A rod of length L at room temperature and uniform area of cross-section A, is made of a metal having

Q: Why is the net change in length zero?

The problem explicitly states that an external compressive force is applied which 'prevents any change in the length of the rod'. Therefore, the net macroscopic change in length is zero.

Q: Can we use $\Delta L = L \alpha \Delta T$ even if the rod doesn't actually expand?

Yes! Think of it as a two-step virtual process: First, the rod expands freely by $\Delta L$ due to heat. Second, the mechanical force compresses it back by that exact same $\Delta L$. The formula calculates the 'prevented' expansion.

Q: Why do we equate thermal strain and mechanical strain?

To keep the net strain zero, the mechanical compression must exactly cancel the thermal expansion. Therefore, their magnitudes must be perfectly equal.

Q: What if the force was tensile instead of compressive?

A tensile force would pull the rod and stretch it further. To counteract expansion and push the rod back to its original length, the force must strictly be compressive.

Q: Does the area of cross-section change during this process?

In standard linear expansion problems, we assume the change in the cross-sectional area is negligible compared to the length, so we treat $A$ as a constant.

Question

A rod of length $L$ at room temperature and uniform area of cross-section $A$, is made of a metal having coefficient of linear expansion $\alpha /^\circ	ext{C}$. It is observed that an external compressive force $F$, is applied on each of its ends, prevents any change in the length of the rod, when its temperature rises by $\Delta T$ K. Young's modulus, $Y$ for this metal is

Options:

(A) $\frac{F}{2A\alpha \Delta T}$

(B) $\frac{F}{A\alpha(\Delta T - 273)}$

(C) $\frac{2F}{A\alpha\Delta T}$

(D) $\frac{F}{A\alpha\Delta T}$

Sigma Percentile · Accepted Answer

This problem is a classic intersection of **Thermodynamics** and **Solid Mechanics**. It beautifully demonstrates how materials respond to temperature changes and how mechanical forces can counteract those natural tendencies.

### Analyzing the Setup
Imagine a metal rod of length $L$, cross-sectional area $A$, and a coefficient of linear expansion $\alpha$. When the temperature of this rod is increased by $\Delta T$, the atoms vibrate more vigorously, pushing each other apart. If the rod were free to expand, its length would increase by an amount $\Delta L$.

According to the laws of thermal expansion, this change in length is given by:

$$\Delta L = L \alpha \Delta T$$

From this, we can define the **thermal strain**, which is the fractional change in length:

$$	ext{Thermal Strain} = \frac{\Delta L}{L} = \alpha \Delta T$$

### The Mechanical Counteraction
However, the problem introduces a twist: the rod is *not* allowed to expand. An external compressive force $F$ is applied at both ends, perfectly preventing any change in length.

What does this mean physically? It means the mechanical force is compressing the rod by the exact same amount $\Delta L$ that the heat is trying to expand it. The net change in length is zero, but internally, the rod is under immense stress.

### Applying Hooke's Law
To relate this compressive force to the material's properties, we invoke **Hooke's Law**. For linear elasticity, Young's modulus $Y$ is defined as the ratio of stress to strain:

$$Y = \frac{	ext{Stress}}{	ext{Strain}}$$

Here, the mechanical stress is the force per unit area:

$$	ext{Stress} = \frac{F}{A}$$

And the mechanical strain required to counteract the thermal expansion is:

$$	ext{Strain} = \frac{\Delta L}{L}$$

Substituting these into Hooke's Law gives us:

$$Y = \frac{F / A}{\Delta L / L}$$

### The Master Equation
Since the mechanical strain must exactly equal the thermal strain to keep the rod's length constant, we can substitute our earlier expression for thermal strain ($\alpha \Delta T$) into the denominator:

$$Y = \frac{F / A}{\alpha \Delta T}$$

### Final Calculation
Now, it's just a matter of simple algebraic rearrangement. We want to find the expression for Young's modulus $Y$. By bringing the area $A$ down to the denominator, we get:

$$Y = \frac{F}{A \alpha \Delta T}$$

This elegant formula tells us exactly how stiff the material must be to generate a specific restoring force under a given temperature change. The correct option is **(d)**.

Select Answer:

Visualized Solution

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$T h er ma l E x p an s i o n$

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$H oo k e^{'} s L a w$

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The Sigma Insight: Thermal Expansion

Analyzing the Setup Imagine a metal rod of length $L$ , cross-sectional area $A$ , and a coefficient of linear expansion $α$

The Mechanical Counteraction However, the problem introduces a twist: the rod is not allowed to expand

Applying Hooke's Law To relate this compressive force to the material's properties, we invoke Hooke's Law

The Master Equation

Final Calculation Now, it's just a matter of simple algebraic rearrangement

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