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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Gravitation
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A particle of mass M is placed at the center of a uniform sperical shell of mass 2M and radius R. The gravitational potential on the surface of the shell is

\[(a)\;\frac{-GM}{R} \quad (b)\;\frac{-3GM}{R} \quad (c)\;\frac{-2GM}{R} \quad (d)\;zero \]
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Potential at the surface of the shell = Potential due to particle + Potential due to shell
$\qquad = \bigg( \large\frac{-GM}{R}\bigg)+\bigg (\large\frac{-2 GM}{R}\bigg)$
$\qquad= \large\frac{-3GM}{R}$
Hence b is the correct answer.


answered Aug 26, 2013 by meena.p
edited Feb 17, 2014 by meena.p

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