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(D) Given,the distance of the bottom of the hole from the surface of the Earth is $d = \frac{R_e}{2}$,where $R_e$ is the radius of the Earth.On the su

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$150$

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(D) Given,the distance of the bottom of the hole from the surface of the Earth is $d = \frac{R_e}{2}$,where $R_e$ is the radius of the Earth.On the surface of the Earth,the weight of the body is $W = mg = 300 \ N$.The acceleration due to gravity at a depth $d$ below the surface is given by the formula $g' = g(1 - \frac{d}{R_e})$.Substituting $d = \frac{R_e}{2}$ into the formula:$g' = g(1 - \frac{R_e/2}{R_e}) = g(1 - \frac{1}{2}) = \frac{g}{2}$.The weight of the body at the bottom of the hole is $W' = mg' = m(\frac{g}{2}) = \frac{mg}{2}$.Substituting $mg = 300 \ N$:$W' = \frac{300}{2} = 150 \ N$.

$A$ hole is drilled halfway to the center of the Earth. $A$ body weighs $300 \ N$ on the surface of the Earth. How much will it weigh at the bottom of the hole (in $N$)?

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