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MHT CET · Physics · Gravitation

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

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

Answer & Solution

Correct Answer

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

Step-by-step Solution

Detailed explanation

\(g_d = g \left(1 - \frac{d}{R}\right)\) \(\frac{g}{n} = g \left(1 - \frac{d}{R}\right)\)