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From a circular ring of mass 'M' and radius 'R'an arc corresponding to a 90∘ sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the centre of the ring and perpendicular to the plane of the ring is 'K' times ′MR 2 ′. Then the value of 'K' is :
- a)3/4
- b)7/8
- c)1/4
- d)1/8
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
From a circular ring of mass M and radius Ran arc corresponding to a 9...
mass per unit length of the ring is
λ = M/2πR
∴ Mass of remaining ring is M′ = λ x 3/4 (2πR)
∴ M′ = 3/4 M.
Moment of inertia of remaining part is
I′ = M′R2 = 3/4 MR2 = KMR2
∴ K = 3/4
λ = M/2πR
∴ Mass of remaining ring is M′ = λ x 3/4 (2πR)
∴ M′ = 3/4 M.
Moment of inertia of remaining part is
I′ = M′R2 = 3/4 MR2 = KMR2
∴ K = 3/4
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From a circular ring of mass M and radius Ran arc corresponding to a 9...
To find the moment of inertia of the remaining part of the ring, we need to subtract the moment of inertia of the removed arc from the moment of inertia of the entire ring.
The moment of inertia of a circular ring about an axis passing through its center and perpendicular to its plane is given by:
I = (1/2) * M * R^2
where M is the mass of the ring and R is the radius of the ring.
Since an arc corresponding to a 90° sector is removed, the remaining part of the ring can be considered as a semicircle. The moment of inertia of a semicircle about an axis passing through its center and perpendicular to its plane is given by:
I_arc = (1/4) * M * R^2
To find the moment of inertia of the remaining part of the ring, we subtract the moment of inertia of the arc from the moment of inertia of the entire ring:
I_remaining = I - I_arc
= (1/2) * M * R^2 - (1/4) * M * R^2
= (2/4) * M * R^2 - (1/4) * M * R^2
= (1/4) * M * R^2
Therefore, the moment of inertia of the remaining part of the ring is (1/4) * M * R^2.
Now, we are given that the moment of inertia of the remaining part of the ring is K times something. Let's call that something I_0.
(1/4) * M * R^2 = K * I_0
Dividing both sides of the equation by I_0, we get:
(1/4) * M * R^2 / I_0 = K
Therefore, K is equal to (1/4) * M * R^2 / I_0.
The moment of inertia of a circular ring about an axis passing through its center and perpendicular to its plane is given by:
I = (1/2) * M * R^2
where M is the mass of the ring and R is the radius of the ring.
Since an arc corresponding to a 90° sector is removed, the remaining part of the ring can be considered as a semicircle. The moment of inertia of a semicircle about an axis passing through its center and perpendicular to its plane is given by:
I_arc = (1/4) * M * R^2
To find the moment of inertia of the remaining part of the ring, we subtract the moment of inertia of the arc from the moment of inertia of the entire ring:
I_remaining = I - I_arc
= (1/2) * M * R^2 - (1/4) * M * R^2
= (2/4) * M * R^2 - (1/4) * M * R^2
= (1/4) * M * R^2
Therefore, the moment of inertia of the remaining part of the ring is (1/4) * M * R^2.
Now, we are given that the moment of inertia of the remaining part of the ring is K times something. Let's call that something I_0.
(1/4) * M * R^2 = K * I_0
Dividing both sides of the equation by I_0, we get:
(1/4) * M * R^2 / I_0 = K
Therefore, K is equal to (1/4) * M * R^2 / I_0.
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From a circular ring of mass M and radius Ran arc corresponding to a 90 sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the centre of the ring and perpendicular to the plane of the ring is K times ′MR2 ′. Then the value of K is :a)3/4b)7/8c)1/4d)1/8Correct answer is option 'A'. Can you explain this answer?
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From a circular ring of mass M and radius Ran arc corresponding to a 90 sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the centre of the ring and perpendicular to the plane of the ring is K times ′MR2 ′. Then the value of K is :a)3/4b)7/8c)1/4d)1/8Correct answer is option 'A'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about From a circular ring of mass M and radius Ran arc corresponding to a 90 sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the centre of the ring and perpendicular to the plane of the ring is K times ′MR2 ′. Then the value of K is :a)3/4b)7/8c)1/4d)1/8Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for From a circular ring of mass M and radius Ran arc corresponding to a 90 sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the centre of the ring and perpendicular to the plane of the ring is K times ′MR2 ′. Then the value of K is :a)3/4b)7/8c)1/4d)1/8Correct answer is option 'A'. Can you explain this answer?.
From a circular ring of mass M and radius Ran arc corresponding to a 90 sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the centre of the ring and perpendicular to the plane of the ring is K times ′MR2 ′. Then the value of K is :a)3/4b)7/8c)1/4d)1/8Correct answer is option 'A'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about From a circular ring of mass M and radius Ran arc corresponding to a 90 sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the centre of the ring and perpendicular to the plane of the ring is K times ′MR2 ′. Then the value of K is :a)3/4b)7/8c)1/4d)1/8Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for From a circular ring of mass M and radius Ran arc corresponding to a 90 sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the centre of the ring and perpendicular to the plane of the ring is K times ′MR2 ′. Then the value of K is :a)3/4b)7/8c)1/4d)1/8Correct answer is option 'A'. Can you explain this answer?.
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