The variation of acceleration due to gravity (g) with distance (r) from the center of the earth is correctly represented by :
(Given R = radius of earth)

Solution

Physics sir · 4 steps

Split the graph at the surface: inside, g grows in a straight line; outside, it falls off as 1/r².

1. Inside the earth. At distance r < R, only the mass inside radius r pulls you, and that mass is Mr³/R³ for a uniform earth. So g = GM/R³ r, which is directly proportional to r: a straight line from the origin, with g = 0 at the centre.
2. Outside the earth. For r ≥ R, the whole mass acts as if it sits at the centre, so g = GM/r². This curve falls off as 1/r² and never reaches zero.

   g(r) = GM/R³ r, r ≤ R ; GM/r², r ≥ R

3. Peak at surface. Both formulas give GM/R² at r = R, so the line and the curve meet there, and g is largest at the surface.
4. So. The correct graph is a straight line rising from zero up to r = R, then a 1/r² fall. That is option (a).

The idea to keep. For a uniform sphere, g is proportional to r inside and to 1/r² outside, and it is largest at the surface. The same shape appears for the field of a uniformly charged solid sphere.

Answer from the NTA final answer key. Solution by PrepWiser.