Hydraulics and Fluid Mechanics in ME The increase of temperature results in increase in viscosity of gas decrease in viscosity of liquid increase in viscosity of liquid decrease in viscosity of gas increase in viscosity of gas decrease in viscosity of liquid increase in viscosity of liquid decrease in viscosity of gas ANSWER DOWNLOAD EXAMIANS APP
Hydraulics and Fluid Mechanics in ME Theoretical power required (in watts) to drive a reciprocating pump is (where w = Specific weight of liquid to be pumped in N/m3, Q = Discharge of the pump in m3/s, Hs = Suction head in meters, and Hd = Delivery head in meters) wQHs wQ (Hs + Hd) wQHd wQ (Hs - Hd) wQHs wQ (Hs + Hd) wQHd wQ (Hs - Hd) ANSWER DOWNLOAD EXAMIANS APP
Hydraulics and Fluid Mechanics in ME A closed tank containing water is moving in a horizontal direction along a straight line at a constant speed. The tank also contains a steel ball and a bubble of air. If the tank is decelerated horizontally, theni) the ball will move to the frontii) the bubble will move to the frontiii) the ball will move to the reariv) the bubble will move to the rear Find out which of the above statements are correct ? (ii) and (iii) (i) and (ii) (i) and (iv) (iii) and (iv) (ii) and (iii) (i) and (ii) (i) and (iv) (iii) and (iv) ANSWER DOWNLOAD EXAMIANS APP
Hydraulics and Fluid Mechanics in ME An opening in the side of a tank or vessel such that the liquid surface with the tank is below the top edge of the opening, is called None of these Orifice Weir Notch None of these Orifice Weir Notch ANSWER DOWNLOAD EXAMIANS APP
Hydraulics and Fluid Mechanics in ME The units of dynamic or absolute viscosity are newton-sec per meter newton-sec² per meter kg sec/meter metres² per sec newton-sec per meter newton-sec² per meter kg sec/meter metres² per sec 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 × (√H3/2 - √H3/2)/Cd × a × √(2g) 2A × (√H₁ - √H₂)/Cd × a × √(2g) 2A × √H₁/Cd × a × √(2g) 2A × √H₂/Cd × a × √(2g) 2A × (√H3/2 - √H3/2)/Cd × a × √(2g) 2A × (√H₁ - √H₂)/Cd × a × √(2g) 2A × √H₁/Cd × a × √(2g) ANSWER DOWNLOAD EXAMIANS APP
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