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. \(OF\) 

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 center of mass of the system at \(t=3\) seconds:
   

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:
1. \(\frac{L}{2}\)
2. \(\frac{L}{4}\)
3. \(\frac{3L}{4}\)
4. \(\frac{L}{3}\)

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. \(L/\)$\sqrt{2}$

Four-point masses each of value \(m\) are placed at the corners of a square ABCD of side \(l\). The moment of inertia of this system about an axis passing through A and parallel to BD will be:

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 coordinates of the centre of mass of a uniform plate of shape as shown in the figure are:

The radius of gyration of a uniform solid sphere about a tangent is:

1. $R\sqrt{\frac{2}{3}}$

2. $R\sqrt{\frac{2}{5}}$

3. $R\sqrt{\frac{5}{3}}$

4. $R\sqrt{\frac{7}{5}}$