# Why doesn't Young's modulus change with length and diameter?

In this question:

The Young modulus of steel is determined using a length of steel wire and is found to have the value $E$. Another experiment is carried out using a wire of the same steel, but of half the length and half the diameter.

What value is obtained for the Young modulus in the 2nd experiment?

I know that the Young's modulus is an intrinsic property of a object. But what I found confusing is that, when I calculated the Young's modulus for the 2nd experiment, I got $2E$. But the answer was $E$, instead of $2E$.

However, my thought kept lying with the equation: $$\text{Young's modulus} = \frac{\text{force}\times\text{length}}{\text{extension}\times\text{area}}$$

Doesn't the change in length and diameter affect the Young's modulus value? How can it be an intrinsic value for a object?

## 1 Answer

From the given information how did you "calculate the Young's modulus for the 2nd experiment"?
You cannot assume that in the second experiment the force and extension are the same as in the first experiment because they are not.
For a given force the extension would be twice in the second experiment as that in the first experiment.

Update in response to a comment

$\text{extension}_1 = \dfrac{\text{force}\times\text{length}}{\text{Young's modulus}\times\text{area}}$

$\text{extension}_2 = \dfrac{\text{force}\times\frac{\text{length}}{2}}{\text{Young's modulus}\times\frac{\text{area}}{4}}=2\times \text{extension}_1$

• For a given amount force extension is half in the 2nd case.Can you please elaborate on this? – Lapmid Jan 31 '17 at 6:48
• @Theasgardian Correction made. – Farcher Jan 31 '17 at 6:53
• I think the extension should be same not twice or half because both lenght and diameter has decreased. – Lapmid Jan 31 '17 at 6:57
• @Theasgardian I have updated my answer. – Farcher Jan 31 '17 at 7:10