
Three rods of the same dimensions are arranged as shown in the figure. They have thermal conductivities \[{K_1},{K_2}\], and \[{K_3}\]. The points P and Q are maintained at different temperatures for the heat to flow at the same rate along PRQ and PQ. Which of the following options is correct?

A. \[{K_3} = \dfrac{1}{2}\left( {{K_1} + {K_2}} \right)\]
B. \[{K_3} = {K_1} + {K_2}\]
C. \[{K_3} = \dfrac{{{K_1}{K_2}}}{{{K_1} + {K_2}}}\]
D. \[{K_3} = 2\left( {{K_1} + {K_3}} \right)\]
Answer
164.1k+ views
Hint:Points P and Q are at different temperatures for maintaining the same rate of flow of heat along PQR and PQ. We have to find the relation connecting the thermal conductivities of three rods with the same dimension. Three rods will have thermal resistance. By connecting thermal resistance and thermal conductivity we can solve the question.
Formula used:
\[Thermal~resistance\; \propto \dfrac{1}{{Thermal~conductance}}\]
\[R = \dfrac{\lambda }{K}\]
Where R is thermal resistance, K is thermal conductivity and λ is the proportionality constant.
Complete answer:
We have three rods of the same dimension and are arranged as shown and they have different thermal conductivities. Two points P and Q are at different temperatures so that the rate of heat flowing through PRQ and PQ is the same. We have to find the relation between thermal conductivity of three rods in this situation.

We have a relation connecting thermal resistance and conductance.
That is, \[R = \dfrac{\lambda }{K}\]
From the figure we can say that resistance of PQ, \[{R_3} = \dfrac{\lambda }{{{K_3}}}\]
Similarly, resistance of PRQ, \[R = {R_1} + {R_2}\]
If we express it in terms of conductivity then the effective resistance will be:
\[R = \dfrac{\lambda }{{{K_1}}} + \dfrac{\lambda }{{{K_2}}}\]
Given that the rate of flow of heat is the same for both PRQ and PQ. It is only possible if resistance of PQ is equal to that of resistance of PRQ.
Mathematically,
\[{R_3} = {R_1} + {R_2}\]
In terms of conductivity, it can be written as:
\[\dfrac{\lambda }{{{K_3}}} = \dfrac{\lambda }{{{K_1}}} + \dfrac{\lambda }{{{K_2}}}\]
On simplifying we get,
\[\dfrac{1}{{{K_3}}} = \dfrac{{{K_2} + {K_1}}}{{{K_2}{K_1}}}\]
Therefore, thermal conductivity of PQ is, \[{K_3} = \dfrac{{{K_2}{K_1}}}{{{K_2} + {K_1}}}\]
Therefore, the answer is option(C)
Note:From the figure it is clear that PR and RQ are connected in series. So, the effective resistance will be the sum of both the resistance. Alternate method to find the solution is we can use the equation for rate of heat flow. Since heat flow through PRQ and PQ is the same we can equate it to find a solution.
Formula used:
\[Thermal~resistance\; \propto \dfrac{1}{{Thermal~conductance}}\]
\[R = \dfrac{\lambda }{K}\]
Where R is thermal resistance, K is thermal conductivity and λ is the proportionality constant.
Complete answer:
We have three rods of the same dimension and are arranged as shown and they have different thermal conductivities. Two points P and Q are at different temperatures so that the rate of heat flowing through PRQ and PQ is the same. We have to find the relation between thermal conductivity of three rods in this situation.

We have a relation connecting thermal resistance and conductance.
That is, \[R = \dfrac{\lambda }{K}\]
From the figure we can say that resistance of PQ, \[{R_3} = \dfrac{\lambda }{{{K_3}}}\]
Similarly, resistance of PRQ, \[R = {R_1} + {R_2}\]
If we express it in terms of conductivity then the effective resistance will be:
\[R = \dfrac{\lambda }{{{K_1}}} + \dfrac{\lambda }{{{K_2}}}\]
Given that the rate of flow of heat is the same for both PRQ and PQ. It is only possible if resistance of PQ is equal to that of resistance of PRQ.
Mathematically,
\[{R_3} = {R_1} + {R_2}\]
In terms of conductivity, it can be written as:
\[\dfrac{\lambda }{{{K_3}}} = \dfrac{\lambda }{{{K_1}}} + \dfrac{\lambda }{{{K_2}}}\]
On simplifying we get,
\[\dfrac{1}{{{K_3}}} = \dfrac{{{K_2} + {K_1}}}{{{K_2}{K_1}}}\]
Therefore, thermal conductivity of PQ is, \[{K_3} = \dfrac{{{K_2}{K_1}}}{{{K_2} + {K_1}}}\]
Therefore, the answer is option(C)
Note:From the figure it is clear that PR and RQ are connected in series. So, the effective resistance will be the sum of both the resistance. Alternate method to find the solution is we can use the equation for rate of heat flow. Since heat flow through PRQ and PQ is the same we can equate it to find a solution.
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