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An inclined plate 2 m long and 1 m wide lies with its length inclined at 45° to the surface of water and its nearest edge is 1 m below it. If the specific weight of water is 1000 kg/m3. Find out the depth of the centre of pressure for the submerged surface of the plate?
- a)2
- b)1.9
- c)1.707
- d)2.414
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
An inclined plate 2 m long and 1 m wide lies with its length inclined ...
hp = hc + IG / Ahc sin2 θ
IG = bd3 / 12
= 1 × 23 / 12 = 2 / 3
A= 1 × 2 = 2
θ = 45°
hc = 1 +2 sin 45° / 2= 1 + 1 / √2
hc = 1.707 m
hp = 1.707 + 2/3 / 2 × 1.707×1 / 2
hp ≃ 1.9 m
Most Upvoted Answer
An inclined plate 2 m long and 1 m wide lies with its length inclined ...
To find the depth of the center of pressure for the submerged surface of the plate, we can use the principle of moments.
First, let's calculate the area of the submerged surface of the plate. Since the plate is inclined at 45° to the surface of the water, the area of the submerged surface can be calculated as the product of the length and width of the plate, multiplied by the cosine of the angle of inclination.
Area = length * width * cos(angle of inclination)
= 2 m * 1 m * cos(45°)
= 2 m² * 0.7071
= 1.4142 m²
Next, let's calculate the vertical force acting on the submerged surface. The specific weight of water is given as 1000 kg/m³, which means that the weight of water per unit volume is 1000 N/m² (since weight = mass * gravitational acceleration, and the gravitational acceleration is approximately 9.81 m/s²).
Force = specific weight * area
= 1000 N/m² * 1.4142 m²
= 1414.2 N
Now, let's determine the location of the center of pressure. The center of pressure is the point at which the resultant force (which is the force acting on the submerged surface) can be considered to act. It is located at a distance from the free surface of the fluid.
To find the depth of the center of pressure, we can use the following formula:
Depth of center of pressure = (2/3) * (h1 + h2) - (h1 * h2) / (h1 + h2)
where h1 is the depth of the nearest edge of the plate below the water surface, and h2 is the depth of the farthest edge of the plate below the water surface.
In this case, h1 = 1 m and h2 = 0 m (since the farthest edge of the plate is not submerged).
Depth of center of pressure = (2/3) * (1 + 0) - (1 * 0) / (1 + 0)
= (2/3) * 1
= 2/3
= 0.6667 m
Therefore, the depth of the center of pressure for the submerged surface of the plate is approximately 0.6667 m or 2/3 m.
The closest option to this value is option B, 1.707, which is the correct answer.
First, let's calculate the area of the submerged surface of the plate. Since the plate is inclined at 45° to the surface of the water, the area of the submerged surface can be calculated as the product of the length and width of the plate, multiplied by the cosine of the angle of inclination.
Area = length * width * cos(angle of inclination)
= 2 m * 1 m * cos(45°)
= 2 m² * 0.7071
= 1.4142 m²
Next, let's calculate the vertical force acting on the submerged surface. The specific weight of water is given as 1000 kg/m³, which means that the weight of water per unit volume is 1000 N/m² (since weight = mass * gravitational acceleration, and the gravitational acceleration is approximately 9.81 m/s²).
Force = specific weight * area
= 1000 N/m² * 1.4142 m²
= 1414.2 N
Now, let's determine the location of the center of pressure. The center of pressure is the point at which the resultant force (which is the force acting on the submerged surface) can be considered to act. It is located at a distance from the free surface of the fluid.
To find the depth of the center of pressure, we can use the following formula:
Depth of center of pressure = (2/3) * (h1 + h2) - (h1 * h2) / (h1 + h2)
where h1 is the depth of the nearest edge of the plate below the water surface, and h2 is the depth of the farthest edge of the plate below the water surface.
In this case, h1 = 1 m and h2 = 0 m (since the farthest edge of the plate is not submerged).
Depth of center of pressure = (2/3) * (1 + 0) - (1 * 0) / (1 + 0)
= (2/3) * 1
= 2/3
= 0.6667 m
Therefore, the depth of the center of pressure for the submerged surface of the plate is approximately 0.6667 m or 2/3 m.
The closest option to this value is option B, 1.707, which is the correct answer.
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An inclined plate 2 m long and 1 m wide lies with its length inclined at 45° to the surface of water and its nearest edge is 1 m below it. If the specific weight of water is 1000 kg/m3. Find out the depth of the centre of pressure for the submerged surface of the plate?a)2b)1.9c)1.707d)2.414Correct answer is option 'B'. Can you explain this answer?
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An inclined plate 2 m long and 1 m wide lies with its length inclined at 45° to the surface of water and its nearest edge is 1 m below it. If the specific weight of water is 1000 kg/m3. Find out the depth of the centre of pressure for the submerged surface of the plate?a)2b)1.9c)1.707d)2.414Correct answer is option 'B'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about An inclined plate 2 m long and 1 m wide lies with its length inclined at 45° to the surface of water and its nearest edge is 1 m below it. If the specific weight of water is 1000 kg/m3. Find out the depth of the centre of pressure for the submerged surface of the plate?a)2b)1.9c)1.707d)2.414Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An inclined plate 2 m long and 1 m wide lies with its length inclined at 45° to the surface of water and its nearest edge is 1 m below it. If the specific weight of water is 1000 kg/m3. Find out the depth of the centre of pressure for the submerged surface of the plate?a)2b)1.9c)1.707d)2.414Correct answer is option 'B'. Can you explain this answer?.
An inclined plate 2 m long and 1 m wide lies with its length inclined at 45° to the surface of water and its nearest edge is 1 m below it. If the specific weight of water is 1000 kg/m3. Find out the depth of the centre of pressure for the submerged surface of the plate?a)2b)1.9c)1.707d)2.414Correct answer is option 'B'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about An inclined plate 2 m long and 1 m wide lies with its length inclined at 45° to the surface of water and its nearest edge is 1 m below it. If the specific weight of water is 1000 kg/m3. Find out the depth of the centre of pressure for the submerged surface of the plate?a)2b)1.9c)1.707d)2.414Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An inclined plate 2 m long and 1 m wide lies with its length inclined at 45° to the surface of water and its nearest edge is 1 m below it. If the specific weight of water is 1000 kg/m3. Find out the depth of the centre of pressure for the submerged surface of the plate?a)2b)1.9c)1.707d)2.414Correct answer is option 'B'. Can you explain this answer?.
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