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

Hint
We should know that viscous force is considered to be directly proportional to the rate at which the fluid velocity is changing in the space. It is defined as the measure of the fluid’s resistance to flow. Based on this concept we have to solve this concept.

Complete step by step answer
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
Let $ r $ is the radius of the sphere and $ {v_t} $ is its terminal speed. Then the weight of sphere is balanced by the buoyant force and viscous force such that:
Weight,
 $ w = mg $
 $ \because p = \dfrac{m}{V} $
 $ m = \dfrac{4}{3}\pi {r^3}\;pg.....(1) $
So,
 $ w = \dfrac{4}{3}\pi {r^3}p $
Buoyant force,
 $ {F_B} = \dfrac{4}{3}\pi {r^3}\;\sigma g.....(2) $
Where $ \sigma $ is density of water.
Viscous force, F = $ 6\pi \eta rvt......(3) $
Where, $ \eta \;is $ viscosity.
From equation (1) (2) and (3)
 $ w = {F_B} + {F_V} $
 $ \dfrac{4}{3}\pi {r^3}pg = \dfrac{4}{3}\pi {r^3}\sigma g + 6\pi \eta rvt $
 $ {V_t} = \dfrac{2}{9}\dfrac{{{r^2}(p - \sigma )g}}{\eta }........(4) $
The rate of production of heat when the sphere attains its terminal velocity is equal to work done by the viscous forces.
 $ W = \dfrac{{dQ}}{{dt}} = {F_V} \times {V_t} $
 $ W = 6\pi \eta r{v_t}^2 $
 $ W = 6\pi \eta r{\left( {\dfrac{2}{9}\dfrac{{\left( {p - \sigma } \right)g}}{\eta }} \right)^2} $
 $ \dfrac{{dQ}}{{dt}} \propto {r^5} $
Hence, the correct answer is Option (A).

Note
We should know when an object will float if the buoyancy force exerted on it by the fluid balances its weight. But from the Archimedes principle we get an idea that the buoyant force is the weight of the fluid displaced. So, in this case for a floating object on a liquid, the weight of the displaced liquid is the weight of the object.