If the value of g (acceleration due to gravit | vedclass

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(D) The acceleration due to gravity $g'$ at a depth $d$ below the Earth's surface is given by the formula: $g' = g \left(1 - \frac{d}{R}\right)$,where

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(D) The acceleration due to gravity $g'$ at a depth $d$ below the Earth's surface is given by the formula: $g' = g \left(1 - \frac{d}{R}\right)$,where $g$ is the acceleration due to gravity at the surface,$d$ is the depth,and $R$ is the radius of the Earth.At the center of the Earth,the depth $d$ is equal to the radius $R$ (i.e.,$d = R$).Substituting $d = R$ into the formula: $g' = g \left(1 - \frac{R}{R}\right) = g(1 - 1) = g(0) = 0$.Therefore,the value of acceleration due to gravity at the center of the Earth is $0 \ m/s^2$.

If the value of $g$ (acceleration due to gravity) at the Earth's surface is $10 \ m/s^2$,what is its value in $m/s^2$ at the center of the Earth,assuming the Earth is a sphere of radius $R$ meters with uniform mass density?

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