A 10 $Ω$ thick wire is stretched so that its length becomes three times then the new resistance of the wire is

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90 $Ω$

30 $Ω$

10 $Ω$

$\frac{10}{9} Ω$

answer is C.

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Detailed Solution

$R^{1} = n^{2} R$
$\begin{array}{l} n = 3 \\ {Rαl}^{2} \\ \frac{R_{1}}{R_{2}} = \frac{1_{1}^{2}}{l_{2}^{2}} \\ \frac{10}{R} = \frac{t^{2}}{9 t^{2}} \\ k = 90 Ω \\ R^{1} = 90 Ω \end{array}$

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