The depth $d$ at which the value of acceleration due to gravity becomes $\frac{1}{n}$ times the value at the surface, is [$R =$ radius of the earth]
The depth $d$ at which the value of acceleration due to gravity becomes $\frac{1}{n}$ times the value at the surface, is [$R =$ radius of the earth]
$\frac{R}{n}$
$R\,\left( {\frac{{n - 1}}{n}} \right)$
$\frac{R}{{{n^2}}}$
$R\,\left( {\frac{n}{{n + 1}}} \right)$
The depth $d$ at which the value of acceleration due to gravity becomes $\frac{1}{n}$ times the value at the surface, is [$R =$ radius of the earth]
$g' = g\left( {1 - \frac{d}{R}} \right)\, \Rightarrow \,\frac{g}{n} = g\left( {1 - \frac{d}{R}} \right)$$ \Rightarrow \,\,d = \left( {\frac{{n - 1}}{n}} \right)\;R$
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