If the linear density of a rod of length L varies as `lambda` = A+xB , the position of its centre of mass is

If linear density of a rod of length 3m varies as lambda = 2 + x, them the position of the centre of gravity of the rod is

The liner density of a rod of length 2m varies as lambda = (2 + 3x) kg/m, where x is distance from end A . The distance of a center of mass from the end B will be

The linear density of a thin rod of length 1m lies as lambda = (1+2x) , where x is the distance from its one end. Find the distance of its center of mass from this end.

Mass is non-uniformly distributed over the rod of length l, its linear mass density varies linearly with length as lamda=kx^(2) . The position of centre of mass (from lighter end) is given by-

A thin rod of length 6 m is lying along the x-axis with its ends at x=0 and x=6m. Its linear density *mass/length ) varies with x as kx^(4) . Find the position of centre of mass of rod in meters.

If the linear mass density of a rod of length 3 m (lying from y=0 to y=3 m) varies as lambda =(2+y)kg/m,then position of the centre of mass of the rod from y=0 is nearly at

The density of a rod of length L varies as rho=A+Bx where x is the distance from the left end. Locate the centre of mass.

Two uniform rods A and B of lengths 5 m and 3m are placed end to end. If their linear densities are 3 kg/m and 2 kg/m, the position of their centre of mass from their interface is

The radius of gyration of a uniform rod of length L about an axis passing through its centre of mass is

A non–uniform thin rod of length L is placed along x-axis as such its one of ends at the origin. The linear mass density of rod is lambda=lambda_(0)x . The distance of centre of mass of rod from the origin is :