$m_{1}$ and $m_{2}$ are two bodies of masses separated by a distance $R$. The distance of the centre of mass of the bodies from the mass $m_{1}$ is

Three rods of the same mass are placed as shown in the figure. The coordinates of the centre of mass of the system are:

The centre of mass of a thin rectangular plate (as shown in the figure) with sides of length $a$ and $b$,whose mass per unit area $(\sigma)$ varies as $\sigma = \frac{\sigma_0 x}{a b}$ (where $\sigma_0$ is a constant),would be . . . . . .

Look at the drawing given in the figure which has been drawn with ink of uniform line-thickness. The mass of ink used to draw each of the two inner circles,and each of the two line segments is $m$. The mass of the ink used to draw the outer circle is $6m$. The coordinates of the centres of the different parts are: outer circle $(0,0)$,left inner circle $(-a, a)$,right inner circle $(a, a)$,vertical line $(0,0)$ and horizontal line $(0,-a)$. The $y$-coordinate of the centre of mass of the ink in this drawing is

Infinite rods of uniform mass density and lengths $L, L/2, L/4, \dots$ are placed one upon another up to infinity as shown in the figure. Find the $x-$ coordinate of the centre of mass.

Two particles whose masses are $10\,kg$ and $30\,kg$ and their position vectors are $\hat{i} + \hat{j} + \hat{k}$ and $-\hat{i} - \hat{j} - \hat{k}$ respectively would have the centre of mass at: