What is the depth at which the value of acceleration due to gravity becomes 1 n , times the value at the…

What is the depth at which the value of acceleration due to gravity becomes 1n, times the value at the surface of the earth? (radius of the earth =R)
  1. R n2
  2. R(n-1)n
  3. Rn(n-1)
  4. Rn

Solution

The value of the acceleration due to gravity when the body moves below the surface of the earth at the depth, d in terms of the radius of the earth, and acceleration due to gravity at the surface of the earth, then at depth:

\(g_{\text {eff }}=g\left(1-\frac{d}{R}\right) \Rightarrow \frac{g}{n}=g\left(1-\frac{d}{R}\right) \Rightarrow d=\frac{(n-1) R}{n}\).

Asked in: NEET 2020 (Phase 2)

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