From a uniform wire,two circular loops are ma | vedclass

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(D) Let the mass per unit length of the wire be $\lambda$.The circumference of loop $P$ is $2\pi r$,so its mass is $M_P = \lambda(2\pi r)$.The moment

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$2$

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(D) Let the mass per unit length of the wire be $\lambda$.The circumference of loop $P$ is $2\pi r$,so its mass is $M_P = \lambda(2\pi r)$.The moment of inertia of loop $P$ about an axis passing through its center and perpendicular to its plane is $I_P = M_P r^2 = \lambda(2\pi r)r^2 = 2\pi\lambda r^3$.The circumference of loop $Q$ is $2\pi(nr)$,so its mass is $M_Q = \lambda(2\pi nr) = n M_P$.The moment of inertia of loop $Q$ about a similar axis is $I_Q = M_Q (nr)^2 = (n M_P)(n^2 r^2) = n^3 M_P r^2 = n^3 I_P$.Given that $I_Q = 8 I_P$,we have $n^3 I_P = 8 I_P$.Therefore,$n^3 = 8$,which gives $n = 2$.

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