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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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