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Answer: B. \(\dfrac{MR^2}{2}\)

Answer: (b) \(\dfrac{MR^2}{2}\)

A disc can be treated as a set of thin concentric rings. A ring of radius \(x\) and width \(dx\) has mass \(dm = \dfrac{M}{\pi R^2}\,2\pi x\,dx\) and MI \(dm\,x^2\). Integrating, \(I = \dfrac{2M}{R^2}\displaystyle\int_0^R x^3\,dx = \dfrac{2M}{R^2}\cdot\dfrac{R^4}{4} = \dfrac{MR^2}{2}\). (\(MR^2/4\) is the MI about a diameter.)

Doubtly AI · 29 Oct 2026 AI-generated · not yet verified Report
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