A ring is made of a wire having a resistance $R_{0} = 12 Ω$ Find the points A and B, as shown in the figure, at which a current carrying conductor should be connected so that the resistance R of the subcircuit between these points is equal $to \frac{8}{3} Ω .$

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$\frac{l_{1}}{l_{2}} = \frac{1}{2}$

$\frac{l_{1}}{l_{2}} = \frac{5}{8}$

$\frac{l_{1}}{l_{2}} = \frac{1}{3}$

$\frac{l_{1}}{l_{2}} = \frac{3}{8}$

answer is A.

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Detailed Solution

$Let resistance of the two parts be R_{1} and R_{2} then R_{1} + R_{2} = R_{0} = 12 as effective resistance between A and B is \frac{8}{3} Ω \frac{R_{1} R_{2}}{R_{1} + R_{2}} = \frac{8}{3} \Rightarrow \frac{R_{1} R_{2}}{12} = \frac{8}{3} \Rightarrow R_{1} R_{2} = \frac{8}{3} \times 12 = 32 Ω from the two expressions R_{1} + R_{2} = 12 & R_{1} R_{2} = 32 it can be given that R_{1} = 4 Ω, R_{2} = 8 Ω; \frac{R_{1}}{R_{2}} = \frac{4}{8} = \frac{1}{2} \Rightarrow \frac{l_{1}}{l_{2}} = \frac{1}{2} (∵ R \propto l)$