Assuming the earth to be a sphere of uniform mass density, the weight of a body at a depth $d=\frac{R}{2}$ from the surface of earth, if its weight on the surface of earth is 200 N, will be:

(Given R = radius of earth)

<p>The gravitational field strength, or equivalently, the weight of an object, decreases linearly from its surface value to zero at the center of a sphere of uniform mass density. This is because only the mass inside the radius at which the object is located contributes to the gravitational force at that location. </p> <p>If ( $d = \frac{R}{2}$ ) is the depth below the surface of the Earth, then the radius of the sphere contributing to the gravitational force at that depth is ( $R - d = R - \frac{R}{2} = \frac{R}{2}$ ).</p> <p>Since the gravitational force (or weight) decreases linearly with the radius in a sphere of uniform density, the weight of the object at depth ( $d = \frac{R}{2} $) is half its weight at the surface of the Earth.</p> <p>So, if the weight of the body on the surface of the Earth is 200 N, its weight at depth ( $d = \frac{R}{2} $) is half of that, or 100 N.</p> <p>Therefore, the correct answer is 100 N.</p>