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Moment of inertia of a thin uniform rod of length $l$ about an axis passing through point A and perpendicular to the rod is $\frac{1}{4} m l^{2}$ . Then the minimum distance of A from one end of the rod is

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$(\frac{1}{2} - \frac{1}{\sqrt{6}}) l$

$(\frac{1}{\sqrt{6}} - \frac{1}{8}) l$

$(\frac{1}{\sqrt{6}} + \frac{1}{4}) l$

$\frac{l}{\sqrt{6}}$

answer is A.

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Detailed Solution

If x be the distance of A from the centre of the rod (centre of mass), Then by parallel axis theorm

$\frac{1}{4} m l^{2} = \frac{1}{12} m l^{2} + m x^{2} \Rightarrow x^{2} = \frac{l^{2}}{6} \Rightarrow x = \frac{l}{\sqrt{6}}$

Therefore, minimum distance of A from one end $= (\frac{l}{2} - \frac{l}{\sqrt{6}})$

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