Question

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

• A. $Gm[A(\frac{1}{a}-\frac{1}{a+L})+BL]$
• B. $Gm[A(\frac{1}{a+L}-\frac{1}{a})+BL]$
• C. $Gm[A(\frac{1}{a+L}-\frac{1}{a})-BL]$
• D. $Gm[A(\frac{1}{a}-\frac{1}{a+L})-BL]$

Answer 1

The correct option is A $Gm[A(\frac{1}{a}-\frac{1}{a+L})+BL]$

Given $\lambda =(A+B{x}^2),$
taking small element $dm$ of length $dx$ at a distance $x$
from $x=0$

So, $dm=\lambda dx$
$dm=(A+B{x}^2)dx$
$dF=\frac{Gmdm}{x^2}$

$\Rightarrow F={\int }_a^{a+L}\frac{Gm}{x^2}(A+B{x}^2)dx$

$=Gm{[-\frac{A}{x}+Bx]}_a^{a+L}$

$=Gm[A(\frac{1}{a}-\frac{1}{a+L})+BL]$