MHT CET · Physics · Gravitation
The depth at which acceleration due to gravity becomes \(\frac{\mathrm{g}}{\mathrm{n}}\) is ( \(\mathrm{R}=\) radius of earth, \(\mathrm{g}=\) acceleration due to gravity) ( \(\mathrm{n}=\) integer)
- A \(\frac{R(n-1)}{n}\)
- B \(\frac{\mathrm{R}(\mathrm{n}+1)}{\mathrm{n}}\)
- C \(\frac{\mathrm{R}(\mathrm{n}-1)^2}{\mathrm{n}}\)
- D \(\frac{\mathrm{R}(\mathrm{n}+1)^2}{\mathrm{n}}\)
Answer & Solution
Correct Answer
(A) \(\frac{R(n-1)}{n}\)
Step-by-step Solution
Detailed explanation
\( \mathrm{g}' = \mathrm{g} \left(1 - \frac{\mathrm{d}}{\mathrm{R}}\right) \) \( \frac{\mathrm{g}}{\mathrm{n}} = \mathrm{g} \left(1 - \frac{\mathrm{d}}{\mathrm{R}}\right) \)
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