A wire one square centimeter area of cross-section is stretched by a force to double its original length. Calculate the force. Young's modulus = 90 GPa. Assume that the wire does not break.

Explanation

To solve the problem step by step, we will use the relationship between Young's modulus, stress, and strain.

### Step 1: Understand the given information
- Area of cross-section (A) = 1 cm² = \(1 \times 10^{-4}\) m² (convert to SI units)
- Young's modulus (Y) = 90 GPa = \(90 \times 10^{9}\) Pa (convert to SI units)
- The wire is stretched to double its original length, so \(L' = 2L\).

### Step 2: Calculate the strain
Strain (ε) is defined as the change in length divided by the original length.

\[ \text{Strain} (ε) = \frac{\Delta L}{L} = \frac{L' - L}{L} = \frac{2L - L}{L} = \frac{L}{L} = 1 \]

### Step 3: Calculate the stress
Stress (σ) is defined as the force (F) applied per unit area (A).

\[ \text{Stress} (σ) = \frac{F}{A} \]

### Step 4: Relate Young's modulus to stress and strain
Young's modulus (Y) is defined as the ratio of stress to strain.

\[ Y = \frac{σ}{ε} \]

From this, we can express stress in terms of Young's modulus and strain:

\[ σ = Y \cdot ε \]

### Step 5: Substitute the values
Substituting the values we have:

\[ σ = 90 \times 10^{9} \, \text{Pa} \cdot 1 = 90 \times 10^{9} \, \text{Pa} \]

### Step 6: Calculate the force
Now, we can substitute the stress back into the stress formula to find the force:

\[ F = σ \cdot A \]
\[ F = (90 \times 10^{9} \, \text{Pa}) \cdot (1 \times 10^{-4} \, \text{m}^2) \]
\[ F = 90 \times 10^{5} \, \text{N} = 9 \times 10^{6} \, \text{N} \]

### Final Answer
The force required to stretch the wire to double its original length is \(9 \, \text{MN}\) (Mega Newtons).

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