Question

Two cylinders P and Q have the same length and diameter and are made of different materials having thermal conductivities in the ratio 2:3. These two cylinders are combined to make a cylinder. One end of P is kept at \[{100^0}C\] and another end of Q at \[{0^0}C\]. Find the temperature at the interface of P and Q.
A. \[{30^0}C\]
B. \[{40^0}C\]
C. \[{50^0}C\]
D. \[{60^0}C\]

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 temperature of the junction in contact the formula is,
\[\theta = \dfrac{{{K_1}{\theta _1} + {K_2}{\theta _2}}}{{{K_1} + {K_2}}}\]
Where, K is the thermal conductivity and \[{\theta _1},{\theta _2}\] is the temperature of two cylinders.

Complete step by step solution:
If two cylinders P and Q have the same length and diameter and are made of different materials having thermal conductivities in the ratio of 2:3. These two cylinders are combined to make a cylinder. One end of P is kept at \[{100^0}C\] and another end of Q at \[{0^0}C\]. We need to find the temperature at the interface of P and Q.

The temperature of the junction in contact with the formula is,
\[\theta = \dfrac{{{K_1}{\theta _1} + {K_2}{\theta _2}}}{{{K_1} + {K_2}}}\]
Here, \[{K_1} = 2K\], \[{K_2} = 3K\], . That is, \[\dfrac{{{K_1}}}{{{K_2}}} = \dfrac{2}{3}\]
And \[{\theta _1} = {100^0}C\] and \[{\theta _2} = {0^0}C\]
Substitute the value in the above equation we obtain,
\[\theta = \dfrac{{2K \times 100{K_1} + 3K \times 0}}{{2K + 3K}} \\ \]
\[\Rightarrow \theta = \dfrac{{200K}}{{5K}} \\ \]
\[\therefore \theta = {40^0}C\]
Therefore, the temperature at the interface of P and Q is \[{40^0}C\].

Hence, option B is the correct answer.

Note:Here in the given problem it is important to remember the equation for the temperature of the junction. Using the formula, we can easily find the solution.