WBJEE · Physics · Gravitation
The height vertically above the earth's surface at which the acceleration due to gravity becomes \(1 \%\) of its value at the surface is ( \(\mathrm{R}\) is the radius of the Earth)
- A \(8 \mathrm{R}\)
- B \(9 \mathrm{R}\)
- C \(10 \mathrm{R}\)
- D \(20 \mathrm{R}\)
Answer & Solution
Correct Answer
(B) \(9 \mathrm{R}\)
Step-by-step Solution
Detailed explanation
\[ \begin{aligned} & \text { Hints : } g^{\prime}=\frac{g}{\left(1+\frac{h}{\mathrm{R}}\right)^2} \Rightarrow \frac{g}{100}=\frac{g}{\left(1+\frac{h}{\mathrm{R}}\right)^2} \\ & 1+\frac{h}{\mathrm{R}}=10 \Rightarrow \frac{h}{\mathrm{R}}=9, h=9 \mathrm{R} \end{aligned} \]
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