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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, then the effective resistance of all the pieces is

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$\frac{101 R}{200}$

$\frac{R}{2}$

$\frac{201 R}{200}$

answer is C.

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Detailed Solution

$\begin{aligned} R^{1} = \frac{R}{20} each; R^{11} = 10 \cdot R^{1} = 10 \cdot \frac{R}{20} = \frac{R}{2} \\ R^{111} = \frac{R^{1}}{10} = \frac{R}{200}; R_{eff.} = \frac{R}{2} + \frac{R}{200} = \frac{101 R}{200} \end{aligned}$

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