Three rods each of mass m and length l are joined together to form an equilateral triangle. The moment of inertia of the system about an axis passing through the centre of mass of the system and perpendicular to the plane of the triangle is

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$\frac{1}{2} {ml}^{2}$

$\frac{5}{12} {ml}^{2}$

$\frac{1}{3} {ml}^{2}$

$\frac{7}{12} {ml}^{2}$

answer is A.

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Detailed Solution

MI of each rod about an axis passing through its centre and perpendicular to its length

$= \frac{{ml}^{2}}{12}$

$d = \frac{1}{2 \sqrt{3}} . l$

MI of each rod about required axis of rotation

(in figure through c)

$I_{c} = \frac{{ml}^{2}}{12} + m {(\frac{l}{2 \sqrt{3}})}^{2} = \frac{{ml}^{2}}{6}$

for the system MI $= 3 \cdot \frac{{ml}^{2}}{6} = \frac{{ml}^{2}}{2}$

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