When a load of 2 kg is applied to a wire of length 5 m, the tension is 0.25 mm. If the radius of the wire is 1 mm, the wire is bent

1. Reflexive power

2. Strain

3. Find the Young's modulus of the material the wire is made of.

Let's tackle this problem step by step. Here's how we can approach each part:

### Given Information:
- **Load (W)** = 2 kg 
- **Length of the wire (L)** = 5 m = 5000 mm
- **Elongation (ΔL)** = 0.25 mm
- **Radius of the wire (r)** = 1 mm
- **Acceleration due to gravity (g)** = 9.8 m/s² (to calculate the force from the weight)

We can now break down the problem into its components:

### 1. Reflexive Power

Reflexive power is not a common physical term in the context of material mechanics, elasticity, or wire tension. It's possible this is a misinterpretation or unrelated to this type of problem. If you meant something else, please clarify.

### 2. Strain

Strain is a measure of deformation of the material. It is given by the ratio of change in length (elongation) to the original length.

\[
\text{Strain} = \frac{\Delta L}{L}
\]

Substituting the values:

\[
\text{Strain} = \frac{0.25}{5000} = 0.00005
\]

So, the strain in the wire is **0.00005**.

### 3. Young's Modulus (E)

Young's modulus is a measure of the stiffness of a material. It is calculated using the formula:

\[
E = \frac{\text{Stress}}{\text{Strain}}
\]

**Stress** is the force applied per unit area. To calculate the force:

\[
\text{Force} = \text{Weight} = m \cdot g = 2 \cdot 9.8 = 19.6 \, \text{N}
\]

The **cross-sectional area** of the wire is given by:

\[
A = \pi r^2 = \pi (1 \, \text{mm})^2 = \pi \, \text{mm}^2 = 3.14 \, \text{mm}^2 = 3.14 \times 10^{-6} \, \text{m}^2
\]

Now, **Stress** is the force divided by the cross-sectional area:

\[
\text{Stress} = \frac{\text{Force}}{A} = \frac{19.6}{3.14 \times 10^{-6}} = 6.24 \times 10^6 \, \text{Pa}
\]

Finally, the **Young's Modulus (E)** is:

\[
E = \frac{6.24 \times 10^6}{0.00005} = 1.248 \times 10^{11} \, \text{Pa}
\]

Thus, the **Young's modulus** of the material is approximately **\(1.25 \times 10^{11} \, \text{Pa}\)**.

---

### Summary of Results:
- **Strain**: 0.00005
- **Young's Modulus**: \(1.25 \times 10^{11} \, \text{Pa}\)

Would you like more details on any of these calculations?

### 5 Related Questions:
1. What is the physical significance of Young's modulus in material science?
2. How would the strain change if the load were doubled?
3. What happens to the wire if the Young's modulus is lower than calculated?
4. How is stress different from pressure in mechanical terms?
5. How would the elongation change if the radius of the wire was halved?

### Tip:
Young's modulus is a fundamental property that indicates how easily a material deforms under tension. High values mean the material is stiffer and more resistant to deformation.

When a load of 2 kg is applied to a wire of length 5 m, the tension is 0.25 mm. If the radius of the wire is 1 mm, the wire is bent. 1. Reflexive power 2. Strain 3. Find the Young's modulus of the material the wire is made of.

Solve a physics problem involving a 2 kg load applied to a 5 m wire, with a radius of 1 mm, that experiences a tension of 0.25 mm. Learn how to calculate the strain and Young's modulus using key formulas and principles of elasticity, perfect for students in Grades 11-12.

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