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Physics · PYQ NEET UG 2016(1)
हिन्दीFrom a disc of radius \( R \) and mass \( M \), a circular hole of diameter \( R \), whose rim passes through the centre is cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis, passing through the centre?
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Correct answer · Option B
\( 13 MR^2/32 \)
Explanation
Mass of removed disc is \( M' = M/4 \), radius \( R' = R/2 \). Using parallel axis theorem, moment of inertia of the removed part about the centre is \( I' = \frac{1}{2}M'(R/2)^2 + M'(R/2)^2 = \frac{3}{8}M'(R/2)^2 \). Subtracting this from original \( I = \frac{1}{2}MR^2 \) gives \( 13 MR^2/32 \).
