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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, the effective resistance of all the pieces is
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The correct answer is:
$\frac{101 R}{200}$
Resistance of each piece $=\frac{R}{20} \Omega$
Equivalent resistance of 10 such pieces connected in series
$$
R_1=10 \times \frac{R}{20}=\frac{R}{2}
$$
Similarly, equivalent resistance of 10 such pieces connected in parallel
$$
\begin{aligned}
\frac{1}{R_2} & =\frac{1}{R / 20}+\ldots 10 \text { times } \\
& =\frac{10}{R / 20}=\frac{200}{R} \\
R_2 & =\frac{R}{200}
\end{aligned}
$$
Now, $R_1$ and $R_2$ are connected in series so total resistance $=R_1+R_2$
$$
\begin{aligned}
& =\frac{R}{2}+\frac{R}{200} \\
& =\frac{100 R+R}{200} \\
& =\frac{101 R}{200}
\end{aligned}
$$
Equivalent resistance of 10 such pieces connected in series
$$
R_1=10 \times \frac{R}{20}=\frac{R}{2}
$$
Similarly, equivalent resistance of 10 such pieces connected in parallel
$$
\begin{aligned}
\frac{1}{R_2} & =\frac{1}{R / 20}+\ldots 10 \text { times } \\
& =\frac{10}{R / 20}=\frac{200}{R} \\
R_2 & =\frac{R}{200}
\end{aligned}
$$
Now, $R_1$ and $R_2$ are connected in series so total resistance $=R_1+R_2$
$$
\begin{aligned}
& =\frac{R}{2}+\frac{R}{200} \\
& =\frac{100 R+R}{200} \\
& =\frac{101 R}{200}
\end{aligned}
$$
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