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STD 11 - 7. gravitation — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 7. gravitation4 Marks
MCQ
The height at which the acceleration due to gravity becomes $\frac{g}{9}$ (where $g$ = the acceleration due to gravity on the surface of the earth) in terms of $R$, the radius of the earth, is
  • ✓
    $2R$
  • B
    $\frac{R}{{\sqrt 2 }}$
  • C
    $\;\frac{R}{2}$
  • D
    $\;\sqrt {2R} $
✓

Answer

Correct option: A.
$2R$
a
We know that $\frac{{g'}}{g} = \frac{{{R^2}}}{{{{\left( {R + h} \right)}^2}}}$

$\therefore \frac{{g/9}}{g} = {\left[ {\frac{R}{{R + h}}} \right]^2}\,\,\,\therefore h = 2R$

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