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⚛️ Physics  ·  Gravitation  ·  NEET & JEE

A body weighs 100 N at the Earth surface. At a depth equal to one-fourth of the Earth radius, its weight is:

Answer: 75 N.

  • A 75 N
  • B 90 N
  • C 50 N
  • D 25 N

Correct answer: A. 75 N

Explanation: At depth d, g' = g(1 - d/R) = g(1 - 1/4) = 0.75g. So the weight is 0.75 × 100 = 75 N.

Variation of g with Height and Depthgg_surfaceAbove surface: g ∝ 1/r² (falls off curving down)Below surface: g ∝ r (falls off LINEARLY)Earth's surface (r = R)centre (r=0): g=0g is MAXIMUM exactly at the surface - it decreases in both directions, but by different laws

g is maximum at Earth's surface; going up, it falls off as 1/r² (inverse-square); going down, it falls off linearly with depth (since only the mass enclosed within radius r contributes), reaching zero at the centre.

Concept context

Universal gravitation, orbital mechanics, gravitational potential, and escape velocity.

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