Gravitation Question 333

Question: The depth d at which the value of acceleration due to gravity becomes $ [\frac{1}{n}] $ times the value at the surface of the earth, is [R = radius of the earth]

Options:

A) $ [\frac{R}{n}] $

B) $ [R\left( \frac{n-1}{n} \right)] $

C) $ [\frac{R}{n^{2}}] $

D) $ [R\left( \frac{n}{n+1} \right)] $

Show Answer

Answer:

Correct Answer: B

Solution:

  • $ [g’=g\left( 1-\frac{d}{R} \right)\Rightarrow \frac{g}{n}=g\left( 1-\frac{d}{R} \right)] $ $ [\Rightarrow d\left( \frac{n-1}{n} \right)R] $


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