Ratio of radius of gyration of a hollow sphere to that of a solid cylinder of equal mass, for moment of Inertia about their diameter axis AB as shown in figure is $\sqrt\frac{8} {x }$. The value of x is:

Question

Two iron solid discs of negligible thickness have radii $R_1$ and $R_2$ and moment of inertia $I_1$ and $I_2$, respectively. For $R_2 = 2R_1$, the ratio of $I_1$ and $I_2$ would be $1/x$, where $x =$ _____.

Answer 1

The moment of inertia of a hollow sphere about its diameter axis is defined as: $ I_{\text{sphere}} = \frac{2}{3} MR^2 = M k_1^2 $, where $k_1$ represents the hollow sphere's radius of gyration.

The moment of inertia of a solid cylinder about its diameter axis is expressed as: $ I_{\text{cylinder}} = \frac{1}{12} M(4R^2) + \frac{1}{4} MR^2 + M(2R)^2 = \frac{67}{12} MR^2 = M k_2^2 $.

The ratio of the radii of gyration is calculated as: $ \frac{k_1}{k_2} = \sqrt{\frac{\frac{2}{3}}{\frac{67}{12}}} = \sqrt{\frac{8}{67}} $.

Consequently, $x$ equals 67.

Ratio of radius of gyration of a hollow sphere to that of a solid cylinder of equal mass, for moment of Inertia about their diameter axis AB as shown in figure is \(\sqrt\frac{8} {x }\). The value of x is:

The Correct Option is C

Solution and Explanation

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