Hydraulics and Fluid Mechanics in ME In order to avoid tendency of separation at throat in a Venturimeter, the ratio of the diameter at throat to the diameter of pipe should be 1/8 to 1/4 1/16 to 1/8 1/3 to 1/2 1/4 to 1/3 1/8 to 1/4 1/16 to 1/8 1/3 to 1/2 1/4 to 1/3 ANSWER DOWNLOAD EXAMIANS APP
Hydraulics and Fluid Mechanics in ME When the flow in an open channel is gradually varied, the flow is said to be Unsteady uniform flow Steady non-uniform flow Unsteady non-uniform flow Steady uniform flow Unsteady uniform flow Steady non-uniform flow Unsteady non-uniform flow Steady uniform flow ANSWER DOWNLOAD EXAMIANS APP
Hydraulics and Fluid Mechanics in ME Stanton diagram is a semi-log plot of friction factor against Reynolds number log-log plot of friction factor against Reynolds number log-log plot of relative roughness against Reynolds number semi-log plot of friction factor against relative roughness semi-log plot of friction factor against Reynolds number log-log plot of friction factor against Reynolds number log-log plot of relative roughness against Reynolds number semi-log plot of friction factor against relative roughness ANSWER DOWNLOAD EXAMIANS APP
Hydraulics and Fluid Mechanics in ME The specific speed of a centrifugal pump, delivering 750 liters of water per second against a head of 15 meters at 725 r.p.m., is 24.8 r.p.m. 248 r.p.m 82.4 r.p.m. 48.2 r.p.m 24.8 r.p.m. 248 r.p.m 82.4 r.p.m. 48.2 r.p.m ANSWER DOWNLOAD EXAMIANS APP
Hydraulics and Fluid Mechanics in ME The kinematic viscosity of an oil (in stokes) whose specific gravity is 0.95 and viscosity 0.011 poise, is 0.0116 stoke 0.116 stoke 0.0611 stoke 0.611 stoke 0.0116 stoke 0.116 stoke 0.0611 stoke 0.611 stoke 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 × √H₁/Cd × a × √(2g) 2A × (√H₁ - √H₂)/Cd × a × √(2g) 2A × √H₂/Cd × a × √(2g) 2A × (√H3/2 - √H3/2)/Cd × a × √(2g) 2A × √H₁/Cd × a × √(2g) 2A × (√H₁ - √H₂)/Cd × a × √(2g) 2A × √H₂/Cd × a × √(2g) 2A × (√H3/2 - √H3/2)/Cd × a × √(2g) ANSWER DOWNLOAD EXAMIANS APP
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