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Q. Two identical solid copper spheres of radius $ R $ are placed in contact with each other. The gravitational attraction between them is proportional to:

KEAMKEAM 2005

Solution:

$ F=\frac{GMM}{{{R}^{2}}} $ $ =\frac{G\left( \frac{4}{3}\pi {{R}^{3}}\rho \right)\left( \frac{4}{3}\pi {{R}^{3}}\rho \right)}{{{R}^{2}}} $
$ \therefore $ $ F\propto {{R}^{4}} $