b
(b) Let the temperature of junction be \(\theta\). Since roads \(B\) and \(C\) are parallel to each other (because both having the same temperature difference).

Hence given figure can be redrawn as follows
( \(\frac{Q}{t} = \frac{{({\theta _1} - {\theta _2})}}{R}\) and \({\left( {\frac{Q}{t}} \right)_{AB}} = {\left( {\frac{Q}{t}} \right)_{BC}}\)

==> \(\frac{{(90 - \theta )}}{{R/2}} = \frac{{(\theta - 0)}}{R}\)

==> \(180 - 2\theta = \theta \)

==>\(\theta = 60^\circ C\)