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A piece of metal of specific gravity 13.6 is placed in mercury of specific gravity 13.6, what fraction of it volume is under mercury
- a)the metal piece will simply float over the mercury
- b)the metal piece will be immersed in mercury by half
- c)whole of the metal piece will be immersed with its top surface just at mercury level
- d)metal piece will sink to the bottom
- e)none of the above
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
A piece of metal of specific gravity 13.6 is placed in mercury of spec...
Solution:
We can use the Archimedes' principle to solve this problem. According to this principle, any object immersed in a fluid experiences an upward force called the buoyant force, which is equal to the weight of the displaced fluid. The object will float if the buoyant force is equal to or greater than its weight, and sink otherwise.
Given data:
Specific gravity of metal, Sg1 = 13.6
Specific gravity of mercury, Sg2 = 13.6
Let us assume the volume of the metal piece to be V.
Applying Archimedes' principle, we have:
Buoyant force on the metal piece = weight of the displaced mercury
⇒ V × ρ2g = V × ρ1g
where g is the acceleration due to gravity, ρ1 is the density of the metal and ρ2 is the density of mercury.
Dividing both sides by Vg, we get:
ρ2/ρ1 = (V/V) = 1
Since the specific gravity of the metal is equal to that of mercury, we can conclude that the volume of the metal piece that is under mercury is equal to its total volume. Therefore, the whole of the metal piece will be immersed with its top surface just at mercury level.
Hence, the correct answer is option 'C'.
We can use the Archimedes' principle to solve this problem. According to this principle, any object immersed in a fluid experiences an upward force called the buoyant force, which is equal to the weight of the displaced fluid. The object will float if the buoyant force is equal to or greater than its weight, and sink otherwise.
Given data:
Specific gravity of metal, Sg1 = 13.6
Specific gravity of mercury, Sg2 = 13.6
Let us assume the volume of the metal piece to be V.
Applying Archimedes' principle, we have:
Buoyant force on the metal piece = weight of the displaced mercury
⇒ V × ρ2g = V × ρ1g
where g is the acceleration due to gravity, ρ1 is the density of the metal and ρ2 is the density of mercury.
Dividing both sides by Vg, we get:
ρ2/ρ1 = (V/V) = 1
Since the specific gravity of the metal is equal to that of mercury, we can conclude that the volume of the metal piece that is under mercury is equal to its total volume. Therefore, the whole of the metal piece will be immersed with its top surface just at mercury level.
Hence, the correct answer is option 'C'.
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Community Answer
A piece of metal of specific gravity 13.6 is placed in mercury of spec...
Because of very high surface tension only, metal piece could not immerse into the mercury. Hence the top surface of the metal piece at same level of the mercury surface. Since the buoyant force and the weight of the metal piece are same, and the height of centre of buoyancy and height of centre of gravity are at same level.
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A piece of metal of specific gravity 13.6 is placed in mercury of specific gravity 13.6, what fraction of it volume is under mercurya)the metal piece will simply float over the mercuryb)the metal piece will be immersed in mercury by halfc)whole of the metal piece will be immersed with its top surface just at mercury leveld)metal piece will sink to the bottome)none of the aboveCorrect answer is option 'C'. Can you explain this answer?
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