Gravitation Question 24

Question: The depth d at which the value of acceleration due to gravity becomes $ [\frac{1}{n}] $ times the value at the surface, is [R = radius of the earth][MP PMT 1999; Kerala PMT 2005]

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)] $

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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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