In a composite rod,two rods of different lengths and the same cross-sectional area are joined end to end. If $K$ is the equivalent coefficient of thermal conductivity of the composite rod,then $\left( \frac{\ell_1 + \ell_2}{K} \right)$ is equal to:

In a composite rod,when two rods of different lengths $l_1$ and $l_2$ and of the same cross-sectional area are joined end to end,if $K$ is the effective coefficient of thermal conductivity,then the value of $(l_1 + l_2)/K$ is

Two metal plates of equal thickness have thermal conductivities $K_1$ and $K_2$ respectively. They are joined together to form a single plate. The equivalent thermal conductivity of this composite plate is ..........

$A$ composite block is made of slabs $A, B, C, D$ and $E$ of different thermal conductivities (given in terms of a constant $K$) and sizes (given in terms of length,$L$) as shown in the figure. All slabs are of same width. Heat $Q$ flows only from left to right through the blocks. Then in steady state:
$(A)$ Heat flow through $A$ and $E$ slabs are same.
$(B)$ Heat flow through slab $E$ is maximum.
$(C)$ Temperature difference across slab $E$ is smallest.
$(D)$ Heat flow through $C =$ Heat flow through $B +$ Heat flow through $D$.

As per the given figure,two plates $A$ and $B$ of thermal conductivity $K$ and $2K$ are joined together to form a compound plate. The thickness of the plates are $4.0 \,cm$ and $2.5 \,cm$ respectively,and the area of cross-section is $120 \,cm^{2}$ for each plate. If the equivalent thermal conductivity of the compound plate is $\left(1+\frac{5}{\alpha}\right) K$,then the value of $\alpha$ will be . . . . . .

Two identical square rods of metal are welded end to end as shown in figure $(a)$. Assume that $10 \, cal$ of heat flows through the rods in $2 \, min$. Now the rods are welded as shown in figure $(b)$. The time it would take for $10 \, cal$ to flow through the rods now,is ........ $\min$.