A wire of resistance 18 ohms is drawn until its radius reduces to $\frac{1}{2}$ th of its original radius then the resistance of the wire is

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288 $Ω$

72 $Ω$

388 $Ω$

188 $Ω$

answer is C.

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Detailed Solution

$Rα \frac{1}{r^{4}} m$ = (constant)
$\begin{array}{l} \frac{R_{1}}{R_{2}} = \frac{r_{2}^{4}}{r_{1}^{4}} \\ \frac{18}{R_{2}} = {(\frac{r}{2 r})}^{4} \\ R_{2} = 18 \times 16 = 288 Ω \end{array}$

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