A Straight Rod Of Length L Extends From X=A To X=L +A. The Gravitational Force Is Exerted On A Point Mass 'm' At X=0, If The Mass Per Unit Length Of The Rod Is A+Bx2is Given By.

\( dm = (A +B x^{2}) dx \) \( df = \frac{GMdm}{x^{2}} \) \( \Rightarrow F = \int_{a}^{a+L} \frac{GM}{x^{2}} (A + B x^{2}) dx \) \( \Rightarrow GM = [-\frac{A}{x} + Bx]_{a}^{a} + L \) \( Gm [A (\frac{1}{a} – \frac{1}{a + L})] + BL \)

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