Problem 875 - JEE Physics

Question

The length of a wire is \(1.0 \mathrm{~m}\) and the area of cross-section is \(1.0 \times 10^{-2} \mathrm{~cm}^{2}\). If the work done for increase in length by \(0.2 \mathrm{~cm}\) is \(0.4\) joule. Then what is the young's modulus? Of material of the wire ? (A) \(2.0 \times 10^{10}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\) (B) \(4.0 \times 10^{10}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\) (C) \(2.0 \times 10^{11}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\) (D) \(4.0 \times 10^{11}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\)

Step-by-Step Solution

Answer

The Young's modulus of the material of the wire is approximately \(2.0 \times 10^{11}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\).

Step 1: Find stress

The work done (W) is given as 0.4 J. We can use the formula for work: Work = Force × Distance. Let's find the force applied on the wire, F: 0.4 J = F × 0.2 cm Now, convert the distance from cm to meters: 0.2 cm = 0.002 m So, we get: 0.4 J = F × 0.002 m Solve for the force, F: F = 0.4 J / 0.002 m = 200 N Now, we need to find the stress. Stress is given by: Stress = Force / Area The area of cross-section is given as 1.0 × 10^{-2} cm². We need to convert it to square meters: 1.0 × 10^{-2} cm² = 1.0 × 10^{-6} m² Now, we can find the stress on the wire: Stress = 200 N / (1.0 × 10^{-6} m²) = 2.0 × 10^8 N/m²

Step 2: Find strain

Now, we need to find the strain on the wire. Strain is given by: Strain = Change in length / Original length Change in length = 0.2 cm = 0.002 m (already converted to meters) Original length = 1.0 m Now, find the strain: Strain = 0.002 m / 1.0 m = 0.002

Step 3: Find Young's modulus

Young's modulus (Y) is given by: Young's modulus = Stress / Strain Now, we can find the Young's modulus of the wire: Y = (2.0 × 10^8 N/m²) / 0.002 = 1.0 × 10^{11} N/m² From the given options, the closest value is: (C) \(2.0 \times 10^{11}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\) So, the answer is (C).

Key Concepts

Stress and Strain

Stress and strain are essential concepts in understanding how materials deform under various forces.
Stress is defined as the force applied over a specific area. Its formula is:

• Stress = Force / Area

For example, if you apply a force of 200 N over an area of 1.0 × 10^{-6} m², the stress is 2.0 × 10^{8} N/m².
This calculation shows the intensity of the internal forces within the material. Strain, on the other hand, measures the deformation of a material. It is defined as the ratio of the change in length to the original length:

• Strain = Change in Length / Original Length

In our exercise, if a wire initially 1.0 m long is stretched by 0.002 m, the strain is calculated to be 0.002. This indicates a small deformation relative to its original length. Understanding how stress and strain relate is crucial in predicting how a material will behave under pressure, guiding engineers in designing everything from small electronic devices to massive bridges.

Work and Energy

In physics, work and energy are closely related to the changes witnessed in a material when force is applied.
Work is defined as the product of force and the distance over which it acts:

• Work = Force × Distance

This formula helps us calculate the energy used to deform a material.
For instance, in our problem, the work done to stretch the wire by 0.2 cm was 0.4 joules.
By knowing the amount of work done, you can determine the stored potential energy in the wire.
The concept of work done links to energy, particularly the type of energy we refer to as potential energy in mechanical contexts.
Understanding work is key for those designing systems where energy transformation and efficiency are critical, such as in engines and machinery setups.

Material Properties

Material properties determine how different materials respond to forces and deformations. One key property, concerning our exercise, is Young's modulus.
This measures a material's stiffness and is represented by:

• Young's Modulus (Y) = Stress / Strain

Young's modulus provides insight into whether a material is elastic or rigid.
In our problem, the calculated Young's modulus is 2.0 × 10^{11} N/m², suggesting a material with significant resistance to deformation. A higher Young's modulus indicates a stiffer material. These properties are crucial for selecting materials in construction and manufacturing, ensuring they will perform well in specific applications.
Engineers need to understand these properties to prevent structural failures and to optimize material performance under various load conditions.