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 \(\frac{(\mathrm{n}-1)}{\mathrm{nR}}\)
- C \(\frac{\mathrm{Rn}}{(\mathrm{n}-1)}\)
- D \(\frac{\mathrm{n}}{\mathrm{R}(\mathrm{n}-1)}\)
Answer & Solution
Correct Answer
(A) \(\frac{R(n-1)}{n}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \mathrm{g}^{\prime}=\mathrm{g}\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right) \\ & \frac{\mathrm{g}}{\mathrm{n}}=\mathrm{g}\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right) \\ & \mathrm{d}=\frac{\mathrm{R}(\mathrm{n}-1)}{\mathrm{n}}\end{aligned}\)
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