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Physics Question on mechanical properties of fluid

A rain drop of radius 0.3 mm0.3 \,mm falls through air with a terminal velocity of 1 m s−11 \,m \,s^{-1} . The viscosity of air is 18×10−518 \times 10^{-5 } poise. The viscous force on the rain drop is

A

1.018×10−21.018 \times 10^{-2} dyne

B

2.018×10−22.018 \times 10^{-2} dyne

C

3.018×10−23.018 \times 10^{-2} dyne

D

4.018×10−24.018 \times 10^{-2} dyne

Answer

1.018×10−21.018 \times 10^{-2} dyne

Explanation

Solution

Here, r=0.3 mm=0.03 cmr = 0.3\, mm = 0.03 \,cm , v=1 m s−1=100 cm s−1v = 1 \,m\, s^{-1} = 100 \,cm \,s^{-1} , η=18×10−5\eta=18\times10^{-5} poise. According to stokes law, force of viscosity on rain drop is F=6πηrvF=6\pi\eta rv =6×3.142×18×10−5×0.03×100=6\times3.142\times18\times10^{-5}\times0.03\times100 dyne =1.018×10−2=1.018\times10^{-2} dyne