b
(b) For no current flow between \(C\) and \(D\) \({\left( {\frac{Q}{t}} \right)_{AC}} = {\left( {\frac{Q}{t}} \right)_{CB}}\)

==> \(\frac{{{K_1}A({\theta _A} - {\theta _C})}}{l} = \frac{{{K_2}A({\theta _C} - {\theta _B})}}{l}\)

==> \(\frac{{{\theta _A} - {\theta _C}}}{{{\theta _C} - {\theta _B}}} = \frac{{{K_2}}}{{{K_1}}}\)...\((i) \)

Also \({\left( {\frac{Q}{t}} \right)_{AD}} = {\left( {\frac{Q}{t}} \right)_{DB}}\)

==>\(\frac{{{K_3}A({\theta _A} - {\theta _D})}}{l} = \frac{{{K_4}A({\theta _D} - {\theta _B})}}{l}\)

==>\(\frac{{{\theta _A} - {\theta _D}}}{{{\theta _D} - {\theta _B}}} = \frac{{{K_4}}}{{{K_3}}}\) ..\(.(ii)\) 

It is given that \({\theta _C} = {\theta _D},\) hence from equation \((i)\) and \((ii)\) we get \(\frac{{{K_2}}}{{{K_1}}} = \frac{{{K_4}}}{{{K_3}}}\) ==> \({K_1}{K_4} = {K_2}{K_3}\)