Problem 900 - JEE Physics

Question

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) \(19.6 \times 10^{20}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\)

Step-by-Step Solution

Answer

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

Step 1: 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

Step 2: 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)

Step 3: 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) \(19.6 \times 10^{10}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\)

Key Concepts

Stress and Strain

When we talk about stress and strain, we are delving into how materials behave under the influence of external forces.
Stress is a measure of force applied over an area. It's calculated by taking the force acting on an object, such as a rod, and dividing it by the cross-sectional area of that object. For example, if a mass hangs from a rod, the gravitational force on that mass creates stress in the rod. You can express stress using the formula: \( \text{Stress} = \frac{\text{Force}}{\text{Area}} \), which is measured in units of N/m² (Pascals).

Strain, on the other hand, measures how much an object deforms compared to its original size. It's the change in length divided by the original length, often written as a simple ratio. Strain has no units because it's a proportion, like stretching a rubber band and comparing the stretched length to the original length. It's expressed as: \( \text{Strain} = \frac{\Delta L}{L} \).
Understanding these concepts helps us predict material behavior under stress, which is essential in engineering and materials science.

Mechanical Properties of Materials

Materials have different mechanical properties that describe how they react under various forces. Young's Modulus is a key property that indicates a material's elasticity. It tells us how easily a material can stretch or compress and still return to its original form once the force is removed. This modulus is essential when designing anything from bridges to bungee cords.

Aside from Young's Modulus, other important mechanical properties include:

• Tensile Strength: The maximum stress that materials can withstand while being stretched or pulled before breaking.
• Compressive Strength: The capacity of a material to withstand loads pushing it together.
• Hardness: A measure of a material's resistance to deformation or scratching.

These properties are crucial for engineers to consider, ensuring that structures and materials perform reliably under everyday loadings without failing.

Elasticity

Elasticity is the ability of a material to undergo deformation and return to its original shape once the force is removed. It's a fundamental concept when dealing with materials, especially in contexts like manufacturing and civil engineering.

Elasticity is typically described by Hooke's Law, which states that the deformation (or strain) of an elastic material is directly proportional to the applied force (or stress), as long as the material's elastic limit is not surpassed. Mathematically, this can be represented as \( F = k \cdot x \), where \(F\) is the force applied, \(k\) is the stiffness or spring constant, and \(x\) is the amount of deformation.

Materials only behave elastically up to their yield point. Beyond this point, they may permanently deform, even after the load is removed. Knowing this limit ensures that materials can be used effectively and safely in various applications without unexpected deformation.

Physics Engineering Formulas

In physics and engineering, formulas are the backbone calculations that allow us to solve problems and design systems accurately. These formulas help us understand relationships between different physical quantities and predict behaviors.

Let's look at some fundamental physics engineering formulas that are often used in material science and mechanics:

• Hooke's Law: Describes the linear relationship between the force exerted on a spring and its change in length: \( F = k \cdot x \).
• Newton's Second Law: Relates force, mass, and acceleration as \( F = m \cdot a \).
• Young's Modulus Formula: It relates stress and strain to determine how elastic a material is: \( Y = \frac{\text{Stress}}{\text{Strain}} \).

Mastering these formulas is essential for solving engineering problems efficiently, allowing students and professionals to accurately predict material behaviors and design successful engineering solutions.