Question

When the length of a wire having cross section area \(10^{-6} \mathrm{~m}^{2}\) is streatched by \(0.1 \%\) then tension in it is \(100 \mathrm{~N}\). Young's modulus of material wire is \(\ldots \ldots \ldots\) (A) \(10^{12}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\) (B) \(10^{2}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\) (C) \(10^{10}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\) (D) \(10^{11}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\)

Short answer

The Young's modulus of the material of the wire is \(10^{11} (\mathrm{N} / \mathrm{m}^{2})\).

Step-by-step solution

Calculate the stress

First, we need to find the stress in the wire. Stress is defined as the force per unit area, so we can calculate it using the formula: \[stress = \frac{force}{area}\]. The force applied to the wire is the tension (100 N), and the cross-sectional area of the wire is \(10^{-6} m^2\). To calculate the stress: \[stress = \frac{100 N}{10^{-6} m^2} = 10^8\ (\mathrm{N} / \mathrm{m}^{2})\].

Calculate the strain

Next, we need to find the strain in the wire. Strain is the ratio of the change in length (the stretch) to the original length. We are given that the wire has been stretched by 0.1%. Since strain is unitless, we can write the strain as the decimal equivalent: \[strain = \frac{0.1}{100} = 0.001\].

Find Young's modulus

Now that we have calculated the stress and strain, we can use the formula for Young's modulus: \[Y = \frac{stress}{strain}\]. Plugging in the values we found in steps 1 and 2: \[Y = \frac{10^8\ (\mathrm{N} / \mathrm{m}^{2})}{0.001} = 10^{11}\ (\mathrm{N} / \mathrm{m}^{2})\]. So, the Young's modulus of the material of the wire is \(10^{11} (\mathrm{N} / \mathrm{m}^{2})\), which corresponds to option (D).

Key concepts

Understanding Stress and Strain

When we talk about materials under load, two important concepts come up: stress and strain. Stress gives us an idea of the internal force within a material when an external force is applied. Think of it as the load or pressure happening inside. It's calculated as force divided by area, \[stress = \frac{force}{area}\].
In our problem, the stress in the wire was determined by dividing the tension of 100 N by the cross-sectional area of \(10^{-6} \, m^2\). This yielded a stress \(10^8 \, (\mathrm{N} / \mathrm{m}^{2})\).
Strain, on the other hand, tells us about how much a material deforms under stress. It is the ratio of the change in length to the original length of the material. This means, if you stretch a wire by a certain percentage, that's your strain. In this exercise, a 0.1% stretch translates into a strain of 0.001 when converted into decimal form.

• Stress measures force intensity within the material.
• Strain indicates deformation level compared to its original size.
• Both are crucial to understanding material behavior under force.

Exploring Tension in Materials

Tension is a type of stress that specifically refers to a pulling force, which tries to elongate a material. Imagine pulling on a rope until it stretches; the force you apply creates tension. In our scenario, the wire experiences tension that amounts to 100 N.
This tension causes the wire to stretch by 0.1%, signifying that the wire is under tensile stress.

• Tension is a pulling force trying to lengthen materials.
• The amount of tension directly impacts the elongation observed.
• Understanding tension is key to designing materials that can withstand stretching forces.

When dealing with engineering or construction, it's vital to know how much tension a material can endure before altering its shape or failing. This understanding, combined with knowledge of stress and strain, helps engineers ensure structures are safe and durable.

Elasticity in Materials and Young's Modulus

Materials tend to revert to their original shape after a stress, thanks to a property known as elasticity. Young's Modulus is a helpful measure of this elasticity. It shows how stiff a material is by comparing how much it stretches or compresses under a certain stress.
To calculate Young's Modulus, we use the formula: \[Y = \frac{stress}{strain}\].
For the wire in the problem, we calculated a stress of \(10^8 \, (\mathrm{N} / \mathrm{m}^{2})\) and a strain of 0.001. Therefore, its Young's Modulus is \(10^{11} \, (\mathrm{N} / \mathrm{m}^{2})\), which tells us the wire is quite stiff and resistant to deformation.

• Elasticity refers to a material's ability to resume its shape after deformation.
• Young's Modulus quantifies this elasticity, showing material rigidity.
• An understanding of elasticity is crucial when selecting materials for load-bearing and flexible applications.

By knowing Young's Modulus, engineers can predict how a material like the wire will behave under different forces. This assists in making informed decisions about material suitability for various applications.