1.

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)`

Answer» Correct Answer - C
Moment of inertia of a uniform cylinder of length `l and radius R` about its perpendicular bisector is
`I = (ml^(2))/(12) + (mR^(2))/(4) = (m)/(4)((l^(2))/(3)+R^(2))` ..(i)
and `m = pi R^(2)l.rho`
`:. R^(2) = (m)/(pi//rho)`
From(i), `I = (m)/(4) ((l^(2))/(3)+(m)/(pi lrho))`
`I` will be minimum, when
`(dI)/(dl) = 0`
`(d)/(dl) [(m)/(4) ((l^(2))/(3)+(m)/(pi lrho))] = 0`
`(m)/(4) ((2l^(2))/(3)+(m)/(pil^(2)rho)) = 0`
or `(2l)/(3) = (m)/(pil^(2)rho) = (piR^(2)l rho)/(pi l^(2)rho) = (R^(2))/(l)`
or `(l^(2))/(R^(2)) = (3)/(2) , (l)/(R ) = sqrt((3)/(2))`


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