
A hot metallic sphere of radius r radiates heat. Its rate of cooling is
A. independent of r
B. proportional to r
C. proportional to ${{r}^{2}}$
D. proportional to $\dfrac{1}{r}$
Answer
164.1k+ views
Hint: In this question, we are given a hot metallic sphere which is of radius r and we have to find its rate of cooling. For this we use the formula of rate of cooling and then by comparing the rate of cooling with radius, we find that the rate of cooling is inversely proportional to $\dfrac{1}{r}$.
Formula Used:
We use the formula of rate of cooling which is
${{R}_{c}}=\dfrac{d\theta }{dt}=\dfrac{A\varepsilon \sigma {{T}^{4}}}{mc}$
Where $\varepsilon $ is the emissivity of the body, $\sigma $ is the Stefan- Boltzmann constant, A is the surface of the body and T is the absolute temperature of the body.
Complete step by step solution:
We have given a hot metallic sphere of radius r. And we have to find the rate of cooling. We know the formula of rate of cooling is,
${{R}_{c}}=\dfrac{d\theta }{dt}=\dfrac{A\varepsilon \sigma ({{T}_{1}}^{4}-{{T}_{0}}^{4})}{mc}$
So from the above equation, we conclude that
$\dfrac{d\theta }{dt}\propto \dfrac{A}{V}\propto \dfrac{{{r}^{2}}}{{{r}^{3}}}$
Hence $\dfrac{d\theta }{dt}\propto \dfrac{{{r}^{2}}}{{{r}^{3}}}$
Simplifying further, we get
This means $\dfrac{d\theta }{dt}\propto \dfrac{1}{r}$
Thus from the above expression, we conclude that the rate of cooling is proportional to $\dfrac{1}{r}$.
Thus, option D is the correct answer.
Note: Radiation is a form of transfer of heat through electromagnetic radiation. This radiation carries heat energy in the form of photons from one place to another. A metallic body does the same when it gets heated. It does not require a medium of propagation.
Formula Used:
We use the formula of rate of cooling which is
${{R}_{c}}=\dfrac{d\theta }{dt}=\dfrac{A\varepsilon \sigma {{T}^{4}}}{mc}$
Where $\varepsilon $ is the emissivity of the body, $\sigma $ is the Stefan- Boltzmann constant, A is the surface of the body and T is the absolute temperature of the body.
Complete step by step solution:
We have given a hot metallic sphere of radius r. And we have to find the rate of cooling. We know the formula of rate of cooling is,
${{R}_{c}}=\dfrac{d\theta }{dt}=\dfrac{A\varepsilon \sigma ({{T}_{1}}^{4}-{{T}_{0}}^{4})}{mc}$
So from the above equation, we conclude that
$\dfrac{d\theta }{dt}\propto \dfrac{A}{V}\propto \dfrac{{{r}^{2}}}{{{r}^{3}}}$
Hence $\dfrac{d\theta }{dt}\propto \dfrac{{{r}^{2}}}{{{r}^{3}}}$
Simplifying further, we get
This means $\dfrac{d\theta }{dt}\propto \dfrac{1}{r}$
Thus from the above expression, we conclude that the rate of cooling is proportional to $\dfrac{1}{r}$.
Thus, option D is the correct answer.
Note: Radiation is a form of transfer of heat through electromagnetic radiation. This radiation carries heat energy in the form of photons from one place to another. A metallic body does the same when it gets heated. It does not require a medium of propagation.
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
