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)`.

Options

A. `10m`
B. `20m`
C. `12m`
D. `5m`

Answer

B

Explanation

The velocity attained by the sphere is falling freely from a height `h` is
`v=sqrt(2gh)` .(i)
this is the terminal velocity of the sphere in water. Hence by Stokes' 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)`
`=20(m)/(s)`
From equation (i). we have
`h=(v^2)/(2g)=(20xx20)/(2xx10)=20cm`