The dependence of acceleration due to gravity `g` on the distance `r` from the centre of the earth, assumed to be a sphere of radius `R` of uniform density is as shown in Fig. below:


The correct figure is

Options

A. 4
B. 1
C. 2
D. 3

Answer

A

Explanation

The acceleration due to gravity at a depth d below the surface of earth is
` g. = (GM)/(R^2) (1 - d/R) = g (1 - d/R)`
g.=0 at d = R
i.e., acceleration due to gravity is zero at the centre of earth. Thus, the variation in value of g with r is
For, r > R,
` g. = (g)/( (1 + h/R)^2) = (gR^2)/(r^2) rArr g. prop (1)/(r^2)`
Here, R+h=r
For r < R,` g.=g(1- d/R) = (gr)/( R)`
Here R - d - r ` rArr g. prop r`
Therefore, the variation of g with distance from centre of the earth will be as shown in the figure.