The value of tension in a long thin metal wire has been changed from T1 to T2. The lengths of the me

Q: Why do we assume the area of cross-section $A$ remains constant?

For small elongations within the elastic limit, the lateral contraction (Poisson's effect) is negligible compared to the longitudinal extension. Therefore, the cross-sectional area $A$ is treated as a constant for simplicity.

Q: Can we use this formula if the wire is stretched beyond its elastic limit?

No. Hooke's Law ($Y = \text{Stress}/\text{Strain}$) is only valid within the proportional limit of the material. Beyond this limit, the relationship between stress and strain is no longer linear.

Q: Why did we divide the two equations instead of subtracting them?

Dividing the equations is a powerful algebraic trick to eliminate multiplicative constants like $A$ and $Y$ in one step. Subtracting them would leave the constants in the equation, making it harder to solve for $l_0$ without knowing their values.

Q: What if the tension is applied horizontally instead of vertically?

The orientation doesn't matter as long as the tension $T$ is uniform throughout the wire. Hooke's law applies to the internal restoring force, which equals the applied tension regardless of whether the wire is horizontal or vertical.

Q: Does the original length depend on the material of the wire?

The actual physical original length $l_0$ is a property of the specific wire you have. However, in this mathematical derivation, the material properties ($Y$) canceled out, meaning this formula for finding $l_0$ works for any material as long as it obeys Hooke's law.

Question

The value of tension in a long thin metal wire has been changed from $T_1$ to $T_2$. The lengths of the metal wire at two different values of tension $T_1$ and $T_2$ are $l_1$ and $l_2$, respectively. The actual length of the metal wire is

Options:

(A) $\frac{T_2 l_1 - T_1 l_2}{T_2 - T_1}$

(B) $\frac{T_1 l_1 - T_2 l_2}{T_1 - T_2}$

(C) $\frac{l_1 + l_2}{2}$

(D) $\sqrt{T_1 T_2 l_1 l_2}$

Sigma Percentile · Accepted Answer

### Visualizing the Setup
Imagine a long, thin metal wire hanging from a ceiling. Let's say its natural, unstretched length is $l_0$. Now, we pull it with a tension $T_1$, and it stretches to a new length $l_1$. If we pull it harder with a tension $T_2$, it stretches even more to a length $l_2$. Our goal is to find that original, unstretched length $l_0$.

### The Master Equation: Hooke's Law
To connect tension and length, we use Hooke's Law. Young's modulus $Y$ is defined as the ratio of stress to strain. Stress is the tension $T$ divided by the cross-sectional area $A$, and strain is the change in length $\Delta l$ divided by the original length $l_0$.

$$Y = \frac{	ext{Stress}}{	ext{Strain}} = \frac{T/A}{\Delta l / l_0} = \frac{T l_0}{A \Delta l}$$

Remember, the change in length $\Delta l$ is just the final length $l$ minus the original length $l_0$. So, we can rewrite the equation as:

$$\Delta l = l - l_0 = \frac{T l_0}{A Y}$$

### Setting Up the Equations
Let's apply this master equation to our two specific cases. When the tension is $T_1$, the final length is $l_1$. So the change in length is $l_1 - l_0$. Rearranging our formula, we get:

$$l_1 - l_0 = \frac{T_1 l_0}{A Y} \quad 	ext{--- (Equation 1)}$$

Similarly, for the second case, when the tension is $T_2$, the length becomes $l_2$. The change in length is $l_2 - l_0$. Following the exact same logic, we get:

$$l_2 - l_0 = \frac{T_2 l_0}{A Y} \quad 	ext{--- (Equation 2)}$$

### Eliminating the Unknown Constants
Now, look closely at these two equations. They both have this messy term $\frac{l_0}{A Y}$. Since the wire is the same in both cases, its cross-sectional area $A$ and Young's modulus $Y$ are constants. To eliminate them, let's simply divide Equation 1 by Equation 2.

$$\frac{l_1 - l_0}{l_2 - l_0} = \frac{\frac{T_1 l_0}{A Y}}{\frac{T_2 l_0}{A Y}}$$

The constants cancel out beautifully, leaving us with a very clean ratio:

$$\frac{l_1 - l_0}{l_2 - l_0} = \frac{T_1}{T_2}$$

### Final Calculation
We are almost there. We just need to isolate $l_0$. Let's cross-multiply to get rid of the fractions.

$$T_2 (l_1 - l_0) = T_1 (l_2 - l_0)$$

Expanding the brackets, we have:

$$T_2 l_1 - T_2 l_0 = T_1 l_2 - T_1 l_0$$

Finally, let's bring all the terms containing $l_0$ to one side. Moving things around, we get:

$$T_1 l_0 - T_2 l_0 = T_1 l_2 - T_2 l_1$$

Factoring out $l_0$:

$$l_0 (T_1 - T_2) = T_1 l_2 - T_2 l_1$$

Dividing by $(T_1 - T_2)$, and multiplying the numerator and denominator by $-1$ to match the given options, we get our final answer:

**$$l_0 = \frac{T_2 l_1 - T_1 l_2}{T_2 - T_1}$$**

This matches option (a) perfectly. This problem elegantly demonstrates how taking ratios can swiftly eliminate unknown constants in physics problems.

Select Answer:

Visualized Solution

\text{Visualizing the Setup}

\text{Hooke's Law}

\text{Equation for State 1}

\text{Equation for State 2}

\text{Eliminating Constants}

\text{Algebraic Manipulation}

\text{Final Answer}

\text{The Way Forward}

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

Visualizing the Setup Imagine a long, thin metal wire hanging from a ceiling

The Master Equation

Setting Up the Equations Let's apply this master equation to our two specific cases

Eliminating the Unknown Constants Now, look closely at these two equations

Final Calculation We are almost there

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