The moment of inertia of a uniform cylinder of length `l and radius R` about its perpendicular bisector is `I`. What is the ratio `l//R` such that the moment of inertia is minimum ?
A
1
B
`(3)/sqrt(2)`
C
`sqrt((3)/(2))`
D
`sqrt(3)/(2)`
Text Solution
Verified by Experts
The correct Answer is:
C
`I=(ml^(2))/(12)+(mR^(2))/(4)` or `1=(m)/(4)((l^(3))/(3)+R^(2))` .....(1) Also `m=piR^(2)lp` `rArrR^(2)=(m)/(pilp)` Put in equation (1) `I=(m)/(4)((l^(2))/(3)+(m)/(pilp))` For maxium `&` minima `(dI)/(dl)=(m)/(4)((2l)/(3)-(m)/(pil^(2)p))=0` `rArr(2l)/(3)=(m)/(pil^(2)p)rArr(2l)/(3)=(piR^(2)lp)/(pil^(2)p)` or `(2l)/(3)=(R^(2))/(l) rArr (l)/(R)=sqrt(3)/(2)`
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