The depth d at which the value of acceleratio | vedclass

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(C) 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 th

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

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(C) 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 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}$$\frac{d}{R} = \frac{n-1}{n}$$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 earth's surface is $(R = \text{radius of the earth})$

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