Rotational Motion

JEE Main level, numerical answer, medium

47 of 280

Question 47 of 280

Moment of Inertia (M.I.) of four bodies having same mass MM and radius 2R2R are as follows
I1={I}_{1}= M.I. of solid sphere about its diameter
I2={I}_{2}= M.I. of solid cylinder about its axis
I3={I}_{3}= M.I. of solid circular disc about its diameter
I4={I}_{4}= M.I. of thin circular ring about its diameter
If 2(I2+I3)+I4=xI12\left({I}_{2}+{I}_{3}\right)+{I}_{4}=x{I}_{1} then the value of xx will be _____ .

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Answer: 5

∵2(I2+I3)+I4=xI1∵2\left({I}_{2}+{I}_{3}\right)+{I}_{4}=x{I}_{1}
Now
For solid sphere about its diameter I1=25MR2{I}_{1}=\dfrac{2}{5}M{R}^{2},
For solid cylinder about its axis I2=12MR2{I}_{2}=\dfrac{1}{2}M{R}^{2},
For circular disc about its diameter I3=14MR2{I}_{3}=\dfrac{1}{4}M{R}^{2}
And for circular ring about its diameter I4=12MR2{I}_{4}=\dfrac{1}{2}M{R}^{2}
⇒2(MR22+MR24)+MR22=x×25MR2\Rightarrow 2\left(\dfrac{M{R}^{2}}{2}+\dfrac{M{R}^{2}}{4}\right)+\dfrac{M{R}^{2}}{2}=x\times \dfrac{2}{5}M{R}^{2}
⇒2(34)+12=x×25\Rightarrow 2\left(\dfrac{3}{4}\right)+\dfrac{1}{2}=x\times \dfrac{2}{5}
⇒2=x×25\Rightarrow 2=x\times \dfrac{2}{5}
⇒x=5\Rightarrow x=5

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