A thin rod of negligible mass and area of cross-section 4 * 10^-6 m^2, suspended vertically from one

Q: Why is the mass of the rod considered negligible? Doesn't its own weight cause stretching?

Yes, in a real rod, its own weight would cause a non-uniform stress distribution (maximum at the top, zero at the bottom). By specifying 'negligible mass', the problem simplifies the stress to be perfectly uniform throughout the rod, caused solely by the suspended mass $M$.

Q: Why do we equate the thermal contraction to the mechanical stretch? Aren't they happening at the same time?

Because the rod is 'prevented from contracting', its net change in length is zero. Even though cooling and stretching happen together, the net effect is $\Delta l_{\text{net}} = \Delta l_{\text{thermal}} + \Delta l_{\text{mechanical}} = 0$. This allows us to equate their absolute magnitudes.

Q: Can I use the formula $U = \frac{1}{2} \times \text{Stress} \times \text{Strain}$ directly for energy?

Be careful! The formula $U = \frac{1}{2} \times \text{Stress} \times \text{Strain}$ gives the energy *density* (energy per unit volume). To get the total energy, you must multiply this density by the total volume of the rod ($A \times L$).

Q: Does the temperature change affect Young's Modulus of the rod?

In reality, Young's Modulus does vary slightly with temperature. However, for standard JEE problems, we assume the elastic properties like $Y$ remain constant over the given temperature range unless specified otherwise.

Q: What would happen if we attached a mass greater than 40 kg?

If $M > 40\text{ kg}$, the mechanical stretching force would exceed the thermal contraction force, and the rod would experience a net elongation (it would become longer than $0.5\text{ m}$).

Question

A thin rod of negligible mass and area of cross-section $4 	imes 10^{-6} 	ext{ m}^2$, suspended vertically from one end, has a length of $0.5 	ext{ m}$ at $100^\circ	ext{C}$. The rod is cooled to $0^\circ	ext{C}$, but prevented from contracting by attaching a mass at the lower end. Find (a) this mass and (b) the energy stored in the rod, given for the rod. Young's modulus $= 10^{11} 	ext{ N/m}^2$, coefficient of linear expansion $= 10^{-5} 	ext{ K}^{-1}$ and $g = 10 	ext{ m/s}^2$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are holding a thin metal rod suspended vertically from a rigid ceiling. At a high temperature of $100^\circ	ext{C}$, the rod hangs in a relaxed state.

Now, we cool the rod down to $0^\circ	ext{C}$. Naturally, the atoms in the metal lose kinetic energy, their average separation decreases, and the rod attempts to **contract** thermally.

However, we prevent this contraction completely by attaching a mass $M$ at the lower end. This mass pulls the rod downwards, creating a **mechanical stretch** that perfectly counteracts the thermal shrinkage.

This is a classic coupling of **thermal expansion** and **elasticity** (Young's Modulus). Let's break down the physics step-by-step.

### Calculating Thermal Contraction

First, let's determine how much the rod would have contracted if it were free to shrink. The change in length due to temperature change is given by the linear thermal expansion formula:

$$\Delta l_1 = L \alpha \Delta	heta$$

Here, the initial length $L = 0.5	ext{ m}$, the coefficient of linear expansion $\alpha = 10^{-5}	ext{ K}^{-1}$, and the temperature change $\Delta	heta = 0^\circ	ext{C} - 100^\circ	ext{C} = -100	ext{ K}$.

Substituting these values:

$$\Delta l_1 = 0.5 	imes 10^{-5} 	imes (-100) = -0.5 	imes 10^{-3}	ext{ m}$$

This negative sign indicates a **decrease in length** of $0.5	ext{ mm}$.

### Calculating Mechanical Elongation

To keep the length constant, the attached mass $M$ must stretch the rod by an equal and opposite amount, $\Delta l_2 = 0.5 	imes 10^{-3}	ext{ m}$.

We relate the stretching force $F$ to the elongation $\Delta l_2$ using **Young's Modulus** $Y$:

$$Y = \frac{	ext{Stress}}{	ext{Strain}} = \frac{F/A}{\Delta l_2 / L}$$

Rearranging this formula for the mechanical elongation $\Delta l_2$ gives:

$$\Delta l_2 = \frac{F L}{A Y}$$

The stretching force is simply the weight of the suspended mass, $F = M g = 10 M$.

Substituting the given values ($A = 4 	imes 10^{-6}	ext{ m}^2$, $Y = 10^{11}	ext{ N/m}^2$, and $L = 0.5	ext{ m}$):

$$\Delta l_2 = \frac{(10 M)(0.5)}{(4 	imes 10^{-6})(10^{11})}$$

Simplifying the denominator:

$$(4 	imes 10^{-6}) 	imes 10^{11} = 4 	imes 10^5	ext{ N}$$

Thus, the elongation is:

$$\Delta l_2 = \frac{5 M}{4 	imes 10^5} = 1.25 	imes 10^{-5} M$$

### Finding the Required Mass

Since the net change in length is zero, the mechanical elongation must exactly equal the magnitude of the thermal contraction:

$$\Delta l_2 = |\Delta l_1|$$

$$1.25 	imes 10^{-5} M = 0.5 	imes 10^{-3}$$

Solving for $M$:

$$M = \frac{0.5 	imes 10^{-3}}{1.25 	imes 10^{-5}} = 40	ext{ kg}$$

Therefore, a **mass of $40	ext{ kg}$** must be attached to prevent the rod from contracting.

### Calculating Stored Elastic Energy

When the rod is stretched, work is done to pull the atoms apart against their metallic bonds. This work is stored in the rod as **elastic potential energy** $U$.

The formula for the elastic potential energy stored in a stretched wire is:

$$U = \frac{1}{2} k (\Delta l)^2$$

where $k = \frac{A Y}{L}$ is the equivalent spring constant of the rod, and $\Delta l$ is the elongation.

Let's calculate $k$ first:

$$k = \frac{(4 	imes 10^{-6}) 	imes 10^{11}}{0.5} = 8 	imes 10^5	ext{ N/m}$$

Now, substitute $k$ and $\Delta l = 0.5 	imes 10^{-3}	ext{ m}$ into the energy equation:

$$U = \frac{1}{2} 	imes (8 	imes 10^5) 	imes (0.5 	imes 10^{-3})^2$$

$$U = (4 	imes 10^5) 	imes (0.25 	imes 10^{-6})$$

$$U = 0.1	ext{ J}$$

Thus, the **elastic potential energy stored in the rod is $0.1	ext{ J}$**.

Visualized Solution

Visualizing the Physical System

Thermal Contraction Formula

Calculating Thermal Contraction $Δ l_{1}$

Mechanical Stretch via Young's Modulus

Substituting Mechanical Parameters

Simplifying $Δ l_{2}$

Applying the Constraint: $Δ l_{net} = 0$

Solving for Mass $M$

Elastic Potential Energy Formula

Raw Setup of Energy Calculation

Calculating Energy $U$

The Way Forward

The Sigma Insight: Young's Modulus, Bulk Modulus and Modulus of Rigidity

Analyzing the Setup

Calculating Thermal Contraction

Calculating Mechanical Elongation

Finding the Required Mass

Calculating Stored Elastic Energy

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