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Chapter 7: Problem 845

A stress of \(3.18 \times 10^{8} \mathrm{Nm}^{2}\) is applied to steel rod of length \(1 \mathrm{~m}\) along its length. Its young's modulus is $2 \times 10^{11}\left(\mathrm{~N} / \mathrm{m}^{2}\right)$ Then what is the elongation produced in the rod in \(\mathrm{mm}\) ? (A) \(3.18\) (B) \(6.36\) (C) \(5.18\) (D) \(1.59\)

Short Answer

Expert verified
The elongation produced in the rod is \(1.59 \mathrm{mm}\).

Step by step solution

01

Write down Hooke's Law formula

Hooke's Law relates stress (σ), Young's modulus (E), and strain (ε) in the following way: \(σ = Eε\)
02

Calculate the strain

Using Hooke's Law formula, we can find the strain (ε) produced in the steel rod: \(\epsilon = \frac{\sigma}{E}\) Plug in the given values for stress and Young's modulus: \( ε = \frac{3.18 \times 10^{8} \mathrm{Nm}^{2}}{2 \times 10^{11}\left(\mathrm{N} / \mathrm{m}^{2}\right)}\)
03

Solve for strain

Calculate the strain (ε) by dividing the stress by the Young's modulus: \(ε = \frac{3.18 \times 10^{8}}{2 \times 10^{11}}\) After the calculation, we get: \(ε = 1.59 \times 10^{-3}\)
04

Calculate elongation

Since strain (ε) is dimensionless and represents the ratio of elongation produced by the initial length, we can use it to find the elongation (ΔL) of the steel rod: \(ΔL = ε \times L\) Plug in the given length of the steel rod and the calculated strain: \(ΔL = (1.59 \times 10^{-3}) \times (1 \mathrm{m})\)
05

Convert elongation to millimeters

Calculate the elongation (ΔL) produced in the rod in meters: \(ΔL = 1.59 \times 10^{-3} \mathrm{m}\) To convert the elongation to millimeters, multiply by 1000: \(ΔL = 1.59 \times 10^{-3} \mathrm{m} \times 1000\) Which gives us the final answer in millimeters: \(ΔL = 1.59 \mathrm{mm}\) Therefore, the elongation produced in the rod is 1.59 mm, and the correct answer is (D).

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