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)`.

Answer

`20m`

Explanation

The velocity attained by the sphere in falling freely from a height `h` is
`v=sqrt(2gh) ……(i)`
This is the terminal velocity of the pshere in water. Hence by Stokes's law, we have
`v=(2)/(9)(r^(2)(rho-sigma)g)/(eta)`
where `r` is the radius of the sphere, `rho` is the density of the amterial of the sphere
`sigma(=1.0xx10^(3)kg//m^(3))` is the density of water and `eta` is coefficient of viscosity of water
`:. v=(2xx(1.0xx10^(-3))^(2)(1.0xx10^(4)-1.0xx10^(3))xx10)/(9xx1.0xx10^(-3))`
`=20m//s`
from equation `(i)`, we ahve
`h=(v^(2))?(2g)=(20xx20)/(2xx10)=20m`