## Problem Statement: Calculation of Tension Force in a Wire### Given Data:- **Length of the Wire (L):** 2.5 m- **Diameter of the Wire:** 0.4 cm (Radius = 0.2 cm)- **Extension of the Wire (ΔL):** 0.35 cm- **Young's Modulus (E):** \(2 \times 10^{11} \, \text{N/m}^2\)### Question:What is the tension force in the wire shown below if it stretches 0.35 cm when tightened if Young's modulus for the wire is \(2 \times 10^{11} \, \text{N/m}^2\)?### Options:a) 1374 N b) 1978 N c) 2693 N d) 3517 N ### Diagram Explanation:The diagram provided illustrates a wire of length 2.5 meters, with one end fixed and the other end where the tension force is applied causing the extension. The diameter of the wire is 0.4 cm, and it shows a circular cross-section to emphasize the diameter for calculation of the area.### Solution Steps:1. **Calculate the Cross-Sectional Area (A) of the Wire:** Given radius (r) = 0.2 cm = 0.002 m \[ A = \pi r^2 = \pi (0.002)^2 = \pi \times 4 \times 10^{-6} \approx 1.2566 \times 10^{-5} \, \text{m}^2 \]2. **Use Hooke's Law for Stress and Strain:** \[ \text{Strain} = \frac{\Delta L}{L} = \frac{0.35 \, \text{cm}}{250 \, \text{cm}} = \frac{0.0035}{2.5} = 0.0014 \] \[ \text{Stress} = E \times \text{Strain} = 2 \times 10^{11} \times 0.0014 = 2.8 \times 10^8 \, \text{N/m}^2 \]3. **Calculate the Tension

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What is the tension force in the wire shown below if it stretches 0.35 cm when tightened if young's modulus for the wire is 2 × 1011 N/m² 2.5 m 0.4 cm a) 1374 N b) 1978 N c) 2693 N d) 3517N

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ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

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Mechanical Properties of Solids

There are three types of matter- solid, liquid and gas. Out of them, solids have a rigid shaped and size. The intermolecular forces of attractions in solids are greatest. Mechanical properties are the physical properties of a material which it shows, when it is provided with certain amount of force or load. These properties are to be kept in mind for various purposes. When a force is applied to an object, the object may or may not get deformed depending upon its elastic property. But every time the object develops a restoring force against the force applied. This is commonly known as stress. If the object changes its dimensions due to the force applied it develops strain ( which is the ratio of change in dimension to the original dimension).

Question

## Problem Statement: Calculation of Tension Force in a Wire

### Given Data:
- **Length of the Wire (L):** 2.5 m
- **Diameter of the Wire:** 0.4 cm (Radius = 0.2 cm)
- **Extension of the Wire (ΔL):** 0.35 cm
- **Young's Modulus (E):** \(2 \times 10^{11} \, \text{N/m}^2\)

### Question:
What is the tension force in the wire shown below if it stretches 0.35 cm when tightened if Young's modulus for the wire is \(2 \times 10^{11} \, \text{N/m}^2\)?

### Options:
a) 1374 N
b) 1978 N
c) 2693 N
d) 3517 N

### Diagram Explanation:
The diagram provided illustrates a wire of length 2.5 meters, with one end fixed and the other end where the tension force is applied causing the extension. The diameter of the wire is 0.4 cm, and it shows a circular cross-section to emphasize the diameter for calculation of the area.

### Solution Steps:
1. **Calculate the Cross-Sectional Area (A) of the Wire:**

Given radius (r) = 0.2 cm = 0.002 m

\[
A = \pi r^2 = \pi (0.002)^2 = \pi \times 4 \times 10^{-6} \approx 1.2566 \times 10^{-5} \, \text{m}^2
\]

2. **Use Hooke's Law for Stress and Strain:**

\[
\text{Strain} = \frac{\Delta L}{L} = \frac{0.35 \, \text{cm}}{250 \, \text{cm}} = \frac{0.0035}{2.5} = 0.0014
\]

\[
\text{Stress} = E \times \text{Strain} = 2 \times 10^{11} \times 0.0014 = 2.8 \times 10^8 \, \text{N/m}^2
\]

3. **Calculate the Tension

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