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A circular disc of radius r and thickness r/6 has moment of inertia I about an axis passing through its centre and perpendicular to its plane. It is melted and recasted to a solid sphere. The moment of inertia of the sphere about its diameter as axis of rotation is
- a)I/2
- b)2I/11
- c)I/10
- d)I/5
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
A circular disc of radius r and thickness r/6 has moment of inertia I ...
Volume oc circular disc = surface of area x thickness
The mass of the circular disc and sphere are the same
where R is the radius of the sphere,
Moment of interia of a circluar disc
Moment of interia of solid sphere
The mass of the circular disc and sphere are the same
where R is the radius of the sphere,
Moment of interia of a circluar disc
Moment of interia of solid sphere
Most Upvoted Answer
A circular disc of radius r and thickness r/6 has moment of inertia I ...
Solution:
Given that,
The moment of inertia of a circular disc of radius r and thickness r/6 about an axis passing through its centre and perpendicular to its plane is I.
Let's find the moment of inertia of the disc about an axis perpendicular to its plane passing through a point on its edge.
The perpendicular distance of any point on the edge from the axis is r/2.
Using perpendicular axis theorem,
I = (Ixx + Iyy)
Where Ixx is the moment of inertia about an axis passing through its centre and perpendicular to its plane and Iyy is the moment of inertia about an axis perpendicular to its plane passing through a point on its edge.
Ixx = (1/2)MR^2 + (1/12)M(R/3)^2 = (7/12)MR^2
Iyy = (1/2)MR^2
Therefore,
I = (7/12)MR^2 + (1/2)MR^2 = (11/12)MR^2
When the disc is melted and recasted to a solid sphere of the same radius, the mass of the sphere will be the same as that of the disc.
Let's find the moment of inertia of the sphere about its diameter as axis of rotation.
Using the formula for the moment of inertia of a solid sphere about its diameter,
I = (2/5)MR^2
Substituting M = (12/11)I/R^2, we get
I = (2/5) * (12/11)I * R^2 / R^2 = 24/55 * I
Therefore, the moment of inertia of the sphere about its diameter as axis of rotation is I/5.
Hence, option D is the correct answer.
Given that,
The moment of inertia of a circular disc of radius r and thickness r/6 about an axis passing through its centre and perpendicular to its plane is I.
Let's find the moment of inertia of the disc about an axis perpendicular to its plane passing through a point on its edge.
The perpendicular distance of any point on the edge from the axis is r/2.
Using perpendicular axis theorem,
I = (Ixx + Iyy)
Where Ixx is the moment of inertia about an axis passing through its centre and perpendicular to its plane and Iyy is the moment of inertia about an axis perpendicular to its plane passing through a point on its edge.
Ixx = (1/2)MR^2 + (1/12)M(R/3)^2 = (7/12)MR^2
Iyy = (1/2)MR^2
Therefore,
I = (7/12)MR^2 + (1/2)MR^2 = (11/12)MR^2
When the disc is melted and recasted to a solid sphere of the same radius, the mass of the sphere will be the same as that of the disc.
Let's find the moment of inertia of the sphere about its diameter as axis of rotation.
Using the formula for the moment of inertia of a solid sphere about its diameter,
I = (2/5)MR^2
Substituting M = (12/11)I/R^2, we get
I = (2/5) * (12/11)I * R^2 / R^2 = 24/55 * I
Therefore, the moment of inertia of the sphere about its diameter as axis of rotation is I/5.
Hence, option D is the correct answer.
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A circular disc of radius r and thickness r/6 has moment of inertia I about an axis passing through its centre and perpendicular to its plane. It is melted and recasted to a solid sphere. The moment of inertia of the sphere about its diameter as axis of rotation isa)I/2b)2I/11c)I/10d)I/5Correct answer is option 'D'. Can you explain this answer?
Question Description
A circular disc of radius r and thickness r/6 has moment of inertia I about an axis passing through its centre and perpendicular to its plane. It is melted and recasted to a solid sphere. The moment of inertia of the sphere about its diameter as axis of rotation isa)I/2b)2I/11c)I/10d)I/5Correct answer is option 'D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A circular disc of radius r and thickness r/6 has moment of inertia I about an axis passing through its centre and perpendicular to its plane. It is melted and recasted to a solid sphere. The moment of inertia of the sphere about its diameter as axis of rotation isa)I/2b)2I/11c)I/10d)I/5Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A circular disc of radius r and thickness r/6 has moment of inertia I about an axis passing through its centre and perpendicular to its plane. It is melted and recasted to a solid sphere. The moment of inertia of the sphere about its diameter as axis of rotation isa)I/2b)2I/11c)I/10d)I/5Correct answer is option 'D'. Can you explain this answer?.
A circular disc of radius r and thickness r/6 has moment of inertia I about an axis passing through its centre and perpendicular to its plane. It is melted and recasted to a solid sphere. The moment of inertia of the sphere about its diameter as axis of rotation isa)I/2b)2I/11c)I/10d)I/5Correct answer is option 'D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A circular disc of radius r and thickness r/6 has moment of inertia I about an axis passing through its centre and perpendicular to its plane. It is melted and recasted to a solid sphere. The moment of inertia of the sphere about its diameter as axis of rotation isa)I/2b)2I/11c)I/10d)I/5Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A circular disc of radius r and thickness r/6 has moment of inertia I about an axis passing through its centre and perpendicular to its plane. It is melted and recasted to a solid sphere. The moment of inertia of the sphere about its diameter as axis of rotation isa)I/2b)2I/11c)I/10d)I/5Correct answer is option 'D'. Can you explain this answer?.
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