
There are two hollow spheres made of different materials, one with double the radius and one-fourth wall thickness of the other are filled with ice. If the time taken for melting the ice completely in a larger sphere is 25 minutes and that for a smaller sphere is 16 minutes, find the ratio of the thermal conductivity of the larger sphere to the smaller sphere.
A. 4:5
B. 5:4
C. 25:1
D. 1:25
Answer
164.4k+ 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. Here, using the formula for heat flow we are going to find the solution.
Formula Used:
To find the heat flow the formula is,
\[\dfrac{Q}{t} = KA\dfrac{{\Delta T}}{L}\]
Where,
A is a cross-sectional area.
\[\Delta T\] is the temperature difference between two ends of the metal.
L is the length of the metal plate.
K is the thermal conductivity.
Complete step by step solution:
The quantity of heat is nothing but how much the heat flow per unit of time in a given material and is given by,
\[Q = KA\dfrac{{\left( {\Delta T} \right)t}}{L}\]
\[\Rightarrow K = \dfrac{{QL}}{{A\left( {\Delta T} \right)t}}\]
Since Q and \[\Delta T\] are the same for both spheres.
Then,
\[K \propto \dfrac{L}{{At}}\]
\[\Rightarrow K \propto \dfrac{L}{{\pi {r^2}t}}\]
\[\Rightarrow K \propto \dfrac{L}{{{r^2}t}}\]
For larger spheres,
\[{K_{larger}} = \dfrac{{{L_l}}}{{{r_l}^2{t_l}}}\]
For smaller spheres,
\[{K_{smaller}} = \dfrac{{{L_S}}}{{{r_S}^2{t_S}}}\]
Here, the ratio of the thermal conductivity of the larger sphere to the smaller sphere is
\[\dfrac{{{K_{larger}}}}{{{K_{smaller}}}} = \dfrac{{\dfrac{{{L_l}}}{{{r_l}^2{t_l}}}}}{{\dfrac{{{L_S}}}{{{r_S}^2{t_S}}}}} \\ \]
\[\Rightarrow \dfrac{{{K_{larger}}}}{{{K_{smaller}}}} = \dfrac{{{L_l}}}{{{r_l}^2{t_l}}} \times \dfrac{{{r_S}^2{t_S}}}{{{L_S}}}\]
\[\dfrac{{{K_{larger}}}}{{{K_{smaller}}}} = \dfrac{{{L_l}}}{{{L_S}}} \times \dfrac{{{r_S}^2}}{{{r_l}^2}} \times \dfrac{{{t_S}}}{{{t_l}}}\]
Here, \[{r_l} = 2{r_S}\], \[{L_l} = \dfrac{1}{4}{L_S}\], \[{t_l} = 25\min \] and \[{t_S} = 16\min \]
Substitute the value of in the above equation, then,
\[\dfrac{{{K_{larger}}}}{{{K_{smaller}}}} = \left( {\dfrac{1}{4}} \right) \times {\left( {\dfrac{1}{2}} \right)^2} \times \dfrac{{16}}{{25}} \\ \]
\[\Rightarrow \dfrac{{{K_{larger}}}}{{{K_{smaller}}}} = \left( {\dfrac{1}{4}} \right) \times \left( {\dfrac{1}{4}} \right) \times \dfrac{{16}}{{25}} \\ \]
\[\Rightarrow \dfrac{{{K_{larger}}}}{{{K_{smaller}}}} = \dfrac{1}{{16}} \times \dfrac{{16}}{{25}} \\ \]
\[\Rightarrow \dfrac{{{K_{larger}}}}{{{K_{smaller}}}} = \dfrac{1}{{25}}\]
That is,
\[{K_{larger}}:{K_{smaller}} = 1:25\]
Therefore, the ratio of the thermal conductivity is 1:25.
Hence, option A is the correct answer.
Note: Do not get confused with the formula for thermal conductivity and the rate of heat transfer, these two are related to each other. The thermal conductivity of a material depends on its temperature, density of material etc.
Formula Used:
To find the heat flow the formula is,
\[\dfrac{Q}{t} = KA\dfrac{{\Delta T}}{L}\]
Where,
A is a cross-sectional area.
\[\Delta T\] is the temperature difference between two ends of the metal.
L is the length of the metal plate.
K is the thermal conductivity.
Complete step by step solution:
The quantity of heat is nothing but how much the heat flow per unit of time in a given material and is given by,
\[Q = KA\dfrac{{\left( {\Delta T} \right)t}}{L}\]
\[\Rightarrow K = \dfrac{{QL}}{{A\left( {\Delta T} \right)t}}\]
Since Q and \[\Delta T\] are the same for both spheres.
Then,
\[K \propto \dfrac{L}{{At}}\]
\[\Rightarrow K \propto \dfrac{L}{{\pi {r^2}t}}\]
\[\Rightarrow K \propto \dfrac{L}{{{r^2}t}}\]
For larger spheres,
\[{K_{larger}} = \dfrac{{{L_l}}}{{{r_l}^2{t_l}}}\]
For smaller spheres,
\[{K_{smaller}} = \dfrac{{{L_S}}}{{{r_S}^2{t_S}}}\]
Here, the ratio of the thermal conductivity of the larger sphere to the smaller sphere is
\[\dfrac{{{K_{larger}}}}{{{K_{smaller}}}} = \dfrac{{\dfrac{{{L_l}}}{{{r_l}^2{t_l}}}}}{{\dfrac{{{L_S}}}{{{r_S}^2{t_S}}}}} \\ \]
\[\Rightarrow \dfrac{{{K_{larger}}}}{{{K_{smaller}}}} = \dfrac{{{L_l}}}{{{r_l}^2{t_l}}} \times \dfrac{{{r_S}^2{t_S}}}{{{L_S}}}\]
\[\dfrac{{{K_{larger}}}}{{{K_{smaller}}}} = \dfrac{{{L_l}}}{{{L_S}}} \times \dfrac{{{r_S}^2}}{{{r_l}^2}} \times \dfrac{{{t_S}}}{{{t_l}}}\]
Here, \[{r_l} = 2{r_S}\], \[{L_l} = \dfrac{1}{4}{L_S}\], \[{t_l} = 25\min \] and \[{t_S} = 16\min \]
Substitute the value of in the above equation, then,
\[\dfrac{{{K_{larger}}}}{{{K_{smaller}}}} = \left( {\dfrac{1}{4}} \right) \times {\left( {\dfrac{1}{2}} \right)^2} \times \dfrac{{16}}{{25}} \\ \]
\[\Rightarrow \dfrac{{{K_{larger}}}}{{{K_{smaller}}}} = \left( {\dfrac{1}{4}} \right) \times \left( {\dfrac{1}{4}} \right) \times \dfrac{{16}}{{25}} \\ \]
\[\Rightarrow \dfrac{{{K_{larger}}}}{{{K_{smaller}}}} = \dfrac{1}{{16}} \times \dfrac{{16}}{{25}} \\ \]
\[\Rightarrow \dfrac{{{K_{larger}}}}{{{K_{smaller}}}} = \dfrac{1}{{25}}\]
That is,
\[{K_{larger}}:{K_{smaller}} = 1:25\]
Therefore, the ratio of the thermal conductivity is 1:25.
Hence, option A is the correct answer.
Note: Do not get confused with the formula for thermal conductivity and the rate of heat transfer, these two are related to each other. The thermal conductivity of a material depends on its temperature, density of material etc.
Recently Updated Pages
Environmental Chemistry Chapter for JEE Main Chemistry

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

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
