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Q.

A solid sphere of radius R gravitationally attracts a particle placed at 3R from its centre with a force F1 . Now a spherical cavity of radius  R2 is made in the sphere (as shown in figure) and the force becomes F2 . The value of F1:F2 is:

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a

50:41

b

25:36

c

41:50

d

36:25

answer is A.

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Detailed Solution

F1=GMm(3R)2=GMm9R2………..(1)F2=GMm9R2−GM8m5R22⇒F2=GMm9R2−GMm50R2=GMmR219−150=41450×GMmR2…………..(2)Dividing (1) & (2) ⇒F1F2=GMm9R241450GMmR2=5041
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