Problem 863 - JEE Physics

Question

An aluminum rod [Young's modulus \(=7 \times 10^{9}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\) has a breaking strain of \(0.2 \%\) what is the minimum cross-sectional area of the rod in order to support a load of \(10^{4}\) Newtons? (A) \(1 \times 10^{-2} \mathrm{~m}^{2}\) (B) \(1.4 \times 10^{-3} \mathrm{~m}^{2}\) (C) \(3.5 \times 10^{-3} \mathrm{~m}^{2}\) (D) \(7.1 \times 10^{-4} \mathrm{~m}^{2}\)

Step-by-Step Solution

Answer

The minimum cross-sectional area of the rod in order to support a load of \(10^{4}\) Newtons is (D) \(7.1 \times 10^{-4} \mathrm{m}^{2}\).

Step 1: Understand stress, strain, and Young's Modulus relationship.

The relationship between stress (σ), strain (ε), and Young's modulus (Y) is given by the formula: \(σ = Yε\) where σ is the stress in N/m² or Pa, ε is the strain (which is unitless), and Y is the Young's modulus in N/m² or Pa.

Step 2: Calculate the stress in the rod.

Given the breaking strain as \(0.2\%\), first, convert it to the decimal form: \(ε = 0.2 \% = 0.002\) Now using the Young's modulus, which is given as \(7 \times 10^{9} \mathrm{N/m}^{2}\), we can calculate the stress using the formula: \(σ = Yε\) \(σ = (7 \times 10^{9} \mathrm{N/m}^{2})(0.002)\) \(σ = 1.4 \times 10^{7} \mathrm{N/m}^{2}\)

Step 3: Calculate the required cross-sectional area.

We need to find the minimum cross-sectional area to support a load of \(10^{4}\) Newtons. We can use the formula for stress: \(σ = \cfrac{F}{A}\) Where F is the force in Newtons and A is the cross-sectional area in m². Rearranging the formula to find A: \(A = \cfrac{F}{σ}\) Now, substitute the values of F and σ: \(A = \cfrac{10^{4} \mathrm{N}}{1.4 \times 10^{7} \mathrm{N/m}^{2}}\) \(A = 7.14 \times 10^{-4} \mathrm{m}^{2}\) Comparing this value to the given options, we can round it up to the closest option: (A) \(1 \times 10^{-2} \mathrm{m}^{2}\) (B) \(1.4 \times 10^{-3} \mathrm{m}^{2}\) (C) \(3.5 \times 10^{-3} \mathrm{m}^{2}\) **(D) \(7.1 \times 10^{-4} \mathrm{m}^{2}\)** So, the minimum cross-sectional area of the rod in order to support a load of \(10^{4}\) Newtons is **(D) \(7.1 \times 10^{-4} \mathrm{m}^{2}\)**.

Key Concepts

Stress and Strain

When it comes to understanding the behavior of materials under force, the terms stress and strain are essential. Stress refers to the internal force per unit area within a material that arises because of externally applied forces. Mathematically, it’s expressed as the force (F) applied divided by the cross-sectional area (A), so \( \sigma = \frac{F}{A} \). It is typically measured in pascals (Pa).Strain, on the other hand, is a measure of how much an object deforms under stress. It’s the ratio of change in the dimension (say, length) to the original dimension and is typically expressed as a percentage. Strain is dimensionless because it’s a ratio.Young’s Modulus (Y), which was given in the exercise, is the constant that describes the relationship between stress and strain for a linear elastic material. It’s defined as the ratio of stress to strain: \( \sigma = Y \varepsilon \). Young's Modulus provides insights into a material's stiffness, with larger values indicating stiffer material.

Aluminum Rod

The aluminum rod in the exercise serves as an example for studying stress and strain in a real-world scenario. Aluminum is a favoured material in engineering due to its considerable mechanical strength and relatively low weight.Often used in applications ranging from construction to aerospace, aluminum's mechanical properties, such as its Young’s Modulus, determine how it responds to stresses applied to it. In this case, the Young's Modulus was given as \( 7 \times 10^9 \, \mathrm{N/m^2} \), which aids in determining the stress the material can endure before yielding or breaking.Using aluminum is particularly beneficial when looking for a material that is easy to work with, resistant to corrosion, and capable of sustaining tensile stress without significant deformation.

Cross-Sectional Area

The cross-sectional area of an object plays a critical role in determining its strength and ability to withstand loads. In the context of the exercise, the task was to find the minimum cross-sectional area of an aluminum rod needed to support a specific load safely.The formula for stress \( \sigma = \frac{F}{A} \) can be rearranged to find the cross-sectional area \( A = \frac{F}{\sigma} \). Thus, knowing the applied force and permissible stress allows us to calculate the minimum required area.This calculation is essential in engineering, ensuring that materials can handle the loads they are subjected to without failure. In the exercise, we calculated the area to find \( 7.1 \times 10^{-4} \, \mathrm{m^2} \), the option nearest to the theoretical requirement for supporting a force of \( 10^4 \, \mathrm{N} \).

Mechanical Properties

Mechanical properties are pivotal in understanding how materials behave under various types of force. These properties include yield strength, tensile strength, and modulus of elasticity, among others.Young's Modulus is one of these key mechanical properties and is crucial for predicting how materials deform under stress. It helps engineers and designers select the appropriate material for projects, ensuring durability and safety.In addition to Young’s Modulus, the context of mechanical failure involves understanding the breaking point or breaking strain of materials. Given the breaking strain of \( 0.2\% \) in this exercise, it tells us the extent to which the aluminum rod can stretch before failure. Mechanical properties guide the design decisions, ensuring that materials not only support the desired functionality but maintain that under intended loads.