Combination are in series, so heat current is same for all Rods
\(\left(\frac{\Delta Q }{\Delta t }\right)_{ AB }=\left(\frac{\Delta Q }{\Delta t }\right)_{ BC }=\left(\frac{\Delta Q }{\Delta t }\right)_{ CD }=\) Heat current
\(\frac{(100-70) K _{1} A }{\ell}=\frac{(70-20) K _{2} A }{\ell}=\frac{(20-0) K _{3} A }{\ell}\)
\(30 K _{1}=50 K _{2}=20 K _{3}\)
\(3 K _{1}=2 K _{3}\)
\(\frac{K_{1}}{K_{3}}=\frac{2}{3}=2: 3\)
\(5 K _{2}=2 K _{3}\)
\(\frac{ K _{2}}{ K _{3}}=\frac{2}{5}=2: 5\)
(આપેલ : બરફની ગલનગુપ્ત ઉષ્મા $=3.36 \times 10^5\; J kg ^{-1}$)