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^(-12)` and density of water `= 1.0 xx 10^(3) kg//m^(3)`.

Explanation

The velocity attained by the sphere in falling freely from a height `h` is
`v=sqrt(2gh)`…i
This is the thermal velocity of the sphere in water. Hence by Stoke'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 material of the sphere.
`sigma(=1.0xx10^(3)kg//m^(3))` is the density of water and `eta` is coefficient of vicosity of water.
`:. v=(2xx(1.0xx10^(-3))^(2)(1.0xx10^(4)-1.0xx10^(3))xx10)/(9xx1.0xx10^(-3))=20m//s`
From eqn i we have `h=(v^(2))/(2g)=(20xx20)/(2xx10)=20m`