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$Y = \dfrac{{Stress}}{{Strain}}$

Elastic Potential Energy is given by the formula,

$U = \int {dW} $

It is given in the question that,

Natural length of the string is l

Cross – sectional area of the string is A

Young’s Modulus of the string is Y

Force acting on the string is F

According to Mohr’s Law,

$Y = \dfrac{{Stress}}{{Strain}}$ ……….(1)

Where,

Y is the Young’s Modulus

Also,

$Stress = \dfrac{F}{A}$

Where,

F is the Tension in the String

And,

\[Strain = \dfrac{{\Delta l}}{l}\]

Inserting the values of stress and strain in equation 1,

We get,

$ = > Y = \dfrac{{\dfrac{F}{A}}}{{\dfrac{{\Delta l}}{l}}}$

$ = > Y = \dfrac{{Fl}}{{A\Delta l}}$

$ = > F = \dfrac{{YA\Delta l}}{l}$ ……..(2)

Elastic Potential Energy is given by the formula,

$U = \int {dW} $

Where dW is the work done

In the above formula,

$dW = F.dx$

Where,

dx is the change in length of the string due to the applied tension

Inserting the value of dW in above equation,

We get,

$ = > U = \int {F.dx} $

Inserting the value of F in the above equation and the limits would be from 0 to $\Delta l$ (Since, initially the string was unstretched so initial limit is 0 and finally the string is stretch by $\Delta l$ so the final limit is $\Delta l$)

We get,

$ = > U = \int\limits_0^{\Delta l} {\dfrac{{YAx}}{l}.dx} $

$ = > U = \dfrac{{YA}}{l}\int\limits_0^{\Delta l} {x.dx} $

$ = > U = \dfrac{{YA}}{l}[\dfrac{{{x^2}}}{2}]_0^{\Delta l}$

$ = > U = \dfrac{{YA}}{l}\dfrac{{{{(\Delta l)}^2}}}{2}$ …………(3)

From Equation 2,

$ = > F = \dfrac{{YA\Delta l}}{l}$

$ = > \Delta l = \dfrac{{Fl}}{{AY}}$

Inserting the value of $\Delta l$ in Equation 3,

We get,

$ = > U = \dfrac{{YA}}{l}\dfrac{{{{(\dfrac{{Fl}}{{AY}})}^2}}}{2}$

$ = > U = \dfrac{{{F^2}l}}{{2AY}}$

Comparing the above equation with the one given in the question,

$ = > U = \dfrac{{{F^2}l}}{{2AY}} = \dfrac{{{F^2}l}}{{xAY}}$

Clearly, we can see from the above expression that,

$x = 2$