MHT CET · Physics · Gravitation
The depth at which acceleration due to gravity becomes \(\frac{\mathrm{g}}{2 \mathrm{n}}\) is \((\mathrm{R}=\) radius of earth, \(\mathrm{g}=\) acceleration due to gravity on earth's surface, \(\mathrm{n}\) is integer)
- A \(\frac{\mathrm{R}(1-2 \mathrm{n})}{\mathrm{n}}\)
- B \(\frac{\mathrm{R}(1-\mathrm{n})}{2 \mathrm{n}}\)
- C \(\frac{\mathrm{R}(\mathrm{n}-1)}{\mathrm{n}}\)
- D \(\frac{\mathrm{R}(2 \mathrm{n}-1)}{2 \mathrm{n}}\)
Answer & Solution
Correct Answer
(D) \(\frac{\mathrm{R}(2 \mathrm{n}-1)}{2 \mathrm{n}}\)
Step-by-step Solution
Detailed explanation
The gravitational acceleration at depth is given as \(g_d=g\left[1-\frac{d}{R}\right]\) Given \(\mathrm{g}_{\mathrm{d}}=\frac{\mathrm{g}}{2 \mathrm{n}}\)
\(\begin{aligned} \therefore \quad \frac{g}{2 n} & =g\left[1-\frac{d}{R}\right] \\ \frac{d}{R} & =1-\frac{1}{2 n} \\ d & =\left[\frac{2 n-1}{2 n}\right] R\end{aligned}\)
\(\begin{aligned} \therefore \quad \frac{g}{2 n} & =g\left[1-\frac{d}{R}\right] \\ \frac{d}{R} & =1-\frac{1}{2 n} \\ d & =\left[\frac{2 n-1}{2 n}\right] R\end{aligned}\)
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