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In Millikan's oil drop experiment , what is the terminal speed of an uncharged drop of radius 2.0xx10^(-5)mand density 1.2xx10^(3)kgm^(-3) . Take the viscosity of air at the temperature of the experiment to be 1.8xx10^(-5) Pa.s . Howmuch is the viscous force on the drop at thatspeed ?Neglect buoyancy of the drop due to air .

Answer» <html><body><p></p>Solution :Here radius of <a href="https://interviewquestions.tuteehub.com/tag/drop-443018" style="font-weight:bold;" target="_blank" title="Click to know more about DROP">DROP</a> `r=2.0xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a>)m` <br/>Density of oil`rho=1.2xx10^(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)kgm^(-3)`<br/>Density of air (dropis of air ) `rho_(o)=0`<br/>The <a href="https://interviewquestions.tuteehub.com/tag/coefficient-920926" style="font-weight:bold;" target="_blank" title="Click to know more about COEFFICIENT">COEFFICIENT</a> of viscosity of oil, <br/>`<a href="https://interviewquestions.tuteehub.com/tag/eta-446786" style="font-weight:bold;" target="_blank" title="Click to know more about ETA">ETA</a>=1.8xx10^(-5)` Pas<br/>Terminal velocityfrom Stock.s law ,<br/>`v_(t)=(2)/(9)(r^(2)g)/(eta)(rho-rho_(o))`<br/>`=(2r^(2)gQ)/(9eta)`<br/>`=(2xx4xx10^(-10)xx9.8xx1.2xx10^(3))/(9xx1.8xx10^(-5))`<br/>`=5.807xx10^(-2)`<br/>`v_(t)=5.8cms^(-1)`<br/>Viscous force on drop from Stoke.s law <br/>`F_(v)=6pietarv_(t)`<br/>`=6xx3.14xx1.8xx10^(-5)xx2xx10^(-5)xx5.8xx10^(-2)`<br/>`=393.34xx10^(-12)`<br/> `=3.93xx10^(-10)N`</body></html>


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