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Physics Question on mechanical properties of fluid

What will be the approximate terminal velocity of a rain drop of diameter 1.8?10−3 m 1.8 ? 10^{-3}\, m, when density of rain water ≈103 kgm−3\approx 10^3\, kgm^{-3} and the co-efficient of viscosity of air ≈1.8?10−5 Nsm−2\approx 1.8 ? 10^{-5}\, Nsm^{-2} ? (Neglect buoyancy of air).

A

49 ms−149\, ms^{-1}

B

98 ms−198\, ms^{-1}

C

392 ms−1392 \,ms^{-1}

D

980 ms−1980 \,ms^{-1}

Answer

98 ms−198\, ms^{-1}

Explanation

Solution

The correct option is(B): 98 ms −1.

Terminal velocity, v=29r2(p−σ)ηgv=\frac{2}{9} r^{2} \frac{(p-\sigma)}{\eta} g
Neglecting buoyancy effect of the fluid,
v=29⋅ρηr2gv=\frac{2}{9} \cdot \frac{\rho}{\eta} r^{2} g
Putting the given values, we get
v=29×103×(0.9×10−3)21.8×10−5×9.8=98 ms−1v=\frac{2}{9} \times \frac{10^{3} \times\left(0.9 \times 10^{-3}\right)^{2}}{1.8 \times 10^{-5}} \times 9.8=98 \,ms ^{-1}