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

As we know, moment of inertia of a solid cylinder about an axis which is perpendicular bisector
`I=(mR^(2))/(4)+(ml^(2))/(12)`
`I=(m)/(4)[R^(2)+(l^(2))/(3)]`
Let V = Volume of cylinder = `piR^(2)l`
`=(m)/(4)[(V)/(pil)+(l^(2))/(3)]implies(dI)/(dl)=(m)/(4)[(-V)/(pil^(2))+(2l)/(3)]=0`
`(V)/(pil^(2))=(2l)/(3)impliesV=(2pil^(3))/(3)`
`piR^(2)l=(2pil^(3))/(3)implies(l^(2))/(R^(2))=(3)/(2)or,(l)/(R)=sqrt((3)/(2))`