Height at which weight of a body becomes 1/16^{th} , its weight on surface of earth (radius R ), is

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7.Gravitation

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The height at which the weight of a body becomes $1/16^{th}$, its weight on the surface of earth (radius $R$), is

A

$3R$

B

$4R$

C

$5R$

D

$15R$

Solution

$\frac{g_{h}}{g}=\frac{R^{2}}{(R+h)^{2}}$

$\frac{\mathrm{g} / 16}{\mathrm{g}}=\frac{\mathrm{R}^{2}}{(\mathrm{R}+\mathrm{h})^{2}}$

$\mathrm{h}=3 \mathrm{R}$

Standard 11

Physics

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