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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Gravitation
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The work done in slowly lifting a body from earth's surface to a height R (radius of earth) is equal to two times the work done in lifing the same body from earth's surface to height h there h is equal to

\[(a)\;\frac{R}{4} \quad (b)\;\frac{R}{3} \quad (c)\;\frac{R}{6} \quad (d)\;\frac{R}{2} \]
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Work done = increase in gravitational potential energy
$W_1 =\large\frac{mg R}{1+\Large\frac{R}{R}}=\large\frac{mgR}{2}$
$\bigg[ \Delta u =\large\frac{mgh}{1+\Large\frac{h}{R}}\bigg]$ Where 'h' height above the earth surface
Given $2W_2=W_1$
Where $W_2$ = work done in lifting
through a height 'h'
$\large\frac{mgR}{2} =\frac{2mgh}{1+\Large\frac{h}{R}}$
Solving we get,
$h= \large\frac{R}{3}$
Hence b is the correct answer.


answered Aug 29, 2013 by meena.p
edited Feb 18, 2014 by meena.p

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