A body weighed 250 ~N on the surface assuming the earth to be a...

At a depth $d$ below the surface of the earth, $$\begin{aligned}& m g^{\prime}=G \frac{M^{\prime} m}{(R-d)^2}\\ & M^{\prime}=\frac{4}{3} \pi(R-d)^3 \rho \end{aligned}$$$$\begin{aligned} \Rightarrow \quadg^{\prime} & =\frac{G M^{\prime}}{(R-d)^2}\\ & =\frac{4}{3}G \pi(R-d) \rho \end{aligned}$$If $\quad d=\frac{R}{2}$ then g^{\prime}=$\frac{4}{3}$G $\pi \rho=\frac{2 \pi G R \rho}{3}$On the surface of the earth $$\begin{aligned}g & =\frac{G m}{R V}=\frac{4}{3} \piG R \rho \\ \therefore \quad g^{\prime} & =\frac{g}{2} \end{aligned}$$\therefore$The body weighed$250 \mathrm{~N}$on the surface of the earth would weigh$\frac{1}{2} \times 250=125 \mathrm{~N}$, half way down towards the centre of the earth.