Q111 / 150 · MHT CET 2021 (21 Sep Shift 1) · Physics

When the value of acceleration due to gravity ' $\mathrm{g}$ ' become $\frac{\mathrm{g}}{3}$ above surface of earth at height ' $h$ ' then relation between ' $h$ ' and ' $\mathrm{R}$ ' is $$ (\mathrm{R}=\text { radius of earth) } $$

  • A
    $\mathrm{h}=\frac{\mathrm{R}}{\sqrt{3}-1}$
  • B
    $\mathrm{h}=\frac{\sqrt{3}}{\mathrm{R}}$
  • C
    $\mathrm{h}=(\sqrt{2}-1) \mathrm{R}$
  • Correct answer
    D
    $\mathrm{h}=(\sqrt{3}-1) \mathrm{R}$

Solution

$$ \begin{aligned} & \mathrm{g}^{\prime}=\mathrm{g} \frac{\mathrm{R}^2}{(\mathrm{R}+\mathrm{h})^2}=\frac{\mathrm{g}}{3} \\ & \frac{1}{3}=\frac{\mathrm{R}^2}{(\mathrm{R}+\mathrm{h})^2} \\ & \frac{1}{\sqrt{3}}=\frac{\mathrm{R}}{\mathrm{R}+\mathrm{h}} \\ & \mathrm{h}=(\sqrt{3}-1) \mathrm{R} \end{aligned} $$

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

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