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Physics Question on mechanical properties of fluid

A rain drop of radius 0.3 mm0.3\, mm has a terminal velocity of 1 m/s1\, m / s and the viscosity of 1 m/s1\, m / s and the viscosity of air is 18×10−518 \times 10^{-5} poise. The viscous force on the drop is:

A

16.95×10−9 N16.95\times 10^{-9}\,N

B

1.695×10−9 N1.695\times 10^{-9}\,N

C

10.17×10−9 N10.17\times 10^{-9}\,N

D

101.17×10−9 N101.17\times 10^{-9}\,N

Answer

101.17×10−9 N101.17\times 10^{-9}\,N

Explanation

Solution

For a drop of radius rr and terminal velocity vv,
the viscous force is given by F=6πηrvF=6 \pi \eta r v
where η\eta is coefficient of viscosity.
Putting the numerical values from the question, we have
η=18×10−5\eta=18 \times 10^{-5} poise =18×10−6kg/m−s=18 \times 10^{-6} kg / m - s
r=0.3mm=0.3×10−3m,v=1m/sr=0.3 mm =0.3 \times 10^{-3} m , v=1 m / s
∴F=6×3.14×18×10−6×0.3×10−3×1\therefore F=6 \times 3.14 \times 18 \times 10^{-6} \times 0.3 \times 10^{-3} \times 1
=101.74×10−9N=101.74 \times 10^{-9} N