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Physics Question on mechanical properties of fluid

A drop of water of radius 0.0015 mm0.0015 \,mm is falling in air. If the coefficient of viscosity of air is 2.0×10−5 kg/ms2.0 \times 10^{-5} \,kg / ms, the terminal velocity of the drop will be (The density of water =1.0×10 kg/m3=1.0 \times 10\, kg / m ^{3} and g=10 m/s2g=10\, m / s ^{2} )

A

1.0×10−4 m/s1.0 \times 10^{-4} \,m / s

B

2.0×10−4 m/s2.0 \times 10^{-4}\, m / s

C

2.5×10−4 m/s2.5 \times 10^{-4} \,m / s

D

5.0×10−4 m/s5.0 \times 10^{-4} \,m / s

Answer

2.5×10−4 m/s2.5 \times 10^{-4} \,m / s

Explanation

Solution

Radius r=0.0015mm=15×10−7 mr=0.0015 mm =15 \times 10^{-7} \,m Coefficient of viscosity =2.0×10−5 kg/ms=2.0 \times 10^{-5} \,kg / ms Density of water =1×103 kg m/m3=1 \times 10^{3} \,kg\,m / m ^{3} Terminal velocity v=29r2ρgηv=\frac{2}{9} \frac{r^{2} \rho g}{\eta} v=29×(15×10−7)2×1×103×102×10−5v=\frac{2}{9} \times \frac{\left(15 \times 10^{-7}\right)^{2} \times 1 \times 10^{3} \times 10}{2 \times 10^{-5}} v=2.5×10−4 m/sv=2.5 \times 10^{-4} \,m / s (\left(\right. The density of water =1.0×103 kg/m3=1.0 \times 10^{3}\, kg / m ^{3} and g=10 m/s2)\left.g =10 \,m / s ^{2}\right)