Question

A small sphere falls from rest in a viscous liquid. due to friction, heat is produced. find the relation between the rate of production of heat and the radius of sphere at terminal velocity

Answer 1

The forces acting on the sphere are :
Weight = 4/3πr³рg downwards
Buoyant force = 4/3πr³бg upwards
Viscous force = 6πηrν upwards
The sphere attains the terminal velocity v₍t₎ when the resultant force acting on it is zero. i-e,
4/3πr³рg = 4/3πr³бg + 6πηrv₍t₎
Solving above equation gives us the terminal velocity,
v₍t₎ = 2r²(р-б)g / 9η
The rate of heat generation is equal to the rate of work done by the viscous force which in turn is equal to its power. Thus,
dQ/dt = (6πηrv₍t₎) v₍t₎ = 8π(р-б)²g²r⁵ / 27η

Answer 2

Given :r is radius of sphere

vt=terminal velocity.

When a sphere is in liquid, it experience following  forces:

its weight=W=mg=(4/3)πr³ρg [d=m/v]

Buoyancy force=4/3πr³σg

Viscous fore upwards =6πηrvt

where η is coefficient of viscosity of liquid.

Sphere attains its terminal velocity when forces acting on it are balanced:

(4/3)πr³ρg =4/3πr³σg+6πηrvt

Solving the above equation for terminal velcoity we get :

vt=2r²(ρ-σ)g/9η
Rate of heat generation is equal to rate of work done[power]] by viscous forces .

dQ/dt=(6πηrvt)vt=8π(ρ-σ)² g² r⁵/27η