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A wire of resistance 50 $Ω$ is cut into six equal parts and they are bundled together side by side to form a thicker wire. The resistance of the bundle is

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$\frac{50}{9}$ $Ω$

$\frac{25}{18}$ $Ω$

$\frac{18}{25}$ $Ω$

$\frac{9}{12.5}$ $Ω$

answer is D.

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Detailed Solution

$\begin{array}{l} l^{1} = \frac{l}{6}, A^{'} = 6 A \\ \frac{R^{'}}{R} = \frac{l^{1}}{l} \times \frac{A}{A} = \frac{l}{6 \times l} \times \frac{A}{6 A} = \frac{1}{36} \\ R^{'} = \frac{R}{36} = \frac{50}{36} = \frac{25}{18} Ω \end{array}$

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