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A uniform circular disc of radius $R$ and mass $M$ has a circular hole of radius $R/2$ cut out, **centred** at distance $R/2$ from the centre of the disc. Moment of inertia of the remaining lamina about an axis through the **original centre**, perpendicular to the plane, is:
A$\tfrac{13}{32}MR^2$
B$\tfrac{1}{2}MR^2$
C$\tfrac{15}{32}MR^2$
D$\tfrac{1}{4}MR^2$
Answer & Solution
Correct answer: A. $\tfrac{13}{32}MR^2$
Full disc: $I_{\text{full}} = MR^2/2$. Removed mass (disc of radius $R/2$ has area 1/4 of original, so mass $M/4$). Its MoI about its OWN centre = $(M/4)(R/2)^2/2 = MR^2/32$. Shift to original centre (parallel axis): $+ (M/4)(R/2)^2 = MR^2/16$. So $I_{\text{removed}} = MR^2/32 + MR^2/16 = 3MR^2/32$. Remaining: $I = MR^2/2 - 3MR^2/32 = 16MR^2/32 - 3MR^2/32 = 13MR^2/32$.
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