
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
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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.
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