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A solid sphere of radius R gravitationally attracts a particle placed at 3R from its centre with a force $F_{1}$ . Now a spherical cavity of radius $(\frac{R}{2})$ is made in the sphere (as shown in figure) and the force becomes $F_{2}$ . The value of $F_{1} : F_{2}$ is:

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detailed solution

Correct option is A

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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