A straight rod of length L has one of its end at the origin and the other at `X=L`. If the mass per unit length of the rod is given by `Ax` where A is constant, where is its centre of mass?

A straight of length L extends from x=a to x=L+a. the gravitational force it exerts on a point mass 'm' at x=0, if the mass per unit length of the rod is A+Bx^(2), is given by :

A straight of length L extends from x=a to x=L+a. the gravitational force it exerts on a point mass 'm' at x=0, if the mass per unit length of the rod is A+Bx^(2), is given by :

A light rod of length l has two masses m_1 and m_2 attached to its two ends. The moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass is.

A rod of mass m and length L is non- uniform.The mass of rod per unit length (mu)], varies as "(mu)=bx" ,where b is a constant.The , centre of mass of the ,rod is given by

A straight rod of length 3l units slides with its ends A,B always on the x and y axes respectively,then the locus of the centriod of Delta OAB is

The centre of mass of a non uniform rod of length L whose mass per unit length lambda = (K x^(2))/(L) , where K is a constant and x is the distance from one end is