At \(t=0,\) the positions of the two blocks are shown. There is no external force acting on the system. Find the coordinates of the centre of mass of the system (in SI units) at \(t=3\) seconds.
       

1.	\((1,0)\)	2.	\((3,0)\)
3.	\((4.5,0)\)	4.	\((2.25,0)\)

A uniform square plate \(ABCD\) has a mass of \(10\) kg. If two point masses of \(5\) kg each are placed at the corners \(C\) and \(D\) as shown in the adjoining figure, then the centre of mass shifts to the mid-point of:
            
1. \(OH\)

2. \(DH\)

3. \(OG\)

4. \(OF\) 

The mass per unit length of a non-uniform rod of length \(L\) is given by \(\mu =λx^{2}\) where \(\lambda\) is a constant and \(x\) is the distance from one end of the rod. The distance between the centre of mass of the rod and this end is:

A particle rotating on a circular path of the radius \(\frac{4}{\pi}~\text{m}\) at \(300\) rpm reaches \(600\) rpm in \(6\) revolutions. If the angular velocity increases at a constant rate, find the tangential acceleration of the particle:
1. \(10\) m/s²
2. \(12.5\) m/s²
3. \(25\) m/s²
4. \(50\) m/s²

The center of mass of a system of particles does not depend upon:

A particle is moving with a constant velocity along a line parallel to the positive x-axis. The magnitude of its angular momentum with respect to the origin is:

A rigid body rotates with an angular momentum of \(L.\) If its kinetic energy is halved, the angular momentum becomes:
1. \(L\)
2. \(L/2\)
3. \(2L\)
4. \(L/\)$\sqrt{2}$

A rod is falling down with constant velocity \(V_0\) as shown. It makes contact with hinge A and rotates around it. The angular velocity of the rod just after the moment when it comes in contact with hinge A is: