A ring is made of a wire having a resistance $R_0 = 12 \,\,\Omega$. 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 sub circuit between these points is equal to $\frac{8}{3}\,\Omega$.
AIPMT 2012, Diffcult
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Let $x$ be resistance per unit length of the wire. Then,

The resistance of the upper portion is

$R_{1}=x l_{1}$

The resistance of the lower portion is

$R_{2}=x l_{2}$

Equivalent resistance between $A$ and $B$ is

$R = \frac{{{R_1}{R_2}}}{{{R_1} + {R_2}}} = \frac{{\left( {x{l_1}} \right)(x{l_2})}}{{x{l_1} + x{l_2}}}$

$\frac{8}{3}=\frac{x l_{1} l_{2}}{l_{1}+l_{2}}$ or $\frac{8}{3}=\frac{x l_{1} l_{2}}{l_{2}\left(\frac{l_{1}}{l_{2}}+1\right)}$ or $\frac{8}{3}=\frac{x_{1}}{\left(\frac{l_{1}}{l_{2}}+1\right)}$        ......$(i)$

Also $\quad R_{0}=x l_{1}+x l_{2}$

${12=x\left(l_{1}+l_{2}\right)}$

${12=x l_{2}\left(\frac{l_{1}}{l_{2}}+1\right)}$        ....$(ii)$

Divide $(i)$ by $(ii),$ we get

$\frac{{\frac{8}{3}}}{{12}} = \frac{{\frac{{x{l_1}}}{{\left( {\frac{{{l_1}}}{{{l_2}}} + 1} \right)}}}}{{x{l_2}\left( {\frac{{{l_1}}}{{{l_2}}} + 1} \right)}}{\rm{ or }}\frac{8}{{36}} = \frac{{{l_1}}}{{{l_2}{{\left( {\frac{{{l_1}}}{{{l_2}}} + 1} \right)}^2}}}$

$\left(\frac{l_{1}}{l_{2}}+1\right)^{2} \frac{8}{36}=\frac{l_{1}}{l_{2}}$ or $\left(\frac{l_{1}}{l_{2}}+1\right)^{2} \frac{2}{9}=\frac{l_{1}}{l_{2}}$

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

$\therefore$       $2(y+1)^{2}=9 y$ or $2 y^{2}+2+4 y=9 y$

or    $ 2 y^{2}-5 y+2=0$

Solving this quadratic equation, we get

$y=\frac{1}{2}$  or  $2 \therefore \frac{l_{1}}{l_{2}}=\frac{1}{2}$

art

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