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A straight rod of length L extends from x = a to x = L + a. The gravitational force it exerts on a point mass 'm' at x = 0, if the mass per unit length of the rod is A + Bx², is given by:

1. Gm [A(1/a - 1/(a+L)) - BL]
2. Gm [A(1/a - 1/(a+L)) + BL]
3. Gm [A(1/(a+L) - 1/a) + BL]
4. Gm [A(1/(a+L) - 1/a) - BL]

Correct answer: Gm [A(1/a - 1/(a+L)) + BL]

Solution

F = Gm * integral_a^{a+L} (A+Bx^2)/x^2 dx = Gm[-A/x + Bx] from a to a+L = Gm[A(1/a - 1/(a+L)) + BL]; the B term integrates to +BL.

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