A straight rod of length $\mathrm{L}$ extends from $x=a$ to $x=\mathrm{L}+\mathrm{a}$. The gravitational force it exerts on point mass ' $\mathrm{m}$ ' at $x=0$, if the mass per unit length of the rod is $\mathrm{A}+\mathrm{B} x^{2}$, is given by:

Question:

A straight rod of length $\mathrm{L}$ extends from $x=a$ to $x=\mathrm{L}+\mathrm{a}$.

The gravitational force it exerts on point mass ‘ $\mathrm{m}$ ‘ at $x=0$,

if the mass per unit length of the rod is $\mathrm{A}+\mathrm{B} x^{2}$, is given

by:

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

JEE Main Previous Year Single Correct Question of JEE Main from Physics Gravitation chapter.

JEE Main Previous Year 12 Jan. 2019 I

Correct Option: 4

Solution:

A straight rod of length $\mathrm{L}$ extends from $x=a$ to $x=\mathrm{L}+\mathrm{a}$. The gravitational force it exerts on point mass ‘ $\mathrm{m}$ ‘ at $x=0$, if the mass per unit length of the rod is $\mathrm{A}+\mathrm{B} x^{2}$, is given by: