The depth d at which the value of acceleratio | vedclass

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(B) The acceleration due to gravity at a depth $d$ below the surface of the earth is given by the formula: $g' = g \left( 1 - \frac{d}{R} \right)$.Giv

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$R \left( \frac{n - 1}{n} \right)$

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(B) The acceleration due to gravity at a depth $d$ below the surface of the earth is given by the formula: $g' = g \left( 1 - \frac{d}{R} \right)$.Given that the value of acceleration due to gravity at depth $d$ is $\frac{1}{n}$ times the value at the surface,we have $g' = \frac{g}{n}$.Substituting this into the formula: $\frac{g}{n} = g \left( 1 - \frac{d}{R} \right)$.Dividing both sides by $g$: $\frac{1}{n} = 1 - \frac{d}{R}$.Rearranging the terms to solve for $d$: $\frac{d}{R} = 1 - \frac{1}{n} = \frac{n - 1}{n}$.Therefore,$d = R \left( \frac{n - 1}{n} \right)$.

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].

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