Question

A wire of cross-section \(4 \mathrm{~mm}^{2}\) is stretched by \(0.1 \mathrm{~mm}\) by a certain weight. How far (length) will be wire of same material and length but of area \(8 \mathrm{~mm}^{2}\) stretched under the action of same force. (A) \(0.05 \mathrm{~mm}\) (B) \(0.10 \mathrm{~mm}\) (C) \(0.15 \mathrm{~mm}\) (D) \(0.20 \mathrm{~mm}\)

Short answer

The elongation of the second wire under the same force is \( 0.05\,mm \). So the correct answer is (A).

Step-by-step solution

Write down the given information and the formula for Young's modulus

The given information is as follows: - Initial wire cross-section area: \( A_1 = 4 \,mm^2 \) - Elongation of the initial wire: \( \Delta l_1 = 0.1 \,mm \) - Second wire cross-section area: \( A_2 = 8 \,mm^2 \) - We need to find the elongation of the second wire: \( \Delta l_2 \) The formula for Young's modulus (\( Y \)) is as follows: \( Y = \frac{stress}{strain} \) Stress is defined as the force acting on the object divided by its area, and strain is the change in length divided by the original length: \( Y = \frac{F/A}{\Delta l/L} \) Now we can rearrange this formula to make the change in length, \(\Delta l\), the focus: \( \Delta l = \frac{FL}{YA} \) We know that the wires are made of the same material and are under the same force, so they have the same Young's modulus and the same value for the force.

Understand the relationship between elongation and force for the two wires

Let's express the original elongation of the wire with area \(A_1\): \( \Delta l_1 = \frac{F_1L_1}{YA_1} \) Now let's express the elongation of the second wire with area \(A_2\): \( \Delta l_2 = \frac{F_2L_2}{YA_2} \) Since both wires are under the same force ( \(F_1 = F_2\) ) and assuming they have the same length ( \(L_1 = L_2\) ), we can set up a proportion between these two equations and solve for \( \Delta l_2 \): \( \frac{\Delta l_1}{\Delta l_2} = \frac{F L_1}{Y A_1} \cdot \frac{YA_2}{F L_2} \)

Solve for the elongation of the second wire

After rearranging the equation from Step 2, we only need to plug in the given values to find \( \Delta l_2 \): \( \Delta l_2 = \Delta l_1 \cdot \frac{A_1}{A_2} \) Now substitute the given values: \( \Delta l_2 = 0.1 \,mm \cdot \frac{4 \,mm^2}{8 \,mm^2} \) Calculate the result: \( \Delta l_2 = 0.1 \,mm \cdot \frac{1}{2} = 0.05 \,mm \) The elongation of the second wire under the same force is \( 0.05 \,mm \). So the correct answer is (A).

Key concepts

Understanding Stress

Stress is a measure of the internal forces within a material when an external force is applied. Imagine a rubber band being pulled—stress measures the intensity of that stretch. In mathematical terms, stress (\( \sigma \)) is calculated by dividing the force (\( F \)) exerted on an area by the cross-sectional area (\( A \)) of the material:

• Formula: \( \sigma = \frac{F}{A} \)

This calculation helps determine how a material reacts when subjected to an external load. If the stress exceeds a certain limit, the material may break or deform. This is crucial when designing materials or structures because excessive stress can lead to failure. In our exercise, even though the original wire's cross-section was doubled, the same force was applied, meaning the stress was halved. Accounting for stress is essential to understand why the second wire stretches less when the force is constant but the area increases.

Exploring Strain

Strain is about the change in shape or size of a material under stress. It's crucial to understand that strain is a relative measure, not absolute. It looks at how much a material elongates or compresses compared to its original length. To find strain (\( \epsilon \)), you divide the change in length (\( \Delta l \)) by the original length (\( L \)):

• Formula: \( \epsilon = \frac{\Delta l}{L} \)

This dimensionless quantity gives insight into a material's flexibility or brittleness. In the context of the exercise, the initially given elongation shows the outcome of both stress and material properties in response to a force. Because the wires are of the same length and material, the strain would decrease in the second wire since it now has a larger cross-sectional area.

Proportionality and Young's Modulus

Young's Modulus (\( Y \)) connects stress and strain, offering insights into a material's elasticity. It's a constant for a given material reflecting its ability to endure deformations under external forces while returning to its original shape. Proportionality here refers to the relationship described by Hooke's Law, which states stress (\( \sigma \)) is directly proportional to strain (\( \epsilon \)) as long as the material’s elastic limit is not exceeded.

• Formula: \( Y = \frac{\sigma}{\epsilon} \)

In simpler terms, Young's Modulus helps figure out how stiff or flexible a material is. In our exercise, since both wires are made of the same material and Young’s modulus and force stayed constant, the effect of doubling the area while keeping the material consistent allows us to predict how the second wire behaves. Thus, with the same stress applied, increasing the wire's cross-section reduces the strain, leading to less elongation as observed in the problem solution.