Problem 869 - JEE Physics

Question

A steel wire is stretched with a definite load. If the young's modulus of the wire is \(\mathrm{Y}\). For decreasing the value of \(\mathrm{Y}\). (A) Radius is to be decreased (B) Radius is to be increased (C) Length is to be increased (D) None of the above

Step-by-Step Solution

Answer

(B) Radius is to be increased

Step 1: Define stress and strain

Stress (\(\sigma\)) is given by the force (\(F\)) applied on an object divided by its cross-sectional area (\(A\)), and strain (\(\epsilon\)) is given by the change in length (\(\Delta L\)) divided by the original length (\(L\)): Stress: \[\sigma = \frac{F}{A}\] Strain: \[\epsilon = \frac{\Delta L}{L}\]

Step 2: Establish Young's Modulus relationship

Young's Modulus (Y) is defined as the ratio of stress to strain: \[Y = \frac{\sigma}{\epsilon}\] Combine the stress and strain definitions from Step 1: \[Y = \frac{\frac{F}{A}}{\frac{\Delta L}{L}}\]

Step 3: Calculate the relationship between radius, length, and stress

The cross-sectional area (A) of the wire is given by the formula: \[A=\pi r^2\] where r is the radius of the wire. Substitute this into the equation for stress: \[\sigma = \frac{F}{\pi r^2}\]

Step 4: Analyze the effects of increasing/decreasing radius and length on stress

Using the equation for stress (\(\sigma = \frac{F}{\pi r^2}\)), we can see that: - If the radius (r) is decreased, the stress on the wire will increase. - If the radius (r) is increased, the stress on the wire will decrease. Since the problem asks for a decrease in the effective Young's Modulus, which should result in less stress on the wire, the correct answer is: (B) Radius is to be increased

Key Concepts

Stress

Stress is a fundamental concept when studying materials and their behavior under load. It represents how much force is acting per unit area on an object. Imagine applying pressure to a wire, the force you exert creates stress within the wire.

The formula to calculate stress is fairly straightforward:

• \( \sigma = \frac{F}{A} \)

Where:

• \( \sigma \) stands for stress
• \( F \) is the force applied
• \( A \) represents the cross-sectional area of the object

Essentially, stress tells us how concentrated the force is on a material. If the area is small, even a small force can lead to a large stress. Conversely, spreading the force over a larger area results in less stress.

Strain

Strain goes hand in hand with stress, as it's a measure of the deformation that stress causes. When a material is subjected to stress, it changes shape or length, and strain quantifies this change.

Strain is calculated by the following equation:

• \( \epsilon = \frac{\Delta L}{L} \)

Where:

• \( \epsilon \) is the strain
• \( \Delta L \) is the change in length
• \( L \) is the original length

Strain is dimensionless, meaning it doesn't have units. It's a relative measure describing how much a material stretches or compresses relative to its original size. In our context, it's important to understand that a higher strain indicates a significant length change under stress. This becomes crucial when evaluating the performance of materials like wires under loads.

Cross-Sectional Area

The cross-sectional area plays a vital role in determining how a material responds to force. Essentially, it represents the size of a slice through the material perpendicular to the force direction. In the case of a wire, imagine cutting through, and seeing a circle; that area is what we're talking about.

The mathematical representation for a circular cross-section is:

• \( A = \pi r^2 \)

Where:

• \( A \) is the area
• \( \pi \) is a constant (~3.14159)
• \( r \) is the radius of the wire

The cross-sectional area influences stress because a larger area distributes the force more effectively, reducing stress. For our exercise, increasing the radius of the wire increases the cross-sectional area, which in turn reduces stress. Understanding this concept is crucial in applications where minimizing stress for a particular load is desired, like in the case of elongating a wire.