
A small sphere of radius $r$ , falls from rest in a viscous liquid. As a result, heat is produced due to viscous force. The rate of production of heat when the sphere attains its terminal velocity, is proportional to:
A) ${r^5}$
B) ${r^3}$
C) ${r^4}$
D) ${r^2}$
Answer
170.7k+ views
Hint: Use the formula of the work done since it is equal to the rate of heat production and substitute the formula of the viscous force and the terminal velocity in it to find the relation between the radius of the object and the rate of the heat production.
Formula used:
(1) The formula of the viscous force is given by
$F = 6\pi \eta r{v_t}$
Where $F$ is the viscous force of the liquid, $\eta $ is the dynamic viscosity coefficient and ${v_t}$ is the terminal velocity of the sphere that falls on the viscous liquid.
(2) The formula of the terminal velocity is given by
${v_t} = \dfrac{2}{9}{r^2}\left( {\rho - \sigma } \right)\dfrac{g}{\eta }$
Where $r$ is the radius of the sphere, $\rho $ density of the sphere that falls on the viscous liquid, $\sigma $ is the density of the viscous liquid and $g$ is the acceleration due to gravity.
(3) The formula of the work done is given by
$W = F{v_t}$
Where $W$ is the work done.
Complete step by step solution:
It is given that the
Radius of the small sphere is $r$
Let the terminal velocity of the small sphere is considered as ${v_t}$ .
The rate of the heat production is equal to the work done. Let’s use the formula of the power,
$W = F{v_t}$
Substituting the formula of the viscous force in the above equation, we get
$\Rightarrow W = \dfrac{{dQ}}{{dt}} = 6\pi \eta r{v_t}^2$
Substituting the formula of the terminal velocity in the above equation, we get
$\Rightarrow q = 6\pi \eta r{\left( {\dfrac{2}{9}{r^2}\left( {\rho - \sigma } \right)g\eta } \right)^2}$
By simplifying the above equation, we get
$\Rightarrow \dfrac{{dQ}}{{dt}} \propto {r^5}$
Thus the option (A) is correct.
Note: Viscous force opposes the flow of the liquid. The buoyancy force is the force that is exerted by the liquid on the object that is immersed on it. It provides the tendency to float. Remember the formula of the viscous force, terminal velocity and the work done.
Formula used:
(1) The formula of the viscous force is given by
$F = 6\pi \eta r{v_t}$
Where $F$ is the viscous force of the liquid, $\eta $ is the dynamic viscosity coefficient and ${v_t}$ is the terminal velocity of the sphere that falls on the viscous liquid.
(2) The formula of the terminal velocity is given by
${v_t} = \dfrac{2}{9}{r^2}\left( {\rho - \sigma } \right)\dfrac{g}{\eta }$
Where $r$ is the radius of the sphere, $\rho $ density of the sphere that falls on the viscous liquid, $\sigma $ is the density of the viscous liquid and $g$ is the acceleration due to gravity.
(3) The formula of the work done is given by
$W = F{v_t}$
Where $W$ is the work done.
Complete step by step solution:
It is given that the
Radius of the small sphere is $r$
Let the terminal velocity of the small sphere is considered as ${v_t}$ .
The rate of the heat production is equal to the work done. Let’s use the formula of the power,
$W = F{v_t}$
Substituting the formula of the viscous force in the above equation, we get
$\Rightarrow W = \dfrac{{dQ}}{{dt}} = 6\pi \eta r{v_t}^2$
Substituting the formula of the terminal velocity in the above equation, we get
$\Rightarrow q = 6\pi \eta r{\left( {\dfrac{2}{9}{r^2}\left( {\rho - \sigma } \right)g\eta } \right)^2}$
By simplifying the above equation, we get
$\Rightarrow \dfrac{{dQ}}{{dt}} \propto {r^5}$
Thus the option (A) is correct.
Note: Viscous force opposes the flow of the liquid. The buoyancy force is the force that is exerted by the liquid on the object that is immersed on it. It provides the tendency to float. Remember the formula of the viscous force, terminal velocity and the work done.
Recently Updated Pages
Molarity vs Molality: Definitions, Formulas & Key Differences

Preparation of Hydrogen Gas: Methods & Uses Explained

Polymers in Chemistry: Definition, Types, Examples & Uses

P Block Elements: Definition, Groups, Trends & Properties for JEE/NEET

Order of Reaction in Chemistry: Definition, Formula & Examples

Hydrocarbons: Types, Formula, Structure & Examples Explained

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

Displacement-Time Graph and Velocity-Time Graph for JEE

Uniform Acceleration

JEE Main 2025: Derivation of Equation of Trajectory in Physics

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

Instantaneous Velocity - Formula based Examples for JEE

Other Pages
NCERT Solution for Class 11 Physics Chapter 1 Units and Measurements - 2025-26

NCERT Solution for Class 11 Physics Chapter 2 Motion In A Straight Line - 2025-26

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

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

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

NCERT Solution for Class 11 Physics Chapter 3 Motion In A Plane - 2025-26
