MHT CET · Physics · Rotational Motion

From a disc of radius \(R\), a concentric dircular portion of radius \(r\) is cut out so as to leave an annular disc of mass \(M\). The moment of inertia
of this annular disc about the axis perpendicular to its plane and passing through its centre of gravity is

A \(\frac{1}{2} M\left(R^{2}+r^{2}\right)\)
B \(\frac{1}{2} M\left(R^{2}-r^{2}\right)\)
C \(\frac{1}{2} M\left(R^{4}+r^{4}\right)\)
D \(\frac{1}{2} M^{\prime}\left(R^{4}-r^{4}\right)\)

Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{1}{2} M\left(R^{2}+r^{2}\right)\)

Step-by-step Solution

Detailed explanation

The moment of inertia of this annular disc about the axis perpendicular to its plane will be \(\frac{1}{2} M\left(R^{2}+r^{2}\right)\)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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From a disc of radius \(R\), a concentric dircular portion of radius \(r\) is cut out so as to leave an annular disc of mass \(M\). The moment of inertiaof this annular disc about the axis perpendicular to its plane and passing through its centre of gravity is

Step-by-step Solution

From a disc of radius \(R\), a concentric dircular portion of radius \(r\) is cut out so as to leave an annular disc of mass \(M\). The moment of inertia
of this annular disc about the axis perpendicular to its plane and passing through its centre of gravity is