Question

The young's modulus of a wire of length \(\mathrm{L}\) and radius \(\mathrm{r}\) is \(\mathrm{Y}\left(\mathrm{N} / \mathrm{m}^{2}\right)\). If the length and radius are reduced to \(\mathrm{L} / 2\) and \(\mathrm{r} / 2\). Then what will be its young's modulus? (A) \(\mathrm{Y} / 2\) (B) \(\mathrm{Y}\) (C) \(2 \mathrm{Y}\) (D) \(4 \mathrm{Y}\)

Short answer

The new Young's modulus (Y') is 8 times greater than the original Young's modulus (Y). Therefore, the correct answer is \(8Y\).

Step-by-step solution

Understand the concept of Young's modulus

Young's modulus (Y) is a measure of the stiffness of a material. It is the ratio of stress to strain in a material. Mathematically, it is represented as: \[Y = \frac{F/A}{\Delta L/L}\] where F is the applied force, A is the cross-sectional area, ΔL is the change in length, and L is the original length of the material.

Calculate the cross-sectional area of the wire

The wire has a radius r, so its cross-sectional area A can be calculated as: \[A = \pi r^2\]

Analyze the given condition in the problem

The length and radius of the wire are both reduced by half. So, the new length and radius of the wire are: \[L' = \frac{L}{2}\] \[r' = \frac{r}{2}\]

Calculate the new cross-sectional area and Young's modulus

The new cross-sectional area A' can be calculated as: \[A' = \pi (r')^2\] \[A' = \pi \left( \frac{r}{2} \right)^2\] \[A' = \pi \cdot \frac{r^2}{4}\] \[A' = \frac{A}{4}\] Let Young's modulus for the new wire be represented by Y'. According to the formula for Young's modulus: \[Y = \frac{F/A}{\Delta L/L}\] \[Y' = \frac{F/A'}{\Delta L/L'}\]

Compare the old and new Young's modulus

Divide Y by Y' to see how they are related: \[\frac{Y}{Y'} = \frac{\frac{F/A}{\Delta L/L}}{\frac{F/A'}{\Delta L/L'}}\] \[\frac{Y}{Y'} = \frac{A'}{A} \cdot \frac{L'}{L}\] Using the values we calculated earlier for A' and L': \[\frac{Y}{Y'} = \frac{\frac{A}{4}}{A} \cdot \frac{\frac{L}{2}}{L}\] \[\frac{Y}{Y'} = \frac{1}{4} \cdot \frac{1}{2}\] \[\frac{Y}{Y'} = \frac{1}{8}\] \[Y' = 8Y\] So, the new Young's modulus (Y') is 8 times greater than the original Young's modulus (Y). This means that none of the given options in (A), (B), (C), and (D) is correct. The correct answer is 8Y.

Key concepts

Material Stiffness

Material stiffness is a way to describe how much a material resists deformation when a force is applied. It's like asking how hard it is to stretch, compress, or bend something.

When we talk about rigidity, we use a number called Young's Modulus, represented by the symbol \(Y\). The higher the Young's Modulus, the stiffer the material is. This means that a material with a high Young's Modulus will change shape less when you put the same force on it compared to a material with a lower Young's Modulus.

Young's Modulus is really useful when you're picking materials for projects because it helps you know how a material will react under different stresses. For example, engineers might choose materials with high Young's Modulus for parts that need to hold up under heavy loads.

Stress and Strain

Stress and strain are fundamental concepts used to understand how materials behave when forces are applied.

• **Stress:** This is the force applied on a material divided by the area over which it is applied. It's like the pressure you apply on the material.
• **Strain:** This is how much the material deforms or changes shape divided by its original shape. It’s like measuring how much a rubber band stretches when you pull it.

The relationship between stress and strain is crucial because it defines the Young's Modulus \(Y\). The equation for Young's Modulus is:
\[Y = \frac{\text{Stress}}{\text{Strain}}\]

In simpler terms, when the stress is high, the strain can still be low if the material is stiff, making the Young's Modulus value high. This relationship helps predict how much a material will stretch or compress under different forces.

Cross-sectional Area

The cross-sectional area of a material is a measure of the size of its face exposed to a force. Think about cutting a wire straight through and looking at the circle at its end—that's the cross-section.

For a wire with radius \(r\), the formula to find the cross-sectional area \(A\) is:
\[A = \pi r^2\]

This area is important because it influences how stress is calculated in a material. A larger cross-sectional area will distribute an applied force more, lowering the stress. Conversely, a smaller cross-sectional area means the same force will create more stress.

When the radius is halved, the area doesn’t just halve; it reduces to a quarter because of the square in the formula. This significantly affects calculations involving stress and Young's Modulus—leading to different material behaviors under the altered conditions.