wire of cross-sectional area 1 cm2 and length 1 m is made of a metal of Young's modulus 2x 10'Nm'. Find the amount of work done by a constant external force in increasing its length by 1mm.

To calculate the work done in stretching the wire by 1 mm using a constant external force, we can follow these steps:

### Given dataTo calculate the work done in stretching the wire by 1 mm using a constant external force, we can follow these steps:

### Given 
- Cross-sectional area, \( A = 1 \, \text{cm}^2 = 1 \times 10^{-4} \, \text{m}^2 \)
- Initial length, \( L = 1 \, \text{m} \)
- Extension, \( \Delta L = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \)
- Young's modulus, \( Y = 2 \times 10^{11} \, \text{N/m}^2 \)

### Step 1: Relation between force and extension (Hooke's Law)

From Hooke's law, the force \( F \) applied to stretch the wire is given by:
\[
F = \frac{Y A \Delta L}{L}
\]
Substitute the values:
\[
F = \frac{(2 \times 10^{11}) (1 \times 10^{-4}) (1 \times 10^{-3})}{1}
\]
\[
F = 2 \times 10^4 \, \text{N}
\]

### Step 2: Work done in stretching the wire

The work done \( W \) to stretch the wire by an amount \( \Delta L \) is given by the formula:
\[
W = \frac{1}{2} F \Delta L
\]
Substitute the values of \( F \) and \( \Delta L \):
\[
W = \frac{1}{2} (2 \times 10^4) (1 \times 10^{-3})
\]
\[
W = 10 \, \text{J}
\]

### Final Answer:
The amount of work done by the constant external force in increasing the length of the wire by 1 mm is \( \boxed{10 \, \text{J}} \).

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Would you like more details or have any further questions?

Here are some related questions for practice:
1. How does the Young's modulus of a material affect the work done during stretching?
2. What would be the work done if the extension was doubled to 2 mm?
3. How does the cross-sectional area impact the force required for a given extension?
4. What is the significance of Young's modulus in material science?
5. If the wire was made of a different metal with half the Young's modulus, how would the work change?

**Tip:** Young's modulus provides a measure of how stiff a material is—higher values mean the material is less likely to deform under stress.

A wire of cross-sectional area 1 cm² and length 1 m is made of a metal of Young's modulus 2 x 10¹¹ N/m². Find the amount of work done by a constant external force in increasing its length by 1 mm.

Learn how to calculate the work done in stretching a wire of 1 cm² cross-sectional area and 1 m length, using a metal with a Young's modulus of 2 x 10¹¹ N/m². This detailed step-by-step solution covers key concepts such as Hooke's Law and the Work-Energy Theorem. Suitable for Grades 10-12 physics students.

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