Hydraulics and Fluid Mechanics in ME Choose the correct relationship kinematicviscosity = dynamicviscosity x density gravity = specific gravity x density dynamicviscosity = kinematicviscosity x density specific gravity = gravity x density kinematicviscosity = dynamicviscosity x density gravity = specific gravity x density dynamicviscosity = kinematicviscosity x density specific gravity = gravity x density ANSWER DOWNLOAD EXAMIANS APP
Hydraulics and Fluid Mechanics in ME The speed of a pressure wave through a pipe depends upon the bulk modulus for the fluid the length of pipe the original head the viscosity of fluid the bulk modulus for the fluid the length of pipe the original head the viscosity of fluid ANSWER DOWNLOAD EXAMIANS APP
Hydraulics and Fluid Mechanics in ME A piece weighing 3 kg in air was found to weigh 2.5 kg when submerged in water. Its specific gravity is 5 7 6 1 5 7 6 1 ANSWER DOWNLOAD EXAMIANS APP
Hydraulics and Fluid Mechanics in ME If the atmospheric pressure on the surface of an oil tank (sp. gr. 0.8) is 0.2 kg/cm”, the pressure at a depth of 50 m below the oil surface will be 3 meters of water column 2 meters of water column 6 meters of water Column 5 meters of water column 3 meters of water column 2 meters of water column 6 meters of water Column 5 meters of water column ANSWER DOWNLOAD EXAMIANS APP
Hydraulics and Fluid Mechanics in ME The force exerted by a jet of water (in a direction normal to flow) impinging on a fixed plate inclined at an angle θ with the jet is (waV/g) × sin θ (waV/2g) × sin θ (waV²/g) × sin θ (waV²/2g) × sin 2θ (waV/g) × sin θ (waV/2g) × sin θ (waV²/g) × sin θ (waV²/2g) × sin 2θ ANSWER DOWNLOAD EXAMIANS APP
Hydraulics and Fluid Mechanics in ME A tank of uniform cross-sectional area (A) containing liquid upto height (H1) has an orifice of cross-sectional area (a) at its bottom. The time required to bring the liquid level from H1 to H2 will be 2A × (√H3/2 - √H3/2)/Cd × a × √(2g) 2A × √H₁/Cd × a × √(2g) 2A × √H₂/Cd × a × √(2g) 2A × (√H₁ - √H₂)/Cd × a × √(2g) 2A × (√H3/2 - √H3/2)/Cd × a × √(2g) 2A × √H₁/Cd × a × √(2g) 2A × √H₂/Cd × a × √(2g) 2A × (√H₁ - √H₂)/Cd × a × √(2g) ANSWER DOWNLOAD EXAMIANS APP
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