
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
162.3k+ views
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.
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.
Recently Updated Pages
A steel rail of length 5m and area of cross section class 11 physics JEE_Main

At which height is gravity zero class 11 physics JEE_Main

A nucleus of mass m + Delta m is at rest and decays class 11 physics JEE_MAIN

A wave is travelling along a string At an instant the class 11 physics JEE_Main

The length of a conductor is halved its conductivity class 11 physics JEE_Main

Two billiard balls of the same size and mass are in class 11 physics JEE_Main

Trending doubts
Differentiate between audible and inaudible sounds class 11 physics JEE_Main

JEE Main Eligibility Criteria 2025

NIT Delhi Cut-Off 2025 - Check Expected and Previous Year Cut-Offs

JEE Main Seat Allotment 2025: How to Check, Documents Required and Fees Structure

JEE Mains 2025 Cut-Off GFIT: Check All Rounds Cutoff Ranks

NIT Durgapur JEE Main Cut-Off 2025 - Check Expected & Previous Year Cut-Offs

Other Pages
JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

Uniform Acceleration

Degree of Dissociation and Its Formula With Solved Example for JEE

Free Radical Substitution Mechanism of Alkanes for JEE Main 2025

The resultant of vec A and vec B is perpendicular to class 11 physics JEE_Main

JEE Advanced 2025 Notes
