A metallic sphere of radius `1.0 xx 10^(-3) m` and density `1.0 xx 10^(4) kg//m^(3)` enters 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 xx 10^(-3) N s//m^(2), g = 10ms^(-2)` and density of water `= 1.0 xx 10^(3) kg//m^(3)`.

Explanation

The velocity of attained by the sphere in falling from a height `h` is
`v = sqrt(2gh) ….(i)`
This is the terminal velocity of the sphere in water. Hence by Stoke's law, we have
`v = (2)/(9)(r^(2)(rho - sigma))/(eta)`
where `r` is the radius of the sphere, rho is the density of the material of the sphere
`sigma (= 1.0 xx 10^(3)kg//m^(3))` is the density of water and `eta` is coefficient of viscosity of water
`:. v = (2 xx (1.0 xx 10^(-3))^(2)(1.0 xx 10^(4) - 1.0 xx 10^(3)) xx 10)/(9 xx 1.0 xx 10^(-2)) = 20 m//s`
from equation `(i)`. we have
`h = (v^(2))/(2g) = (20 xx 20)/(2 xx 10) = 20 m`