Question

Wires A and B have identical lengths and have circular cross-sections. The radius of A is twice the radius of B i.e., \[{r_A} = 2{r_B}\]. For a given temperature difference between the two ends, both wires conduct heat at the same rate. Find the relation between the thermal conductivities.
A. \[{K_A} = 4{K_B}\]
B. \[{K_A} = 2{K_B}\]
C. \[{K_A} = \dfrac{{{K_B}}}{2}\]
D. \[{K_A} = \dfrac{{{K_B}}}{4}\]

Answer 1

Hint: In order to solve this problem we need to understand the thermal conductivity. The rate at which heat is transferred by conduction through a unit cross-section area of a material is known as thermal conductivity.

Formula Used:
To find the rate of flow of heat the formula is,
\[\dfrac{Q}{t} = KA\dfrac{{\Delta T}}{L}\]
Where, A is cross-sectional area, \[\Delta T\] is the temperature difference between two ends, L is the length of the cylinder and K is the thermal conductivity.

Complete step by step solution:
We have two wires A and B having identical lengths and circular cross-sections. The radius of A is twice the radius of B i.e., \[{r_A} = 2{r_B}\]. For a given temperature difference both wires conduct heat at the same rate. We need to find the relation between the thermal conductivities. To find the rate of flow of heat the formula is,
\[\dfrac{Q}{t} = KA\dfrac{{\Delta T}}{L}\]
From this, the value of K can be written as,
\[K = \dfrac{{QL}}{{t\Delta TA}}\]

From this, we can say that the relation between thermal conductivity and the area is,
\[K = \left( {\dfrac{{QL}}{{t\Delta T}}} \right)\dfrac{1}{A}\]
\[\Rightarrow K \propto \dfrac{1}{A}\]
Since we have two wires A and B then,
\[{K_A} \propto \dfrac{1}{{{A_1}}}\]and \[{K_B} \propto \dfrac{1}{{{A_2}}}\]
\[\Rightarrow \dfrac{{{K_A}}}{{{K_B}}} = \dfrac{{{A_2}}}{{{A_1}}}\]
Here, area \[A = \pi {r^2}\] then,
\[\dfrac{{{K_A}}}{{{K_B}}} = \dfrac{{\pi {r_B}^2}}{{\pi {r_A}^2}}\]

From data, \[{r_A} = 2{r_B}\]
Substitute the value in the above equation we get,
\[\dfrac{{{K_A}}}{{{K_B}}} = \dfrac{{{r_B}^2}}{{{{\left( {2{r_B}} \right)}^2}}}\]
\[\Rightarrow \dfrac{{{K_A}}}{{{K_B}}} = \dfrac{1}{4}\]
\[\therefore {K_A} = \dfrac{{{K_B}}}{4}\]
Therefore, the relation between the thermal conductivities is \[{K_A} = \dfrac{{{K_B}}}{4}\].

Hence, option D is the correct answer.

Note:The materials which have a high thermal conductivity are used in heat sinks whereas materials with low values are used as thermal insulators. In order to measure the thermal conductivities of materials there exist several methods, which are broadly classified into two types of techniques one is transient and other is steady-state techniques.