Question

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:
A. 3 : 4
B. 4 : 3
C. 1 : 2
D. 2 : 1

Answer 1

Hint: In order to deal with this question first we will manipulate the formula of thermal conductivity and find the relation of the rate of heat flow in terms of change in temperature, thermal conductivity and length of the rods. Further we know that the rate of flow of heat for both the rods is the same so we will equate the two values and find the relation.
Formula used- $K = \dfrac{{Ql}}{{A\Delta T}}$

Complete Step-by-Step solution:
Given that:
Length of the rod 1 = 40 cm
Temperature difference at its two ends \[\Delta {T_1} = {80^0}C\]
Length of the rod 2 = 60 cm
Temperature difference at its two ends \[\Delta {T_2} = {90^0}C\]
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:
$K = \dfrac{{Ql}}{{A\Delta T}}$
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 $\Delta T$ is the difference in temperature
From the above formula we have:
$
   \Rightarrow Q = \dfrac{{KA\Delta T}}{l} \\
   \Rightarrow \dfrac{Q}{A} = \dfrac{{K\Delta T}}{l}.........(1) \\
 $
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 \dfrac{{{Q_1}}}{{{A_1}}} = \dfrac{{{Q_2}}}{{{A_2}}}$
Let us use the formula from equation (1)
$
  \because \dfrac{{{Q_1}}}{{{A_1}}} = \dfrac{{{Q_2}}}{{{A_2}}} \\
   \Rightarrow \dfrac{{{K_1}\Delta {T_1}}}{{{l_1}}} = \dfrac{{{K_2}\Delta {T_2}}}{{{l_2}}}........(2) \\
 $
Let us now substitute the values in equation (2)
\[
  \because \dfrac{{{K_1}\Delta {T_1}}}{{{l_1}}} = \dfrac{{{K_2}\Delta {T_2}}}{{{l_2}}} \\
   \Rightarrow \dfrac{{{K_1}}}{{{K_2}}} = \dfrac{{\Delta {T_2}{l_1}}}{{\Delta {T_1}{l_2}}} \\
   \Rightarrow \dfrac{{{K_1}}}{{{K_2}}} = \dfrac{{90 \times 40}}{{80 \times 60}} \\
   \Rightarrow \dfrac{{{K_1}}}{{{K_2}}} = \dfrac{3}{4} \\
 \]
Hence, the ratio of the thermal conductivities of the rods is 3 : 4
So, the correct answer is option A.

Note- Thermal conductivity is the rate at which heat is transmitted through a given material is proportional to the temperature gradient's negative value. And it's also proportional to the area the heat passes into but inversely proportional to the angle between the two isothermal planes.