Hydraulics and Fluid Mechanics in ME The power absorbed (in watts) in overcoming the viscous resistance of a footstep bearing is μ π³ N² R⁴ /1800 t μ π³ N² R² /3600 t μ π³ N² R² /1800 t μ π³ N² R⁴ /3600 t μ π³ N² R⁴ /1800 t μ π³ N² R² /3600 t μ π³ N² R² /1800 t μ π³ N² R⁴ /3600 t ANSWER DOWNLOAD EXAMIANS APP
Hydraulics and Fluid Mechanics in ME The viscosity of a gas increases with increase in temperature decreases with increase in temperature is independent of pressure for very high pressure intensities is independent of temperature increases with increase in temperature decreases with increase in temperature is independent of pressure for very high pressure intensities is independent of temperature ANSWER DOWNLOAD EXAMIANS APP
Hydraulics and Fluid Mechanics in ME The weight per unit volume of a liquid at a standard temperature and pressure is called Specific gravity Mass density None of these Specific weight Specific gravity Mass density None of these Specific weight ANSWER DOWNLOAD EXAMIANS APP
Hydraulics and Fluid Mechanics in ME In a free nappe, The pressure above the nappe is atmospheric The pressure above the nappe is negative The pressure below the nappe is atmospheric The pressure below the nappe is negative The pressure above the nappe is atmospheric The pressure above the nappe is negative The pressure below the nappe is atmospheric The pressure below the nappe is negative ANSWER DOWNLOAD EXAMIANS APP
Hydraulics and Fluid Mechanics in ME Two pipe systems can be said to be equivalent, when the following quantites are same friction factor and diameter friction loss and flow flow and length length and diameter friction factor and diameter friction loss and flow flow and length length and diameter ANSWER DOWNLOAD EXAMIANS APP
Hydraulics and Fluid Mechanics in ME The discharge over a rectangular weir, considering the velocity of approach, is (whereH1 = H + Ha = Total height of water above the weir, H = Height of water over the crest of the weir, and Ha = Height of water due to velocity of approach) (2/3) Cd × L.√2g [H1 - Ha] (2/3) Cd × L. √2g [H15/2 - Ha5/2] (2/3) Cd × L. √2g [H13/2 - Ha3/2] (2/3) Cd × L.√2g [H12 - Ha2] (2/3) Cd × L.√2g [H1 - Ha] (2/3) Cd × L. √2g [H15/2 - Ha5/2] (2/3) Cd × L. √2g [H13/2 - Ha3/2] (2/3) Cd × L.√2g [H12 - Ha2] ANSWER DOWNLOAD EXAMIANS APP
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