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The moment of inertia of a thin uniform rod of mass ' $M$ ' and length ' $L$ ' about an axis passing through a point at a distance $\frac{L}{4}$ from one of its ends and perpendicular to the length of the rod is
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The correct answer is:
$\frac{7 \mathrm{ML}^2}{48}$
The moment of inertia of the rod about an axis passing through the center and perpendicular to its length is given by
$\mathrm{I}_0=\frac{\mathrm{ML}^2}{12}$
Hence by parallel axis theorem,
$\mathrm{I}=\mathrm{I}_0+\mathrm{M}\left(\frac{\mathrm{L}}{4}\right)^2=\frac{\mathrm{ML}^2}{12}+\frac{\mathrm{ML}^2}{16}=\frac{7 \mathrm{ML}^2}{48}$
$\mathrm{I}_0=\frac{\mathrm{ML}^2}{12}$
Hence by parallel axis theorem,
$\mathrm{I}=\mathrm{I}_0+\mathrm{M}\left(\frac{\mathrm{L}}{4}\right)^2=\frac{\mathrm{ML}^2}{12}+\frac{\mathrm{ML}^2}{16}=\frac{7 \mathrm{ML}^2}{48}$
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