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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» <html><body><p></p>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`.<br/><img src="https://doubtnut-static.s.llnwi.net/static/physics_images/KPK_AIO_PHY_XI_P2_C10_E01_052_S01.png" width="80%"/><br/>As shown in <a href="https://interviewquestions.tuteehub.com/tag/figure-987693" style="font-weight:bold;" target="_blank" title="Click to know more about FIGURE">FIGURE</a> a small spherical body of radius r, density `rho` falling in viscous medium of density `sigma`. <br/> Following forces acted on it .<br/> (i) Weight `F_(1)=mg=("volume"<a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a>"density")g`<br/>`=((4)/(3)pir^(3)rho))g`...(1) (In downward)<br/>Where m = mass of spehre <br/> (ii) Buoyant force `F_(2)=m_(o)g`<br/>`=((4)/(3)pir^(3)sigma)g=`(2) (in upward)<br/> where `m_(0)` = mass of liquid having volume of sphere <br/> (iii) According to stokes.law<br/>`F_(v)=6pietarv` ...(3)<br/><a href="https://interviewquestions.tuteehub.com/tag/resultant-1187362" style="font-weight:bold;" target="_blank" title="Click to know more about RESULTANT">RESULTANT</a> force acting on the sphere,<br/>`F_(R)=F_(1)-F_(2)-F_((v))`<br/>`F_(R)=((4)/(3)pir^(3)rho)g-((4)/(3)pir^(3)sigma))g-6pietarv`<br/>`F_(R)=(4)/(3)pir^(3)g(rho-sigma)-6pietarv` ...(4)<br/> (i)Equation of initial acceleration<br/>If mass of sphere is m and initial acceleration `a_(0)` then<br/> `F_(R)=ma_(0)=(4)/(3)pi^(3)rhoa_(0)`<br/>From equation (4),<br/>`((4)/(3)pir^(3)rho)a_(0)=(4)/(3)pir^(3)g(rho-sigma)-6pietarv`<br/>Taking initial velocity of freely falling sphere v=0<br/>`((4)/(3)pir^(3)rho)a_(0)=(4)/(3)pir^(3)g(rho-sigma)`<br/>`thereforerho(a_(0))=g(rho-sigma)`<br/>`thereforea_(0)=((rho-sigma)g)/(rho)` ...(5)<br/> (ii)Terminal velocity :<br/>As the time passes velocity of sphere will increase hence in equation (3), the force `6pietarv` wil increase and resultant force `F_(R)` decreases. <br/>For any one velocity of spehere `F_(R)`becomes zero and hence according to <a href="https://interviewquestions.tuteehub.com/tag/newton-11212" style="font-weight:bold;" target="_blank" title="Click to know more about NEWTON">NEWTON</a> first law , spehre moves with constant velocity . This constant velocity is known as final or terminal velocity `v_(t)` of sphere.<br/>In equation (4) `v=v_(t)andF_(R)=0`,<br/> `therefore(6pietar)v_(t)=(4)/(3)pir^(3)g(rho-sigma)`<br/>`thereforev_(t)=(2)/(9)(r^(2)g(rho-sigma))/(eta)` ...(6)<br/>Hence terminal velocity of small sphere in viscous medium `v_(t)`<br/> (i) Proportional to the square radius of sphere `v_(t)<a href="https://interviewquestions.tuteehub.com/tag/propr-3782068" style="font-weight:bold;" target="_blank" title="Click to know more about PROPR">PROPR</a>^(2)`<br/>(ii) Proportional to the difference of densities of sphere and fluid.<br/> `v_(t)prop(rho-sigma)`<br/>(iii)Inversely proportional to the coefficients of viscosity of fluid `v_(t)prop(1)/(eta)`.<br/> (iv)Upward motion of bubbles produced in fluid :<br/>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.</body></html>


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