3 particles each of mass m are kept at vertices of an equilateral triangle of side L. The

gravitational field at centre due to these particles is

1. zero                                                                       

2. $\frac{3GM}{L^2}$

3. $\frac{9GM}{L^2}$                                                                     

4. $\frac{12}{\sqrt{3}}\frac{GM}{L^2}$

The weight of an object in the coal mine, sea level, at the top of the mountain are  $W_1$,

$W_2$ and $W_3$   respectively, then

1.  $W_1\le W_2\ge W_3$                      

2.  $W_1=W_2=W_3$

3.  $W_1\le W_2\le W_3$                     

4.  $W_1\ge W_2\ge W_3$      

The value of escape velocity on a certain planet is 2 km/s. Then the value of orbital speed

for a satellite orbiting close to its surface is:

1. 12km/s                                                                 

2. 1km/s

3. $\sqrt{2<mspace\; width="1em"></mspace>}<mspace\; width="1em"></mspace>km/s$                                                             

4. $2\sqrt{2}<mspace\; width="1em"></mspace>km/s$

Four particles each of mass M, are located at the vertices of a square with side L. The

gravitational potential due to this at the centre of the square is

1. $-\sqrt{32}\frac{GM}{L}$                                   

2. $-\sqrt{64}\frac{GM}{L^2}$

3. zero                                             

4. $\sqrt{32}\frac{GM}{L}$

The mass of the moon is $7.34\times {10}^{22}kg$ and the radius is $1.74\times {10}^{6}m$. The value of

gravitation field will be

1. $1.45<mspace\; width="1em"></mspace>N/kg$                               

2. $1.55<mspace\; width="1em"></mspace>N/kg$

3. $1.75<mspace\; width="1em"></mspace>N/kg$                               

4.  $1.62<mspace\; width="1em"></mspace>N/kg$

The centripetal force acting on a satellite orbiting round the earth and the gravitational

force of earth acting on the satellite both equal F. The net force on the satellite is

1. Zero                                  

2. F

3. $F\sqrt{2}$                                 

4. 2 F

Reason of weightlessness in a satellite is:

1. Zero gravity

2. Centre of mass

3. Zero reaction force by satellite surface

4. None

Mass M is divided into two parts xM and $\left(1-x\right)M$ . For a given separation, the value of x

for which the gravitational attraction between the two pieces becomes maximum is

1. $\frac{1}{2}$                                 

2. $\frac{3}{5}$

3. 1                                   

4. 2

If earth is supposed to be a sphere of radius R, if $g_{30°}$ is value of acceleration due to gravity at latitude of 30^o and g at the pole, the value of $g-g_{{30}^o}$ is

1. $\frac{1}{4}<mspace\; width="1em"></mspace>{\omega }^{2}R$                      

2. $\frac{3}{4}{\omega }^{2}R$

3. ${\omega }^{2}R$                           

4. $\frac{1}{2}{\omega }^{2}R$

If M the mass of the earth and R its radius, the ratio of the gravitational acceleration and

the gravitational constant is

1. $\frac{R^2}{M}$                       

2. $\frac{M}{R^2}$

3. $M{R}^2$                       

4. $\frac{M}{R}$