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(D) The moment of inertia of a single rod of mass $m$ and length $L$ about an axis through its center is $I = \frac{mL^2}{12}$.For a square frame of f

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$16$

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(D) The moment of inertia of a single rod of mass $m$ and length $L$ about an axis through its center is $I = \frac{mL^2}{12}$.For a square frame of four rods,we use the parallel axis theorem for each rod to find its moment of inertia about the center of the square.The distance from the center of the rod to the center of the square is $d = \frac{L}{2}$.The moment of inertia of one rod about the center of the square is $I_{rod} = I_{cm} + md^2 = \frac{mL^2}{12} + m(\frac{L}{2})^2 = \frac{mL^2}{12} + \frac{mL^2}{4} = \frac{mL^2 + 3mL^2}{12} = \frac{4mL^2}{12} = \frac{mL^2}{3}$.Since there are four such rods,the total moment of inertia $I'$ is $I' = 4 	imes \frac{mL^2}{3} = \frac{4mL^2}{3}$.Since $I = \frac{mL^2}{12}$,we have $mL^2 = 12I$.Substituting this into the expression for $I'$,we get $I' = \frac{4}{3} 	imes (12I) = 16I$.

The moment of inertia of a thin rod about an axis passing through its center and perpendicular to its length $L$ is $I$. What will be the moment of inertia of a square frame formed by four such rods about an axis passing through its center and perpendicular to the plane of the square (in $I$)?

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