A uniform wire of resistance 20 ,Omega having | vedclass

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(A) Let the resistance of the shorter part $MN$ be $x \,\Omega$. Since the total resistance of the wire is $20 \,\Omega$,the resistance of the longer

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(A) Let the resistance of the shorter part $MN$ be $x \,\Omega$. Since the total resistance of the wire is $20 \,\Omega$,the resistance of the longer part $MN$ will be $(20 - x) \,\Omega$. With respect to points $M$ and $N$,these two portions of the wire are connected in parallel. The equivalent resistance $R_{eq}$ is given by the formula: $R_{eq} = \frac{R_1 R_2}{R_1 + R_2} = \frac{x(20 - x)}{x + (20 - x)} = 1.8$ $R_{eq} = \frac{20x - x^2}{20} = 1.8$ $20x - x^2 = 36$ $x^2 - 20x + 36 = 0$ Solving the quadratic equation: $(x - 18)(x - 2) = 0$ So,$x = 18 \,\Omega$ or $x = 2 \,\Omega$. Since we are looking for the shorter section,we take $x = 2 \,\Omega$. Given that the resistance per unit length is $1 \,\Omega/m$,the length of the shorter part is $L = \frac{R}{	ext{resistance per unit length}} = \frac{2 \,\Omega}{1 \,\Omega/m} = 2 \,m$.

$A$ uniform wire of resistance $20 \,\Omega$ having resistance $1 \,\Omega/m$ is bent into 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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