The Young's modulus of steel wire of radius $$r$$ and length $$L$$ is $$Y$$. If the radius $$r$$ and length $$L$$ of the wire are doubled then the value of $$Y$$

Young’s modulus:

Y = (stress) / (strain)

$$Y=\frac{F/A}{\Delta L/L}=\frac{FL}{A\Delta L}$$

Now if:

radius → 2r ⇒ area $$A\propto r^2$$ becomes 4A
length → 2L

Substitute in formula:

$$Y'=\frac{F(2L)}{(4A)(\Delta L')}$$

But extension $$ΔL′$$ also changes such that:

$$\Delta L'=\frac{F(2L)}{Y(4A)}=\frac{1}{2}\Delta L$$

Putting back:

Y′=Y

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