Two rings of radius R and nR made of the same | vedclass

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(D) Let the mass per unit length be $\lambda$. The mass of a ring is $M = \lambda 	imes (2\pi r)$.For the first ring of radius $R$,mass $M_1 = \lambd

Vedclass · Accepted Answer

$2$

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(D) Let the mass per unit length be $\lambda$. The mass of a ring is $M = \lambda 	imes (2\pi r)$.For the first ring of radius $R$,mass $M_1 = \lambda (2\pi R)$. The moment of inertia is $I_1 = M_1 R^2 = \lambda (2\pi R) R^2 = 2\pi \lambda R^3$.For the second ring of radius $nR$,mass $M_2 = \lambda (2\pi nR) = n M_1$. The moment of inertia is $I_2 = M_2 (nR)^2 = (n M_1) (n^2 R^2) = n^3 M_1 R^2 = n^3 I_1$.Given the ratio $\frac{I_1}{I_2} = \frac{1}{8}$.Substituting $I_2 = n^3 I_1$,we get $\frac{I_1}{n^3 I_1} = \frac{1}{n^3} = \frac{1}{8}$.Therefore,$n^3 = 8$,which gives $n = 2$.

Two rings of radius $R$ and $nR$ made of the same material have a ratio of moment of inertia about an axis passing through their centers and perpendicular to their planes as $1:8$. The value of $n$ is (mass per unit length is constant).

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