NEET · Physics · STD 11 - 7. gravitation
What is the depth at which the value of acceleration due to gravity becomes \(\frac{1}{n}\) times the value that at the surface of earth? (radius of earth \(=R\) )
- A \(\frac{R}{n}\)
- B \(\frac{R}{n^{2}}\)
- C \(\frac{R(n-1)}{n}\)
- D \(\frac{R n}{(n-1)}\)
Answer & Solution
Correct Answer
(C) \(\frac{R(n-1)}{n}\)
Step-by-step Solution
Detailed explanation
Value of acceleration due to gravity at depth d, \(g^{\prime}=g\left(1-\frac{d}{R}\right)\)
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