Four identical rods, each of mass m and length l , are joined to form a rigid square frame. The frame lies in the xy $-$ plane , with its centre at the origin and the sides parallel to the x and y axes. Its moment of inertia about:

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One side is $\frac{5}{2} {ml}^{2}$

the x $-$ axis is $\frac{2}{3} {ml}^{2}$

An axis parallel to the z $-$ axis and passing through a corner is $\frac{10}{3} {ml}^{2}$

the z $-$ axis is $\frac{4}{3} {ml}^{2}$

answer is A, B, C.

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Detailed Solution

Due to symmetry $I_{x} = I_{y} = \frac{{mL}^{2}}{12} \times 2 + \{0 + m {(\frac{L}{2})}^{2}\} \times 2 = \frac{2}{3} {mL}^{2}$ and $I_{z} = I_{y} + I_{x} = 2 (\frac{2}{3} {ml}^{2}) = \frac{4}{3} {ml}^{2}$

$I_{c or n e r} = {(I_{z})}_{c . m} + 4 m {(\frac{L}{\sqrt{2}})}^{2} \Rightarrow I_{c or n e r} = \frac{4}{3} m L^{2} + 2 m L^{2} = \frac{10}{3} m L^{2}$