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The moment of inertia of a uniform cylin...

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

`(3)/(sqrt(2))`

B

`sqrt((3)/(2))`

C

`(sqrt(3))/(2)`

D

`1`

Text Solution

Verified by Experts

The correct Answer is:
B
`I=(Ml^(2))/(12)+(MR^(2))/(4)`
= `(Ml^(2))/(12)+(M)/(4)xx(M)/(rho pi l)[ :. M=(piR^(2)l)piimplies(M)/(rho pi l)=R^(2)]`
`(DI)/(dl)=(M)/(12)(2l)-(M^(2))/(4pirho)((1)/(l^(2)))=0`
`(l)/(6)=(M)/(4 rho pi l^(2))`
`l^(3)=(3M)/(2 rho pi)=(3)/(2 prho pi)xxpiR^(2)l rho`
`(l^(2))/(R^(2))=(3)/(2)implies(l)/(R)=sqrt((3)/(2))`
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