Height at which weight of a body becomes frac{1}{9} ^{th} its weight on surface of earth (radius of

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7.Gravitation

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The height at which the weight of a body becomes $\frac{1}{9} ^{th}$ its weight on the surface of earth (radius of earth is $R$)

A

$h= 3\, R$

B

$h = R$

C

$h = \frac{R}{2}$

D

$h = 2R$

Solution

$g^{\prime}=\frac{g}{\left(1+\frac{h}{R}\right)^{2}}, g^{\prime}=\frac{g}{9}$

$\Rightarrow \frac{g}{9}=\frac{g}{\left(1+\frac{h}{R}\right)^{2}} \Rightarrow 1+\frac{h}{R}=3 \Rightarrow h=2 R$

Standard 11

Physics

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