© All Rights reserved @ LearnWithDash
Step-by-Step Solution
1. Understanding the Problem
Two wires are made of the same material (hence having the same Young’s modulus), but they differ in length and radius, and different forces are applied to them. We know the following for the first wire:
Length = $L$
Radius = $r$
Applied force = $f$
Observed extension = $l$
The second wire has:
Length = $2L$
Radius = $2r$
Applied force = $2f$
We want to find the extension of the second wire.
2. Formula for Extension
The extension $\Delta L$ of a wire under tension is given by the formula:
$ \Delta L = \frac{F\,L}{A\,Y},$
where:
$F$ = Applied force
$L$ = Original length of the wire
$A$ = Cross-sectional area of the wire
$Y$ = Young’s modulus of the material
3. Extension of the First Wire
For the first wire with length $L$ and cross-sectional area $A = \pi r^2$ (denoted generally by $A$):
$ l = \frac{f \, L}{A \, Y}.
$
Rearranging to solve for $f$, we get:
$ f = \frac{A \, Y \, l}{L}.
$
4. Extension of the Second Wire
The second wire has:
Length = $2L$
Cross-sectional area = $4A$ (since area $\propto r^2$ and the new radius is $2r$)
Applied force = $2f$
Let the extension of the second wire be $\Delta l_{\text{second}}$. Then, from the extension formula:
$ \Delta l_{\text{second}}
= \frac{(2f)\,(2L)}{(4A)\,Y}
= \frac{4fL}{4AY}
= \frac{fL}{AY}.
$
5. Substitute $f$ from the First Wire
Using $ f = \frac{A\,Y\,l}{L}$ from the first wire, substitute into the expression above:
$ \Delta l_{\text{second}}
= \frac{\left(\frac{A\,Y\,l}{L}\right) \cdot L}{A\,Y}
= \frac{A\,Y\,l}{L} \cdot \frac{L}{A\,Y}
= l.
$
6. Conclusion
The increase in length of the second wire is therefore $l$.
