MHT CET · Physics · Gravitation
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 \(\mathrm{h}=(\sqrt{2}-1) \mathrm{R}\)
- D \(\mathrm{h}=(\sqrt{3}-1) \mathrm{R}\)
Answer & Solution
Correct Answer
(D) \(\mathrm{h}=(\sqrt{3}-1) \mathrm{R}\)
Step-by-step Solution
Detailed explanation
\( \begin{aligned} & \mathrm{g}^{\prime}=\mathrm{g} \frac{\mathrm{R}^2}{(\mathrm{R}+\mathrm{h})^2}=\frac{\mathrm{g}}{3} \\ & \frac{1}{3}=\frac{\mathrm{R}^2}{(\mathrm{R}+\mathrm{h})^2} \\ & \frac{1}{\sqrt{3}}=\frac{\mathrm{R}}{\mathrm{R}+\mathrm{h}} \\ & \mathrm{h}=(\sqrt{3}-1) \mathrm{R} \end{aligned} \)
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