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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 ′MR2 '. 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 ' R ' an arc correspondin...
Given that,
Mass of Ring = M; Radius of Ring = R
Now 90° arc is removed from circular ring, then mass removed = M/4
Mass of remaining portion = 3M/4
Moment of inertia of remaining part = ∫d mr2
So the value of K is 3/4
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From a circular ring of mass ' M and radius ' R ' an arc correspondin...
Given Information:
- Mass of the circular ring = M
- Radius of the circular ring = R
- Moment of inertia of the remaining part of the ring = KMR²
Solution:
Step 1: Calculating the Moment of Inertia of the Original 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 = MR².
- Given that the moment of inertia of the remaining part after removing a 90° sector is KMR².
- Therefore, the moment of inertia of the removed sector is (1-K)MR².
- Since the sector is 90°, its moment of inertia is (1/4)MR².
- Thus, the moment of inertia of the original ring is I = MR² - (1/4)MR² = (3/4)MR².
Step 2: Calculating the Value of K
- Given that the moment of inertia of the remaining part after removing a 90° sector is KMR².
- From step 1, we know that the moment of inertia of the original ring is (3/4)MR².
- Therefore, K = (3/4) / 1 = 3/4.
Therefore, the value of K is 3/4, which corresponds to option 'a'.
- Mass of the circular ring = M
- Radius of the circular ring = R
- Moment of inertia of the remaining part of the ring = KMR²
Solution:
Step 1: Calculating the Moment of Inertia of the Original 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 = MR².
- Given that the moment of inertia of the remaining part after removing a 90° sector is KMR².
- Therefore, the moment of inertia of the removed sector is (1-K)MR².
- Since the sector is 90°, its moment of inertia is (1/4)MR².
- Thus, the moment of inertia of the original ring is I = MR² - (1/4)MR² = (3/4)MR².
Step 2: Calculating the Value of K
- Given that the moment of inertia of the remaining part after removing a 90° sector is KMR².
- From step 1, we know that the moment of inertia of the original ring is (3/4)MR².
- Therefore, K = (3/4) / 1 = 3/4.
Therefore, the value of K is 3/4, which corresponds to option 'a'.
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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 ′MR2 '. Then the value of ' K isa)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 ' 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 ′MR2 '. Then the value of ' K isa)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 ' 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 ′MR2 '. Then the value of ' K isa)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 ' 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 ′MR2 '. Then the value of ' K isa)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 ' 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 ′MR2 '. Then the value of ' K isa)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 ' 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 ′MR2 '. Then the value of ' K isa)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 ' 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 ′MR2 '. Then the value of ' K isa)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 ' 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 ′MR2 '. Then the value of ' K isa)3/4b)7/8c)1/4d)1/8Correct answer is option 'A'. Can you explain this answer?, a detailed solution for 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 ′MR2 '. Then the value of ' K isa)3/4b)7/8c)1/4d)1/8Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of 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 ′MR2 '. Then the value of ' K isa)3/4b)7/8c)1/4d)1/8Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
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