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:
Circle

Updated On: Apr 17, 2025
  • 34
  • 17
  • 67
  • 51
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is C

Solution and Explanation

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.

Was this answer helpful?
4
0

Questions Asked in JEE Main exam

View More Questions