Problem 864 - JEE Physics

Question

Two wires of copper having the length in the ratio \(4: 1\) and their radii are as \(1: 4\) are stretched by the same force. What will be the ratio of longitudinal strain in the two wires? (A) \(1: 16\) (B) \(16: 1\) (C) 1:64 (D) \(64: 1\)

Step-by-Step Solution

Answer

The ratio of longitudinal strain in the two wires is \(16:1\).

Step 1: List the given information

- Length ratio: \(L_1:L_2 = 4:1\) - Radii ratio: \(r_1:r_2 = 1:4\) - Same force F is applied on both wires

Step 2: Define longitudinal strain and Young's modulus (Y)

Longitudinal strain, \(\epsilon\), is the fractional change in length (\(\frac{\Delta L}{L}\)) when the force is applied. Young's modulus (Y) is a property of the material that relates the longitudinal stress and strain: \(Y = \frac{σ}{\epsilon}\) Where σ is the longitudinal stress, given by \(\frac{F}{A}\), and A is the cross-sectional area of the wire.

Step 3: Find the expression for the longitudinal strain

For the two copper wires, the Young's Modulus is the same, as their material is identical. Thus, \(Y_1 = Y_2\) Using \(Y = \frac{σ}{\epsilon}\), \(\frac{σ_1}{\epsilon_1} = \frac{σ_2}{\epsilon_2}\) Then, stress can be written in terms of force and area. Since both wires experience the same force, F, \(\frac{F/A_1}{\epsilon_1} = \frac{F/A_2}{\epsilon_2}\) We simplify this expression to obtain the ratio of \(\epsilon\): \(\frac{\epsilon_1}{\epsilon_2} = \frac{A_2}{A_1}\)

Step 4: Calculate the cross-sectional area ratio

The cross-sectional area of a wire is given by \(A = πr^2\). Using the given radius ratio, we can calculate the area ratio: \(\frac{A_1}{A_2} = \frac{πr_1^2}{πr_2^2} = \frac{(1)^2}{(4)^2} = \frac{1}{16}\)

Step 5: Conclude the ratio of strain

Now, we can use the obtained area ratio to find the ratio of strains: \(\frac{\epsilon_1}{\epsilon_2} = \frac{A_2}{A_1} = \frac{16}{1}\) This corresponds to option (B) \(16:1\). So, the ratio of longitudinal strain in the two wires is \(16:1\).

Key Concepts

Young's Modulus

Young's modulus is a fundamental concept in material science that helps us understand how materials respond to stress and strain. It is named after the British scientist Thomas Young. Young's modulus is a measure of the stiffness of a solid material. It quantifies the relationship between the stress applied to a material and the resulting strain. In simple terms, Young's modulus (denoted as Y) helps us determine how much a material will stretch or compress under a given force. It provides a way to compare the elastic properties of different materials. Mathematically, Young's modulus is defined as:\[Y = \frac{\sigma}{\epsilon}\]where:

• \(\sigma\) (sigma) is the stress applied to the material, measured in Pascals (Pa),
• \(\epsilon\) (epsilon) is the strain, which is the change in length divided by the original length.

Young's modulus is a constant for a given material, meaning that no matter how much you scale up or down the sample size (provided the material remains in its elastic limit), this ratio remains constant. In the context of the exercise, both wires are made of copper, so they have the same Young's modulus.

Cross-Sectional Area

The cross-sectional area of a material is crucial in understanding how it will handle forces applied to it. It refers to the size of the exposed surface when a solid object is sliced through, perpendicular to its length. For cylindrical objects like wires, the cross-sectional area (A) is calculated using the formula for the area of a circle:\[A = \pi r^2\]where

• \(\pi\) is a constant (approximately 3.14159),
• \(r\) is the radius of the circular cross-section.

In the given exercise, the radius of each wire plays a crucial role in determining the cross-sectional area ratio. Since the radius influences the amount of material exposed to the applied force, it directly affects the stress as well as the longitudinal strain experienced by the wire.The ratio of the cross-sectional areas of the two wires, calculated using the radius ratio, provides us insight into how each wire will deform under the same force.

Stress in Materials

Stress is a measure of the internal forces acting within a material when an external force is applied. It tells us how much force is usually applied over a certain area of the material's cross-section. This is fundamentally important in understanding how materials deform under various loads.The formula to calculate stress (σ) is:\[\sigma = \frac{F}{A}\]where

• \(F\) represents the force applied to the material,
• \(A\) is the cross-sectional area over which this force is distributed.

Larger cross-sectional areas typically mean less stress, assuming the same force is applied. This is because the load is spread over a larger area, reducing the force per unit area. Conversely, smaller cross-sectional areas increase the stress, leading to more deformation.In the exercise, both wires are subject to the same force, but due to different radii, they have different cross-sectional areas. This leads to varying levels of stress in each wire, directly influencing the longitudinal strain. Understanding stress in materials helps us predict how and why materials change shape when force is applied.