Question

Moment of inertia of a ring about an axis passing through its centre and ...... to
its plane is...​

Answer 1

Moment of Inertia of ring about an axis passing through its centre and perpendicular to it.

Let mass of ring is 'M' and radius is 'R'

cut an element, dx at circumference of ring.

So, mass of elementary length of ring, dm is given by dm

$dm = (m \div 2\pi \: r)dx \\ \\ i = (dm) {r}^{2} \\ \\ i = (m \div 2\pi \: r)dx \: {r}^{2} \\ \\ i = (mr \div 2\pi)2\pi \: r \\ \\ i = m{r}^{2}$

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