Hydraulics and Fluid Mechanics in ME When an ideal fluid flows past a sphere lowest pressure intensity occurs at front stagnation point highest intensity of pressure occurs around the circumference at right angles to flow lowest pressure intensity occurs at rear stagnation point total drag is zero lowest pressure intensity occurs at front stagnation point highest intensity of pressure occurs around the circumference at right angles to flow lowest pressure intensity occurs at rear stagnation point total drag is zero ANSWER DOWNLOAD EXAMIANS APP
Hydraulics and Fluid Mechanics in ME The property of a liquid which offers resistance to the movement of one layer of liquid over another adjacent layer of liquid, is called Viscosity Surface tension Capillarity Compressibility Viscosity Surface tension Capillarity Compressibility ANSWER DOWNLOAD EXAMIANS APP
Hydraulics and Fluid Mechanics in ME In a reaction turbine, the draft tube is used To prevent air to enter the turbine To run the turbine full To transport water to downstream To increase the head of water by an amount equal to the height of the runner outlet above the tail race To prevent air to enter the turbine To run the turbine full To transport water to downstream To increase the head of water by an amount equal to the height of the runner outlet above the tail race ANSWER DOWNLOAD EXAMIANS APP
Hydraulics and Fluid Mechanics in ME The boundary layer thickness at a distance of l m from the leading edge of a flat plate, kept at zero angle of incidence to the flow direction, is O.l cm. The velocity outside the boundary layer is 25 ml sec. The boundary layer thickness at a distance of 4 m is (Assume that boundary layer is entirely laminar) 0.40 cm 0.10 cm 0.05 cm 0.20 cm 0.40 cm 0.10 cm 0.05 cm 0.20 cm ANSWER DOWNLOAD EXAMIANS APP
Hydraulics and Fluid Mechanics in ME The loss of head at exit of a pipe is (where v = Velocity of liquid in the pipe) 0.375v²/2g 0.5v²/2g v²/2g 0.75v²/2g 0.375v²/2g 0.5v²/2g v²/2g 0.75v²/2g 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 × √H₂/Cd × a × √(2g) 2A × √H₁/Cd × a × √(2g) 2A × (√H₁ - √H₂)/Cd × a × √(2g) 2A × (√H3/2 - √H3/2)/Cd × a × √(2g) 2A × √H₂/Cd × a × √(2g) 2A × √H₁/Cd × a × √(2g) 2A × (√H₁ - √H₂)/Cd × a × √(2g) 2A × (√H3/2 - √H3/2)/Cd × a × √(2g) ANSWER DOWNLOAD EXAMIANS APP
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