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The height at which the acceleration due to gravity becomes $\frac{g}{16}$ (where, $g=$ acceleration due to gravity on the surface of the earth) in terms of $R$ is, if $R$ is the radius of earth.
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Verified Answer
The correct answer is:
$3 R$
As, acceleration due to gravity at height $h$,
$$
\begin{aligned}
& g^{\prime}=\frac{G M}{(R+h)^2}, \\
& \Rightarrow \quad \frac{g}{16}=\frac{G M}{R^2}\left(\frac{R^2}{(R+h)^2}\right)=g \frac{R^2}{(R+h)^2}\left(\because g^{\prime}=\frac{g}{16}\right) \\
& \Rightarrow \quad \frac{1}{4}=\frac{R}{R+h} \Rightarrow h=3 R \\
&
\end{aligned}
$$
$$
\begin{aligned}
& g^{\prime}=\frac{G M}{(R+h)^2}, \\
& \Rightarrow \quad \frac{g}{16}=\frac{G M}{R^2}\left(\frac{R^2}{(R+h)^2}\right)=g \frac{R^2}{(R+h)^2}\left(\because g^{\prime}=\frac{g}{16}\right) \\
& \Rightarrow \quad \frac{1}{4}=\frac{R}{R+h} \Rightarrow h=3 R \\
&
\end{aligned}
$$
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