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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Gravitation
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The magnitude of potential energy per unit mass of object at surface of earth is E. then the escape velocity of the object is

\[(a)\;\sqrt {2E} \quad (b)\;\sqrt {4E^2} \quad (c)\;\sqrt E \quad (d)\;\sqrt {E/2} \]

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1 Answer

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Potential energy per unit mass $=E$
Total potetial energy $=mE$
Which is equal to binding energy
Hence :
$\frac{1}{2}mv_e^2=mE$
$v_e=\sqrt {2E}$
answered Aug 29, 2013 by meena.p
 

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