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(C) Let $M$ be the mass, $R$ be the radius, and $L$ be the length of the uniform solid cylinder.The moment of inertia of the cylinder about its centra

Vedclass · Accepted Answer

$\sqrt{3(2 n-1)}$

Vedclass · Answer

(C) Let $M$ be the mass, $R$ be the radius, and $L$ be the length of the uniform solid cylinder.The moment of inertia of the cylinder about its central axis is $I_1 = \frac{1}{2}MR^2$.The moment of inertia of the cylinder about an axis passing through its center and perpendicular to its length is $I_2 = \frac{MR^2}{4} + \frac{ML^2}{12}$.According to the problem, $I_1 = \frac{1}{n} I_2$, which implies $n I_1 = I_2$.Substituting the expressions: $n \left( \frac{1}{2} MR^2 \right) = \frac{MR^2}{4} + \frac{ML^2}{12}$.Dividing by $M$: $\frac{nR^2}{2} = \frac{R^2}{4} + \frac{L^2}{12}$.Multiply by $12$: $6nR^2 = 3R^2 + L^2$.Rearranging for $L^2$: $L^2 = 6nR^2 - 3R^2 = 3R^2(2n - 1)$.Therefore, $\frac{L^2}{R^2} = 3(2n - 1)$, which gives $\frac{L}{R} = \sqrt{3(2n - 1)}$.

If the moment of inertia of a uniform solid cylinder about the axis of the cylinder is $\frac{1}{n}$ times its moment of inertia about an axis passing through its midpoint and perpendicular to its length, then the ratio of the length and radius of the cylinder is

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