1.

Derive, making use of an integral, the formula for the moment of inertia of a right circular cone about its height.

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Solution :For a CONE (Fig) `DM=(Mr^2dz)/(1//3hR^2), R = (Rz)/(h) , DI = (dm.r^2)/(2)=(3MR^2z^4dz)/(2h^5)`
`I=int_0^h(3MR^2z^4dz)/(2h^5)=(3MR^2)/(2h^5)int_0^hz^4dz.=(3MR^2h^2)/(2h^5xx5)=0.3MR^2`


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