$A$ circular hole of radius $\left(\frac{a}{2}\right)$ is cut out of a circular disc of radius $'a'$ as shown in the figure. The centroid of the remaining circular portion with respect to point $'O'$ will be:

$A$ uniform square plate is shown in the figure. Four identical small squares are removed from its corners. If squares $1, 2,$ and $3$ are removed,where will the center of mass $(C.M.)$ be located?

$A$ uniform square plate of side $a$ has a hole of side $b$ cut out of it as shown in the figure. The hole is centered at $(a/4, a/4)$ from the center of the square plate. The distance of the center of mass of the remaining part from the center of the square plate is:

$A$ circular disc of radius $R$ is removed from one end of a bigger circular disc of radius $2R$. The centre of mass of the new disc is at a distance $\alpha R$ from the centre of the bigger disc. The value of $\alpha$ is

$A$ ring is formed by joining two uniform semi-circular rings $ABC$ and $ADC$. The mass of $ABC$ is thrice that of $ADC$. If the ring is hinged to a fixed support at $A$,it can rotate freely in a vertical plane. Find the value of $\tan \theta$,where $\theta$ is the angle made by the line $AC$ with the vertical in equilibrium.

$A$ circular hole of radius $\frac{R}{4}$ is made in a thin uniform disc having mass $M$ and radius $R$,as shown in the figure. The moment of inertia of the remaining portion of the disc about an axis passing through the point $O$ and perpendicular to the plane of the disc is: