Question

The equivalent thermal conductivity of the compound bar is
$A.\;\dfrac{{{K_1}{K_2}}}{{{K_1} + {K_2}}}$
$B.\;\dfrac{{2{K_1}{K_2}}}{{{K_1} + {K_2}}}$
$C.\;\dfrac{{{K_1}}}{{{K_1} + {K_2}}}$
$D.\;\dfrac{{{K_2}}}{{{K_1} + {K_2}}}$

Answer 1

Hint: The thermal conductivity of a material is a measure of its ability to conduct heat. It is commonly denoted by, $K{\text{ or }}\lambda$. Heat transfer occurs at a lower rate in materials of low thermal conductivity than in materials of high thermal conductivity.

Complete step-by-step solution -
The equation for thermal conductivity is $q = - k\Delta T$, where q is heat flux, k is the thermal conductivity and $\Delta T$ is the thermal gradient. This is also called Fourier's Law for heat conduction. It is generally expressed as a scalar quantity.
The equation $q = - k\Delta T$ can be written as $q = - k\dfrac{{{T_2} - {T_1}}}{L}$ q is directly proportional to the temperature difference and inversely proportional to separation and where q is heat flux, k is the thermal conductivity, ${T_1}$ & ${T_2}$ are temperatures of two surfaces and $L$ is the thickness of the plane.
Let $K$ be the thermal conductivity of the compound bar therefore heat current through the compound bar where length of the bar is $2L$ , ${T_1}$ & ${T_2}$ are temperatures of two surfaces and $L$ is the thickness of the plane, $H$is the heat current, $A$ is the area of the bar, is given by
$H = \dfrac{{KA({T_1} - {T_2})}}{{2L}}$----(i)
We know at steady state, heat current $H$ is equal to heat current through first material is ${H_1}$ and heat current through second material is ${H_2}$.
Therefore, $H = {H_1} = {H_2}$
Therefore,
$\dfrac{{KA({T_1} - {T_2})}}{{2L}}$=$\dfrac{{{K_1}A({T_1} - {T_0})}}{L}$ ----(ii) where ${T_0}$ is the temperature at the joint of both the material or temperature at the junction,
Now we know that temperature at junction is equal to ${T_0} = \dfrac{{{K_1}{T_1} + {K_2}{T_2}}}{{{K_1} + {K_2}}}$
Substituting the value of ${T_0}$ in (ii), we get
$\dfrac{{KA({T_1} - {T_2})}}{{2L}} = \dfrac{{{K_1}A}}{L}\left[ {{T_1} - \left( {\dfrac{{{K_1}{T_1} + {K_2}{T_2}}}{{{K_1} + {K_2}}}} \right)} \right]$
$\to \dfrac{{K({T_1} - {T_2})}}{2} = {K_1}\left[ {{T_1} - \left( {\dfrac{{{K_1}{T_1} + {K_2}{T_2}}}{{{K_1} + {K_2}}}} \right)} \right]$, here $A$ & $L$ are present on both the side and so they will get cancelled.
$\to \dfrac{{K({T_1} - {T_2})}}{2} = {K_1}\left[ {\left( {\dfrac{{{T_1}({K_1} + {K_2}) - ({K_1}{T_1} + {K_2}{T_2})}}{{{K_1} + {K_2}}}} \right)} \right]$
$\to \dfrac{{K({T_1} - {T_2})}}{2} = {K_1}\left[ {\left( {\dfrac{{{K_1}{T_1} + {K_2}{T_1} - {K_1}{T_1} - {K_2}{T_2}}}{{{K_1} + {K_2}}}} \right)} \right]$
$\to \dfrac{{K({T_1} - {T_2})}}{2} = {K_1}\left[ {\left( {\dfrac{{{K_2}{T_1} - {K_2}{T_2}}}{{{K_1} + {K_2}}}} \right)} \right]$
$\to \dfrac{{K({T_1} - {T_2})}}{2} = {K_1}\left[ {\left( {\dfrac{{{K_2}({T_1} - {T_2})}}{{{K_1} + {K_2}}}} \right)} \right]$
$\to \dfrac{K}{2} = {K_1}\left[ {\left( {\dfrac{{{K_2}}}{{{K_1} + {K_2}}}} \right)} \right]$
$\to K = \dfrac{{2{K_1}{K_2}}}{{{K_1} + {K_2}}}$
Hence option B is the correct answer which states that equivalent thermal conductivity of the compound bar is $\dfrac{{2{K_1}{K_2}}}{{{K_1} + {K_2}}}$.

Note: We must understand properties of a compound bar for solving these types of questions. The compound bar consists of strips of dissimilar metals bonded together, and can be used to demonstrate unequal expansion in different metals on heating.