Solution:
The acceleration due to gravity at depth \(d\) is \(g' = g \left(1 - \frac{d}{R}\right)\). Setting \(g' = g/4\), we get \(1 - \frac{d}{R} = \frac{1}{4}\) which gives \(d = \frac{3R}{4}\).
How much deep inside the earth (radius \(R\)) should a man go, so that his weight becomes one-fourth of that on the earth's surface?
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