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(D) The acceleration due to gravity at a depth $d$ below the surface of the earth is given by the formula: $g' = g(1 - d/R)$,where $g$ is the accelera

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(D) The acceleration due to gravity at a depth $d$ below the surface of the earth is given by the formula: $g' = g(1 - d/R)$,where $g$ is the acceleration due to gravity at the surface and $R$ is the radius of the earth.Given that the body is halfway down to the center,the depth $d = R/2$.Substituting this value into the formula:$g' = g(1 - (R/2)/R) = g(1 - 1/2) = g/2$.Since weight $W = mg$,the new weight $W'$ at depth $d$ is $W' = mg' = m(g/2) = W/2$.Given the weight at the surface $W = 200 \; N$,the weight at depth $d = R/2$ is $W' = 200/2 = 100 \; N$.

$A$ body weighs $200 \; N$ on the surface of the earth. How much will it weigh halfway down to the centre of the earth (in $; N$)?

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