The variation of g with depth d below Earth's surface (assuming uniform density) is
A{'text': 'g × (1 + d/R)', 'label': 'A'}
B{'text': 'g × (1 − d/R)', 'label': 'B'}
C{'text': 'g × (R/d)', 'label': 'C'}
D{'text': 'g × (d/R)', 'label': 'D'}
Answer & Solution
Correct answer: B. {'text': 'g × (1 − d/R)', 'label': 'B'}
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: NCERT Class 11 Physics, Ch 7 "Gravitation", §7.4_
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