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Chapter 7: Problem 880

A copper wire of length \(4 \mathrm{~m}\) and area of cross-section $1.2 \mathrm{~cm}^{2}\( is stretched with a force of \)4.8 \times 10^{3} \mathrm{~N}\(. If young's modulus for copper is \)1.2 \times 10^{11}\left(\mathrm{~N} / \mathrm{m}^{2}\right)$. What will be the length increase of the wire? (A) \(1.33 \mathrm{~mm}\) (B) \(1.33 \mathrm{~cm}\) (C) \(2.66 \mathrm{~mm}\) (D) \(2.66 \mathrm{~cm}\)

Short Answer

Expert verified
The length increase of the copper wire is \(\Delta L = 1.33\,\text{mm}\), which corresponds to answer (A).

Step by step solution

01

Write down the given values

First, we need to write down all the given information: - Length of the copper wire, \(L = 4\,\text{m}\) - Area of cross-section, \(A = 1.2\,\text{cm}^2 = 1.2 \times 10^{-4}\,\text{m}^2\) (converted to square meters) - Stretching force, \(F = 4.8 \times 10^3\,\text{N}\) - Young's modulus for copper, \(Y = 1.2 \times 10^{11}\,\text{N/m}^2\) - Length increase of the wire, \(\Delta L = ?\)
02

Rearrange the Young's modulus formula to solve for the length increase

We will rearrange the Young's modulus formula to solve for the length increase, \(\Delta L\). \(Y = \frac{F \cdot L}{A \cdot \Delta L}\) Rearrange for \(\Delta L\): \(\Delta L = \frac{F \cdot L}{A \cdot Y}\)
03

Substitute the given values into the formula

Now, we can substitute all the given values into the rearranged formula: \(\Delta L = \frac{(4.8 \times 10^3\,\text{N}) \cdot (4\,\text{m})}{(1.2 \times 10^{-4}\,\text{m}^2) \cdot (1.2 \times 10^{11}\,\text{N/m}^2)}\)
04

Calculate the length increase

Calculate the length increase, \(\Delta L\): \(\Delta L = \frac{1.92 \times 10^4\,\text{N} \cdot \text{m}}{1.44 \times 10^7\,\text{N} \cdot \text{m}^2} = 1.33 \times 10^{-3}\, \text{m}\) Convert the result into millimeters: \(\Delta L = 1.33 \times 10^{-3}\, \text{m} \cdot \frac{1000\,\text{mm}}{1\,\text{m}} = 1.33\,\text{mm}\) So the length increase of the copper wire is \(\Delta L = 1.33\,\text{mm}\), which corresponds to answer (A).

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