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Physics Question on Viscosity

A rain drop of radius 0.3 mm0.3\, mm has a terminal velocity in air is 1 m/s1\, m/s. The viscosity of air is 8×10−58 \times 10^{-5} poise. The viscous force on it is

A

45.2×10−445.2 \times 10^{-4} dyne

B

101.73×10−5101.73 \times 10^{-5} dyne

C

16.95×10−416.95 \times 10^{-4} dyne

D

16.95×10−516.95 \times 10^{-5} dyne

Answer

45.2×10−445.2 \times 10^{-4} dyne

Explanation

Solution

Radius of drop, r=0.3 mm=0.03 cmr = 0.3\, mm = 0.03\, cm
Terminal velocity, v=1 m/s=100 cm/secv = 1 \,m/s = 100\, cm/sec
Viscosity of air, η=8×10−5\eta=8\times10^{-5} poise
Viscous force, f=6 πηrvf = 6\,\pi\eta rv
F=6×3.14×(8×10−5)×0.03×100=4.42×10−3F=6\times3.14\times\left(8\times10^{-5}\right)\times0.03\times100=4.42\times10^{-3} dyne