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in Physics by (75.1k points)
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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
image
A. `(5)/(9)`
B. `(7)/(8)`
C. `(3)/(4)`
D. `(7)/(9)`

1 Answer

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by (67.6k points)
 
Best 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)`.

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