Fluid Mechanics

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178 of 192

Question 178 of 192

A spherical ball of radius 1mm1mm and density 10.5gcc−110.5g{cc}^{-1} is dropped in glycerine of coefficient of viscosity 9.89.8 poise and density 1.5gcc−11.5g{cc}^{-1}. Viscous force on the ball when it attains constant velocity is 3696×10−xN3696\times {10}^{-x}N. The value of xx is
(Given, g=9.8ms−2andπ=227g=9.8m{s}^{-2}\text{and}\pi =\dfrac{22}{7})

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Answer: 7

Viscous force on the ball when it attains constant velocity i.e., terminal velocity is,
Fv=(mg−FB)(As,a=0){F}_{v}=\left(mg-{F}_{B}\right)\left(\text{As},a=0\right)
Here, FB{F}_{B} is buoyant force.
⇒Fv=Vρbg−Vρgg=Vg(ρb−ρg)=43π(10−3)3×9.8[(10.5−1.5)×103]≈3696×10−7N\Rightarrow {F}_{v}=V{\rho }_{b}g-V{\rho }_{g}g =Vg\left({\rho }_{b}-{\rho }_{g}\right) =\dfrac{4}{3}\pi {\left({10}^{-3}\right)}^{3}\times 9.8\left[\left(10.5-1.5\right)\times {10}^{3}\right] \approx 3696\times {10}^{-7}N
Thus, x=7x=7.

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