If ' $\mathrm{I}$ ' is moment of inertia of a thin circular disc about an axis passing through the tangent…
If ' $\mathrm{I}$ ' is moment of inertia of a thin circular disc about an axis passing through the tangent of the disc and in the plane of disc. The moment of inertia of same circular disc about an axis perpendicular to plane and passing through its centre is
$\frac{4 \mathrm{I}}{5}$
$\frac{2 I}{5}$
$\frac{4 \mathrm{I}}{3}$
$\frac{2 I}{3}$
Solution
M.I. of thin circular disc through the tangent in the plane of the disc is $I=\frac{5}{4} \mathrm{MR}^2$
$\Rightarrow \mathrm{MR}^2=\frac{4}{5} \mathrm{I}$
$\therefore \quad$ M.I. of thin circular disc about an axis perpendicular to plane and passing through its
$\text { centre }=\frac{M^2}{2}=\frac{\left(\frac{4}{5} \mathrm{~L}\right)}{2}=\frac{2 \mathrm{I}}{5}$