
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 sphere attains terminal velocity is proportional to
$\begin{align}
& a){{r}^{5}} \\
& b){{r}^{3}} \\
& c){{r}^{4}} \\
& d){{r}^{2}} \\
\end{align}$
Answer
570.6k+ views
Hint: The power of the sphere when it attains terminal velocity is equal to the rate of production of heat. Terminal velocity can be written in terms of the radius of the sphere. Now, power is equal to the force times the terminal velocity. Force can be written in terms of radius of sphere using stokes formula. Now we can easily relate the rate of heat produced in terms of radius of the sphere.
Formulas used:
$\begin{align}
& P=F{{V}_{T}} \\
& F=6\pi \eta r{{V}_{T}} \\
& {{V}_{T}}=\dfrac{2{{r}^{2}}(\rho -\sigma )}{9\eta } \\
\end{align}$
Complete answer:
The power of the sphere can be equal to the rate of heat produced when it hits the viscous liquid.
$P=F{{V}_{T}}$
Also, from stokes theorem,
$F=6\pi \eta r{{V}_{T}}$,
Now, write the formula of force in the power equation, we get,
$P=6\pi \eta r{{V}_{T}}^{2}$.
Now, the terminal velocity according to stokes law is,
$\begin{align}
& {{V}_{T}}=\dfrac{2{{r}^{2}}(\rho -\sigma )}{9\eta } \\
& {{V}_{T}}\alpha {{r}^{2}} \\
& \\
\end{align}$
As power is also dependence on terminal velocity, we can write,
$P\alpha {{r}^{5}}$
Therefore, the correct option is option a.
Additional information:
When a sphere falls through a viscous liquid, the velocity of the sphere goes on increasing as it travels until it reaches a velocity called terminal velocity. At the terminal velocity, the frictional drag acted on the sphere due to the viscous force of liquid is balanced by the gravitational force and thus, the velocity will be constant after this. When the velocity increases, the viscosity on the sphere increases as the fluid displaces abruptly causing increase in internal resistance for the molecules to collide.
Note:
The viscous force acting on a larger object will be greater than the viscous force acting on the smaller sphere. The stokes law isn’t applicable if the liquid is not streamlined. The body must be rigid and should not change its shape and size due to viscous force.
Formulas used:
$\begin{align}
& P=F{{V}_{T}} \\
& F=6\pi \eta r{{V}_{T}} \\
& {{V}_{T}}=\dfrac{2{{r}^{2}}(\rho -\sigma )}{9\eta } \\
\end{align}$
Complete answer:
The power of the sphere can be equal to the rate of heat produced when it hits the viscous liquid.
$P=F{{V}_{T}}$
Also, from stokes theorem,
$F=6\pi \eta r{{V}_{T}}$,
Now, write the formula of force in the power equation, we get,
$P=6\pi \eta r{{V}_{T}}^{2}$.
Now, the terminal velocity according to stokes law is,
$\begin{align}
& {{V}_{T}}=\dfrac{2{{r}^{2}}(\rho -\sigma )}{9\eta } \\
& {{V}_{T}}\alpha {{r}^{2}} \\
& \\
\end{align}$
As power is also dependence on terminal velocity, we can write,
$P\alpha {{r}^{5}}$
Therefore, the correct option is option a.
Additional information:
When a sphere falls through a viscous liquid, the velocity of the sphere goes on increasing as it travels until it reaches a velocity called terminal velocity. At the terminal velocity, the frictional drag acted on the sphere due to the viscous force of liquid is balanced by the gravitational force and thus, the velocity will be constant after this. When the velocity increases, the viscosity on the sphere increases as the fluid displaces abruptly causing increase in internal resistance for the molecules to collide.
Note:
The viscous force acting on a larger object will be greater than the viscous force acting on the smaller sphere. The stokes law isn’t applicable if the liquid is not streamlined. The body must be rigid and should not change its shape and size due to viscous force.
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