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A wire of resistance $5 \Omega$ is drawn out so that its new length is 3 times its original length. What is the reistance of the new wire?
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The correct answer is:
$45 \Omega$
$\left(\frac{r_1}{r_2}\right)^2=\left(\frac{\ell_2}{\ell_1}\right)=\frac{3 \ell}{\ell}=3$
$\left(\frac{R_2}{R_1}\right)=\frac{\ell_2}{\ell_1} \times \frac{A_1}{A_2}=3 \times\left(\frac{r_1}{r_2}\right)^2=3 \times 3 \Rightarrow R_2=45$
$\left(\frac{R_2}{R_1}\right)=\frac{\ell_2}{\ell_1} \times \frac{A_1}{A_2}=3 \times\left(\frac{r_1}{r_2}\right)^2=3 \times 3 \Rightarrow R_2=45$
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