The length of a metal wire is l1, when the tension in it is T1 and is l2 when the tension is T2. The

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

In reality, as a wire stretches, it gets slightly thinner (due to Poisson's ratio). However, for small elastic deformations typical in these problems, the change in area is negligible and is standardly ignored unless explicitly mentioned.

Q: Can we just subtract the two equations instead of dividing?

If you subtract them, you will get an equation that still contains the unknown constants $Y$ and $A$. Dividing the equations is a strategic mathematical move to completely eliminate these unknowns in one step.

Q: What if the wire is thick and its own weight is not negligible?

If the wire has significant mass, the tension would not be uniform; it would be maximum at the top and zero at the bottom. You would have to set up a differential equation and integrate along the length of the wire to find the total elongation.

Q: Why is the change in length $\Delta l$ written as $l_1 - l_0$?

The variable $l_1$ represents the *total* final length of the wire under tension $T_1$. The strain formula requires the *change* in length, which is the final length minus the original natural length ($l_0$).

Q: Is Young's modulus different for different tensions?

No, Young's modulus is an intrinsic property of the material itself (like density). As long as the applied tension does not stretch the wire beyond its elastic limit, $Y$ remains perfectly constant.

Question

The length of a metal wire is $l_1$, when the tension in it is $T_1$ and is $l_2$ when the tension is $T_2$. The natural length of the wire is

Options:

(A) $\sqrt{l_1 l_2}$

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

(C) $\frac{l_1 T_2 + l_2 T_1}{T_2 + T_1}$

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

Sigma Percentile · Accepted Answer

This problem is a classic application of Hooke's Law and the concept of Young's Modulus. It tests your ability to set up physical equations and manipulate them algebraically to eliminate unknown constants.

### Analyzing the Setup
Imagine a metal wire hanging vertically from a rigid support. When no external force is applied, the wire rests at its natural length, which we will call $l_0$. We are given two distinct states of this wire under tension:

1.  **State 1:** A tension $T_1$ is applied, and the wire stretches to a new total length $l_1$.
2.  **State 2:** A different tension $T_2$ is applied, and the wire stretches to a total length $l_2$.

Our objective is to find an expression for the natural length $l_0$ purely in terms of the known variables: $l_1$, $l_2$, $T_1$, and $T_2$.

### The Master Equation
To connect tension and length, we rely on Young's Modulus ($Y$), which is a property of the material of the wire. Hooke's Law states that within the elastic limit, stress is proportional to strain.

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

Here, $F$ is the tension ($T$), $A$ is the cross-sectional area, $\Delta l$ is the change in length, and $l_0$ is the original natural length. Let's apply this master equation to our two states.

For **State 1**, the change in length is $(l_1 - l_0)$. Substituting this into our formula gives:

$$\frac{T_1}{A} = \frac{Y(l_1 - l_0)}{l_0}$$

Similarly, for **State 2**, the change in length is $(l_2 - l_0)$, yielding:

$$\frac{T_2}{A} = \frac{Y(l_2 - l_0)}{l_0}$$

### Algebraic Manipulation
We now have a system of two equations. Notice that the Young's Modulus $Y$, the cross-sectional area $A$, and the denominator $l_0$ are constants present in both equations. The most elegant way to eliminate these unknown constants is to divide the first equation by the second equation.

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

All the constants cancel out beautifully, leaving us with a clean, purely algebraic relationship:

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

### Final Calculation
To solve for $l_0$, we cross-multiply to remove the fractions:

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

Expanding the brackets on both sides, we get:

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

Our goal is to isolate $l_0$. Let's group all terms containing $l_0$ on the left side of the equation and move the remaining terms to the right side:

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

Factoring out $l_0$ on the left side gives:

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

Finally, dividing both sides by $(T_2 - T_1)$, we arrive at our final expression for the natural length of the wire:

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

This matches option (b). The beauty of this result is that it allows us to determine the natural length of a wire without ever needing to know its material properties or thickness, simply by measuring its length under two different known tensions.

Animated Solution for Physics - Properties of Solids and Liquids: The length of a metal wire is $l_{1}$ , when the tension in it is $T_{1}$ and is $l_{2}$ when the tension is $T_{2}$ . The natural length of the wire is

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Visualized Solution

$Visualizing the Wire States$

$Hooke’s Law$

$Equation for State 1$

$Equation for State 2$

$Eliminating Constants$

$Cross Multiplication$

$Expanding the Brackets$

$Grouping l_{0} Terms$

$Final Answer$

$The Way Forward$

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

Analyzing the Setup Imagine a metal wire hanging vertically from a rigid support

The Master Equation To connect tension and length, we rely on Young's Modulus ( $Y$ ), which is a property of the material of the wire

Algebraic Manipulation We now have a system of two equations

Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Wire States

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Hooke’s Law

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equation for State 2

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Grouping l0​ Terms

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

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