The radius of gyration of a disc of mass 50 g and radius 2.5 cm, about an axis passing through its centre of gravity and perpendicular to the plane is

Options

A. 0.52cm
B. 1.76cm
C. 3.54cm
D. 6.54cm

Answer

B

Explanation

`I= (MR^(2))/(2)= (0.05 xx (2.5 xx 10^(-2))^(2))/(2)`
Now `I= MK^(2)`
`rArr (0.05 xx (2.5 xx 10^(-2))^(2))/(2)= MK^(2)`
`rArr K= (2.5)/(sqrt2)cm= 1.76cm`