For a uniform rectangular sheet shown in the figure, the ratio of moments of inertia about the axes perpendicular to the sheet and passing through O (the centre of mass) and O' (corner point) is :

Consider two uniform discs of the same thickness and different radii R₁ = R and
R₂ = $$\alpha $$R made of the same material. If the ratio of their moments of inertia I₁ and I₂ , respectively, about their axes is I₁ : I₂ = 1 : 16 then the value of $$\alpha $$ is :

A uniform rod of length ‘$$l$$’ is pivoted at one of its ends on a vertical shaft of negligible radius. When the shaft rotates at angular speed $$\omega $$ the rod makes an angle $$\theta $$ with it (see figure). To find $$\theta $$ equate the rate of change of angular momentum (direction going into the paper) $${{m{l^2}} \over {12}}{\omega ^2}\sin \theta \cos \theta $$ about the centre of mass (CM) to the torque provided by the horizontal and vertical forces F_H and F_V about the CM. The value of $$\theta $$ is then such that :

Moment of inertia of a cylinder of mass M, length L and radius R about an axis passing through its centre and perpendicular to the axis of the cylinder is
I = $$M\left( {{{{R^2}} \over 4} + {{{L^2}} \over {12}}} \right)$$. If such a cylinder is to be made for a given mass of a material, the ratio $${L \over R}$$ for it to have minimum possible I is