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A 15 cm x 15 cm x 15cm metal cube
and its Specific gravity 8.6 is submerged
in a two layered liquid, the bottom layer
being mercury and the top layer being
water. The percentage of the volume of
the cube remaining above the interface
will be, approximately?
and its Specific gravity 8.6 is submerged
in a two layered liquid, the bottom layer
being mercury and the top layer being
water. The percentage of the volume of
the cube remaining above the interface
will be, approximately?
Most Upvoted Answer
A 15 cm x 15 cm x 15cm metal cube and its Specific gravity 8.6 is subm...
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Given Data:
- Side length of the metal cube (l) = 15 cm
- Specific gravity of the metal cube = 8.6
- Density of water = 1 g/cm³
- Density of mercury = 13.6 g/cm³
Calculating the mass of the cube:
- Volume of the cube = l^3 = 15 cm x 15 cm x 15 cm = 3375 cm³
- Mass of the cube = Volume x Density = 3375 cm³ x 8.6 g/cm³ = 29025 g = 29.025 kg
Calculating the buoyant force:
- Buoyant force in mercury = Volume submerged in mercury x Density of mercury x g
- Buoyant force in water = Volume submerged in water x Density of water x g
Calculating the volume submerged in each liquid:
- Volume of cube submerged in mercury = Volume of cube x Specific gravity = 3375 cm³ x 8.6 = 29025 cm³
- Volume of cube submerged in water = Volume of cube - Volume submerged in mercury = 3375 cm³ - 29025 cm³ = 3087.5 cm³
Calculating the percentage of volume remaining above the interface:
- Percentage volume remaining = (Volume above interface / Total volume of cube) x 100%
- Volume above interface = Volume of cube - Volume submerged in water = 3375 cm³ - 3087.5 cm³ = 287.5 cm³
- Percentage volume remaining = (287.5 cm³ / 3375 cm³) x 100% ≈ 8.52%
Therefore, approximately 8.52% of the volume of the cube will remain above the interface between mercury and water.
Given Data:
- Side length of the metal cube (l) = 15 cm
- Specific gravity of the metal cube = 8.6
- Density of water = 1 g/cm³
- Density of mercury = 13.6 g/cm³
Calculating the mass of the cube:
- Volume of the cube = l^3 = 15 cm x 15 cm x 15 cm = 3375 cm³
- Mass of the cube = Volume x Density = 3375 cm³ x 8.6 g/cm³ = 29025 g = 29.025 kg
Calculating the buoyant force:
- Buoyant force in mercury = Volume submerged in mercury x Density of mercury x g
- Buoyant force in water = Volume submerged in water x Density of water x g
Calculating the volume submerged in each liquid:
- Volume of cube submerged in mercury = Volume of cube x Specific gravity = 3375 cm³ x 8.6 = 29025 cm³
- Volume of cube submerged in water = Volume of cube - Volume submerged in mercury = 3375 cm³ - 29025 cm³ = 3087.5 cm³
Calculating the percentage of volume remaining above the interface:
- Percentage volume remaining = (Volume above interface / Total volume of cube) x 100%
- Volume above interface = Volume of cube - Volume submerged in water = 3375 cm³ - 3087.5 cm³ = 287.5 cm³
- Percentage volume remaining = (287.5 cm³ / 3375 cm³) x 100% ≈ 8.52%
Therefore, approximately 8.52% of the volume of the cube will remain above the interface between mercury and water.
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A 15 cm x 15 cm x 15cm metal cube and its Specific gravity 8.6 is submerged in a two layered liquid, the bottom layer being mercury and the top layer being water. The percentage of the volume of the cube remaining above the interface will be, approximately?
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A 15 cm x 15 cm x 15cm metal cube and its Specific gravity 8.6 is submerged in a two layered liquid, the bottom layer being mercury and the top layer being water. The percentage of the volume of the cube remaining above the interface will be, approximately? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about A 15 cm x 15 cm x 15cm metal cube and its Specific gravity 8.6 is submerged in a two layered liquid, the bottom layer being mercury and the top layer being water. The percentage of the volume of the cube remaining above the interface will be, approximately? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 15 cm x 15 cm x 15cm metal cube and its Specific gravity 8.6 is submerged in a two layered liquid, the bottom layer being mercury and the top layer being water. The percentage of the volume of the cube remaining above the interface will be, approximately?.
A 15 cm x 15 cm x 15cm metal cube and its Specific gravity 8.6 is submerged in a two layered liquid, the bottom layer being mercury and the top layer being water. The percentage of the volume of the cube remaining above the interface will be, approximately? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about A 15 cm x 15 cm x 15cm metal cube and its Specific gravity 8.6 is submerged in a two layered liquid, the bottom layer being mercury and the top layer being water. The percentage of the volume of the cube remaining above the interface will be, approximately? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 15 cm x 15 cm x 15cm metal cube and its Specific gravity 8.6 is submerged in a two layered liquid, the bottom layer being mercury and the top layer being water. The percentage of the volume of the cube remaining above the interface will be, approximately?.
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