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NEET · Physics · STD 11 - 7. gravitation

What is the depth at which the value of acceleration due to gravity becomes \(\frac{1}{n}\) times the value that at the surface of earth? (radius of earth \(=R\) )

  1. A \(\frac{R}{n}\)
  2. B \(\frac{R}{n^{2}}\)
  3. C \(\frac{R(n-1)}{n}\)
  4. D \(\frac{R n}{(n-1)}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\frac{R(n-1)}{n}\)

Step-by-step Solution

Detailed explanation

Value of acceleration due to gravity at depth d, \(g^{\prime}=g\left(1-\frac{d}{R}\right)\)
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