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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Gravitation
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An object of mass m starts falling towards planet of mass M and Radius R. As it reaches near the surface it passes through a small hole in the planet. The planet is made of two pieces, a sperical shell of negligible thickness and mass $\large\frac{2M}{3}$ and a point mass of $\large\frac{M}{3}$ at center. Change in the force of gravity experienced by the object is

\[(a)\;\frac{2}{3}\frac{GMm}{R^2} \quad (b)\;0 \quad (c)\;\frac{1}{3}\frac{GM^2}{R^2} \quad (d)\;\frac{4}{3}\frac{GMm}{R^2} \]

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1 Answer

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Inside the shell gravitational force is zero.
Force is only due to mass $\large\frac{M}{3}$
Change in force of gravity
$\qquad=\large\frac{GMm}{R^2}-\large\frac{G \Large\frac{M}{3}\large m}{R^2}$
$\qquad=\large\frac{2 GMm}{3 R^2}$
Hence a is the correct answer.
answered Aug 21, 2013 by meena.p
edited Jun 30, 2014 by lmohan717
 

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