Question

A rod has a mass of 100 g and length 1 m. Its radius of gyration perpendicular to its
length and passing through its centre of gravity is
(a)0.2886 m
(b)2.886 m
(C)0.02886 m
(d)0.2688 m
(e)Answer not known​

Answer 1

Answer:

a is correct answer

Formula for com for cylinder at centre and perpendicular to its length is ML²/12

Now relation between com and radius of gyration(K) is:-

ML²/12=MK²

L²/12=K²

1/12=K²

1/√12=K

K=0.2886751346

Answer 2

Answer:

A

Explanation:

This is book sum or you created sum sir???