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 ?

Options

A. 1
B. `(3)/sqrt(2)`
C. `sqrt((3)/(2))`
D. `sqrt(3)/(2)`

Answer

C

Explanation

`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)`