A mass m is at a distance a from one end of a uniform rod of length l and M. The gravitational force on the mass due to the rod is

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GMma+l

GmMal+a

GMma

GmM2l+a

answer is B.

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Detailed Solution

Considering mass per unit length x=ml mass of a small portion dx=λdx at distance x. Force due to small portion Gmλdxx2, Total fore ∫aa+lGmλdxx2F=Gmλ-1xaa+l=Gmλ1a-1a+l=Gmλ1aa+l=GMmaa+l

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