A rod is of 40 cm in length and has a temperature difference of ${80}^{0}C$at its two ends. Another rod B is of length 60 cm and has a temperature difference of ${90}^{0}C$ , but has the same area of cross section. If the rate of flow of heat is the same, then the ratio of their thermal conductivities will be:

To start, we'll alter the thermal conductivity formula to determine the relationship between the rate of heat flow and temperature change, thermal conductivity, and rod length. We will equal the two numbers and discover the relationship since we also know that the rate of heat transfer for both rods is the same.

Length of the rod 1 = 40 cm
Temperature difference at its two end$$\mathrm{\Delta }{T}_1={80}^{0}C$$
Length of the rod 2 = 60 cm
Temperature difference at its two ends$$\mathrm{\Delta }{T}_2={90}^{0}C$$
We know that the Formula for Thermal Conductivity is given as
Every substance has its own capacity for conducting and transferring the heat. The thermal conductivity of a material is explained by the following formula:
$K={\displaystyle \frac{Ql}{A\mathrm{\Delta }T}}$
Where K is the thermal conductivity, Q is the amount of heat transferred through the material, l is the distance between the two isothermal planes, A is the area of the surface in square meters and$\mathrm{\Delta }T$ is the difference in temperature
From the above formula we have:

$$\begin{gathered} \Rightarrow Q=\frac{KA\mathrm{\Delta }T}{l} \\ \Rightarrow \frac{Q}{A}=\frac{K\mathrm{\Delta }T}{l}.........1 \end{gathered}$$
As in both the rods the rate of flow of heat is the same. Also the cross section area is the same. So, we have:
$\Rightarrow {\displaystyle \frac{Q_1}{A_1}}={\displaystyle \frac{Q_2}{A_2}}$
Let us use the formula from equation (1)

$$\begin{gathered} \because {\displaystyle \frac{Q_1}{A_1}}={\displaystyle \frac{Q_2}{A_2}} \\ \Rightarrow {\displaystyle \frac{K_1\mathrm{\Delta }{T}_1}{l_1}}={\displaystyle \frac{K_2\mathrm{\Delta }{T}_2}{l_2}}........(2) \end{gathered}$$
 Let us now substitute the values in equation (2)
∵$$\begin{gathered} \frac{K_1}{K_2}=\frac{ \Delta T_2}{\Delta T_1 } \\ \Rightarrow \frac{K_1}{K_2} = \frac{90\times 40 }{80\times 60 } \\ \Rightarrow K_1:K_2 = 3:4 \end{gathered}$$

So the answer the is 3:4.