The depth d at which the value of acceleration due to gravity becomes 1/n times the value at the surface, is [R = radius of the earth] :
Answer & explanation
Correct answer: option 2
\(g' = g(1-\frac{d}{R})\)
\(\Rightarrow \frac{g}{n} = g(1-\frac{d}{R})\)
\(\Rightarrow \frac{d}{R} = 1 - \frac{1}{n}\)
\(\Rightarrow d = \frac{n-1}{n}R\)