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(D) Let the total mass of each body be $m$. The moment of inertia $(I)$ about an axis passing through the centre of mass and perpendicular to the plan

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Four rods forming a square of side $2a$

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(D) Let the total mass of each body be $m$. The moment of inertia $(I)$ about an axis passing through the centre of mass and perpendicular to the plane is:$1$. For a disc of radius $a$: $I_{	ext{disc}} = \frac{1}{2} m a^2 = 0.5 m a^2$.$2$. For a ring of radius $a$: $I_{	ext{ring}} = m a^2 = 1.0 m a^2$.$3$. For a square lamina of side $L = 2a$: Using the perpendicular axis theorem,$I_z = I_x + I_y$. Since $I_x = I_y = \frac{m L^2}{12}$,we have $I_{	ext{sq}} = \frac{m(2a)^2}{12} + \frac{m(2a)^2}{12} = \frac{4 m a^2}{12} + \frac{4 m a^2}{12} = \frac{8 m a^2}{12} = \frac{2}{3} m a^2 \approx 0.67 m a^2$.$4$. For four rods of length $L = 2a$ forming a square: Each rod has mass $m/4$. The moment of inertia of one rod about the centre of the square is found using the parallel axis theorem: $I_{	ext{rod}} = I_{	ext{cm}} + (m/4)d^2 = \frac{(m/4)(2a)^2}{12} + (m/4)a^2 = \frac{m a^2}{3} + \frac{m a^2}{4} = \frac{7}{12} m a^2$. For four rods,$I_{	ext{total}} = 4 	imes \frac{7}{12} m a^2 = \frac{7}{3} m a^2 \approx 2.33 m a^2$.Comparing the values,the four rods forming a square have the largest moment of inertia.

For the same total mass,which of the following will have the largest moment of inertia about an axis passing through its centre of mass and perpendicular to the plane of the body?

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