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The resistance of a 10m long wire is 10 $Ω$ . Its length is increased by 25% by stretching the wire uniformly. The resistance of wire will change to

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12.5 $Ω$

15.6 $Ω$

16.6 $Ω$

14.5 $Ω$

answer is C.

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Detailed Solution

$\frac{R_{2}}{R_{1}} = {(\frac{l_{2}}{l_{1}})}^{2}$

$\frac{R_{2}}{10} = {(\frac{12.5}{10})}^{2}$

$R_{2} = 15.625 Ω$

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