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 empty the tank completely will be (2AH₁²)/(Cd × a√2g) (2A√H₁)/(Cd × a√2g) (2AH₁)/(Cd × a√2g) (2AH₁3/2)/(Cd × a√2g) (2AH₁²)/(Cd × a√2g) (2A√H₁)/(Cd × a√2g) (2AH₁)/(Cd × a√2g) (2AH₁3/2)/(Cd × a√2g) ANSWER DOWNLOAD EXAMIANS APP
Hydraulics and Fluid Mechanics in ME In a forced vortex motion, the velocity of flow is directly proportional to the square of its radial distance from the axis of rotation directly proportional to its radial distance from axis of rotation inversely proportional to the square of its radial distance from the axis of rotation inversely proportional to its radial distance from the axis of rotation directly proportional to the square of its radial distance from the axis of rotation directly proportional to its radial distance from axis of rotation inversely proportional to the square of its radial distance from the axis of rotation inversely proportional to its radial distance from the axis of rotation ANSWER DOWNLOAD EXAMIANS APP
Hydraulics and Fluid Mechanics in ME One cubic metre of water weighs 250 liters 100 liters 1000 liters 500 liters 250 liters 100 liters 1000 liters 500 liters ANSWER DOWNLOAD EXAMIANS APP
Hydraulics and Fluid Mechanics in ME Cavitation will begin when pressure becomes more than critical pressure the pressure at any location reaches an absolute pressure equal to the saturated vapour pressure of the liquid pressure is increased flow is increased pressure becomes more than critical pressure the pressure at any location reaches an absolute pressure equal to the saturated vapour pressure of the liquid pressure is increased flow is increased ANSWER DOWNLOAD EXAMIANS APP
Hydraulics and Fluid Mechanics in ME The upper surface of a weir over which water flows is known is nappe crest sill weir top nappe crest sill weir top ANSWER DOWNLOAD EXAMIANS APP
Hydraulics and Fluid Mechanics in ME A compound pipe of diameter d1, d2 and d3 having lengths l1, l2 and l3 is to be replaced by an equivalent pipe of uniform diameter d and of the same length (l) as that of the compound pipe. The size of the equivalent pipe is given by l/d⁵ = (l₁/d₁⁵) + (l₂/d₂⁵) + (l₃/d₃⁵) l/d⁴ = (l₁/d₁⁴) + (l₂/d₂⁴) + (l₃/d₃⁴) l/d³ = (l₁/d₁³) + (l₂/d₂³) + (l₃/d₃³) l/d² = (l₁/d₁²) + (l₂/d₂²) + (l₃/d₃²) l/d⁵ = (l₁/d₁⁵) + (l₂/d₂⁵) + (l₃/d₃⁵) l/d⁴ = (l₁/d₁⁴) + (l₂/d₂⁴) + (l₃/d₃⁴) l/d³ = (l₁/d₁³) + (l₂/d₂³) + (l₃/d₃³) l/d² = (l₁/d₁²) + (l₂/d₂²) + (l₃/d₃²) ANSWER DOWNLOAD EXAMIANS APP
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