Problem 901 - JEE Physics

Question

A load \(W\) produces an extension of \(1 \mathrm{~mm}\) in a thread of radius \(r\). Now if the load is made \(4 \mathrm{~W}\) and radius is made \(2 \mathrm{r}\) all other things remaining same the extension will becomes.......... (A) \(4 \mathrm{~mm}\) (B) \(16 \mathrm{~mm}\) (C) \(1 \mathrm{~mm}\) (D) \(0.25 \mathrm{~mm}\)

Step-by-Step Solution

Answer

The new extension becomes \(\boxed{0.25 \, \text{mm}}\).

Step 1: Identify the given information

In the problem, we are provided with the following information: 1. Initial load: \(W\) 2. Initial extension: \(1 \, \text{mm}\) 3. Initial radius: \(r\) We are also given the changes in load and radius: 1. New load: \(4W\) 2. New radius: \(2r\) The problem asks for the new extension after these changes have been made.

Step 2: Understand the Young's modulus formula

The Young's modulus of a material is defined as the ratio of stress to strain. Mathematically, this can be formulated as: \(Y = \frac{F/A}{\Delta L/L}\) Where: \(Y\) - Young's modulus \(F\) - Applied force/load (in our case, it is \(W\)) \(A\) - Cross-sectional area of the thread \(\Delta L\) - Extension \(L\) - Original length of the thread Since the problem assumes all other things remain the same, it implies that \(Y\) and \(L\) remain constant. Thus, our focus will be on the relationship between the load, radius, and extensions.

Step 3: Write the formula for the initial condition

From the given information, for the initial condition, we have: \(Y = \frac{W / \pi r^{2}}{1 / L}\) (Y is constant in this case, we will equate this with the equation of the modified parameters) Now, we have to find an expression for the new condition when the load is changed to \(4W\) and the radius is changed to \(2r\).

Step 4: Write the formula for the new condition

For the new condition, we have: \(Y = \frac{4W / \pi (2r)^{2}}{\Delta L' / L}\) Where: \(\Delta L'\) - The extension in the new condition

Step 5: Equate the Young's modulus for both conditions and solve for the new extension

As the Young's modulus remains constant throughout this process, we can equate the two expressions we derived in Steps 3 and 4: \(\frac{W / \pi r^{2}}{1 / L} = \frac{4W / \pi (2r)^{2}}{\Delta L' / L}\) Now, we solve for the new extension \(\Delta L'\): \(\frac{W}{\pi r^{2}} \cdot \frac{L}{1} = \frac{4W}{\pi (2r)^{2}} \cdot \frac{L}{\Delta L'}\) Simplifying the equation and canceling out the common terms: \(\frac{1}{r^{2}} = \frac{4}{(2r)^{2}} \cdot \frac{1}{\Delta L'}\) \(\frac{1}{r^{2}} = \frac{4}{4r^{2}} \cdot \frac{1}{\Delta L'}\) Now, solve for \(\Delta L'\): \(\Delta L' = \frac{1}{4}\) Thus, the new extension becomes \(\boxed{0.25 \, \text{mm}}\), which corresponds to the option (D).

Key Concepts

Load and Extension

Understanding how a load affects the extension of a material is crucial in the study of mechanics. When a force (or load) is applied to an elastic material, it stretches, resulting in what we call an extension. This happens because the material's atoms are displaced, causing the material to elongate. The relationship between load and extension is linear for small deformations, as described by Hooke's Law.

Here is what you need to know about load and extension:

• When the load increases, the extension typically increases, assuming the material does not break.
• The relationship between them can change when more complicated forces or larger deformations are involved.
• In our example, changing the load from \( W \) to \( 4W \) changes the extension, while considering the changes in other factors such as the radius of the thread.

By understanding how loads affect the extension, engineers and scientists can predict how structures will react under different conditions, ensuring safety and reliability.

Stress and Strain

Stress and strain are fundamental concepts when examining how materials deform under external forces. Stress refers to the internal force per unit area in a material, while strain measures the deformation of the material.

Consider these points about stress and strain:

• Stress is calculated as force divided by the cross-sectional area: \( \sigma = \frac{F}{A} \).
• Strain is defined as the change in length divided by the original length: \( \epsilon = \frac{\Delta L}{L} \).
• Young's Modulus \( Y \) relates stress and strain through the formula: \( Y = \frac{\text{stress}}{\text{strain}} \).
• In this problem, as the load changes while considering the constant Young's modulus, we see how stress and resulting strain adapt with a modified radius.

With these definitions, stress and strain provide a clear way to quantify and predict how various materials will behave under distinctive forces, which is significant in design and material choice.

Cross-sectional Area

The cross-sectional area of a thread or wire directly influences how materials respond to applied loads. It is essentially the shape's face that is perpendicular to the direction of the force.

Some key aspects of cross-sectional area include:

• The area \( A \) of a circular thread is given by the formula \( A = \pi r^2 \), which means the area and, thus, the material's ability to bear loads increase with the square of the radius.
• A larger cross-sectional area means that the material can handle more load before deforming excessively.
• In the example, changing the radius from \( r \) to \( 2r \) increases the cross-sectional area by a factor of four (\( \pi (2r)^2 = 4\pi r^2 \)).
• However, this change impacts stress and subsequently affects the strain and extension derived from it.

Understanding how the cross-sectional area impacts stress-bearing capabilities helps in selecting appropriate materials and dimensions for various engineering applications.