Two uniforms, thin identical rods each of mass M and length L are joined at the middle so as to form a cross.Then:-

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Solution

The moment of inertia for each rod is $I = \frac{m l^{2}}{12}$ , about the middle point and perpendicular to the plane thus making the sum total for two rods as $2 I$ , i.e. $I = \frac{m l^{2}}{6}$

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Two uniforms, thin identical rods each of mass M and length L are joined at the middle so as to form a cross.Then:-

The moment of inertia for each rod is I=ml212 , about the middle point and perpendicular to the plane thus making the sum total for two rods as 2I , i.e. I=ml26

The moment of inertia for each rod is $I = \frac{m l^{2}}{12}$ , about the middle point and perpendicular to the plane thus making the sum total for two rods as $2 I$ , i.e. $I = \frac{m l^{2}}{6}$