An increase in the length of a copper wire due to stretching is 25%, $R_{1}$ is its resistance before stretching and $R_{2}$ is its resistance after stretching, then $R_{1} : R_{2}$ ?

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Solution

The correct option is B
0.2%

Given,

Length of wire after stretching $l_{2}$ = $25$ % = $1.25 l_{1}$

Volume remains constant

$l_{1} A_{1}$ = $l_{2} A_{2}$ = $1.25 l_{1} A_{2}$

$A_{1} = 1.25 A_{2}$

As R = ρ $\frac{L}{A}$

$R_{1}$ = ρ $\frac{l_{1}}{A_{1}}$ , $R_{2}$ = ρ $\frac{l_{2}}{A_{2}}$

$\frac{R_{1}}{R_{2}}$ = $\frac{l_{1} A_{2}}{l_{2} A_{1}}$

= $\frac{l_{1} A_{2}}{1.25 l_{1} \times 1.25 A_{2}}$ = $\frac{1}{1.56}$

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