Q 11-07-189JEE MainJEE Main 2019 (12 Jan, Shift 1)Medium
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 $A + Bx^2$, is given by:
Answer: (D) $Gm\left[A\left(\dfrac1a - \dfrac{1}{a + L}\right) + BL\right]$
An element $dx$ at $x$ has mass $(A + Bx^2)dx$ and attracts $m$ with
$$dF = \frac{Gm(A + Bx^2)}{x^2}dx$$
$$F = Gm\int_a^{a + L}\left(\frac{A}{x^2} + B\right)dx = Gm\left[A\left(\frac1a - \frac{1}{a + L}\right) + BL\right]$$
Solution by Sreeraj P, M.Sc Physics