Given below are two statements:
Statement I : Acceleration due to earth's gravity decreases as you go 'up' or 'down' from earth's surface.
Statement II : Acceleration due to earth's gravity is same at a height 'h' and depth 'd' from earth's surface, if h = d.
In the light of above statements, choose the most appropriate answer from the options given below

Solution

Physics sir · 5 steps

Write both formulas side by side: g falls twice as fast going up as going down, so h = d cannot match.

1. Going up. At height h you move farther from the centre while the whole mass stays below you. So gₕ = gR²/(R+h)², which for small h is about g(1 - 2h/R).
2. Going down. At depth d only the inner sphere of radius R - d pulls you, and the outer shell gives no net force. So g_d = g(1 - d/R), which also decreases, reaching zero at the centre.
3. Check Statement I. Both expressions are less than g for any h > 0 or d > 0. So g is largest at the surface and decreases in both directions. Statement I is correct.

   h = d: gₕ ≈ g(1 - 2h/R) ≠ g_d = g(1 - h/R)

4. Check Statement II. For the same distance, the drop going up (2h/R) is about twice the drop going down (h/R). The two values of g are not equal. In the small-h form they match only when d = 2h. Statement II is incorrect.
5. So. Statement I is correct and Statement II is incorrect, which is option (a).

The idea to keep. Near the surface, g drops by 2h/R going up but only by d/R going down. Equal decrease needs d = 2h, not d = h.

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