The resistance of a wire is $R Ω$ . If it is melted and stretched to $n$ times its original length, its new resistance will be

\frac{R}{n^{2}}

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n R

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\frac{R}{n}

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n^{2} R

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Solution

The correct option is D $n^{2} R$
$R_{1} = \frac{ρ l_{1}}{A_{1}}$
$R_{2} = \frac{ρ l_{2}}{A_{2}}$
$l_{1} A_{1} = l_{2} A_{2} \to$ (Since volume remains constant)
$l_{1} A_{1} = n l_{1} A_{2}$ ( $l_{2} = n l_{1}$ )
$\Rightarrow A_{2} = \frac{A_{1}}{n}$
$\Rightarrow R_{2} = \frac{ρ n^{2} l_{1}}{A_{1}}$
Hence $R_{2} = n^{2} R_{1}$

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The correct option is D n2RR1=ρl1A1 R2=ρl2A2 l1A1=l2A2→ (Since volume remains constant) l1A1=nl1A2 (l2=nl1) ⇒A2=A1n ⇒R2=ρn2l1A1 Hence R2=n2R1

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