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Chapter 7: Problem 900

An iron rod of length \(2 \mathrm{~m}\) and cross-section area of $50 \mathrm{~mm}^{2}\( stretched by \)0.5 \mathrm{~mm}\(, when a mass of \)250 \mathrm{~kg}$ is hung from its lower end. What is young's modulus of the iron rod? (A) \(19.6 \times 10^{10}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\) (B) \(19.6 \times 10^{15}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\) (C) \(19.6 \times 10^{18}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\) (D) \(19.6 \times 10^{20}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\)

Short Answer

Expert verified
The Young's modulus of the iron rod is approximately \(19.6 \times 10^{10}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\).

Step by step solution

01

Calculate Stress

Stress is defined as the force acting on an object divided by the cross-sectional area of the object. In this case, the force acting on the iron rod is due to the mass hung from its lower end. We can calculate the force using gravitational force formula: \(F = m \cdot g\), where m is the mass (250 kg), and g is acceleration due to gravity (approx. 9.81 m/s²). First, calculate the force: F = 250 kg × 9.81 m/s² ≈ 2450 N (Newton) Now, convert the cross-sectional area given in mm² to m²: Area = 50 mm² × (1 m² / 10^6 mm²) = 50 × 10^(-6) m² Next, calculate stress: Stress = Force / Area ≈ 2450 N / 50 × 10^(-6) m² = 4.9 × 10^7 N/m²
02

Calculate Strain

Strain is defined as the relative change in length, which is the elongation (∆L) divided by the initial length (L). We are given the elongation as 0.5 mm, which we must convert to meters, and the initial length as 2 meters. First, convert the elongation to meters: Elongation = 0.5 mm × (1 m / 1000 mm) = 0.5 × 10^(-3) m Now, calculate strain: Strain = Elongation / Length = (0.5 × 10^(-3) m) / (2 m) = 2.5 × 10^(-4)
03

Calculate Young's Modulus

Young's modulus is defined as the ratio of stress to strain. We can use our calculated values for stress and strain to find Young's modulus. Y = Stress / Strain = (4.9 × 10^7 N/m²) / (2.5 × 10^(-4)) ≈ 1.96 × 10^(11) N/m² The closest answer from the given choices is: (A) \(19.6 \times 10^{10}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\)

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