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] $
