Solution:
The depth at which the acceleration due to gravity becomes \( \frac{1}{n} \) times the surface value is:
\[
d = \left( 1 - \frac{1}{n} \right) R
\]
where \( R \) is the radius of the Earth.
The depth at which the value of acceleration due to gravity becomes 1/n times the value at the surface is (R be the radius of the earth) :
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