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(A) The total resistance of the wire is $R = 20 \Omega$. It is divided into $10$ equal parts,so the resistance of each part is $r = \frac{20 \Omega}{1

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$5$

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(A) The total resistance of the wire is $R = 20 \Omega$. It is divided into $10$ equal parts,so the resistance of each part is $r = \frac{20 \Omega}{10} = 2 \Omega$.Two parts are connected in parallel. The equivalent resistance of one such parallel pair is $R_p = \frac{r 	imes r}{r + r} = \frac{2 	imes 2}{2 + 2} = 1 \Omega$.Since there were $10$ parts in total,and we used them in pairs,we have $10 / 2 = 5$ such parallel pairs.These $5$ pairs are then connected in series. The total equivalent resistance is $R_{eq} = 5 	imes R_p = 5 	imes 1 \Omega = 5 \Omega$.

$A$ wire of resistance $20 \Omega$ is divided into $10$ equal parts. $A$ combination of two parts are connected in parallel and so on. Now,the resulting pairs of parallel combinations are connected in series. The equivalent resistance of the final combination is . . . . . . $\Omega$.

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