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Give some practical applications of Stoke's law.

Answer» <html><body><p></p>Solution :The viscous force F acting on a spherical body of <a href="https://interviewquestions.tuteehub.com/tag/radius-1176229" style="font-weight:bold;" target="_blank" title="Click to know more about RADIUS">RADIUS</a> r depends directly on: (i) radius (r) of the sphere <br/> (ii) velocity (v) of the sphere and <br/>(iii) coefficient of viscosity n of the liquid <br/>Therefore `F propeta^(x)r^(y)r^(y) = F k eta^(x)v^2 `where k is a dimensionless constant. . Using dimensions, the above equation can be written as<br/> `[MLT^(-2)]=K [ML^(-1)T^(-1) ] ^(z )xx [L]^(y) xx [LT^(-1) ]^(x)` <br/> On solving, we get x = 1, y = 1 and z = 1. Therefore, F = `k eta rv `<br/> Experimentally, Stoke found that the value of `k = 6 pi` <br/> ` f= 6 pietar v `<br/> This relation is <a href="https://interviewquestions.tuteehub.com/tag/known-534098" style="font-weight:bold;" target="_blank" title="Click to know more about KNOWN">KNOWN</a> as Stoke.s law.,<br/>Practical applications of Stoke.s law Since the raindrops are smaller in size and their terminal velocities are small, <a href="https://interviewquestions.tuteehub.com/tag/remain-1184279" style="font-weight:bold;" target="_blank" title="Click to know more about REMAIN">REMAIN</a> suspended in air in the form of clouds. As they grow up in size, their terminal velocities increase and they start falling in the form of rain. This law explains the following: <br/>(a) Floatation of clouds<br/> (b) Larger raindrops hurt us more than the smaller ones <br/> (c) A man <a href="https://interviewquestions.tuteehub.com/tag/coming-922990" style="font-weight:bold;" target="_blank" title="Click to know more about COMING">COMING</a> down with the <a href="https://interviewquestions.tuteehub.com/tag/help-1018089" style="font-weight:bold;" target="_blank" title="Click to know more about HELP">HELP</a> of a parachute acquires constant terminal velocity.</body></html>


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