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(C) The resistance of each piece is $r = \frac{R}{20}$.There are $20$ pieces in total,so $10$ pieces are used for the series combination and $10$ piec

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$\frac{101 R}{200}$

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(C) The resistance of each piece is $r = \frac{R}{20}$.There are $20$ pieces in total,so $10$ pieces are used for the series combination and $10$ pieces for the parallel combination.For the $10$ pieces connected in series,the equivalent resistance $R_1$ is:$R_1 = 10 	imes r = 10 	imes \frac{R}{20} = \frac{R}{2}$.For the $10$ pieces connected in parallel,the equivalent resistance $R_2$ is:$\frac{1}{R_2} = \frac{1}{r} + \frac{1}{r} + \dots (10 	ext{ times}) = \frac{10}{r} = \frac{10}{R/20} = \frac{200}{R}$.Thus,$R_2 = \frac{R}{200}$.Since the two combinations are joined in series,the total effective resistance $R_{eq}$ is:$R_{eq} = R_1 + R_2 = \frac{R}{2} + \frac{R}{200} = \frac{100R + R}{200} = \frac{101R}{200}$.

$A$ uniform conductor of resistance $R$ is cut into $20$ equal pieces. Half of them are joined in series and the remaining half of them are connected in parallel. If the two combinations are joined in series,the effective resistance of all the pieces is

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