Acceleration due to gravity is ‘g’ on the surface of the earth. The value of acceleration due to gravity at a height of 32 km above earth’s surface is (Radius of the earth = 6400 km)

(1) 0.9 g           

(2) 0.99g

(3) 0.8 g           

(4) 1.01 g

The height of a point vertically above the earth’s surface, at which the acceleration due to gravity becomes \(1\%\) of its value at the surface is: (Radius of the earth = \(R\))
1. \(8R\)
2. \(9R\)
3. \(10R\)
4. \(20R\)

If radius of earth is R then the height h’ at which value of ‘g’ becomes one-fourth is 

(1)$\frac{R}{4}$               

(2)$\frac{3R}{4}$

(3)R                 

(4)$\frac{R}{8}$

Two planets have the same average density but their radii are $R_1$  and $R_2$. If acceleration due to gravity on these planets be $g_1$ and $g_2$ respectively, then

(1)    $\frac{g_1}{g_2}$= $\frac{R_1}{R_2}$                   

(2) $\frac{g_1}{g_2}$=$\frac{R_2}{R_1}$

(3)    $\frac{g_1}{g_2}$= $\frac{{R_1}^2}{{R_2}^2}$                  

(4) $\frac{g_1}{g_2}$= $\frac{{R_1}^3}{{R_2}^3}$

Assume that the acceleration due to gravity on the surface of the moon is 0.2 times the acceleration due to gravity on the surface of the earth. If $R_E$ is the maximum range of a projectile on the earth’s surface, what is the maximum range on the surface of the moon for the same velocity of projection ?

(a) 0.2$R_{\epsilon }$                     

(b) 2$R_{\epsilon }$

(c) 0.5$R_{\epsilon }$                     

(d) 5$R_{\epsilon }$