How much deep inside the earth (radius $R$ ) should a man go, so that his weight becomes one-fourth of that on the earth's surface?

• A

  $\frac{R}{2}$
• Correct answer
  B

  $\frac{3 R}{4}$
• C

  $\frac{R}{4}$
• D

  $\frac{R}{3}$

Solution

Suppose depth inside the earth is $h$.
As we know that, inside the earth $g_i \propto\left(\frac{R-h}{R^3}\right)$ ....(i) On the surface of the earth $g_s \propto \frac{1}{R^2}$ ....(ii) From Eqs. (i) and (ii), we get $\frac{g_i}{g_c}=\frac{(R-h) R^2}{R^3}=\frac{R-h}{R}$ $\Rightarrow \quad \frac{1}{4}=\frac{R-h}{R}$ $\Rightarrow \quad R=4 R-4 h \Rightarrow 3 R=4 h$ $\Rightarrow \quad h=\frac{3}{4} R$.

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