Explanation

(b) Here , $K_{A} = 2K_{B}\ (dx)_{A} = d(dx)_{B}.If\ \theta\ $ is temperature of junction,
$(dT)_{A} = \theta_{A} - \theta,\ (dT)\_ B = \left( \theta - \theta_{B} \right)$
As $\left( \frac{dQ}{dt} \right)_{A} = \left( \frac{dQ}{dt} \right)_{B}$
$\therefore K_{A}A\ \frac{(dT)_{A}}{(dx)_{A}} = K_{B}\frac{A(dT)_{B}}{(dx)_{B}}$
$2K_{B}\left( \theta_{A} - \theta \right) = K_{B}\left( \theta - \theta_{B} \right)$
2$\theta_{A} - 2\theta = \theta - \theta_{B}$
2$\theta_{A} + \theta_{B} = 3\theta$
As $\theta_{A} - \theta_{B}\ \ = 48{^\circ};$ ….(i)
$\theta_{A} = 48 + \theta_{B}$
Put in Eq. (i)
2(48+$\ \theta_{B}) + \theta_{B} = 3\theta\ \ $
96+3$\ \theta_{B} = 3\theta$
96 = 3($\theta - \theta_{B})$
$\mathbf{\therefore\ }\theta - \theta_{B} = 96/3$ =32$℃$