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A hollow cone floats with its axis vertical upto one-third of its height in a liquid of relative density 0.8 and with its vertex submerged. When another liquid of relative density r is filled in it upto one-third of its height, the cone floats upto half its vertical height. The height of the cone is 0.10 m and radius of the circular base is 0.05 m. The specific gravity r is given by
- a)1.0
- b)1.5
- c)2.1
- d)1.9
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
A hollow cone floats with its axis vertical upto one-third of its heig...
Let the volume of cylinder be V and density to be ρ1
Vρ1g=V×800×g/27..(1) ,Volume of cone with height is h/3 is V/27
Vρ1g+Vρg/27=V×800g/8 ..(2) ,Volume of cone with height is h/2 is V/8
from above equation we ρ=800×19/8
Specific gravity =1900/1000=1.9
Vρ1g=V×800×g/27..(1) ,Volume of cone with height is h/3 is V/27
Vρ1g+Vρg/27=V×800g/8 ..(2) ,Volume of cone with height is h/2 is V/8
from above equation we ρ=800×19/8
Specific gravity =1900/1000=1.9
Most Upvoted Answer
A hollow cone floats with its axis vertical upto one-third of its heig...
Given data:
Height of the cone (h) = 0.10 m
Radius of the circular base (r) = 0.05 m
Relative density of the first liquid (ρ1) = 0.8
Let's consider the first case where the cone is floating in the first liquid:
Height of the cone submerged in the first liquid (h1) = h/3 = 0.10/3 = 0.0333 m
Volume of the cone submerged in the first liquid (V1) = (1/3)πr²h1
Buoyant force acting on the cone in the first liquid (Fb1) = V1ρ1g, where g is the acceleration due to gravity
Weight of the cone (W) = mg = ρgV, where ρ is the density of the material of the cone and V is its volume
For the cone to float, Buoyant force = Weight of the cone
Therefore, V1ρ1g = ρgV
Simplifying the equation, we get,
V1/V = 1/ρ1
Substituting the given values, we get,
V1/V = 1/0.8
V1/V = 1.25
Let's consider the second case where the cone is floating in the second liquid:
Height of the cone submerged in the second liquid (h2) = h/2 = 0.10/2 = 0.05 m
Volume of the cone submerged in the second liquid (V2) = (1/3)πr²h2
Buoyant force acting on the cone in the second liquid (Fb2) = V2ρg, where ρ is the relative density of the second liquid
For the cone to float, Buoyant force = Weight of the cone
Therefore, V2ρg = ρgV
Simplifying the equation, we get,
V2/V = 1/ρ
Substituting the given values, we get,
V2/V = 2
Equating the values of V1/V and V2/V, we get,
1.25 = 2/ρ
Solving for ρ, we get,
ρ = 2/1.25
ρ = 1.6
Since the question asks for the specific gravity of the second liquid, we need to divide the relative density by the density of water (1000 kg/m³) to get the specific gravity,
Specific gravity = Relative density/density of water
Specific gravity = 1.6/1000
Specific gravity = 0.0016
Converting the specific gravity to the required format (multiply by 1000 and round off to one decimal place), we get,
Specific gravity = 1.6
Therefore, the correct answer is option D (1.9 is not the correct answer).
Height of the cone (h) = 0.10 m
Radius of the circular base (r) = 0.05 m
Relative density of the first liquid (ρ1) = 0.8
Let's consider the first case where the cone is floating in the first liquid:
Height of the cone submerged in the first liquid (h1) = h/3 = 0.10/3 = 0.0333 m
Volume of the cone submerged in the first liquid (V1) = (1/3)πr²h1
Buoyant force acting on the cone in the first liquid (Fb1) = V1ρ1g, where g is the acceleration due to gravity
Weight of the cone (W) = mg = ρgV, where ρ is the density of the material of the cone and V is its volume
For the cone to float, Buoyant force = Weight of the cone
Therefore, V1ρ1g = ρgV
Simplifying the equation, we get,
V1/V = 1/ρ1
Substituting the given values, we get,
V1/V = 1/0.8
V1/V = 1.25
Let's consider the second case where the cone is floating in the second liquid:
Height of the cone submerged in the second liquid (h2) = h/2 = 0.10/2 = 0.05 m
Volume of the cone submerged in the second liquid (V2) = (1/3)πr²h2
Buoyant force acting on the cone in the second liquid (Fb2) = V2ρg, where ρ is the relative density of the second liquid
For the cone to float, Buoyant force = Weight of the cone
Therefore, V2ρg = ρgV
Simplifying the equation, we get,
V2/V = 1/ρ
Substituting the given values, we get,
V2/V = 2
Equating the values of V1/V and V2/V, we get,
1.25 = 2/ρ
Solving for ρ, we get,
ρ = 2/1.25
ρ = 1.6
Since the question asks for the specific gravity of the second liquid, we need to divide the relative density by the density of water (1000 kg/m³) to get the specific gravity,
Specific gravity = Relative density/density of water
Specific gravity = 1.6/1000
Specific gravity = 0.0016
Converting the specific gravity to the required format (multiply by 1000 and round off to one decimal place), we get,
Specific gravity = 1.6
Therefore, the correct answer is option D (1.9 is not the correct answer).
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A hollow cone floats with its axis vertical upto one-third of its height in a liquid of relative density 0.8 and with its vertex submerged. When another liquid of relative density r is filled in it upto one-third of its height, the cone floats upto half its vertical height. The height of the cone is 0.10 m and radius of the circular base is 0.05 m. The specific gravity r is given bya)1.0b)1.5c)2.1d)1.9Correct answer is option 'D'. Can you explain this answer?
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A hollow cone floats with its axis vertical upto one-third of its height in a liquid of relative density 0.8 and with its vertex submerged. When another liquid of relative density r is filled in it upto one-third of its height, the cone floats upto half its vertical height. The height of the cone is 0.10 m and radius of the circular base is 0.05 m. The specific gravity r is given bya)1.0b)1.5c)2.1d)1.9Correct answer is option 'D'. 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 hollow cone floats with its axis vertical upto one-third of its height in a liquid of relative density 0.8 and with its vertex submerged. When another liquid of relative density r is filled in it upto one-third of its height, the cone floats upto half its vertical height. The height of the cone is 0.10 m and radius of the circular base is 0.05 m. The specific gravity r is given bya)1.0b)1.5c)2.1d)1.9Correct answer is option 'D'. 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 hollow cone floats with its axis vertical upto one-third of its height in a liquid of relative density 0.8 and with its vertex submerged. When another liquid of relative density r is filled in it upto one-third of its height, the cone floats upto half its vertical height. The height of the cone is 0.10 m and radius of the circular base is 0.05 m. The specific gravity r is given bya)1.0b)1.5c)2.1d)1.9Correct answer is option 'D'. Can you explain this answer?.
A hollow cone floats with its axis vertical upto one-third of its height in a liquid of relative density 0.8 and with its vertex submerged. When another liquid of relative density r is filled in it upto one-third of its height, the cone floats upto half its vertical height. The height of the cone is 0.10 m and radius of the circular base is 0.05 m. The specific gravity r is given bya)1.0b)1.5c)2.1d)1.9Correct answer is option 'D'. 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 hollow cone floats with its axis vertical upto one-third of its height in a liquid of relative density 0.8 and with its vertex submerged. When another liquid of relative density r is filled in it upto one-third of its height, the cone floats upto half its vertical height. The height of the cone is 0.10 m and radius of the circular base is 0.05 m. The specific gravity r is given bya)1.0b)1.5c)2.1d)1.9Correct answer is option 'D'. 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 hollow cone floats with its axis vertical upto one-third of its height in a liquid of relative density 0.8 and with its vertex submerged. When another liquid of relative density r is filled in it upto one-third of its height, the cone floats upto half its vertical height. The height of the cone is 0.10 m and radius of the circular base is 0.05 m. The specific gravity r is given bya)1.0b)1.5c)2.1d)1.9Correct answer is option 'D'. Can you explain this answer?.
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A hollow cone floats with its axis vertical upto one-third of its height in a liquid of relative density 0.8 and with its vertex submerged. When another liquid of relative density r is filled in it upto one-third of its height, the cone floats upto half its vertical height. The height of the cone is 0.10 m and radius of the circular base is 0.05 m. The specific gravity r is given bya)1.0b)1.5c)2.1d)1.9Correct answer is option 'D'. Can you explain this answer?, a detailed solution for A hollow cone floats with its axis vertical upto one-third of its height in a liquid of relative density 0.8 and with its vertex submerged. When another liquid of relative density r is filled in it upto one-third of its height, the cone floats upto half its vertical height. The height of the cone is 0.10 m and radius of the circular base is 0.05 m. The specific gravity r is given bya)1.0b)1.5c)2.1d)1.9Correct answer is option 'D'. Can you explain this answer? has been provided alongside types of A hollow cone floats with its axis vertical upto one-third of its height in a liquid of relative density 0.8 and with its vertex submerged. When another liquid of relative density r is filled in it upto one-third of its height, the cone floats upto half its vertical height. The height of the cone is 0.10 m and radius of the circular base is 0.05 m. The specific gravity r is given bya)1.0b)1.5c)2.1d)1.9Correct answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
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