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

20

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 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 (= 1.0 xx 10 ^3 kg // m ^ 3 ) ` is the density of water and `eta ` is coefficient of viscosity of water.
` therefore v = (2xx (1.0 xx 10 ^(-3) ) ^2 (1.0 xx 10 ^(4) - 1.0 xx 10 ^3 ) xx10)/(9 xx1.0 xx 10 ^(-3)) = 20 m//s `
From equation (i), we have ` h = (v ^ 2 )/(2g ) = (20 xx 20)/(2 xx 10) = 20 m `