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 L/4 from one of its ends and perpendicular to the length of the rod is (a) (ML²)/(48) (b) (7ML²)/(48) (c) (5ML²)/(48) (d) (9ML²)/(48)

Answer: (b). Distance of the axis from the centre: d = L/2 - L/4 = L/4. I = (ML²)/(12) + (ML²)/(16) = (7ML²)/(48) So the answer is (b) (7ML²)/(48).

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Moment of Inertia & Radius of Gyration

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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Answer: (b)

Distance of the axis from the centre: $d = \frac{L}{2} - \frac{L}{4} = \frac{L}{4}$ .
$I = \frac{ML^2}{12} + \frac{ML^2}{16} = \frac{7ML^2}{48}$
So the answer is (b) $\frac{7ML^2}{48}$ .

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