A composite rod is made by joining a copper rod, end to end, with a second rod of different material

Q: Why do we add the expansions in the first part?

Because the rods are joined end-to-end and are free to expand. The total change in length is just the sum of how much each individual rod grows.

Q: Why doesn't the length of the individual rods change in the second part?

The problem explicitly states this as a condition. Physically, this happens when the thermal expansion of each material is exactly balanced by the elastic compression caused by the rigid walls pushing back.

Q: Why are the forces equal at the junction?

Newton's Third Law! The junction is in equilibrium, meaning the force exerted by the copper rod on the second rod must be equal and opposite to the force exerted by the second rod on the copper rod.

Q: What happens to the area of cross-section $A$?

Since both rods have the same cross-sectional area, it appears on both sides of our force equation ($F = Y A \alpha \Delta \theta$) and neatly cancels out.

Q: Why do we use $\Delta \theta = 100^\circ\text{C}$?

The temperature is raised from $25^\circ\text{C}$ to $125^\circ\text{C}$. The change in temperature, which drives the thermal expansion, is the difference between these two values: $125 - 25 = 100^\circ\text{C}$.

Question

A composite rod is made by joining a copper rod, end to end, with a second rod of different material but of the same cross-section. At 25°C, the composite rod is 1 m in length, of which the length of the copper rod is 30 cm. At 125°C the length of the composite rod increases by 1.91 mm.
When the composite rod is not allowed to expand by holding it between two rigid walls, it is found that the length of the two constituents do not change with the rise of temperature. Find the Young's modulus and the coefficient of linear expansion of the second rod. (Given, Coefficient of linear expansion of copper = $1.7 	imes 10^{-5}	ext{ per }^\circ	ext{C}$, Young's modulus of copper = $1.3 	imes 10^{11}	ext{ N/m}^2$ )

Sigma Percentile · Accepted Answer

### The Setup: A Tale of Two Rods

Imagine a composite rod, a perfect union of copper and an unknown material, stretching out to a total length of $1	ext{ m}$. The copper segment claims $30	ext{ cm}$ of this length, leaving the remaining $70	ext{ cm}$ to its mysterious partner. As the temperature rises from $25^\circ	ext{C}$ to $125^\circ	ext{C}$, a fiery energy is injected into the system, creating a temperature change $\Delta 	heta = 100^\circ	ext{C}$. This heat will test the physical properties of both materials in two distinct scenarios.

### Phase 1: The Freedom to Expand

In the first scenario, the rod is free to stretch its legs. When heated, the total expansion of the composite rod is simply the sum of the individual expansions of its two parts. We know the fundamental law of linear expansion: $\Delta l = l \alpha \Delta 	heta$.

Therefore, the total change in length is given by:

$$\Delta l = l_1 \alpha_1 \Delta 	heta + l_2 \alpha_2 \Delta 	heta$$

We are given that the total expansion $\Delta l$ is $1.91	ext{ mm}$, or $1.91 	imes 10^{-3}	ext{ m}$. Plugging in the known values for copper ($l_1 = 0.3	ext{ m}$, $\alpha_1 = 1.7 	imes 10^{-5}	ext{ /}^\circ	ext{C}$) and the second rod ($l_2 = 0.7	ext{ m}$), we get:

$$1.91 	imes 10^{-3} = (0.3)(1.7 	imes 10^{-5})(100) + (0.7)(\alpha_2)(100)$$

Solving this elegant linear equation reveals the identity of the second material's expansion coefficient:

$$\alpha_2 = 2.0 	imes 10^{-5}	ext{ /}^\circ	ext{C}$$

### Phase 2: The Unyielding Walls

Now, the plot thickens. The composite rod is trapped between two rigid, immovable walls. As the temperature rises again, the rods desperately want to expand, but the walls push back with an equal and opposite mechanical force. The problem presents a fascinating condition: the length of the *individual* constituents does not change.

This implies a perfect stalemate. The thermal expansion of each rod is exactly nullified by the mechanical compression caused by the walls. For this delicate equilibrium to hold at the junction where the two rods meet, the thermal force exerted by the copper rod must perfectly match the thermal force exerted by the second rod.

### The Grand Finale: Unveiling the Unknowns

The thermal force developed in a constrained rod is given by $F = Y A \alpha \Delta 	heta$. Equating the forces for both rods, we get:

$$F_1 = F_2$$

$$Y_1 A \alpha_1 \Delta 	heta = Y_2 A \alpha_2 \Delta 	heta$$

The cross-sectional area $A$ and the temperature change $\Delta 	heta$ are identical for both rods, allowing us to gracefully cancel them out:

$$Y_1 \alpha_1 = Y_2 \alpha_2$$

We now have a direct bridge to our final unknown, the Young's modulus of the second rod ($Y_2$). Rearranging and substituting the values:

$$Y_2 = \frac{Y_1 \alpha_1}{\alpha_2} = \frac{(1.3 	imes 10^{11})(1.7 	imes 10^{-5})}{2.0 	imes 10^{-5}}$$

$$Y_2 = 1.105 	imes 10^{11}	ext{ N/m}^2$$

And there we have it! By understanding the interplay between thermal expansion and elastic forces, we've successfully decoded the properties of the mysterious second rod.

Visualized Solution

$Initial Setup$

$Free Expansion$

$Substituting Values$

$Calculating α_{2}$

$Constrained Expansion$

$Thermal Force$

$Equating Forces$

$Calculating Y_{2}$

$Final Answer$

The Sigma Insight: Thermal Expansion

The Setup

Phase 1

Phase 2

The Grand Finale

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Initial Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Free Expansion

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating α2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Constrained Expansion

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Thermal Force

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equating Forces

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Y2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer