Question

A particle moves in a potential region given by $U=8x^2-4x+400J.$ Its state of equilibrium will be$-$

• A. 8 m
• B. 10 m
• C. 12 m
• D. 0.25 m

Answer 1

The correct option is D 0.25 m

$U=8x^2-4x+400$
Differentiate with $dx$ we will get $F$
$F=16x-4$
At equilibrium $F=0$
$0=16x-4$
$x=1/4=0.25m$
It’s state of equilibrium will be at $x=0.25m$

Hence,

option $\left(D\right)$ is correct answer.