Question

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:

• A. 34
• B. 17
• C. 67
• D. 51

Answer 1

Answer: (C)

67

Answer 2

The moment of inertia of a hollow sphere about its diameter axis is given by:

$I_{\text{sphere}} = \frac{2}{3} MR^2 = M k_1^2$

where $k_1$ is the radius of gyration of the hollow sphere.

The moment of inertia of a solid cylinder about its diameter axis is:

$I_{\text{cylinder}} = \frac{1}{12} M(4R^2) + \frac{1}{4} MR^2 + M(2R)^2 = \frac{67}{12} MR^2 = M k_2^2$

Calculating the ratio of the radii of gyration:

$\frac{k_1}{k_2} = \sqrt{\frac{\frac{2}{3}}{\frac{67}{12}}} = \sqrt{\frac{8}{67}}$

Hence, the value of $x$ is 67.