MHT CET · Physics · Gravitation
The height above the earth's surface at which the acceleration due to gravity becomes \(\left(\frac{1}{n}\right)\) times the value at the surface is ( \(R=\) radius of earth)
- A \(\frac{R}{\sqrt{n}}\)
- B \(\mathrm{R} \cdot \sqrt{\mathrm{n}}\)
- C \((\sqrt{\mathrm{n}}+1) \mathrm{R}\)
- D \((\sqrt{\mathrm{n}}-1) \mathrm{R}\)
Answer & Solution
Correct Answer
(D) \((\sqrt{\mathrm{n}}-1) \mathrm{R}\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{ll} & \text { Given: } \frac{\mathrm{g}^{\prime}}{\mathrm{g}}=\frac{1}{\mathrm{n}} \\ & \frac{1}{\mathrm{n}}=\frac{\mathrm{R}^2}{(\mathrm{R}+\mathrm{h})^2} \Rightarrow \frac{\mathrm{R}}{\mathrm{R}+\mathrm{h}}=\frac{1}{\sqrt{\mathrm{n}}} \\ \therefore \quad & \sqrt{\mathrm{n}} \mathrm{R}=\mathrm{R}+\mathrm{h} \\ \therefore \quad & \mathrm{R}(\sqrt{\mathrm{n}}-1)=\mathrm{h}\end{array}\)
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