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Physics - Rotation Question with Solution | TestHub

PhysicsRotationCalculation of M.I.Medium2 minPYQ_2021

PhysicsMediumsingle choice

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 $M R^{2}$ . Then the value of $K$ is___NEET 2021

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Answer:

Solution:

mass per unit length of the ring is

$λ = \frac{M}{2 π R}$

$∴$ Mass of remaining ring is $M^{'} = λ \times \frac{3}{4} (2 π R)$

$∴ M^{'} = \frac{3}{4} M$

Moment of inertia of remaining part is

$I^{'} = M^{'} R^{2} = \frac{3}{4} M R^{2} = K M R^{2}$

$∴ K = \frac{3}{4}$

Stream:NEETSubject:PhysicsTopic:RotationSubtopic:Calculation of M.I.

⏱ 2mℹ️ Source: PYQ_2021

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