It gives maximum discharge for a given cross-sectional area and bed slope
It has minimum wetted perimeter
It involves lesser excavation for the designed amount of discharge
All of the above
D. All of the above
Continuity equation
Bernoulli's equation
Pascal's law
Archimedess principle
Critical flow
Turbulent flow
Tranquil flow
Torrential flow
Volumetric strain
Volumetric index
Compressibility
Adhesion
Lesser
Greater
Same
None of these
Meta center
Center of pressure
Center of buoyancy
Center of gravity
Maximum
Minimum
Zero
Nonzero finite
Venturimeter
Orifice meter
Pitot tube
All of these
Effects
Does not effect
Both A and B
None of these
w × Q × H
w × Q × hf
w × Q (H - hf)
w × Q (H + hf)
Pressure of liquid
Discharge of liquid
Pressure difference between two points in a channel
Pressure difference between two points in a pipe
Gauge pressure
Absolute pressure
Positive gauge pressure
Vacuum pressure
Double
Four times
Eight times
Sixteen times
Centroid of the volume of fluid vertically above the body
Centre of the volume of floating body
Center of gravity of any submerged body
Centroid of the displaced volume of fluid
It has low vapour pressure
It is clearly visible
It has low surface tension
It can provide longer column due to low density
Supersonics, as with projectile and jet propulsion
Full immersion or completely enclosed flow, as with pipes, aircraft wings, nozzles etc.
Simultaneous motion through two fluids where there is a surface of discontinuity, gravity forces, and wave making effect, as with ship's hulls
All of the above
Pressure
Discharge
Velocity
Volume
p = T × r
p = T/r
p = T/2r
p = 2T/r
Narrow crested weir
Broad crested weir
Ogee weir
Submerged weir
d/6
d/4
d/2
d
Same
More
Less
None of these
Velocity of flow at the required point in a pipe
Pressure difference between two points in a pipe
Total pressure of liquid flowing in a pipe
Discharge through a pipe
Plus
Minus
Divide
Multiply
Q = (2/3) Cd × b × √(2g) × (H2 - H1)
Q = (2/3) Cd × b × √(2g) × (H2 1/2 - H1 1/2)
Q = (2/3) Cd × b × √(2g) × (H2 3/2 - H1 3/2)
Q = (2/3) Cd × b × √(2g) × (H2 2 - H1 2)
Inertia force
Viscous force
Gravity force
Pressure force
(2/3) Cd × b × √(2gH)
(2/3) Cd × b × √(2g) × H
(2/3) Cd × b × √(2g) × H3/2
(2/3) Cd × b × √(2g) × H2
Centre of pressure
Centre of gravity
Centre of buoyancy
Metacentre
The head loss for all the pipes is same
The total discharge is equal to the sum of discharges in the various pipes
The total head loss is the sum of head losses in the various pipes
Both (A) and (B)
Negligible
Same as buoyant force
Zero
None of the above
Less than unity
Unity
Between 1 and 6
None of these
Remains constant
Increases
Decreases
Depends upon mass of liquid