Question

Two copper wires of length $L$ and $2L$ have radii $r$ and $2r$ respectively. If they are subjected to the same tension force, the ratio of their extension ($\Delta L_1 / \Delta L_2$) is:

• A. $2 : 1$
• B. $1 : 2$
• C. $1 : 1$
• D. $4 : 1$

Hint:

Use proportionality directly: $\Delta L \propto \frac{L}{r^2}$. Doubling the length doubles the extension, but doubling the radius quarter-sizes it. Net change is $2 \times \frac{1}{4} = \frac{1}{2}$, meaning the second wire stretches half as much.

Answer 1

The Correct Option is A

Step 1: Recall the stretch formula.
When a wire is pulled by a force $F$, it stretches by \[ \Delta L = \frac{FL}{AY} = \frac{FL}{\pi r^2 Y} \] Here $L$ is length, $r$ is radius, $A=\pi r^2$ is area, and $Y$ is Young's modulus of the material.

Step 2: Note what is the same.
Both wires are copper, so they share the same $Y$. The pulling force $F$ is also the same. So the stretch only depends on length and radius. \[ \Delta L \propto \frac{L}{r^2} \]

Step 3: Write the first wire's stretch.
First wire has length $L$ and radius $r$. \[ \Delta L_1 \propto \frac{L}{r^2} \]

Step 4: Write the second wire's stretch.
Second wire has length $2L$ and radius $2r$. Its area is $(2r)^2 = 4r^2$. \[ \Delta L_2 \propto \frac{2L}{4r^2} = \frac{1}{2}\cdot\frac{L}{r^2} \]

Step 5: Take the ratio.
Divide the first by the second. \[ \frac{\Delta L_1}{\Delta L_2} = \frac{\frac{L}{r^2}}{\frac{1}{2}\frac{L}{r^2}} = 2 \]

Step 6: State the result.
The thinner, shorter wire stretches twice as much. \[ \boxed{\Delta L_1 : \Delta L_2 = 2 : 1} \]