Please solve this question with explanation of the solution.... Share with your friends Share 0 Shivam Kumar answered this Dear Student, Moment of Inertia of the cylinder along the diameter of the end = 14MR2+13ML2here L=6RSo,I=14MR2+13M(6R)2=494MR2As I=MRg2 Rg=Radius of gyrationMRg2=494MR2So,Rg=494R2=72R=3.5R Regards, 0 View Full Answer