Two particles each of mass m are placed at points A and C such that AB =BC =L as shown in fig. The gravitational force on a third particle placed at D at a distance I metre on the perpendicular bisector of the line AC, is

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Gm22L2 along BD

Gm22L2 along DB

Gm2L2 along AC

none of these

answer is B.

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Detailed Solution

Here F=Gm2(2L)2=Gm22L2The resultant force on mass m placed at point D is F' = 2 F cos 45o long DB=2Gm22L2×12=Gm22L2 along DB

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