A wire of $1 Ω$ has a length of $1 m$ . It is stretched till its length increases by $25 %$ . The percentage change in resistance to the nearest integer is:

$25 %$

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$12.5 %$

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$76 %$

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$56 %$

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Solution

The correct option is D
$56 %$

Step 1. Given data
Resistance of wire before stretching, $R_{1} = 1 Ω$
Length of wire before stretching, $l_{1} = 1 m$ .
It is stretched till its length increases by
$l_{2} = l_{1} + 25 % of I_{1} l_{2} = 1 + \frac{25}{100} \times 1 l_{2} = 1 + 0.25 l_{2} = 1.25 m$ [ $l_{2}$ length of the wire after stretching]
Step 2. Calculating area after stretching,
Resistance is directly proportional to the length and inversely proportional to the area
We know, $R \propto \frac{l}{A}$ (where, $R$ is the resistance, $l$ is the length of the wire, and $A$ is the cross-sectional area of the wire).
For stretched or compressed wire, volume remains constant,
Area $\times$ length $=$ constant [As volume is constant, volume is the product of the area and length.]
$A_{1} l_{1} = A_{2} l_{2} A_{2} = \frac{A_{1} l_{1}}{l_{2}}$ [ $A_{1}$ is the area before stretching and $A_{2}$ is the area after stretching]
Step 3. Calculating percentage change in resistance,
Now, the ratio of the resistance before stretching, $R_{1}$ and resistance after stretching, $R_{2}$
$\frac{R_{1}}{R_{2}} = \frac{l_{1}}{l_{2}} (\frac{A_{2}}{A_{1}}) \frac{R_{1}}{R_{2}} = \frac{l_{1}}{l_{2}} (\frac{A_{1} l_{1}}{A_{1} l_{2}}) By substituting the value of A_{2}$
$\frac{R_{1}}{R_{2}} = \frac{{I_{1}}^{2}}{{I_{2}}^{2}} \frac{R_{1}}{R_{2}} = \frac{1^{2}}{{(1.25)}^{2}} [R_{1} = 1 Ω] R_{2} = 1.5625 Ω$
So, the percentage change in resistance, $R %$
$R % = \frac{R_{2} - R_{1}}{R_{1}} \times 100 R % = \frac{1.5625 - 1}{1} \times 100 R % = 56.25 %$
$R % = 56 %$ [Approximate value]
Therefore, percentage ( $%$ ) increase will be $56 %$
Hence, option D is correct.

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