Question 164: Gravitation

Question

A solid sphere of radius R gravitationally attracts a particle placed at 3R from its centre with a force F
1
. Now a spherical cavity of radius $\left( {{R \over 2}} \right)$ is made in the sphere (as shown in figure) and the force becomes F
2
. The value of F
1
: F
2
is
![](https://imagex.cdn.examgoal.net/1klse24x4/4859a6d7-a609-43ae-b23b-345e599fe470/4caf4e80-7b8b-11eb-a157-49f87e3afb79/file-1klse24x5-307w.jpg)

Accepted Answer

${g_1} = {{GM} \over {{{(3R)}^2}}} = {{GM} \over {9{R^2}}}$
${g^2} = {{GM} \over {9{R^2}}} - {{G\left( {{M \over 8}} ight)} \over {{{\left( {3R - {R \over 2}} ight)}^2}}}$
$ = {{GM} \over {9{R^2}}} - {{GM} \over {{R^2}50}} = {{41} \over {9 	imes 50}}{{GM} \over {{R^2}}}$
${{{g_1}} \over {{g_2}}} = {{41} \over {50}}$
$ 	herefore $ $ {{{F_1}} \over {{F_2}}} = {{m{g_1}} \over {m{g_2}}} = {{41} \over {50}}$

Answer

36 : 25

Answer

41 : 50

Answer

50 : 41

Answer

25 : 36

Gravitation

A solid sphere of radius R gravitationally attracts a particle placed at 3R from its centre with a force F 1 . Now a spherical cavity of radius \left( {{R \over 2}} \right) is made in the sphere (as shown in figure) and the force becomes F 2 . The value of F 1 : F 2 is

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