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A rectangular block (6m × 3m × 1.2m) is immersed in sea water with depth of immersion 0.8m in water. If the centre of gravity is 0.6 m above the bottom of the block, determine the metacentric height. The density for sea water = 1025 kg/m3
- a)0.42 m
- b)0.84 m
- c)0.5 m
- d)0.74 m
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
A rectangular block (6m × 3m × 1.2m) is immersed in sea wa...
X 4m x 3m) is floating in water with the longer side parallel to the water surface. The density of the block is 1500 kg/m^3. What is the height of the block above the water surface?
The weight of the block can be calculated using the formula:
Weight = Density x Volume x Gravity
The volume of the block can be calculated using the formula:
Volume = Length x Width x Height
Given:
Length = 6m
Width = 4m
Height = ?
Using the volume formula:
Volume = 6m x 4m x Height = 24m^2 x Height
The weight of the block can be calculated as:
Weight = 1500 kg/m^3 x (24m^2 x Height) x 9.8 m/s^2
Since the block is floating, the weight of the block is equal to the buoyant force acting on it. The buoyant force can be calculated using the formula:
Buoyant force = Density of water x Volume of displaced water x Gravity
Since the block is floating, the volume of displaced water is equal to the volume of the block. So:
Buoyant force = Density of water x (24m^2 x Height) x 9.8 m/s^2
Since the buoyant force is equal to the weight of the block, we can set the two equations equal to each other:
1500 kg/m^3 x (24m^2 x Height) x 9.8 m/s^2 = Density of water x (24m^2 x Height) x 9.8 m/s^2
The density of water cancels out on both sides of the equation, leaving:
1500 kg/m^3 x 24m^2 x Height = 24m^2 x Height
Simplifying further:
1500 kg/m^3 x Height = Height
1500 kg/m^3 = 1
Height = 1m
Therefore, the height of the block above the water surface is 1 meter.
The weight of the block can be calculated using the formula:
Weight = Density x Volume x Gravity
The volume of the block can be calculated using the formula:
Volume = Length x Width x Height
Given:
Length = 6m
Width = 4m
Height = ?
Using the volume formula:
Volume = 6m x 4m x Height = 24m^2 x Height
The weight of the block can be calculated as:
Weight = 1500 kg/m^3 x (24m^2 x Height) x 9.8 m/s^2
Since the block is floating, the weight of the block is equal to the buoyant force acting on it. The buoyant force can be calculated using the formula:
Buoyant force = Density of water x Volume of displaced water x Gravity
Since the block is floating, the volume of displaced water is equal to the volume of the block. So:
Buoyant force = Density of water x (24m^2 x Height) x 9.8 m/s^2
Since the buoyant force is equal to the weight of the block, we can set the two equations equal to each other:
1500 kg/m^3 x (24m^2 x Height) x 9.8 m/s^2 = Density of water x (24m^2 x Height) x 9.8 m/s^2
The density of water cancels out on both sides of the equation, leaving:
1500 kg/m^3 x 24m^2 x Height = 24m^2 x Height
Simplifying further:
1500 kg/m^3 x Height = Height
1500 kg/m^3 = 1
Height = 1m
Therefore, the height of the block above the water surface is 1 meter.
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Community Answer
A rectangular block (6m × 3m × 1.2m) is immersed in sea wa...
Metacentric height (GM):
GM = I/V - BG
Note:
1) M.O.I. (I) of top view is calculated about water surface about y-y axis.
2) V = Volume of body submerged in water.
Now,
∴ I = 13.5 m4
Now,
V = 6 × 3 × 0.8 = 14.4 m3
BG = AG – AB – 0.6 – 0.4 = 0.2 m
Thus,
GM = 13.5/14.4 - 0.2
∴ GM = 0.7375 m
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A rectangular block (6m × 3m × 1.2m) is immersed in sea water with depth of immersion 0.8m in water. If the centre of gravity is 0.6 m above the bottom of the block, determine the metacentric height. The density for sea water = 1025 kg/m3a)0.42 mb)0.84 mc)0.5 md)0.74 mCorrect answer is option 'D'. Can you explain this answer?
Question Description
A rectangular block (6m × 3m × 1.2m) is immersed in sea water with depth of immersion 0.8m in water. If the centre of gravity is 0.6 m above the bottom of the block, determine the metacentric height. The density for sea water = 1025 kg/m3a)0.42 mb)0.84 mc)0.5 md)0.74 mCorrect answer is option 'D'. 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 A rectangular block (6m × 3m × 1.2m) is immersed in sea water with depth of immersion 0.8m in water. If the centre of gravity is 0.6 m above the bottom of the block, determine the metacentric height. The density for sea water = 1025 kg/m3a)0.42 mb)0.84 mc)0.5 md)0.74 mCorrect answer is option 'D'. 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 A rectangular block (6m × 3m × 1.2m) is immersed in sea water with depth of immersion 0.8m in water. If the centre of gravity is 0.6 m above the bottom of the block, determine the metacentric height. The density for sea water = 1025 kg/m3a)0.42 mb)0.84 mc)0.5 md)0.74 mCorrect answer is option 'D'. Can you explain this answer?.
A rectangular block (6m × 3m × 1.2m) is immersed in sea water with depth of immersion 0.8m in water. If the centre of gravity is 0.6 m above the bottom of the block, determine the metacentric height. The density for sea water = 1025 kg/m3a)0.42 mb)0.84 mc)0.5 md)0.74 mCorrect answer is option 'D'. 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 A rectangular block (6m × 3m × 1.2m) is immersed in sea water with depth of immersion 0.8m in water. If the centre of gravity is 0.6 m above the bottom of the block, determine the metacentric height. The density for sea water = 1025 kg/m3a)0.42 mb)0.84 mc)0.5 md)0.74 mCorrect answer is option 'D'. 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 A rectangular block (6m × 3m × 1.2m) is immersed in sea water with depth of immersion 0.8m in water. If the centre of gravity is 0.6 m above the bottom of the block, determine the metacentric height. The density for sea water = 1025 kg/m3a)0.42 mb)0.84 mc)0.5 md)0.74 mCorrect answer is option 'D'. Can you explain this answer?.
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A rectangular block (6m × 3m × 1.2m) is immersed in sea water with depth of immersion 0.8m in water. If the centre of gravity is 0.6 m above the bottom of the block, determine the metacentric height. The density for sea water = 1025 kg/m3a)0.42 mb)0.84 mc)0.5 md)0.74 mCorrect answer is option 'D'. Can you explain this answer?, a detailed solution for A rectangular block (6m × 3m × 1.2m) is immersed in sea water with depth of immersion 0.8m in water. If the centre of gravity is 0.6 m above the bottom of the block, determine the metacentric height. The density for sea water = 1025 kg/m3a)0.42 mb)0.84 mc)0.5 md)0.74 mCorrect answer is option 'D'. Can you explain this answer? has been provided alongside types of A rectangular block (6m × 3m × 1.2m) is immersed in sea water with depth of immersion 0.8m in water. If the centre of gravity is 0.6 m above the bottom of the block, determine the metacentric height. The density for sea water = 1025 kg/m3a)0.42 mb)0.84 mc)0.5 md)0.74 mCorrect answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
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