Question

# A particle moves in a potential region given by $U=8{x}^{2}âˆ’4x+400$. Its state of equilibrium will be

A

$0.25M$

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B

$8M$

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C

$10M$

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D

$12M$

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Solution

## The correct option is A $0.25M$Step 1. Given Data,$U=8{x}^{2}âˆ’4x+400$Step 2. Formula Used,Let the force be $F$.$\frac{dU}{dx}=F$Step 3. Calculating the state of equilibrium,Let the force be $F$,$\frac{dU}{dx}=F$Differentiating the equation we get,$U=8{x}^{2}âˆ’4x+400\phantom{\rule{0ex}{0ex}}â‡’\frac{dU}{dx}=16xâˆ’4$Therefore, we get,$F=16xâˆ’4$In equilibrium condition, $F=0$,$16xâˆ’4=0$$âˆ´x=0.25\phantom{\rule{0ex}{0ex}}m$Therefore the state of equilibrium of the particle will be $0.25m$.Hence, option A is the correct option.

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