A uniform wire of resistance $20\,ohm$ having resistance $1\Omega /m$ is bent in the adjoining form of a circle as shown in the figure. If the equivalent resistance between $M$ and $N$ is $1.8 \,\Omega $, then the length of the shorter section is ................ $m$
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Let the resistance of the shorter part $\mathrm{MN}$ be $\mathrm{x}$. Total resistance is $20 \,\Omega .$ Hence, the resistance of longer $\mathrm{MN}$ part will be $(20-x)$. With respect to $\mathrm{M}$ and $\mathrm{N},$ the two portions are connected in parallel. Hence,
$R_{e q}=\frac{(20-x) x}{(20-x)+x}=1.8$
Solving, we get; $\mathrm{x}=2\, \Omega$
The resistance per unit length is $1\, \Omega / \mathrm{m} .$ So, length of shorter part $=2 \mathrm{\,m}.$
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