The depth d at which the value of acceleration due to gravity becomes 1 n times the value…
Physics · JEE Advanced · NTA Exams — Gravitation
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]
- \(\frac{R}{n}\)
- \(R\left(\frac{n-1}{n}\right)\)
- \(\frac{\mathrm{R}}{\mathrm{n}^{2}}\)
- \(R\left(\frac{n}{n+1}\right)\)
Answer
(B) R ( n-1 n )
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