From a uniform disc of radius R, an equilateral triangle of side `sqrt(3)R` is cut as shown in the figure. The new position of centre of mass is :

From a uniform disc of radius R, an equilibatered triangle of side sqrt3R is cut as shown. The new position of centre of mass is:

The figure shows a disc of radius 3R from which a circular hole of radius R is cut as shown in the figure. The distance of the centre of mass of the remaining object from the point O is

The figure shows a disc of radius 3R from which a circular hole of radius R is cut as shown in the figure. The distance of the centre of mass of the remaining object from the point O is

From a uniform disc of radius R, a small disc of radius (R)/(2) is cut and removed as shown in the diagram. Find the center of mass of the remaining portion of the disc.

From a uniform circular disc of radius R and mass 9M, a small disc of radius (R )/(3) is removed as shown in the figure. The moment of intertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through centre of the disc is

A thin disc of mass 9M and radius R is given.Now a disc of radius (R/3) is cut as shown in figure.The new position of C .m from orgin will be :-

From a hemisphere of radius R a cone of base radius R/w and height R is cut as shown in figure. Find the height of centre of mass of the remaining object.