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The height at which the weight of a body becomes ${\frac{1}{16}}^{th}$ , its weight on the surface of earth (radius $R$), is
$5R$
$15R$
$3R$
$4R$
Solution
Accleration due to gravity at a height $h$ from the suface of earth is
$g' = \frac{g}{{{{\left( {1 + \frac{h}{R}} \right)}^2}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,…\left( i \right)$
Where g is the acceleration due to gravity at the surface of earth and $R$ is the radius of earth.
Multiplying by $m\,\, (mass\,\, of\,\ the\,\, body)$ on both sides in $(i)$, we get
$mg' = \frac{{mg}}{{{{\left( {1 + \frac{h}{R}} \right)}^2}}}$
$\therefore $ Weight of body at height $h,\, w'\, =\, mg'$
Weight of body at surface of earth, $W\,=\,mg$
According to question, $W' = \frac{1}{{16}}W$
$\therefore \frac{1}{{16}} = \frac{1}{{{{\left( {1 + \frac{h}{R}} \right)}^2}}}$
${\left( {1 + \frac{h}{R}} \right)^2} = 16\,\,or\,\,1 + \frac{h}{R} = 4$
$or\,\frac{h}{R} = 3\,\,or\,\,h = 3R$.
