d
ધારો કે \(x\) એ એકમ લંબાઈ દીઠ અવરોધે છે.

તો સમતુલ્ય અવરોધ \(R = \,\,\frac{{{R_1}{R_2}}}{{{R_1}\, + \,{R_2}}}\)

\( \Rightarrow \,\frac{8}{3}\,\, = \,\,\frac{{(x{{l}_1})\,(x{{l}_2})}}{{x{{l}_1}\, + \,x{{l}_2}}}\)

\( \Rightarrow \,\frac{8}{3}\,\, = \,\,x\,\frac{{{{l}_1}\,{{l}_2}}}{{{{l}_1}\, + \,{{l}_2}}}\)

\( \Rightarrow \,\frac{8}{3}\,\, = \,\,x\,\frac{{{{l}_1}}}{{\frac{{{{l}_1}}}{{{{l}_2}}}\, + \,1}}......(1)\)

અને \({R_0}\,\, = \,x{{l}_1}\, + \,x{{l}_2}\,\)  \( \Rightarrow \,12\,\,\, = \,\,x\,({{l}_1}\, + \,{{l}_2})\) 

\( \Rightarrow \,\,12\,\, = x{{l}_2}\,\left( {\frac{{{{l}_1}}}{{{{l}_2}}}\, + \,1} \right).....(2)\)

સમીકરણ \({\text{(1)}}\) અને  \({\text{(2)}}\) નો ગુણોતર લેતા;

\(\,\frac{{\frac{8}{3}}}{{\frac{{12}}{1}}}\,\,\, = \,\,\frac{{\frac{{x{{l}_1}}}{{\left( {\frac{{{{l}_1}}}{{{{l}_2}}}\, + \,1} \right)}}}}{{x{l_2}\,\left( {\frac{{{{l}_1}}}{{{{l}_2}}}\, + \,1} \right)}}\,\, = \,\,\frac{{{{l}_1}}}{{{l_2}\,{{\left( {\frac{{{{l}_1}}}{{{{l}_2}}}\, + \,1} \right)}^2}}}\)

\( \Rightarrow \,{\left( {\frac{{{{l}_1}}}{{{{l}_2}}}\, + \,1} \right)^2}\, \times \,\frac{8}{{36}}\,\, = \,\,\frac{{{{l}_1}}}{{{{l}_2}}}\)

\( \Rightarrow \,({y^2}\, + \,1\, + \,2y)\, \times \,\frac{8}{{36}}\,\, = \,\,y\,\) [ જ્યા \({y\,\, = \,\,\frac{{{{l}_1}}}{{{{l}_2}}}\,}\)]

\( \Rightarrow \,8{y^2}\, + \,8\, + \,16y\,\, = \,\,36y\)

\( \Rightarrow \,8{y^2}\, + \,8\, + \,16y\,\, = \,\,36y\)

\( \Rightarrow \,8{y^2}\, - \,20y\, + \,8\,\, = \,\,0\)

સમીકરણ ઉકેલતા \({\text{y  =  1/2 }}\) અથવા \(\,{\text{2}}\)

\(\,\therefore \,y\,\, = \,\,\frac{{{{l}_1}}}{{{{l}_2}}}\,\, = \,\,\frac{1}{2}\) અથવા \({\text{2}}\)