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Rotational Motion question

2021 · Q161
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Rotational Motion question

2021 · Q161

NEETPhysicsRotational MotionMCQ+4 / −1
From a circular ring of mass 'M' and radius 'R' an arc corresponding to a 90∘^\circ∘ sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the center of the ring and perpendicular to the plane of the ring is 'K' times 'MR2'. Then the value of 'K' is :
  1. A
    18{1 \over 8}81​
  2. B
    34{3 \over 4}43​
  3. C
    78{7 \over 8}87​
  4. D
    14{1 \over 4}41​
View written solutionFree

Correct answer: B

Given that,

Mass of Ring = M; Radius of Ring = R

Now 90° arc is removed from circular ring, then

Mass removed = M4{M \over 4}4M​

∴\therefore∴ Mass of remaining portion = 3M4{{3M} \over 4}43M​

NEET 2021 Physics - Rotational Motion Question 14 English Explanation


I = MR2

I' = 3M4{{3M} \over 4}43M​R2

I' = 34{3 \over 4}43​I

∴\therefore∴ K = 34{3 \over 4}43​

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