JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Gravitation

Take a small element of the rod of length $dx$ at distance $x$ from the point mass. Its mass is $dm=(A+B{x}^2)dx$, and the gravitational field it sets up at the origin is $\frac{Gdm}{x^2}=G\left(\frac{A}{x^2}+B\right)dx$. Integrate along the rod from $x=a$ to $x=a+L$: the $\frac{A}{x^2}$ part integrates to $A\left(\frac{1}{a}-\frac{1}{a+L}\right)$ and the constant $B$ part integrates to $BL$. Multiplying the field by the point mass $m$, the force is $F=Gm\left[A\left(\frac{1}{a}-\frac{1}{a+L}\right)+BL\right]$.