Animated Solution for Physics - Properties of Solids and Liquids: The area of cross-section of a railway track is $0.01{\text{ m}}^2$. The temperature variation is $10^{\circ }\text{C}$. Coefficient of linear expansion of material of track is $10^{-5}{/}^{\circ }\text{C}$. The energy stored per metre in the track is ...... $\text{J/m}$. (Take, Young's modulus of material of track is $10^{11}{\text{ Nm}}^{-2}$)

Imagine a long, continuous railway track baking under the scorching summer sun. As the temperature rises, the metal naturally wants to expand. But there's a catch—the track is firmly bolted down and constrained between rigid supports. Because it cannot physically lengthen, the track experiences immense internal forces. This phenomenon is known as thermal stress, and it turns the track into a giant, invisible spring storing elastic potential energy.

To understand how much energy is trapped inside, we first need to figure out how much the track wants to expand. If the track were free to move, its change in length $\Delta l$ would be given by the formula $\Delta l=l\alpha \Delta T$, where $\alpha$ is the coefficient of linear expansion and $\Delta T$ is the temperature change.

Because the track is constrained, it is effectively being compressed back to its original length by the supports. The fractional change in length, or thermal strain, is therefore:

When a material is subjected to stress and strain, it stores elastic potential energy. The energy stored per unit volume, known as energy density ($u$), is a fundamental concept in the mechanical properties of solids. It is given by the equation:

Using Hooke's Law, where $\text{Stress}=Y\times \text{Strain}$ ($Y$ being Young's modulus), we can rewrite the energy density entirely in terms of strain:

The question specifically asks for the energy stored per meter of the track, which is the energy per unit length ($U^{\prime }$). Since the total volume of the track is its cross-sectional area ($A$) multiplied by its length ($l$), we can find the energy per unit length by simply multiplying the energy density by the area:

Substituting our expression for energy density, we get our master equation:

Now, it's time to plug in the numbers. We are given $Y=10^{11}{\text{ N/m}}^2$, $\alpha =10^{-5}{/}^{\circ }\text{C}$, $\Delta T=10^{\circ }\text{C}$, and $A=0.01{\text{ m}}^2$ (which is $10^{-2}{\text{ m}}^2$).

First, let's calculate the thermal strain:

Squaring the strain gives us $10^{-8}$. Now, we substitute everything into our master equation:

Combining the powers of 10 ($11-8-2=1$), we get:

The final answer is 5 J/m. This elegant result shows that for every single meter of this railway track, a temperature rise of just $10^{\circ }\text{C}$ stores 5 Joules of energy, ready to be unleashed if the constraints ever fail!