A point \(P\) lies on the axis of a ring of mass \(M\) and radius \(a\) at a distance \(a\) from its centre \(C\). A small particle starts from \(P\) and reaches \(C\) under gravitational attraction. Its speed at \(C\) will be:
1. \(\sqrt{\frac{2 GM}{a}}\)
2. \(\sqrt{\frac{2 GM}{a} \left(1 - \frac{1}{\sqrt{2}}\right)}\)
3. \(\sqrt{\frac{2 GM}{a} \left(\sqrt{2} - 1\right)}\)
4. zero

If g is the acceleration due to gravity on the earth's surface, the gain in the potential energy of an object of mass m raised from the surface of earth to a height equal to the radius of the earth R, is 

1. $\frac{1}{2}mgR$

2. 2 mgR

3. mgR

4. $\frac{1}{4}mgR$

The change in the potential energy, when a body of mass \(m\) is raised to a height \(nR\) from the Earth's surface is: (\(R\) = Radius of the Earth)
1. \(mgR\left(\frac{n}{n-1}\right)\)
2. \(nmgR\)
3. \(mgR\left(\frac{n^2}{n^2+1}\right)\)
4. \(mgR\left(\frac{n}{n+1}\right)\)

If the mass of the earth is M, the radius is R and the gravitational constant is G, then work done to take

1 kg mass from earth surface to infinity will be:

1.  $\sqrt{\frac{GM}{2R}}$                      

2. $<mspace\; width="1em"></mspace>\frac{GM}{R}$

3.   $\sqrt{\frac{2GM}{R}}$                

4.   $\frac{GM}{2R}$

Two identical satellites are at R and 7R away from earth surface, the wrong statement is (R = Radius of earth)

1. Ratio of total energy will be 4

2. Ratio of kinetic energies will be 4

3. Ratio of potential energies will be 4

4. Ratio of total energy will be 4 but ratio of potential and kinetic energies will be 2