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(A) The moment of inertia of a ring of mass $M$ and radius $r$ about an axis passing through its center and perpendicular to its plane is $I = Mr^2$.S

Vedclass · Accepted Answer

$2$

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(A) The moment of inertia of a ring of mass $M$ and radius $r$ about an axis passing through its center and perpendicular to its plane is $I = Mr^2$.Since the rings are made of the same material and have the same cross-sectional area,their mass is proportional to their circumference ($M = \lambda \cdot 2\pi r$,where $\lambda$ is the linear mass density).For the first ring: $M_1 = \lambda \cdot 2\pi R$ and $I_1 = M_1 R^2 = (\lambda \cdot 2\pi R) R^2 = 2\pi \lambda R^3$.For the second ring: $M_2 = \lambda \cdot 2\pi (nR)$ and $I_2 = M_2 (nR)^2 = (\lambda \cdot 2\pi nR) (nR)^2 = 2\pi \lambda n^3 R^3$.The ratio of the moments of inertia is given as $\frac{I_1}{I_2} = \frac{1}{8}$.Substituting the expressions: $\frac{2\pi \lambda R^3}{2\pi \lambda n^3 R^3} = \frac{1}{n^3} = \frac{1}{8}$.Therefore,$n^3 = 8$,which gives $n = 2$.

Two rings of the same material have radii $R$ and $nR$ respectively. The ratio of their moments of inertia about an axis passing through their centers and perpendicular to their planes is $1 : 8$. The value of $n$ is:

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