InterviewSolution
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InterviewSolution
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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 figure. The particle with cavity now applies a gravitational force `F_(2)` on same particle placed at `P`. The radio `F_(2)//F_(1)` will be |
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Answer» If `M` and `m` are the masses of solid sphere and particle respectively and `R` is the radius of the sphere, then gravitational force of attraction acting on the particle at `P` is, `F_(1) = GM m//4 R^(2)`. Mass of sphere of radius `R//2 = (4)/(3) pi (R//2)^(3) rho = (M)/(8)` The gravitational force of attraction `(F_(2))` acting on the particle due to sphere with cavity = force due to soild sphere - force due to the sphere creating the cavity assumed to be present at that position alone So, `F_(2) = (GM m)/((2R)^(2)) - (G(M//8)m)/((3R//2)^(2)) = (7)/(36) (GM m)/(R^(2))` Hence, `(F_(2))/(F_(1)) = (7)/(36) (GM m)/(R^(2))((GM m)/(4R^(2))) = (7)/(9)` |
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