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Moment of inertia of a thin uniform rod of mass M and length L about an axis through its centre perpendicular to its length is
AML²/3
BML²/6
CML²/2
DML²/12
Answer & Solution
Correct answer: D. ML²/12
1. Treat the rod as a continuous line distribution of mass with linear density λ = M/L.
2. Integrate r² dm from −L/2 to L/2: I = ∫(−L/2 to L/2) x² (M/L) dx.
3. Evaluating gives I = (M/L)(L³/12) = ML²/12.
_Source: Maharashtra Balbharati Std XII Physics, Ch 1 "Rotational Dynamics" §1.6_
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