Question

An area of a cross-section of rubber string is \(2 \mathrm{~cm}^{3}\). Its length is doubled when stretched with a linear force of \(2 \times 10^{5}\) dynes. What will be young's modulus of the rubber in dynes? (A) \(4 \times 10^{5}\) (B) \(1 \times 10^{5}\) (C) \(2 \times 10^{5}\) (D) \(1 \times 10^{4}\)

Short answer

The short answer based on the step-by-step solution is: The Young's modulus of the rubber is \(1 \times 10^5 \, \text{dynes/cm}^2\). The correct answer is (B) \(1 \times 10^{5}\).

Step-by-step solution

Write down the given values

The area of the cross-section is given as 2 cm². The force applied is given as \(2 \times 10^5\) dynes. When the rubber is stretched, its length doubles.

Calculate the stress

Stress is defined as force per unit area. In this case, we have: Stress = Force / Area of cross-section Stress = \(2 \times 10^5 \, \text{dynes} / 2 \, \text{cm}^2\) Stress = \(10^5 \, \text{dynes/cm}^2\)

Calculate the strain

Strain is defined as the change in length divided by the original length. Since the length of the rubber doubles when it is stretched, the change in length is equal to the initial length: Strain = Change in length / Original length Strain = 1 (since the change and original lengths are equal)

Determine Young's modulus

Now we can determine the Young's modulus, which is defined as the ratio of stress to strain: Young's modulus (Y) = Stress / Strain Y = \(10^5 \, \text{dynes/cm}^2\ / 1\) Y = \(1 \times 10^5 \, \text{dynes/cm}^2\) The Young's modulus of the rubber is \(1 \times 10^5 \, \text{dynes/cm}^2\). The correct answer is (B) \(1 \times 10^{5}\).

Key concepts

Stress

Stress is a fundamental concept in understanding how materials react to external forces. Simply put, stress is the amount of force applied per unit area of a material. It's essential when analyzing how an object like a rubber string deforms. In the original exercise, the force applied was

• Force = \(2 \times 10^5\) dynes

for an area of

• Area = 2 \(\text{cm}^2\).

To find the stress, we use the formula:\[ ext{Stress} = \frac{\text{Force}}{\text{Area}}\]By substituting the values, we find the stress is:\[ \text{Stress} = \frac{2 \times 10^5\, \text{dynes}}{2\, \text{cm}^2} = 10^5\, \text{dynes/cm}^2\]This calculation highlights the relationship between force, area, and the resulting stress experienced by the material.

Strain

Strain describes how much a material deforms as a response to stress. It is a measure of the relative change in shape or size under the influence of stress. Unlike stress, which is a measure of force over an area, strain is a unit-less measure, often expressed as a ratio or percentage. In our exercise, the original length doubles upon stretching, implying:

• Change in length = Original length
• Therefore, Strain = \(\frac{\text{Change in length}}{\text{Original length}} = 1\)

This means that the rubber stretches entirely in proportion to its original length, making strain equal to 1. Understanding strain is crucial as it reflects the degree of deformation a material undergoes, giving insights into its ductility and endurance.

Elasticity

Elasticity is the property of a material that enables it to return to its original shape after the removal of the force causing the deformation. Rubber is a perfect example of an elastic material due to its ability to stretch significantly and bounce back after the stretch. Young's modulus is a measure of elasticity. It quantifies a material's tendency to deform under stress. High elasticity is illustrated by a lower Young's modulus, meaning minimal stress is needed for deformation. Conversely, materials with a higher Young's modulus require significant force to deform, signifying less elasticity. In the exercise, we calculated Young's modulus for the rubber using the given stress and strain.

Linear Force

Linear force is applied in a straight line and directly influences how a material like rubber stretches or compresses. In the original exercise, the rubber is stretched by a linear force of \(2 \times 10^5\) dynes.This force acts along the length of the rubber, and due to its linear nature, it uniformly affects its elongation. Understanding linear force is pivotal because it dictates the manner of deformation and is vital in calculating stress, which in turn is needed to evaluate the material's responses like elasticity and strength.