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JEEMAIN and NEET
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Physics
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Class11
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Gravitation
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The height to which the acceleration due to gravity becomes $\large\frac{g}{9}$ (where g= acceleration due to gravity at surface of earth ) in terms of $R$, is
\[(a)\;\frac{R}{\sqrt 2} \quad (b)\;\frac{R}{2} \quad (c)\;\sqrt 2 R \quad (d)\;2R \]
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