Two loops P and Q of radii R_1 and R_2 are ma | vedclass

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(B) The moment of inertia of a loop of mass $M$ and radius $R$ is given by $I = MR^2$. Since the loops are made of the same material and wire,let $\la

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$3$

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(B) The moment of inertia of a loop of mass $M$ and radius $R$ is given by $I = MR^2$. Since the loops are made of the same material and wire,let $\lambda$ be the mass per unit length. The mass $M$ of a loop is $M = \lambda \cdot (2\pi R)$. Substituting this into the expression for $I$: $I = (\lambda \cdot 2\pi R) \cdot R^2 = 2\pi\lambda R^3$. Thus,$I \propto R^3$. Therefore,the ratio of the moments of inertia is: $\frac{I_P}{I_Q} = \left(\frac{R_1}{R_2}\right)^3$. Given $\frac{I_P}{I_Q} = 27$,we have: $27 = \left(\frac{R_1}{R_2}\right)^3$. Taking the cube root on both sides: $\frac{R_1}{R_2} = \sqrt[3]{27} = 3$. So,the ratio $R_1 / R_2$ is $3:1$.

Two loops $P$ and $Q$ of radii $R_1$ and $R_2$ are made from uniform metal wire of the same material. If $I_P$ and $I_Q$ are the moments of inertia of loops $P$ and $Q$ respectively,then the ratio $R_1 / R_2$ is (Given $I_P / I_Q = 27$) (in $:1$)

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