Question

The potential energy of a particle in a central field has the form $U ( r )=\frac{1}{r^2}-\frac{1}{r}$, where ' $r ^{\prime}$ is the distance form the centre of the field. The magnitude of the maximum attractive force in Newton is

• A. $\frac{1}{27}$
• B. $\frac{1}{9}$
• C. $\frac{1}{3}$
• D. 1

Answer 1

Answer: (A)

Given, equation of potential energy,
$U(r)=\frac{1}{r^2}-\frac{1}{r}$
As we know that, attractive force,
$F =-\frac{d U}{d r} $
$F =-\frac{d}{d r}\left(\frac{1}{r^2}-\frac{1}{r}\right) $
$ =-\frac{d}{d r}\left(r^{-2}-r^{-1}\right) $
$ =-\left(-2 r^{-3}+r^{-2}\right) $
$ =\left(2 r^{-3}-r^{-2}\right) \ldots \text { (i) }$
Now, for $F_{\max }$ i.e. maximum value of force,
$\frac{d F}{d r}=0 $
$\frac{d}{d r}\left(2 r^{-3}-r^{-2}\right)=0 $
$-6 r^{-4}+2 r^{-3}=0 $
$2 r^{-3}=6 r^{-4} $
$\therefore 1=3 r^{-1} \Rightarrow r=3$Substituting in Eq. (i), we get
$F=\frac{2}{r^3}-\left.\frac{1}{r^2}\right|_{r=3}$
$ =\frac{2}{3^2}-\frac{1}{3^2}$
$ =\frac{2}{27}-\frac{1}{9}$
$ =\frac{2}{27}-\frac{3}{27}$
$ =-\frac{1}{27}$