Angular acceleration of a body of mass 50 kg under the action of a torque of magnitude 500 Nm is 25 `rad//s^(2)`. Then radius of gyration of the body about its axis of rotation is nearly equal to

Options

A. 0.2 m
B. 0.6 m
C. 0.4 m
D. 1 m

Answer

B

Explanation

`tau = I prop`
`I = (tau)/(prop)=(500)/(25)=20`
`K = sqrt((I)/(M))=sqrt((20)/(50))=sqrt(0.4) = 0.6 m`.