Moment of inertia of the isosceles right angled triangle ABC about the side BC is I. Then its moment of inertia about side AC is

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$\frac{I}{3}$

$\frac{I}{2}$

$\frac{2 I}{3}$

$\frac{I}{\sqrt{2}}$

answer is A.

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Detailed Solution

$\begin{array}{l} I = I_{B C} = \frac{1}{6} . m {(A B)}^{2} = \frac{1}{6} m a^{2} [A B = B C = a] \\ ∴ I_{A C} = \frac{1}{6} . m . {(A B \sin 45^{o})}^{2} = (\frac{1}{6} . m . a^{2}) \frac{1}{2} = \frac{I}{2} \end{array}$

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