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. $\sqrt{\frac{2\mathrm{GM}}{\mathrm{a}}}$

2. $\sqrt{\frac{2\mathrm{GM}}{\mathrm{a}}\left(1-\frac{1}{\sqrt{2}}\right)}$

3. $\sqrt{\frac{2\mathrm{GM}}{\mathrm{a}}\left(\sqrt{2}-1\right)}$

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

Kepler's second law regarding constancy of the areal velocity of a planet is a consequence of the law of conservation of:

1. Energy

2. Linear momentum

3. Angular momentum

4. Mass

A projectile fired vertically upwards with a speed v escapes from the earth. If it is to be fired at 45$°$ to the horizontal, what should be its speed so that it escapes from the earth?

1.  v

2.  $\frac{\mathrm{v}}{\sqrt{2}}$

3.  $\sqrt{2}\mathrm{v}$

4.  2v

Magnitude of potential energy (U) and time period (T) of a satellite are related to each other as:

1. $T^2$ $\alpha$ $\frac{1}{U^3}$

2. $T$ $\alpha$ $\frac{1}{U^3}$

3. $T^2$ $\alpha$ $U^3$

4. $T^2$ $\alpha$ $\frac{1}{U^2}$

Two bodies of masses m and 4m are placed at a distance r. The gravitational potential at a point on the line joining them where the gravitational field is zero is

1.  $-\frac{5\mathrm{Gm}}{\mathrm{r}}$

2.  $-\frac{6\mathrm{Gm}}{\mathrm{r}}$

3.  $-\frac{9\mathrm{Gm}}{\mathrm{r}}$

4.  0

If \(A\) is the areal velocity of a planet of mass \(M,\) then its angular momentum is:

In planetary motion, the areal velocity of the position vector of a planet depends on the angular velocity ($\omega$) and the distance of the planet from the sun (r). The correct relation for areal velocity is:

1. $\frac{dA}{dt}$ $\alpha$ $\omega r$

2. $\frac{dA}{dt}$ $\alpha$ ${\omega }^{2}r$

3. $\frac{dA}{dt}$ $\alpha$ $\omega r^2$

4. $\frac{dA}{dt}$ $\alpha$ $\sqrt{\omega r}$

A planet is moving in an elliptical orbit. If T, V, E, and L stand, respectively, for its kinetic energy, gravitational potential energy, total energy and angular momentum about the center of the orbit, then:

The gravitational potential difference between the surface of a planet and 10 m above is 5 J/kg. If the gravitational field is supposed to be uniform, the work done in moving a 2 kg mass from the surface of the planet to a height of 8 m is

1.  2J

2.  4J

3.  6J

4.  8J