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A container of large surface area is filled with liquid of density r. A cubical block of side edge a and mass M is floating in it with four-fifth of its volume submerged. If a coin of mass m is placed gently on the top surface of the block is just submerged. M is
- a)4m/5
- b)m/5
- c)4m
- d)5m
Correct answer is option 'C'. Can you explain this answer?
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A container of large surface area is filled with liquid of density r. ...
Understanding the Floating Block
When a block floats in a liquid, it displaces a volume of liquid equal to its weight. In this scenario, we have a cubical block of side edge "a" and mass "M," floating in a liquid of density "r." Since four-fifths of the block's volume is submerged, we can denote the submerged volume as:
- Volume of the block = a^3
- Submerged volume = (4/5) * a^3
The weight of the displaced liquid equals the weight of the block:
- Weight of displaced liquid = r * (4/5) * a^3 * g
- Weight of the block = M * g
Thus, we have:
- M * g = r * (4/5) * a^3 * g
From this, we can simplify to find M:
- M = r * (4/5) * a^3
Adding the Coin
When a coin of mass "m" is placed on top of the block, it causes the block to sink further. Since the coin just causes the block to be fully submerged, the total weight now includes both the block and the coin:
- Total weight = (M + m) * g
Now, the total displaced volume must equal the weight of the total system:
- Displaced volume = a^3 (since the block is fully submerged)
The weight of the displaced liquid is now:
- Weight of displaced liquid = r * a^3 * g
Equating the weights gives:
- (M + m) * g = r * a^3 * g
Cancelling "g" from both sides, we get:
- M + m = r * a^3
Finding "M"
From our earlier equation:
- M = r * (4/5) * a^3
Substituting this into the new equation:
- (r * (4/5) * a^3) + m = r * a^3
Rearranging yields:
- m = r * a^3 * (1 - 4/5) = r * a^3 / 5
Now substituting this back into our expression for M:
- M = 4m
Thus, the correct answer is:
The Correct Answer
- M = 4m (Option C)
When a block floats in a liquid, it displaces a volume of liquid equal to its weight. In this scenario, we have a cubical block of side edge "a" and mass "M," floating in a liquid of density "r." Since four-fifths of the block's volume is submerged, we can denote the submerged volume as:
- Volume of the block = a^3
- Submerged volume = (4/5) * a^3
The weight of the displaced liquid equals the weight of the block:
- Weight of displaced liquid = r * (4/5) * a^3 * g
- Weight of the block = M * g
Thus, we have:
- M * g = r * (4/5) * a^3 * g
From this, we can simplify to find M:
- M = r * (4/5) * a^3
Adding the Coin
When a coin of mass "m" is placed on top of the block, it causes the block to sink further. Since the coin just causes the block to be fully submerged, the total weight now includes both the block and the coin:
- Total weight = (M + m) * g
Now, the total displaced volume must equal the weight of the total system:
- Displaced volume = a^3 (since the block is fully submerged)
The weight of the displaced liquid is now:
- Weight of displaced liquid = r * a^3 * g
Equating the weights gives:
- (M + m) * g = r * a^3 * g
Cancelling "g" from both sides, we get:
- M + m = r * a^3
Finding "M"
From our earlier equation:
- M = r * (4/5) * a^3
Substituting this into the new equation:
- (r * (4/5) * a^3) + m = r * a^3
Rearranging yields:
- m = r * a^3 * (1 - 4/5) = r * a^3 / 5
Now substituting this back into our expression for M:
- M = 4m
Thus, the correct answer is:
The Correct Answer
- M = 4m (Option C)
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A container of large surface area is filled with liquid of density r. A cubical block of side edge a and mass M is floating in it with four-fifth of its volume submerged. If a coin of mass m is placed gently on the top surface of the block is just submerged. M isa)4m/5b)m/5c)4md)5mCorrect answer is option 'C'. Can you explain this answer?
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A container of large surface area is filled with liquid of density r. A cubical block of side edge a and mass M is floating in it with four-fifth of its volume submerged. If a coin of mass m is placed gently on the top surface of the block is just submerged. M isa)4m/5b)m/5c)4md)5mCorrect answer is option 'C'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A container of large surface area is filled with liquid of density r. A cubical block of side edge a and mass M is floating in it with four-fifth of its volume submerged. If a coin of mass m is placed gently on the top surface of the block is just submerged. M isa)4m/5b)m/5c)4md)5mCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A container of large surface area is filled with liquid of density r. A cubical block of side edge a and mass M is floating in it with four-fifth of its volume submerged. If a coin of mass m is placed gently on the top surface of the block is just submerged. M isa)4m/5b)m/5c)4md)5mCorrect answer is option 'C'. Can you explain this answer?.
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