In the figure,one-fourth part of a uniform disc of radius $R$ is shown. The distance of the centre of mass of this object from the centre $O$ is:

To find the centre of mass of a rigid body,why is it not possible to know $\sum m_i \vec{r}_i$ for all the particles?

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 . . . . . .

Two bodies of mass $1\,kg$ and $3\,kg$ have position vectors $\hat{i}+2\hat{j}+\hat{k}$ and $-3\hat{i}-2\hat{j}+\hat{k}$ respectively. The magnitude of the position vector of the centre of mass of this system will be equal to the magnitude of which of the following vectors?

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 $(3, 3, 3)$?

Two objects of masses $200 \text{ g}$ and $500 \text{ g}$ possess velocities $10 \hat{i} \text{ m/s}$ and $(3 \hat{i} + 5 \hat{j}) \text{ m/s}$ respectively. The velocity of their centre of mass in $\text{m/s}$ is: