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CGPET 2021 Solved Paper

Section: Physics

Question : 4 of 150

Marks: +1, -0

M

l

Solution:

Moment of inertia of a rod of mass

M

and length

l

about an axis perpendicular to it passing through its centre of mass,

I_{CM} = \frac{Ml^2}{12}

According to parallel axes theorem, moment of inertia of rod perpendicular to it and passing through one end,

I=I_{CM} + M\left(\frac{l}{2}\right)^2

= \frac{Ml^2}{12} + \frac{Ml^2}{4} = \frac{Ml^2}{3}

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