
The lengths and radii of two rods made of the same material are in the ratios 1:2 and 2:3 respectively; If the temperature difference between the ends for the two rods is the same, then in the steady state, find the ratio of the amount of heat flowing per second through them.
A. 1:3
B. 4:3
C. 8:9
D. 3:2
Answer
163.8k+ views
Hint:In order to solve this problem we need to understand the rate of heat transfer. The rate of flow of heat is the amount of heat that is transferred per unit of time temperature. A temperature gradient is defined as the change in temperature over a specified distance between two isothermal surfaces or two points in a material.
Formula Used:
To find the heat flow the formula is,
\[\dfrac{Q}{t} = KA\dfrac{{\Delta T}}{L}\]
Where, A is cross-sectional area of metal rod, \[\Delta T\] is temperature difference between two ends of the metal rod, L is length of the metal rod and K is thermal conductivity.
Complete step by step solution:
If the lengths and radii of two rods made of the same material are in the ratios 1:2 and 2:3 and the temperature difference between the ends for the two rods is the same, then in the steady state, we need to find the ratio of the amount of heat flowing per second through them. The quantity of heat is nothing but how much the heat flow per unit of time in a given material and is given by,
\[\dfrac{Q}{t} = KA\dfrac{{\Delta T}}{L}\]
Given that, \[\dfrac{{{L_1}}}{{{L_2}}} = \dfrac{1}{2}\] and \[\dfrac{{{r_1}}}{{{r_2}}} = \dfrac{2}{3}\]
The rate of flow of heat of shorter rod and the longer rod is,
\[{\left( {\dfrac{Q}{t}} \right)_1} = \dfrac{{K\left( {\pi {r_1}^2} \right)\Delta T}}{{{L_1}}}\] and
\[{\left( {\dfrac{Q}{t}} \right)_2} = \dfrac{{K\left( {\pi {r_2}^2} \right)\Delta T}}{{{L_2}}}\]
Then, the ratio of the amount of heat flowing per second through them is,
\[\dfrac{{{{\left( {\dfrac{Q}{t}} \right)}_1}}}{{{{\left( {\dfrac{Q}{t}} \right)}_2}}} = \dfrac{{\dfrac{{K\left( {\pi {r_1}^2} \right)\Delta T}}{{{L_1}}}}}{{\dfrac{{K\left( {\pi {r_2}^2} \right)\Delta T}}{{{L_2}}}}}\]
Since the temperature difference between the ends for the two rods is the same then,
\[\dfrac{{{{\left( {\dfrac{Q}{t}} \right)}_1}}}{{{{\left( {\dfrac{Q}{t}} \right)}_2}}} = \dfrac{{\dfrac{{\left( {{r_1}^2} \right)}}{{{L_1}}}}}{{\dfrac{{\left( {{r_2}^2} \right)}}{{{L_2}}}}}\]
\[\Rightarrow \dfrac{{{{\left( {\dfrac{Q}{t}} \right)}_1}}}{{{{\left( {\dfrac{Q}{t}} \right)}_2}}} = \dfrac{{\left( {{r_1}^2} \right)}}{{{L_1}}} \times \dfrac{{{L_2}}}{{\left( {{r_2}^2} \right)}} \\ \]
\[\Rightarrow \dfrac{{{{\left( {\dfrac{Q}{t}} \right)}_1}}}{{{{\left( {\dfrac{Q}{t}} \right)}_2}}} = \dfrac{{{r_1}^2}}{{{r_2}^2}} \times \dfrac{{{L_2}}}{{{L_1}}}\]
Substitute the value of \[\dfrac{{{L_1}}}{{{L_2}}} = \dfrac{1}{2}\]and \[\dfrac{{{r_1}}}{{{r_2}}} = \dfrac{2}{3}\] in above equation, then,
\[\Rightarrow \dfrac{{{{\left( {\dfrac{Q}{t}} \right)}_1}}}{{{{\left( {\dfrac{Q}{t}} \right)}_2}}} = {\left( {\dfrac{2}{3}} \right)^2} \times \dfrac{2}{1} \\ \]
\[\Rightarrow \dfrac{{{{\left( {\dfrac{Q}{t}} \right)}_1}}}{{{{\left( {\dfrac{Q}{t}} \right)}_2}}} = \dfrac{4}{9} \times \dfrac{2}{1} \\ \]
\[\therefore \dfrac{{{{\left( {\dfrac{Q}{t}} \right)}_1}}}{{{{\left( {\dfrac{Q}{t}} \right)}_2}}} = \dfrac{8}{9}\]
Therefore, the ratio of the amount of heat flowing per second through them is 8:9.
Hence, option C is the correct answer.
Note:Here in the given problem it is important to remember the equation for the heat flow in the rod. Using the formula, we can easily find the solution. Moreover, thermal conductivity refers to a material's capacity to conduct/transfer heat. It is commonly represented by the sign 'k’. Thermal resistivity is the reciprocal of this parameter.
Formula Used:
To find the heat flow the formula is,
\[\dfrac{Q}{t} = KA\dfrac{{\Delta T}}{L}\]
Where, A is cross-sectional area of metal rod, \[\Delta T\] is temperature difference between two ends of the metal rod, L is length of the metal rod and K is thermal conductivity.
Complete step by step solution:
If the lengths and radii of two rods made of the same material are in the ratios 1:2 and 2:3 and the temperature difference between the ends for the two rods is the same, then in the steady state, we need to find the ratio of the amount of heat flowing per second through them. The quantity of heat is nothing but how much the heat flow per unit of time in a given material and is given by,
\[\dfrac{Q}{t} = KA\dfrac{{\Delta T}}{L}\]
Given that, \[\dfrac{{{L_1}}}{{{L_2}}} = \dfrac{1}{2}\] and \[\dfrac{{{r_1}}}{{{r_2}}} = \dfrac{2}{3}\]
The rate of flow of heat of shorter rod and the longer rod is,
\[{\left( {\dfrac{Q}{t}} \right)_1} = \dfrac{{K\left( {\pi {r_1}^2} \right)\Delta T}}{{{L_1}}}\] and
\[{\left( {\dfrac{Q}{t}} \right)_2} = \dfrac{{K\left( {\pi {r_2}^2} \right)\Delta T}}{{{L_2}}}\]
Then, the ratio of the amount of heat flowing per second through them is,
\[\dfrac{{{{\left( {\dfrac{Q}{t}} \right)}_1}}}{{{{\left( {\dfrac{Q}{t}} \right)}_2}}} = \dfrac{{\dfrac{{K\left( {\pi {r_1}^2} \right)\Delta T}}{{{L_1}}}}}{{\dfrac{{K\left( {\pi {r_2}^2} \right)\Delta T}}{{{L_2}}}}}\]
Since the temperature difference between the ends for the two rods is the same then,
\[\dfrac{{{{\left( {\dfrac{Q}{t}} \right)}_1}}}{{{{\left( {\dfrac{Q}{t}} \right)}_2}}} = \dfrac{{\dfrac{{\left( {{r_1}^2} \right)}}{{{L_1}}}}}{{\dfrac{{\left( {{r_2}^2} \right)}}{{{L_2}}}}}\]
\[\Rightarrow \dfrac{{{{\left( {\dfrac{Q}{t}} \right)}_1}}}{{{{\left( {\dfrac{Q}{t}} \right)}_2}}} = \dfrac{{\left( {{r_1}^2} \right)}}{{{L_1}}} \times \dfrac{{{L_2}}}{{\left( {{r_2}^2} \right)}} \\ \]
\[\Rightarrow \dfrac{{{{\left( {\dfrac{Q}{t}} \right)}_1}}}{{{{\left( {\dfrac{Q}{t}} \right)}_2}}} = \dfrac{{{r_1}^2}}{{{r_2}^2}} \times \dfrac{{{L_2}}}{{{L_1}}}\]
Substitute the value of \[\dfrac{{{L_1}}}{{{L_2}}} = \dfrac{1}{2}\]and \[\dfrac{{{r_1}}}{{{r_2}}} = \dfrac{2}{3}\] in above equation, then,
\[\Rightarrow \dfrac{{{{\left( {\dfrac{Q}{t}} \right)}_1}}}{{{{\left( {\dfrac{Q}{t}} \right)}_2}}} = {\left( {\dfrac{2}{3}} \right)^2} \times \dfrac{2}{1} \\ \]
\[\Rightarrow \dfrac{{{{\left( {\dfrac{Q}{t}} \right)}_1}}}{{{{\left( {\dfrac{Q}{t}} \right)}_2}}} = \dfrac{4}{9} \times \dfrac{2}{1} \\ \]
\[\therefore \dfrac{{{{\left( {\dfrac{Q}{t}} \right)}_1}}}{{{{\left( {\dfrac{Q}{t}} \right)}_2}}} = \dfrac{8}{9}\]
Therefore, the ratio of the amount of heat flowing per second through them is 8:9.
Hence, option C is the correct answer.
Note:Here in the given problem it is important to remember the equation for the heat flow in the rod. Using the formula, we can easily find the solution. Moreover, thermal conductivity refers to a material's capacity to conduct/transfer heat. It is commonly represented by the sign 'k’. Thermal resistivity is the reciprocal of this parameter.
Recently Updated Pages
Uniform Acceleration - Definition, Equation, Examples, and FAQs

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

Important Questions for CBSE Class 11 Physics Chapter 1 - Units and Measurement
