Two loops A and B of radii R_1 and R_2 are ma | vedclass

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(C) The moment of inertia of a loop (ring) about its central axis is given by $I = MR^2$.Thus,$I_A = M_1 R_1^2$ and $I_B = M_2 R_2^2$ ... $(i)$Since t

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$1: 3$

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(C) The moment of inertia of a loop (ring) about its central axis is given by $I = MR^2$.Thus,$I_A = M_1 R_1^2$ and $I_B = M_2 R_2^2$ ... $(i)$Since the loops are made from a uniform wire,the mass $M$ is proportional to the circumference $(2 \pi R)$.Let $m$ be the mass per unit length. Then $M_1 = 2 \pi R_1 m$ and $M_2 = 2 \pi R_2 m$.Therefore,$\frac{M_1}{M_2} = \frac{R_1}{R_2}$ ... (ii)Substituting (ii) into the ratio of moments of inertia:$\frac{I_A}{I_B} = \frac{M_1 R_1^2}{M_2 R_2^2} = \left(\frac{M_1}{M_2}\right) \left(\frac{R_1}{R_2}\right)^2 = \left(\frac{R_1}{R_2}\right) \left(\frac{R_1}{R_2}\right)^2 = \left(\frac{R_1}{R_2}\right)^3$.Given $\frac{I_A}{I_B} = 27$,we have $\left(\frac{R_1}{R_2}\right)^3 = 27$.Taking the cube root on both sides,$\frac{R_1}{R_2} = 3$.Therefore,$\frac{R_2}{R_1} = \frac{1}{3}$.

Two loops $A$ and $B$ of radii $R_1$ and $R_2$ are made from a uniform wire. If the moment of inertia of $A$ is $I_A$ and that of $B$ is $I_B$,then find the ratio $R_2 / R_1$ given that $I_A / I_B = 27$.

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