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Q.

A solid sphere of uniform density and radius R applies a gravitational force of  attraction equal to F1 on a particle placed atP, 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 sphere with cavity now applies an gravitational force F2 on same particle placed at P. The ratio F2/ F1 will be

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a

1 /2

b

7/9

c

3

d

7

answer is B.

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Detailed Solution

Gravitational force due to solid sphere,  F1=GMm(2R)2 where M and mare masses of the solid sphere and particle respectivelyand R is the radius of the sphere. The gravitational force on particle due to sphere with cavity = force due to solid sphere creating cavity, assumed to be present above at that position.i, e, F2=GMm4R2-G(M/8)m(3R/2)2=736GMmR2So, F2F1=7GMm36R2/GMm4R2=79
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