The ratio of the accelerations due to gravity at the bottom of a deep mine and that on the surface of the earth is 978/980. Find the depth of the mine if the density of the earth is uniform throughout and the radius of the earth is 6300 km.

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25.38 km

12.86 km

90.9 km

13.0 km

answer is A.

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Detailed Solution

$\begin{array}{l} g_{d} = g (1 - \frac{d}{R}) \Rightarrow \frac{g_{d}}{g} = 1 - \frac{d}{R} \Rightarrow \frac{978}{980} = 1 - \frac{d}{R} \\ \frac{d}{R} = 1 - \frac{978}{980} = \frac{2}{980} = \frac{1}{490} \\ d = \frac{R}{490} = \frac{6300}{490} = 12 .86 km \end{array}$

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