Question

Moment of inertia of a disc of mass $M$ and radius ' $R$ ' about any of its diameter is $\frac{M^2}{4}$. The moment of inertia of this disc about an axis normal to the disc and passing through a point on its edge will be, $\frac{x}{2}M{R}^2$. The value of $x$ is __

Answer 1

Answer: 3

$I=I_{cm}+M{d}^2$
$=\frac{M{R}^2}{2}+M{R}^2$
$=\frac{3}{2}M{R}^2$
$x=3$