The adjacent graph shows the extension (Δl) of a wire of length 1 m suspended from the top of a roof

Q: Why does the graph pass through the origin?

The graph passes through the origin because when there is no load applied ($W = 0$), there is no extension in the wire ($\Delta l = 0$). This is a direct consequence of Hooke's Law, which states that deformation is directly proportional to the applied force within the elastic limit.

Q: What is the most common mistake students make in this problem?

The most common mistake is forgetting the scale factor of $10^{-4}$ on the vertical axis. If you read the extension values as simply $1$ and $4$ instead of $1 \times 10^{-4}$ and $4 \times 10^{-4}$, your calculated Young's Modulus will be off by four orders of magnitude, leading to an incorrect option.

Q: How would the graph change if we used a thicker wire of the same material?

If the wire is thicker, its cross-sectional area $A$ is larger. Since the slope of the graph is $\frac{L}{YA}$, a larger $A$ will decrease the slope. The line on the graph would be flatter (less steep) because a thicker wire is stiffer and stretches less for the same applied load.

Q: Can we use any two points on the line to calculate the slope?

Yes, because the graph is a straight line, its slope is constant everywhere. You can choose any two points on the line. However, choosing points that lie exactly on the grid intersections (like $20\text{ N}$ and $80\text{ N}$) makes the reading more accurate and the calculations simpler.

Q: What does a higher Young's Modulus physically represent?

Young's Modulus is a measure of a material's stiffness. A higher Young's Modulus means the material is stiffer and more resistant to elastic deformation. For example, steel has a much higher Young's Modulus than rubber, meaning it requires a much larger force to stretch by the same amount.

Question

The adjacent graph shows the extension ($\Delta l$) of a wire of length $1	ext{ m}$ suspended from the top of a roof at one end and with a load $W$ connected to the other end. If the cross-sectional area of the wire is $10^{-6}	ext{ m}^2$, calculate from the graph the Young's modulus of the material of the wire.

Options:

(A) $2 	imes 10^{11}	ext{ N/m}^2$

(B) $2 	imes 10^{-11}	ext{ N/m}^2$

(C) $3 	imes 10^{12}	ext{ N/m}^2$

(D) $2 	imes 10^{13}	ext{ N/m}^2$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are conducting an experiment in a physics lab. You hang a metal wire of length $L = 1	ext{ m}$ from a rigid ceiling and attach various weights to its free end.

As you add more weight, the wire stretches. You carefully measure this tiny stretch, $\Delta l$, and plot it against the applied load, $W$.

The resulting graph is a beautiful, straight line passing through the origin. This linear relationship is a direct manifestation of **Hooke's Law** within the elastic limit of the material.

$$	ext{Stress} \propto 	ext{Strain}$$

Let's write this down mathematically:

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

Here, the tension force $F$ in the wire is exactly equal to the suspended load $W$, and $A$ is the cross-sectional area of the wire.

Our goal is to find the Young's Modulus, $Y$, of the wire's material using only the information provided in the graph.

### The Master Equation

To connect our physical formula to the graph, we need to express the vertical axis variable ($\Delta l$) in terms of the horizontal axis variable ($W$). Let's rearrange our Hooke's Law equation:

$$\Delta l = \left(\frac{L}{Y A}ight) W$$

This equation is in the standard form of a straight line passing through the origin:

$$y = m x$$

where:
- $y = \Delta l$ (the vertical axis)
- $x = W$ (the horizontal axis)
- $m = \frac{L}{Y A}$ (the slope of the line)

This is our conceptual breakthrough! The slope of the line on the graph is inversely proportional to the Young's Modulus of the material. Therefore, if we can calculate the slope of this line, we can easily solve for $Y$:

$$Y = \frac{L}{A 	imes 	ext{Slope}}$$

### Calculating the Slope

Let's look closely at the graph to find the slope. We need to choose two convenient points on the line that are easy to read.

Let's pick:
- **Point 1:** At $W_1 = 20	ext{ N}$, the extension is $\Delta l_1 = 1 	imes 10^{-4}	ext{ m}$.
- **Point 2:** At $W_2 = 80	ext{ N}$, the extension is $\Delta l_2 = 4 	imes 10^{-4}	ext{ m}$.

Now, we compute the change in both variables:

$$\Delta W = 80 - 20 = 60	ext{ N}$$

$$\Delta(\Delta l) = (4 - 1) 	imes 10^{-4} = 3 	imes 10^{-4}	ext{ m}$$

The slope ($m$) is the ratio of the change in the vertical coordinate to the change in the horizontal coordinate:

$$	ext{Slope} = \frac{\Delta(\Delta l)}{\Delta W} = \frac{3 	imes 10^{-4}	ext{ m}}{60	ext{ N}} = 0.5 	imes 10^{-5}	ext{ m/N}$$

### Final Calculation

Now that we have the slope, we can substitute all our known values into our rearranged formula for Young's Modulus:
- Length of the wire, $L = 1	ext{ m}$
- Cross-sectional area, $A = 10^{-6}	ext{ m}^2$
- Slope of the graph, $	ext{Slope} = 0.5 	imes 10^{-5}	ext{ m/N}$

Let's plug these in:

$$Y = \frac{1}{10^{-6} 	imes 0.5 	imes 10^{-5}}$$

$$Y = \frac{1}{0.5 	imes 10^{-11}}$$

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

This matches **Option (a)** perfectly!

This value is highly realistic—it is, in fact, the Young's Modulus of steel, one of the most common materials tested in such experiments.

Select Answer:

Visualized Solution

Understanding the Physical Setup and the Graph

Connecting Hooke's Law and Young's Modulus

Expressing Extension in Terms of Load

Calculating the Slope from the Graph

Solving for Young's Modulus $Y$

Substituting Values to Find $Y$

Exploring Variations and Key Takeaways

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

Analyzing the Setup

The Master Equation

Calculating the Slope

Final Calculation

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