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When the water surface coincides with the top edge of a rectangular vertical gate 40 m (wide) x 3 m (deep), then the depth of centre of pressure is
- a)1m
- b)1.5 m
- c)2 m
- d)2.5 m
Correct answer is option 'C'. Can you explain this answer?
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Given data:
Width of gate (b) = 40 m
Depth of gate (d) = 3 m
Depth of centre of pressure = ?
To find: Depth of centre of pressure
Concept:
The pressure on the surface of the gate is given by the formula,
P = ρgh
Where,
ρ = Density of water
g = acceleration due to gravity
h = Depth of water above the gate
The total force acting on the gate is given by,
F = P × A
Where,
A = Area of gate
The centre of pressure is the point where the total force acting on the gate can be considered to act.
The depth of the centre of pressure can be calculated as,
Depth of centre of pressure = (2/3) × Depth of water above the gate
Calculation:
Area of gate (A) = b × d = 40 × 3 = 120 m²
Density of water (ρ) = 1000 kg/m³
Acceleration due to gravity (g) = 9.81 m/s²
When the water surface coincides with the top edge of the gate, the depth of water above the gate is equal to the depth of the gate, i.e., h = d = 3 m
Therefore,
Pressure (P) = ρgh = 1000 × 9.81 × 3 = 29430 N/m²
Total force (F) = P × A = 29430 × 120 = 3531600 N
Depth of centre of pressure = (2/3) × Depth of water above the gate
= (2/3) × 3 = 2 m
Therefore, the depth of the centre of pressure is 2 m.
Hence, option (c) is the correct answer.
Width of gate (b) = 40 m
Depth of gate (d) = 3 m
Depth of centre of pressure = ?
To find: Depth of centre of pressure
Concept:
The pressure on the surface of the gate is given by the formula,
P = ρgh
Where,
ρ = Density of water
g = acceleration due to gravity
h = Depth of water above the gate
The total force acting on the gate is given by,
F = P × A
Where,
A = Area of gate
The centre of pressure is the point where the total force acting on the gate can be considered to act.
The depth of the centre of pressure can be calculated as,
Depth of centre of pressure = (2/3) × Depth of water above the gate
Calculation:
Area of gate (A) = b × d = 40 × 3 = 120 m²
Density of water (ρ) = 1000 kg/m³
Acceleration due to gravity (g) = 9.81 m/s²
When the water surface coincides with the top edge of the gate, the depth of water above the gate is equal to the depth of the gate, i.e., h = d = 3 m
Therefore,
Pressure (P) = ρgh = 1000 × 9.81 × 3 = 29430 N/m²
Total force (F) = P × A = 29430 × 120 = 3531600 N
Depth of centre of pressure = (2/3) × Depth of water above the gate
= (2/3) × 3 = 2 m
Therefore, the depth of the centre of pressure is 2 m.
Hence, option (c) is the correct answer.
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When the water surface coincides with the top edge of a rectangular vertical gate 40 m (wide) x 3 m (deep), then the depth of centre of pressure isa)1mb)1.5 mc)2 md)2.5 mCorrect answer is option 'C'. Can you explain this answer?
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