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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Gravitation

The height to which the acceleration due to gravity becomes $\large\frac{g}{9}$ (where g= acceleration due to gravity at surface of earth ) in terms of $R$, is

\[(a)\;\frac{R}{\sqrt 2} \quad (b)\;\frac{R}{2} \quad (c)\;\sqrt 2 R \quad (d)\;2R \]

1 Answer

$\large\frac{g'}{g}=\large\frac{R^2}{(R+h)^2}$
$\large\frac{\Large\frac{g}{9}}{g}=\bigg(\large\frac{R}{R+h}\bigg)^2$
$\large\frac{R}{R+h}=\frac{1}{3}$
$h= 2R$
Hence a is the correct answer.

 

answered Aug 21, 2013 by meena.p
edited Feb 17, 2014 by meena.p
 

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