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For a wire of mass 1.2 kg and length 2 metres its moment of inertia about an axes passing through its centre of mass and perpendicular to its length is

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$0.1 {kg m}^{2}$

$0.2 {kg m}^{2}$

$0.3 {kg m}^{2}$

$0.4 {kg m}^{2}$

answer is D.

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Detailed Solution

$I = \frac{{ML}^{2}}{12} = \frac{1.2 (2^{2})}{12} = 0.4$

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