Please solve this question with explanation of the solution....

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,

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