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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Gravitation
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A uniform ring of mass $M$ and radius $R$ is placed directly above a uniform sphere of mass $8M$ and of radius $R$. The center of the ring is at a distance of $d= \sqrt 3 R$ from the center of the sphere. The gravitational attraction between the sphere and ring is

\[(a)\;\frac{8GM^2}{R^2} \quad (b)\;\frac{2GM^2}{\sqrt R^2} \quad (c)\;\frac{3GM^2}{2R^2} \quad (d)\;\frac{\sqrt 3 GM^2}{R^2} \]

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1 Answer

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Gravitational field due to ring at a distance
$d= \sqrt 3 R$ on its axis is
$E=\large\frac{GMd}{(R^2+d^2)^{3/2}}$
$\quad=\large\frac{{\sqrt 3}GM}{8R^2}$
Force on the sphere = mass of sphere $ \times $ E
$\quad=8ME$
$\quad=\large \frac{\sqrt 3 \;GM^2}{R^2}$
Hence d is the correct answer.

 

answered Aug 30, 2013 by meena.p
edited Feb 18, 2014 by meena.p
 

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