The height at which the acceleration due to gravity becomes \( \fra...
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The height at which the acceleration due to gravity becomes \( \frac{\mathrm{g}}{9} \) ( where \( \mathrm{g}= \) the acceleration due to gravity on the surface of the earth) in terms of \( \mathrm{R} \), the radius of the earth, is -
(A) \( \frac{R}{\sqrt{2}} \)
(B) \( R / 2 \)
(C) \( \sqrt{2} \mathrm{R} \)
\( \mathrm{P} \)
W
(D) \( 2 \mathrm{R} \)
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