Option: D
NA
- Bodies in flotation to be in stable equilibrium the necessary and sufficient condition is that the centre of gravity is located below the ____.
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Located below the metacentre.
Correct Option: C
Located below the metacentre.
- A hinged gate of length 5 m, inclined at 30° with the horizontal and with water mass on its left, is shown in figure below. Density of water is 1000 kg/m³. The minimum mass of the gate in kg per unit width (perpendicular to the plane of paper), required to keep it closed 5 m

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Given data
l = 5 m
θ = 30°
ρ = 1000 kg/m³
h = 2.5sin30° = 1.25m
A = bl = 1 × 5 = 5 m²IG = bl³ = 1 × (5)³ = 10.41 m4 12 12
F = ρgAh = 1000 × 9.81 × 5 × 1.25 = 61312.5Nh* = h + IG . sin² θ Ah = 1.25 + 10.41 . sin² 30° = 1.67 m 5 × 1.25
Taking moment about hingeF = h* = W × 2.5 × cos30° sin 30° 61312.5 × 1.66 = Mg × 2.165 0.5
M = 9623kgCorrect Option: D
Given data
l = 5 m
θ = 30°
ρ = 1000 kg/m³
h = 2.5sin30° = 1.25m
A = bl = 1 × 5 = 5 m²IG = bl³ = 1 × (5)³ = 10.41 m4 12 12
F = ρgAh = 1000 × 9.81 × 5 × 1.25 = 61312.5Nh* = h + IG . sin² θ Ah = 1.25 + 10.41 . sin² 30° = 1.67 m 5 × 1.25
Taking moment about hingeF = h* = W × 2.5 × cos30° sin 30° 61312.5 × 1.66 = Mg × 2.165 0.5
M = 9623kg
- Consider a frictionless, massless and leak-proof plug blocking a rectangular hole of dimensions 2R × L at the bottom of an open tank as shown in the figure. The head of the plug has the shape of a semi-cylinder of radius R. The tank is filled with a liquid of density ρup to the tip of the plug. The gravitational acceleration is g. Neglect the effect of the atmospheric pressure.

The force F required to hold the plug in its position is
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Total volume ⇒ Vtotal = 2R × L × R = 2 R²L ...(i)
Volume : V1 = πR2 × L....(ii) 2
F = Weight of fluit =
[Vtotal – V1] ρg= 
2R²L - πR²L 
ρg 2 F = 2GρR²gL 
1 - π 
4
Correct Option: A
Total volume ⇒ Vtotal = 2R × L × R = 2 R²L ...(i)
Volume : V1 = πR2 × L....(ii) 2
F = Weight of fluit =
[Vtotal – V1] ρg= 
2R²L - πR²L 
ρg 2 F = 2GρR²gL 
1 - π 
4
- A jet of water issues from a nozzle with a velocity 20 m/s and it impinges normally on a flat plate moving away from it at 10 m/s. The crosssectional area of the jet is 0.01 m², and the density of water = 1000 kg/m³. The force developed on the plate is

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The force developed on the plate,
Fx = ρa(vj - u)² = 1000 × 0.01(20 10)²
Fx = 1000 NCorrect Option: A
The force developed on the plate,
Fx = ρa(vj - u)² = 1000 × 0.01(20 10)²
Fx = 1000 N