Question

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 1

Viscous liquid creates viscous force when a body falls into liquid that opposes the motion of the body. The dissipation heat depends on viscous force and terminal velocity. Write formulas of viscous force and terminal velocity in terms of radius of sphere and find the value of r proportional to heat rate when body achieves terminal velocity.

Complete answer:
Let r be the radius of the sphere and $v_{t}$ be the terminal speed.
We have to find the rate of production of heat when the sphere attains its terminal velocity is equal to the work that is to be done by viscous force. It is given by-
$\dfrac{dQ}{dt} = F_{v} \times v_{t}$
Where, $v_{t}$ is the terminal velocity.
$F_{v}$ is the viscous force and it is given by-
$F_{v}=6\pi \eta r v_{t}$
Put this value of viscous force in the heat rate expression.
$\dfrac{dQ}{dt}= 6\pi \eta r v_{t}^{2}$
Terminal velocity is directly proportional to the square of radius.
$v_{t} \propto r^{2}$
So, write the heat rate in terms of proportional to r.
$\dfrac{dQ}{dt} \propto r \times r^{4}$
$\dfrac{dQ}{dt} \propto r^{5}$

Option (a) is correct.

Additional Information:
When the frictional force opposing the falling body in a viscous medium and the buoyant force collectively equal the sphere's weight leads to a terminal velocity as upward and the downward force becomes equal, the sphere will stop accelerating and maintain the final velocity it reaches.

Note:
Since no net force acts on the system, the body moves with a consistent velocity known as terminal velocity. When there is viscous force, it tends to oppose the motion and cause heat to be lost in the surroundings. The rate of heat loss depends directly on terminal velocity. As terminal velocity increases, the rate of dissipation of heat increases too.