What is the depth at which the value of accel | vedclass

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Question

(C) The value of 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

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

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(C) The value of acceleration due to gravity at a depth $d$ below the surface of the Earth is given by the formula:$g' = g \left(1 - \frac{d}{R}\right)$where $g$ is the acceleration due to gravity at the surface and $R$ is the radius of the Earth.According to the problem,the value of acceleration at depth $d$ is $\frac{1}{n}$ times the value at the surface:$g' = \frac{g}{n}$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)$

What is the depth at which the value of acceleration due to gravity becomes $\frac{1}{n}$ times the value at the surface of the Earth? (radius of Earth $= R$)

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