Q137 / 180 · COMEDK 2023 (Memory Based) · Physics

Starting from the centre of the earth having radius $R$, the variation of $g$ (acceleration due to gravity) is shown by

  • A
    Figure
  • Correct answer
    B
    Figure
  • C
    Figure
  • D
    Figure

Solution

Acceleration due to gravity at a depth $d$ below the surface of the earth is given by $\begin{aligned} g_{\text {depth }} & =g_{\text {surface }}\left(1-\frac{d}{R}\right) \\ & =g_{\text {surface }}\left[\frac{R-d}{R}\right]=d_{\text {surface }}\left(\frac{r}{R}\right) \end{aligned}$ Also, for a point at height $h$ above surface, $g_{\text {height }}=g_{\text {surface }}\left[\frac{R^2}{(R+h)^2}\right]$ Therefore, we can say that value of $g$ increases from centre to maximum at the surface and then decreases as depicted in graph (b).

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

Relevant Questions