Moment of Inertia (M.I.) of four bodies having same mass M and radius 2 R are as follows I 1 = M.I. of solid sphere about its diameter I 2 = M.I. of solid cylinder about its axis I 3 = M.I. of solid circular disc about its diameter I 4 = M.I. of thin circular ring about its diameter If 2 I 2 + I 3 + I 4 = x I 1 then the value of x will be _____ .

Step-by-step solution

∵   2 I 2 + I 3 + I 4 = x I 1 Now For solid sphere about its diameter I 1 = 2 5 M R 2 , For solid cylinder about its axis I 2 = 1 2 M R 2 , For circular disc about its diameter I 3 = 1 4 M R 2 And for circular ring about its diameter I 4 = 1 2 M R 2 ⇒ 2 M R 2 2 + M R 2 4 + M R 2 2 = x × 2 5 M R 2 ⇒ 2 3 4 + 1 2 = x × 2 5 ⇒ 2 = x × 2 5 ⇒ x = 5