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(c) $F = G\frac{{M 	imes M}}{{{R^2}}} = \frac{{G\, 	imes \,{{\left( {\frac{4}{3}\pi {R^3}ho } ight)}^2}}}{{{{(2R)}^2}}} \Rightarrow F \propto \f

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${R^4}$

Vedclass · Answer

(c) $F = G\frac{{M 	imes M}}{{{R^2}}} = \frac{{G\, 	imes \,{{\left( {\frac{4}{3}\pi {R^3}ho } ight)}^2}}}{{{{(2R)}^2}}} \Rightarrow F \propto \frac{{{R^6}}}{{{R^2}}} \propto {R^4}$

Two identical spheres are placed in contact with each other. The force of gravitation between the spheres will be proportional to ($R =$ radius of each sphere)

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