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(C) The moment of inertia of a circular loop of mass $M$ and radius $R$ about its central axis is $I = MR^2$.Let $m$ be the mass per unit length of th

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$2$

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(C) The moment of inertia of a circular loop of mass $M$ and radius $R$ about its central axis is $I = MR^2$.Let $m$ be the mass per unit length of the wire. Then the masses of the loops are $M_P = 2\pi R_1 m$ and $M_Q = 2\pi R_2 m$.The moments of inertia are $I_P = M_P R_1^2 = (2\pi R_1 m) R_1^2 = 2\pi m R_1^3$ and $I_Q = M_Q R_2^2 = (2\pi R_2 m) R_2^2 = 2\pi m R_2^3$.Taking the ratio: $\frac{I_P}{I_Q} = \frac{2\pi m R_1^3}{2\pi m R_2^3} = \left(\frac{R_1}{R_2}\right)^3$.Given $\frac{I_P}{I_Q} = \frac{1}{8}$,we have $\left(\frac{R_1}{R_2}\right)^3 = \frac{1}{8}$.Taking the cube root on both sides: $\frac{R_1}{R_2} = \frac{1}{2}$.Therefore,$\frac{R_2}{R_1} = 2$.

Two circular loops $P$ and $Q$ are made from a uniform wire. The radii of $P$ and $Q$ are $R_1$ and $R_2$ respectively. The moments of inertia about their own axis are $I_{P}$ and $I_{Q}$ respectively. If $\frac{I_{P}}{I_{Q}}=\frac{1}{8}$,then $\frac{R_2}{R_1}$ is

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