Problem 915 - JEE Physics

Question

When a force is applied on a wire of uniform cross-sectional area \(3 \times 10^{-6} \mathrm{~m}^{2}\) and length \(4 \mathrm{~m}\), the increase in length is \(1 \mathrm{~mm}\). what will be energy stored in it ? [ \(\mathrm{Y}=2 \times 10^{11}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\) (A) \(62.50 \mathrm{~J}\) (B) \(0.177 \mathrm{~J}\) (C) \(0.075 \mathrm{~J}\) (D) \(0.150 \mathrm{~J}\)

Step-by-Step Solution

Answer

The energy stored in the wire is \(0.075 J\).

Step 1: List the given information and required formula.

We are given the following: 1. Cross-sectional area (A) = \(3 \times 10^{-6} m^2\) 2. Length of the wire (L) = 4 m 3. Increase in length, ∆L = 1 mm = \(1 \times 10^{-3} m\) 4. Young's modulus (Y) = \(2 \times 10^{11} N/m^2\) We need to find the energy stored (U) in the wire. The formula to calculate energy is: \(U = \frac{1}{2} \times \frac{F^2}{Y} \times \frac{\Delta L}{A}\)

Step 2: Calculate the applied force (F) on the wire.

From Young's modulus, we can find the applied force using the formula: \(Y = \frac{F \times L}{A \times \Delta L}\) Now, we solve for F: \(F = \frac{Y \times A \times \Delta L}{L}\) Plugging in the given values: \(F = \frac{(2 \times 10^{11} N/m^2)(3 \times 10^{-6} m^2)(1 \times 10^{-3} m)}{4 m}\) Calculating the value of F: \(F = 15 \times 10^5 N\)

Step 3: Calculate the energy stored in the wire.

Now we can use the formula for the potential energy stored in the wire: \(U = \frac{1}{2} \times \frac{F^2}{Y} \times \frac{\Delta L}{A}\) Plugging in the values we've found and the given values: \(U = \frac{1}{2} \times \frac{(15 \times 10^5 N)^2}{(2 \times 10^{11} N/m^2)} \times \frac{(1 \times 10^{-3} m)}{(3 \times 10^{-6} m^2)}\) Calculating the value of U: \(U = 0.075 J\) Therefore, the energy stored in the wire is \(0.075 J\), which corresponds to answer choice (C).

Key Concepts

Young's Modulus

Young's Modulus is an essential concept in understanding how materials respond to stress. It measures the stiffness of a material, defined as the ratio of tensile stress to tensile strain. In simple terms, it tells us how much a material will stretch or compress when a force is applied.
Mathematically, it's expressed as:

• \( Y = \frac{Stress}{Strain} = \frac{F \cdot L}{A \cdot \Delta L} \)

where:

• \( Y \) is Young's Modulus.
• \( F \) is the force applied on the material.
• \( L \) is the original length of the material.
• \( A \) is the cross-sectional area.
• \( \Delta L \) is the change in length.

The higher the value of Young's Modulus, the more rigid the material is. For example, metals like steel have high Young's Moduli making them less deformable under stress. Understanding this helps in designing structures that need to withstand high stresses.

Elastic Potential Energy

Elastic Potential Energy (EPE) is the energy stored in elastic materials as a result of their deformation. When materials like wires or springs are stretched or compressed, they store energy that can do work later when the force is removed.
The formula to calculate the elastic potential energy stored in a material is:

• \( U = \frac{1}{2} F \Delta L \)

Using Young's Modulus, EPE can also be expressed as:

• \( U = \frac{1}{2} \frac{F^2}{Y} \frac{\Delta L}{A} \)

This formula considers both Young’s Modulus and the material’s physical properties, like length and cross-sectional area. It helps in understanding the energy stored due to elastic deformation.
So when a wire is stretched, calculating the EPE helps in assessing how much work can be potentially performed by releasing this energy.

Deformation of Materials

Deformation is the change in the shape or size of an object due to an applied force. When a force acts upon a material, it undergoes deformation, which can be elastic or plastic. Elastic deformation is reversible, meaning the object returns to its original shape once the force is removed. In contrast, plastic deformation is permanent.
In the context of Young's Modulus, deformation is calculated as:

• \( \Delta L = \frac{F \cdot L}{Y \cdot A} \)

Understanding deformation is crucial in fields like engineering and material science to predict how materials behave under various conditions.
Elastic deformation allows objects like bridges, buildings, and machinery to withstand natural forces like wind or earthquakes without sustaining permanent damage. By understanding the deformation characteristics, engineers can choose the right materials and design structures that ensure safety and durability.