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Chapter 7: Problem 892

On applying a stress of $20 \times 10^{8}\left(\mathrm{~N} / \mathrm{m}^{2}\right)$ the length of a perfect elastic wire is doubled. What will be its Young's modulus ? (A) \(40 \times 10^{8}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\) (B) \(20 \times 10^{8}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\) (C) \(10 \times 10^{8}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\) (D) \(5 \times 10^{8}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\)

Short Answer

Expert verified
The Young's modulus of the elastic wire is (B) \(20 \times 10^{8}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\).

Step by step solution

01

Determine the stress

The stress applied on the elastic wire is given in the problem: \(20 \times 10^{8}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\).
02

Determine the strain

Since the length of the wire is doubled, the change in length is equal to the original length. Therefore, the strain is equal to 1.
03

Use the Young's modulus formula

Young's modulus (Y) is defined as the ratio of stress to strain: \[Y = \frac{stress}{strain}\]
04

Insert the values and solve for Y

Plug in the values of stress and strain we've determined: \[Y = \frac{20 \times 10^{8} \left(\mathrm{N} / \mathrm{m}^{2}\right)}{1}\] Y = \(20 \times 10^{8} \left(\mathrm{N} / \mathrm{m}^{2}\right)\) So the correct answer is (B) \(20 \times 10^{8}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\).

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