Length of conducting wire is increased by 100%. The change in the resistance of the wire will be:-

50 %

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100 %

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200 %

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300 %

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Solution

The correct option is D 300 %
Let the initial length of wire= $l_{1}$
Final length of wire after stretching $l - 2 = l_{1} + (100 / 100) l_{1}$
$= 2 l_{1}$

$l_{2} / l_{1} = 2$ ;

Area of initial wire $= A_{1}$
Area of final wire $= A_{2}$
Volume of wire remains constant.
$l_{1} A_{1} = l_{2} A_{2}$
$A_{2} = l_{1} A_{1} / l_{2}$
As we know that
$R = ρ l / A$
$R_{1} / R_{2} = (l_{1} / A_{1}) / (l_{2} / A_{2})$ ,
$= l_{1} A_{2} / l_{2} A_{1} .$
$= l_{1} (l_{1} A_{1} / l_{2}) / l_{2} A_{1}$ ,
$R_{1} / R_{2} = (l_{1} / l_{2})^{2} = 0.25$ ; (since $l_{2} / l_{1} = 2$
$R_{2} = 4 R_{1}$ ;
$(Δ R / R) \times 100 = (R_{2} - R_{1} / R_{1}) \times 100 = 300$
The percentage increase in resistance is $300$

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length of a conducting wire is increased by 100% by stretching it.the change in the resistance of the wire is?

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