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A solid sphere of uniform density and radius R applies a gravitational force of attraction equal to F_(1) on a particle placed at P, distance 2R from the centre O of the sphere. A spherical cavity of radius R//2 is now made in the sphere as shown in given figure. The sphere with cavity now applies a gravitational force F_(2) on same particle placed at P. The ratio F_(2)//F_(1) will be |
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Answer» `1//2` `F_(1)=(GMm)/((2R)^(2))`, where M and m are MASS of the solid sphere and particle respectively and R is the radius of the sphere. The gravitational force on particle due to sphere with cavity= force due to solid sphere-force due to sphere creating cavity, ASSUMED to be present above at that position. i.e., `F_(2)=(GMm)/(4R^(2))-(G(M//8)m)/((3R//2)^(2))=(7)/(36)(GM m)/(R^(2))` So, `(F_(2))/(F_(1))=(7 GMm)/(36R^(2))//((GM m)/(4 R^(2)))=(7)/(9)` |
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