The position vector of a particle in a circular motion about the origin sweeps out equal area in equal time. Its

If the density of the material of a square plate and a circular plate shown in the figure is same, the centre of mass of the composite system will be:

An isolated particle of mass $m$ is moving in a horizontal plane $\left(X-Y\right)$ along the $X-\mathrm{axis}$, at a certain height above the ground. It suddenly explodes into two fragments of masses $\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$ and $\raisebox{1ex}{$3m$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$. An instant later, the smaller fragment is at $y=+15 \mathrm{cm}$. The larger fragment at this instant is at:

The centre of mass of a system of particles does not depend on

Three identical metal balls, each of radius $\mathrm{r}$ are placed touching each other in a horizontal surface, such that an equilateral triangle is formed, when centre of the three balls are joined. The centre of mass of the system is located at

Two blocks of masses $10 \mathrm{kg}, 4 \mathrm{kg}$ are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An Impulse gives a velocity of $14 \mathrm{m}/\mathrm{s}$ to the heavier block in the direction of the lighter block. The velocity of the centre of mass is: