First slide

Viscosity

Question

A metallic sphere of radius $1.0 \times 10^{- 3} m and density 1.0 \times 10^{4} kg / m^{3}$ enter a tank of water, after a free fall through a distance of h in the earth's gravitational field, if its velocity remains unchanged after entering water, determine the value of h. Given: coefficient of viscosity of water = $1.0 \times 10^{- 3} N - s / m^{2}, g = 10 m / s^{2} and density of water = 1.0 \times 10^{3} kg / m^{3} .$

Moderate

A

10 m
B

20 m
C

12 m
D

5 m

Solution

The velocity attained by the sphere in falling freely from a height h is
v = $\sqrt{2 gh} - - - - - (i)$
This is the terminal velocity of the sphere in water. Hence by Stoke's law, we have
v = $\frac{2}{9} \frac{r^{2} (ρ - σ) g}{η}$
where r is the radius of the sphere, $ρ$ is the density of the material of the sphere
$σ (= 1.0 \times 10^{3} kg / m^{3})$ is the density of water and $η$ is coefficient of viscosity of water.
v = $\frac{2 \times {(1.0 \times 10^{- 3})}^{2} (1.0 \times 10^{4} - 1.0 \times 10^{3}) \times 10}{9 \times 1.0 \times 10^{- 3}}$
= 20 m/s
from equation (i), we have
h = $\frac{v^{2}}{2 g} = \frac{20 \times 20}{2 \times 10} = 20 m$

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