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Step-by-Step Solution
Step 1: Identify the Known Parameters
• Initial resistance of the wire, $R_0 = 1\,\Omega$
• Initial length of the wire, $l_0 = 1\,\text{m}$
• Final length of the wire, $l_1 = 1.25\,\text{m}$
• The volume of the wire remains constant when it is stretched.
Step 2: Express the Relationship for Area
Since the volume of the wire remains constant, we have
$A_0\,l_0 = A_1\,l_1 \quad \Rightarrow \quad A_1 = \dfrac{A_0\,l_0}{l_1}.$
Here, $A_0$ and $A_1$ are the initial and final cross-sectional areas respectively.
Step 3: Recall the Formula for Resistance
The resistance $R$ of a conductor of resistivity $p$, length $l$, and cross-sectional area $A$ is given by
$R = \dfrac{p\,l}{A}.$
Step 4: Relate Initial and Final Resistances
Let the initial resistivity be $p$, initial length $l_0$, and initial area $A_0$. We then write:
$R_0 = \dfrac{p\,l_0}{A_0}, \quad R_1 = \dfrac{p\,l_1}{A_1}.$
Using $A_1 = \frac{A_0\,l_0}{l_1}$, substitute into the expression for $R_1$:
$R_1 = \dfrac{p\,l_1}{\frac{A_0\,l_0}{l_1}} = \dfrac{p\,l_1^2}{A_0\,l_0}.$
Therefore, dividing $R_1$ by $R_0$ gives:
$\dfrac{R_1}{R_0} = \dfrac{\dfrac{p\,l_1^2}{A_0\,l_0}}{\dfrac{p\,l_0}{A_0}} \;=\; \dfrac{l_1^2}{l_0^2}.$
Hence,
$R_1 = R_0 \times \dfrac{l_1^2}{l_0^2}.$
Step 5: Substitute Numerical Values
Given $l_0 = 1\,\text{m}$ and $l_1 = 1.25\,\text{m}$, then
$ \dfrac{l_1}{l_0} = \dfrac{1.25}{1} = 1.25 \quad \Rightarrow \quad
\left(\dfrac{l_1}{l_0}\right)^2 = 1.25^2 = 1.5625. $
Thus,
$R_1 = 1\,\Omega \times 1.5625 = 1.5625\,\Omega.$
Step 6: Calculate the Percentage Change in Resistance
The percentage change in resistance is
$\dfrac{R_1 - R_0}{R_0} \times 100\% = \dfrac{1.5625 - 1}{1} \times 100\% = 56.25\%. $
Rounded to the nearest integer, the percentage change is $56\%.$
Final Answer
The percentage change in resistance, to the nearest integer, is 56%.
