Q. A straight rod of length $L$ extends from $x = a$ to $x = L + a$. Find the gravitational force it, exerts on a point mass $m$ at $x = 0$ if the linear density of rod $\mu = A + Bx^{2}$
A) $Gm\left\lbrack \frac{A}{a} + BL \right\rbrack$
B) $Gm\left\lbrack A\left( \frac{1}{a} - \frac{1}{a + L} \right) + BL \right\rbrack$
C) $Gm\left\lbrack BL + \frac{A}{a + L} \right\rbrack$
D) $Gm\left\lbrack BL - \frac{A}{a} \right\rbrack$
Explanation
(b) $\therefore dF = \frac{Gm(\mu dx)}{x^{2}}$
[image:8_image67.emf]
$F = Gm\ \int_{a}^{a + L}{(A + Bx)\frac{dx}{x^{2}}}$
$F = Gm\left\lbrack A\left( \frac{1}{a} - \frac{1}{a + L} \right) + BL \right\rbrack$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.