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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 viscousmedium. (iii) Explain : Upward motion of bubbles produced in fluid. |
Answer» Solution :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. |
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