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Question

The moment of inertia of a uniform rod of length 2l and mass m about an axis (XX) passing through its centre and inclined at an angle α=30 is

A
ml212
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B
ml23
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C
ml26
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D
ml24
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Solution

The correct option is A ml212


Mass per unit length of the rod λ=m2l,
dm=λdx
The perpendicular distance of the element from the given axis is xsinα

The desired moment of inertia is

dI=(dm)(xsinα)2
I=x=+lx=ldI
=ll(m2l.dx)(xsinα)2
=ml23sin2α, Putting the value of α=30o we get,

I=ml212

Hence option A is the correct answer

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