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Step-by-Step Solution
Step 1: Identify the physical law involved
The extension of a wire under a given force is related to its Young’s modulus y , cross-sectional area A , and original length l . The relationship can be written as
y = \frac{\text{Stress}}{\text{Strain}} = \frac{\frac{F}{A}}{\frac{\Delta l}{l}} \quad \Longrightarrow \quad y \cdot \frac{\Delta l}{l} = \frac{F}{A}.
Rearranging, we get
l = \frac{A \, y \, \Delta l}{F}.
Step 2: Write down the known data
Wire A is stretched by \Delta l_{A} = 2 \text{ mm} under a force F = 2 \text{ N} .
Wire B is stretched by \Delta l_{B} = 4 \text{ mm} under the same force F = 2 \text{ N} .
The radius of wire B is r_{B} = 4 \, r_{A} , where r_{A} is the radius of wire A.
Both wires are made of the same material, so their Young’s modulus y is the same.
Step 3: Express the lengths using the formula
For a wire that extends from its original length l by \Delta l under a force F , with cross-sectional area A , we have:
l = \frac{A \, y \, \Delta l}{F}.
Thus,
l_{A} = \frac{A_{A} \, y \, \Delta l_{A}}{F}, \quad l_{B} = \frac{A_{B} \, y \, \Delta l_{B}}{F}.
Here,
A_{A} = \pi \, r_{A}^{2}, \quad A_{B} = \pi \, r_{B}^{2} = \pi \cdot (4 r_{A})^{2} = 16 \,\pi \, r_{A}^{2}.
Step 4: Form the ratio l_{A} : l_{B}
We want the ratio
\frac{l_{A}}{l_{B}}
= \frac{\frac{A_{A} \, y \, \Delta l_{A}}{F}}{\frac{A_{B} \, y \, \Delta l_{B}}{F}}
= \frac{A_{A} \, \Delta l_{A}}{A_{B} \, \Delta l_{B}}.
Substitute A_{A} = \pi \, r_{A}^{2} and A_{B} = 16 \, \pi \, r_{A}^{2} :
\frac{l_{A}}{l_{B}}
= \frac{\pi \, r_{A}^{2} \cdot \Delta l_{A}}{16 \, \pi \, r_{A}^{2} \cdot \Delta l_{B}}
= \frac{\Delta l_{A}}{16 \, \Delta l_{B}}.
Given \Delta l_{A} = 2 \text{ mm} and \Delta l_{B} = 4 \text{ mm} , we have
\frac{l_{A}}{l_{B}}
= \frac{2}{16 \cdot 4}
= \frac{2}{64}
= \frac{1}{32}.
Step 5: Final answer
The ratio of the lengths l_{A} : l_{B} is 1 : 32 . Hence if a : b is l_{A} : l_{B} and a/b = 1/x , then x = 32 .
