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Two uniform rods of mass ‘m’ and length ‘l’ form a cross, moment of inertia of cross about an axis parallel to CD passing through A is

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

$\frac{5 m l^{2}}{12}$

$\frac{7}{12} m l^{2}$

$\frac{m l^{2}}{6}$

answer is A.

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Detailed Solution

moment of inertia of rod AB about an axis passing through A and parallel to CD $\frac{m l^{2}}{3}$ and that for rod CD is $\frac{m l^{2}}{4}$ so total moment of inertia is $\frac{7 m l^{2}}{12}$

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