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Home  >>  EAMCET  >>  Physics
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The gravitational force acting on a particle, due to a solid sphere of uniform density and radius $R$, at a distance of $3R$ from the center of the sphere is $F_1$. A spherical hole of radius $(R/2)$ is now made in the sphere as shown in the figure. The sphere with hole now exerts a force $F_2$ on the same particle. Ratio of $F_1$ to $F_2$ is :

$(a)\;\frac{50}{41} \\ (b)\;\frac{41}{50} \\(c)\; \frac{41}{25} \\(d)\;\frac{25}{41}$

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1 Answer

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$(a)\;\large\frac{50}{41}$
answered Nov 7, 2013 by pady_1
 

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