InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
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 A. `(5)/(9)`B. `(7)/(8)`C. `(3)/(4)`D. `(7)/(9)` |
|
Answer» Correct Answer - D `F_(1)=(GMm)/((2R)^(2))=(GMm)/(4R^(2))` `F_(2)=F_(1)-F_("cavity")=(GMm)/(4R^(2))-(G((M)/(8))(m))/(((3)/(2)R)^(2))` `rArr F_(2)=(7GMn)/(36 R^(2))rArr (F_(2))/(F_(1))=(7)/(9)`. |
|