$CASE-1$
$\frac{(60-50)}{4 \min }=k(55-t)$ $(1)$
Again, $CASE- 2$
$\frac{(40-30)}{8 \min }=k(35-t)$ $(2)$
$\frac{(1)}{(2)}=\frac{\frac{10}{4}}{\frac{10}{8}}=\frac{(55-t)}{(35-t)}$
$\Rightarrow 2(35-t)=55-t$
$\Rightarrow 70-2 t=55-t$
$\Rightarrow t=15^{\circ} \mathrm{C}$


In the experiment $I$ : a copper rod is used and all ice melts in $20$ minutes.
In the experiment $II$ : a steel rod of identical dimensions is used and all ice melts in $80$ minutes.
In the experiment $III$ : both the rods are used in series and all ice melts in $t_{10}$ minutes.
In the experiment $IV$ : both rods are used in parallel and all ice melts in $t_{20}$ minutes.