Two point masses $m$ and $M$ are separated by a distance $L$. The distance of the centre of mass of the system from $m$ is

Particles of masses $m, 2m, 3m, \ldots, nm$ grams are placed on the same line at distances $l, 2l, 3l, \ldots, nl$ cm from a fixed point. The distance of the centre of mass of the particles from the fixed point in centimetres is:

The variation of density of a solid cylindrical rod of cross-sectional area $\alpha$ and length $L$ is given by $\rho = \rho_0 \frac{x^2}{L^2}$, where $x$ is the distance from one end of the rod. The position of its centre of mass from that end $(x=0)$ is:

Three identical spheres,each of mass $2 M$,are placed at the corners of a right-angled triangle with mutually perpendicular sides equal to $4 \ m$ each. Taking the point of intersection of these two sides as the origin,the magnitude of the position vector of the centre of mass of the system is $\frac{4 \sqrt{2}}{x}$,where the value of $x$ is . . . . . .

The center of mass of three particles of masses $10 \ kg, 20 \ kg$,and $30 \ kg$ is at $(0, 0, 0)$. Where should a fourth particle of mass $40 \ kg$ be placed so that the center of mass of the system shifts to $(3, 3, 3)$?

What do you mean by mass element $dm$?