MHT CET · Physics · Rotational Motion
A thin uniform rod has mass 'M' and length 'L'. The moment of inertia about an axis perpendicular to it and passing through the point at a distance \(\frac{\mathrm{L}}{3}\) from one of its ends, will be
- A \(\frac{\mathrm{ML}^{2}}{12}\)
- B \(\frac{7}{8} \mathrm{ML}^{2}\)
- C \(\frac{\mathrm{ML}^{2}}{9}\)
- D \(\frac{\mathrm{ML}^{2}}{3}\)
Answer & Solution
Correct Answer
(C) \(\frac{\mathrm{ML}^{2}}{9}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} \mathrm{I} &=\mathrm{I}_{0}+\mathrm{Mh}^{2} \\ \mathrm{~h} &=\frac{\mathrm{L}}{2}-\frac{\mathrm{L}}{3}=\frac{\mathrm{L}}{6} \\ \mathrm{I} &=\frac{\mathrm{ML}^{2}}{12}+\mathrm{Mh}^{2} \\ &=\frac{\mathrm{ML}^{2}}{12}+\frac{\mathrm{ML}^{2}}{36} \\ &=\frac{4 \mathrm{ML}^{2}}{36}=\frac{\mathrm{ML}^{2}}{9} \end{aligned}\)


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