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Physics Question on mechanical properties of fluid

Neglecting the density of air, the terminal velocity obtained by a raindrop of radius 0.3 mm0.3\, mm falling through air of viscosity 1.8×10−5 N s m−21.8 \times 10^{-5}\, N \,s \,m ^{-2} will be

A

10.9 m s−110.9\, m\, s ^{-1}

B

7.48 ms−17.48\, m s ^{-1}

C

3.7 ms−13.7\, m s ^{-1}

D

12.8 ms−112.8\, m s ^{-1}

Answer

10.9 m s−110.9\, m\, s ^{-1}

Explanation

Solution

The terminal velocity of the spherical raindrop of radius rr is given by vt=2r2ρg9ηv_{t} = \frac{2r^{2}\rho g}{9\eta} where ρ\rho is the density of water and η\eta the viscosity of air. Substituting r=0.3 mmr = 0.3\, mm =0.3�10−3 m,ρ=103 kg/m3,g=9.8 m/s2= 0.3 � 10^{-3} \,m, \rho = 10^{3} \,kg/m^{3}, g = 9.8 \,m/s^{2} and η=1.8�10−5 Ns/m2\eta = 1.8 � 10^{-5} \,N s/m^{2}, we get vt=2×(0.3)2×10−3×9.89×1.8×10−5v_{t} = \frac{2\times\left(0.3\right)^{2}\times10^{-3}\times9.8}{9\times1.8\times10^{-5}} =10.9 m s−1= 10.9 \,m\,s^{-1}