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