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An equilateral triangle ABC is fabricated by three identical rods each of mass m and length l. Then the moment of inertia of the triangular frame about an axis passing through vertex A and perpendicular to the plane of the triangular frame is

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$\frac{4}{3} m l^{2}$

$\frac{3}{4} m l^{2}$

$\frac{2}{3} m l^{2}$

$\frac{3}{2} m l^{2}$

answer is D.

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Detailed Solution

$I_{A} = I_{A B} + I_{B C} + I_{A C}$

Now $A M = l \sin 60^{o} = \frac{\sqrt{3} l}{2}$

$∴ I_{A} = \frac{1}{3} m l^{2} + [\frac{1}{12} m l^{2} + m {(\frac{\sqrt{3} l}{2})}^{2}] + \frac{1}{3} m l^{2}$

$= \frac{2}{3} m l^{2} + \frac{5}{6} m l^{2} = \frac{3}{2} m l^{2}$

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