Question

If the length of a wire is made double and radius is halved of its respective values. Then, the Young’s modulus of the material of the wire will :

• A. Remain same
• B. Become 8 times its initial value
• C. Become \(\frac 14^{th}\) of its initial value
• D. Become 4 times its initial value

Answer 1

The Correct Option is A

The Young’s modulus is defined as the ratio of stress to strain in a material and is a measure of the stiffness of the material. It is a property that is intrinsic to the material and does not depend on the dimensions of the material's shape or size.

Let's dive into the problem:

The Young's modulus \( Y \) is given by:

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

Where:

• \(F\) is the force applied,
• \(A\) is the cross-sectional area,
• \(\Delta L\) is the change in length,
• \(L\) is the original length.

Now, let's consider the changes in the wire:

1. The length of the wire is doubled: \(L' = 2L\)
2. The radius is halved: If the original radius is \(r\), the new radius is \(r' = \frac{r}{2}\)

The new cross-sectional area \(A'\) becomes:

\(A' = \pi (r')^2 = \pi \left(\frac{r}{2}\right)^2 = \frac{\pi r^2}{4} = \frac{A}{4}\)

At this point, it is essential to remember that Young's modulus is a material property and does not depend on the material’s dimensions or shape. Thus, regardless of changes in length or cross-sectional area, the intrinsic stiffness or Young’s modulus of the material remains the same.

Consequently, the correct answer is:

Remain same