$A$ uniform circular disc has radius $r$. $A$ square portion of diagonal $r$ is cut from it. The centre of mass of the remaining disc from the centre of the disc 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.

$A$ disc of mass $M$ with uniform surface mass density $\sigma$ is shown in the figure. The centre of mass of the quarter disc (the shaded area) is at the position $(\frac{x}{3} \frac{R}{\pi}, \frac{x}{3} \frac{R}{\pi})$,where $R$ is the radius of the disc and $x$ is ....... . (Round off to the nearest integer)

From a circular disc of radius $R$,a square is cut out with a radius as its diagonal. The center of mass of the remaining part is at a distance (from the centre) of:

$A$ 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:

$A$ solid cone is placed on a horizontal surface has height $h$,radius $R$ and apex angle $\theta$ as shown. If the gravitational potential energy of the cone does not change as the position of the cone is changed from figure $(A)$ to figure $(B)$,then,