First slide

Viscosity

Question

A drop of water of radius 0.0015 mm is falling in air. If the coefficient of viscosity of air is $1.8 \times 10^{- 5} kg / (m - s),$ what will be the terminal velocity of the drop? (density of
water = $1.0 \times 10^{3} kg / m^{2} and g = 9.8 m / s^{2}$ ) Density of air can be neglected.

Moderate

A

$2.72 \times 10^{- 4} m / s$
B

$4.72 \times 10^{- 4} m / s$
C

$5.28 \times 10^{- 4} m / s$
D

$1.36 \times 10^{- 4} m / s$

Solution

By Stokes'law , the terminal velocity of a water drop of radius r is given by
$v = \frac{2}{9} \frac{r^{2} (ρ - σ) g}{η}$
where $ρ$ is the density of water, $σ$ is the density of air and $η$ the coefficient of viscosity of air. Here $σ$ is negligible and r = 0.0015 mm = 1.5 x $10^{- 3}$ mm: 1.5 x $10^{- 6}$ m. Substituting the values:
$v = \frac{2}{9} \times \frac{{(1.5 \times 10^{- 6})}^{2} \times (1.0 \times 10^{3}) \times 9.8}{1.8 \times 10^{- 5}}$
= 2.72 $\times$ $10^{- 4}$ m/s

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