A long cylinderical rod is welded to a thin circular disc of diameter 0.5 m at a point on its circumference.
The rod is in the same plane as that of the disc and forms a tangent to the disc.
The radius of gyration of the disc about the rod (in m ) is

Options

A. `1/4`
B. `sqrt(5/8)`
C. `1/2`
D. `2sqrt2`

Answer

B

Explanation

As we know moment of inertia of disc tangents to the rim and parallel to diameter ,
`I = 5/4 MR^2`
Now, radius of gyration,
`K = sqrt(I/M) = sqrt((5/4 MR^2)/M) = sqrt(5/4) R`
As , given diameter of disc , R = 0.5 m
So, ` k= sqrt(5/4) xx 0.5 rArr k = sqrt(5/8)`