The moment of inertia of a thin rod about an axis passing through its mid point and perpendicular to the rod is 2400 g cm². The length of the 400 g rod is nearly: (a) 8.5 cm (b) 17.5 cm (c) 20.7 cm (d) 72.0 cm

Answer: (a). I = (ML²)/(12) ⇒ L² = 12I/M L² = (12 × 2400)/(400) = 72 cm² L = √(72) ≈ 8.5 cm So the answer is (a) 8.5 cm.

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Moment of Inertia & Radius of Gyration

The moment of inertia of a thin rod about an axis passing through its mid point and perpendicular to the rod is $2400\text{ g cm}^{2}$ . The length of the $400\text{ g}$ rod is nearly:

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Answer: (a)

$I = \frac{ML^2}{12} \Rightarrow L^2 = \frac{12I}{M}$
$L^2 = \frac{12 \times 2400}{400} = 72\ \text{cm}^2$
$L = \sqrt{72} \approx 8.5\ \text{cm}$
So the answer is (a) 8.5 cm.

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