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

Short answer

(B) Radius is to be increased

Step-by-step solution

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: \[\epsilon = \frac{\Delta L}{L}\]

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: \[Y = \frac{\frac{F}{A}}{\frac{\Delta L}{L}}\]

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: \[\sigma = \frac{F}{\pi r^2}\]

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.