InterviewSolution
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InterviewSolution
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A solid sphere of uniform density and radius `R` applies a gravitational force 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 shows 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 A. `1//2`B. `7//9`C. `3`D. `7` |
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Answer» Correct Answer - B `C rarr cavity`, `T rarr` Total, `R rarr` remaining `F_(1) = (GMm)/((2R)^(2))`..(i) `F_(R ) = F_(2) = F_(T) - F_(C) = F_(1) - F_(C)` or `F_(2) = (GMm)/((2R)^(2)) - (G((M)/(8)) m)/((3R//2)^(2))` or `F_(2) = (14 GMm)/(72 R^(2))` From Eqs. (i) and (ii) we get, `(F_(2))/(F_(1)) = (7)/(9)` |
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