MHT CET · Physics · Gravitation
The height ' \(h\) ' above the earth's surface at which the value of acceleration due to gravity ( g ) becomes \(\left(\frac{\mathrm{g}}{3}\right)\) is ( \(\mathrm{R}=\) radius of the earth)
- A \((\sqrt{3}+1) \mathrm{R}\)
- B \((\sqrt{3}-1) R\)
- C \(\sqrt{3} R\)
- D \(3 \sqrt{R}\)
Answer & Solution
Correct Answer
(B) \((\sqrt{3}-1) R\)
Step-by-step Solution
Detailed explanation
We know that,
\(\begin{aligned}
& \mathrm{g}_{\mathrm{h}}=\mathrm{g}\left(\frac{\mathrm{R}}{\mathrm{R}+\mathrm{h}}\right)^2 \\
& \text { For } g_h=\frac{g}{3} \\
& \frac{g}{3}=g\left(\frac{R}{R+h}\right)^2 \\
& \frac{1}{\sqrt{3}}=\frac{\mathrm{R}}{\mathrm{R}+\mathrm{h}} \\
& \sqrt{3} \mathrm{R}=\mathrm{R}+\mathrm{h} \\
& \therefore \quad h=(\sqrt{3}-1) \mathrm{R}
\end{aligned}\)
\(\begin{aligned}
& \mathrm{g}_{\mathrm{h}}=\mathrm{g}\left(\frac{\mathrm{R}}{\mathrm{R}+\mathrm{h}}\right)^2 \\
& \text { For } g_h=\frac{g}{3} \\
& \frac{g}{3}=g\left(\frac{R}{R+h}\right)^2 \\
& \frac{1}{\sqrt{3}}=\frac{\mathrm{R}}{\mathrm{R}+\mathrm{h}} \\
& \sqrt{3} \mathrm{R}=\mathrm{R}+\mathrm{h} \\
& \therefore \quad h=(\sqrt{3}-1) \mathrm{R}
\end{aligned}\)
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