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The variation of g with depth d below Earth's surface (assuming uniform density) is

Ag × (1 + d/R)
Bg × (1 − d/R)
Cg × (R/d)
Dg × (d/R)
Answer & Solution
Correct answer: B. g × (1 − d/R)
1. Inside a spherical Earth with uniform density, only mass within radius (R − d) attracts the object. 2. That enclosed mass scales as (R − d)³. 3. Combined with the 1/(R − d)² dependence, effective g_d = g × (R − d)/R. 4. So g_d = g × (1 − d/R). At the centre (d = R), g = 0. _Source: syllabus Physics, Ch 7 "Gravitation", §7.4_
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