A body weighed $250 \mathrm{~N}$ on the surface assuming the earth to be a sphere of uniform mass density, how much would it weigh half way down to the centre of the earth?

• A

  $195 \mathrm{~N}$
• B

  $240 \mathrm{~N}$
• Correct answer
  C

  $125 \mathrm{~N}$
• D

  $210 \mathrm{~N}$.

Solution

At a depth $d$ below the surface of the earth, $$ \begin{aligned} & m g^{\prime}=G \frac{M^{\prime} m}{(R-d)^2} \\ & M^{\prime}=\frac{4}{3} \pi(R-d)^3 \rho \end{aligned} $$ $$ \begin{aligned} \Rightarrow \quad g^{\prime} & =\frac{G M^{\prime}}{(R-d)^2} \\ & =\frac{4}{3} G \pi(R-d) \rho \end{aligned} $$ 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 $$ \begin{aligned} g & =\frac{G m}{R V}=\frac{4}{3} \pi G R \rho \\ \therefore \quad g^{\prime} & =\frac{g}{2} \end{aligned} $$ $\therefore$ 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.

Hence, the correct answer is option: 'C'.