State stokes'law .By using it deduce the expression for :
(i) Initial acceleration of smooth sphere and
(ii) Equation of terminal velocity of sphere falling freely through the viscous medium.
(iii) Explain : Upward motion of bubbles produced in fluid.

Explanation

Scientist Stokes. said that , viscous force `F_(V)` on small spherical solid body of radius r and moving with velocity v through a viscous medium of large dimensions having coefficient of viscosity `eta` is `6pietarv`.

As shown in figure a small spherical body of radius r, density `rho` falling in viscous medium of density `sigma`.
Following forces acted on it .
(i) Weight `F_(1)=mg=("volume"xx"density")g`
`=((4)/(3)pir^(3)rho))g`...(1) (In downward)
Where m = mass of spehre
(ii) Buoyant force `F_(2)=m_(o)g`
`=((4)/(3)pir^(3)sigma)g=`(2) (in upward)
where `m_(0)` = mass of liquid having volume of sphere
(iii) According to stokes.law
`F_(v)=6pietarv` ...(3)
Resultant force acting on the sphere,
`F_(R)=F_(1)-F_(2)-F_((v))`
`F_(R)=((4)/(3)pir^(3)rho)g-((4)/(3)pir^(3)sigma))g-6pietarv`
`F_(R)=(4)/(3)pir^(3)g(rho-sigma)-6pietarv` ...(4)
(i) Equation of initial acceleration
If mass of sphere is m and initial acceleration `a_(0)` then
`F_(R)=ma_(0)=(4)/(3)pi^(3)rhoa_(0)`
From equation (4),
`((4)/(3)pir^(3)rho)a_(0)=(4)/(3)pir^(3)g(rho-sigma)-6pietarv`
Taking initial velocity of freely falling sphere v=0
`((4)/(3)pir^(3)rho)a_(0)=(4)/(3)pir^(3)g(rho-sigma)`
`thereforerho(a_(0))=g(rho-sigma)`
`thereforea_(0)=((rho-sigma)g)/(rho)` ...(5)
(ii) Terminal velocity :
As the time passes velocity of sphere will increase hence in equation (3), the force `6pietarv` wil increase and resultant force `F_(R)` decreases.
For any one velocity of spehere `F_(R)` becomes zero and hence according to Newton first law , spehre moves with constant velocity . This constant velocity is known as final or terminal velocity `v_(t)` of sphere.
In equation (4) `v=v_(t)andF_(R)=0`,
`therefore(6pietar)v_(t)=(4)/(3)pir^(3)g(rho-sigma)`
`thereforev_(t)=(2)/(9)(r^(2)g(rho-sigma))/(eta)` ...(6)
Hence terminal velocity of small sphere in viscous medium `v_(t)`
(i) Proportional to the square radius of sphere `v_(t)propr^(2)`
(ii) Proportional to the difference of densities of sphere and fluid.
`v_(t)prop(rho-sigma)`
(iii) Inversely proportional to the coefficients of viscosity of fluid `v_(t)prop(1)/(eta)`.
(iv) Upward motion of bubbles produced in fluid :
The denisty `(rho)` of air bubble produced in the fluid is less then the density of fluid `(sigma)` .Hence `(rholtsigma)`, so terminal velocity is negative and hence bubble would travel in upward direction.