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(B) Let the resistances of the two segments of the ring be $R_1$ and $R_2$. Since the segments are in parallel,the equivalent resistance $R$ is given

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$\frac{l_1}{l_2} = \frac{1}{2}$

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(B) Let the resistances of the two segments of the ring be $R_1$ and $R_2$. Since the segments are in parallel,the equivalent resistance $R$ is given by:$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} \Rightarrow R = \frac{R_1 R_2}{R_1 + R_2} = \frac{8}{3}\,\Omega$.Given the total resistance $R_0 = R_1 + R_2 = 12\,\Omega$,we have:$R_1 R_2 = \frac{8}{3} 	imes 12 = 32$.Now,we use the identity $(R_2 - R_1)^2 = (R_1 + R_2)^2 - 4R_1 R_2$:$(R_2 - R_1)^2 = 12^2 - 4(32) = 144 - 128 = 16$.Thus,$R_2 - R_1 = 4\,\Omega$.Solving the system $R_1 + R_2 = 12$ and $R_2 - R_1 = 4$,we get $2R_2 = 16 \Rightarrow R_2 = 8\,\Omega$ and $R_1 = 4\,\Omega$.Since resistance is proportional to length $(R \propto l)$,the ratio of lengths is $\frac{l_1}{l_2} = \frac{R_1}{R_2} = \frac{4}{8} = \frac{1}{2}$.

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