The length of a metal wire is `l_1` when the tensionin it is `T_1 and is l_2` when the tension is `T_2`. The natural length of the wire is

Options

A. `(l_1+l_2)/2`
B. `sqrt(l_1l_2)`
C. `(l_1T_2-l_2T_1)/(T_2-T_1)`
D. `(l_1T_2+l_2T_1)/(T_2+T_1)`

Answer

To find the natural length of the metal wire, we can use the relationship between tension, length, and Young's modulus. Here’s a step-by-step solution:

### Step 1: Understand the relationship between stress, strain, and Young's modulus
According to Hooke's Law, stress is proportional to strain. The formula for stress (σ) is given by:
\[ \sigma = \frac{T}{A} \]
where \(T\) is the tension and \(A\) is the cross-sectional area of the wire. The strain (ε) is defined as:
\[ \epsilon = \frac{\Delta L}{L_0} \]
where \(\Delta L\) is the change in length and \(L_0\) is the original (natural) length of the wire.

### Step 2: Write the equations for the two states of the wire
For the first state of the wire (length \(L_1\) and tension \(T_1\)):
\[ \text{Strain}_1 = \frac{L_1 - L_0}{L_0} \]
Thus, the Young's modulus \(Y\) can be expressed as:
\[ Y = \frac{\sigma_1}{\epsilon_1} = \frac{\frac{T_1}{A}}{\frac{L_1 - L_0}{L_0}} \implies Y = \frac{T_1 L_0}{A (L_1 - L_0)} \]

For the second state of the wire (length \(L_2\) and tension \(T_2\)):
\[ \text{Strain}_2 = \frac{L_2 - L_0}{L_0} \]
Thus, the Young's modulus \(Y\) can also be expressed as:
\[ Y = \frac{\sigma_2}{\epsilon_2} = \frac{\frac{T_2}{A}}{\frac{L_2 - L_0}{L_0}} \implies Y = \frac{T_2 L_0}{A (L_2 - L_0)} \]

### Step 3: Set the two expressions for Young's modulus equal
Since the Young's modulus is the same for both states (as the material is the same), we can set the two equations equal to each other:
\[ \frac{T_1 L_0}{L_1 - L_0} = \frac{T_2 L_0}{L_2 - L_0} \]

### Step 4: Cross-multiply to eliminate fractions
Cross-multiplying gives:
\[ T_1 L_0 (L_2 - L_0) = T_2 L_0 (L_1 - L_0) \]

### Step 5: Expand and rearrange the equation
Expanding both sides:
\[ T_1 L_0 L_2 - T_1 L_0^2 = T_2 L_0 L_1 - T_2 L_0^2 \]
Rearranging gives:
\[ T_1 L_0 L_2 - T_2 L_0 L_1 = T_1 L_0^2 - T_2 L_0^2 \]

### Step 6: Factor out \(L_0\)
Factoring out \(L_0\) from the right side:
\[ T_1 L_0 L_2 - T_2 L_0 L_1 = L_0 (T_1 - T_2) \]

### Step 7: Solve for the natural length \(L_0\)
Rearranging gives:
\[ L_0 (T_1 - T_2) = T_1 L_2 - T_2 L_1 \]
Thus, we can solve for \(L_0\):
\[ L_0 = \frac{T_1 L_2 - T_2 L_1}{T_1 - T_2} \]

### Final Answer:
The natural length of the wire is:
\[ L_0 = \frac{T_1 L_2 - T_2 L_1}{T_1 - T_2} \]