Moment of Inertia (M.I.) of four bodies having same mass M and radius 2R are as follows I_1= M.I. of solid sphere about its diameter I_2= M.I. of…

Question

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 _____ .

MARKS Content Team · Accepted Answer

Correct answer: 5. ∵ 2I_2+I_3+I_4=xI_1 Now For solid sphere about its diameter I_1=2/5MR^2, For solid cylinder about its axis I_2=1/2MR^2, For circular disc about its diameter I_3=1/4MR^2 And for circular ring about its diameter I_4=1/2MR^2 ⇒2(MR^2)/2+(MR^2)/4+(MR^2)/2=x×2/5MR^2 ⇒23/4+1/2=x×2/5 ⇒2=x×2/5 ⇒x=5

Solution

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