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

\(Gm\left[ {A\left( {\frac{1}{a} - \frac{1}{{a + L}}} \right) - BL} \right]\)

\(Gm\left[ {A\left( {\frac{1}{{a + L}} - \frac{1}{a}} \right) - BL} \right]\)

\(Gm\left[ {A\left( {\frac{1}{{a + L}} - \frac{1}{a}} \right) + BL} \right]\)

\(Gm\left[ {A\left( {\frac{1}{a} - \frac{1}{{a + L}}} \right) + BL} \right]\)

A straight rod of length L extends from x = a to x = L + a. The gravit

1.	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
A.	\(Gm\left[ {A\left( {\frac{1}{{a + L}} - \frac{1}{a}} \right) - BL} \right]\)
B.	\(Gm\left[ {A\left( {\frac{1}{{a + L}} - \frac{1}{a}} \right) + BL} \right]\)
C.	\(Gm\left[ {A\left( {\frac{1}{a} - \frac{1}{{a + L}}} \right) + BL} \right]\)
D.	\(Gm\left[ {A\left( {\frac{1}{a} - \frac{1}{{a + L}}} \right) - BL} \right]\)
Answer» D. \(Gm\left[ {A\left( {\frac{1}{a} - \frac{1}{{a + L}}} \right) - BL} \right]\)

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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

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