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(A) Let the initial temperature be $T_0 = 20 ^o C$. At this temperature,the wire is unstrained,so the strain $\epsilon = 0$ and the elastic energy den

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(A) Let the initial temperature be $T_0 = 20 ^o C$. At this temperature,the wire is unstrained,so the strain $\epsilon = 0$ and the elastic energy density $u = 0$.When the temperature decreases by $\Delta T = (T_0 - T)$,the wire tends to contract by an amount $\Delta L = L \alpha \Delta T$,where $L$ is the length and $\alpha$ is the coefficient of linear expansion.Since the wire is clamped,it is stretched by this same amount $\Delta L$ to maintain its length between the walls.The strain is $\epsilon = \frac{\Delta L}{L} = \alpha \Delta T = \alpha (T_0 - T)$.The elastic energy density is given by $u = \frac{1}{2} Y \epsilon^2$,where $Y$ is Young's modulus.Substituting the expression for strain,we get $u = \frac{1}{2} Y \alpha^2 (T_0 - T)^2$.This equation represents a parabola opening upwards with its vertex at $T = T_0 = 20 ^o C$ and $u = 0$.As $T$ decreases from $20 ^o C$,$(T_0 - T)$ increases,and $u$ increases quadratically.Comparing this with the given options,the graph in image $86-$a110 shows $u$ increasing as $T$ decreases from $20 ^o C$ following a parabolic path,which matches our derived relation.

$A$ metal wire is clamped between two vertical walls. At $20 ^o C$,the unstrained length of the wire is exactly equal to the separation between the walls. If the temperature of the wire is decreased,the graph between elastic energy density $(u)$ and temperature $(T)$ of the wire is:

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