Question

Small spheres of diameter 1.01 mm fall through 20°C water with a terminal speed of 1.20 cm/s. Calculate the density of the spheres. The viscosity of water is 1.00 × 10^-3 Pa·s. (Enter your answer to at least the nearest kg/m³.)

Answer 1

Given

diameter of sphere is d = 1.01 mm , radius r = 0.505 *10^-3 m

terminal speed v = 1.20 cm/s = 0.012 m/s

viscosity of water is 1.00 × 10-3 Pa·s

using Stoke's law the force is (drag) F = 6 pi *eta*r*V

where eta is viscosity of water and r is radius of the sphere , v is terminal speed of sphere

and the force F = mg = RHO*V*g = (rho- sigma)(4/3)(pi*r^3)*g

where rho is density fo the spheres , sigma is density of water and V is volume of the spheres

(rho- sigma)(4/3)(pi*r^3)*g = 6 pi *eta*r*V

(rho- sigma)(4/3)(r^2)*g = 6*eta*V

substituting the values  

(rho - 1000)(4/3)(0.505 *10^-3)^2*9.8 = 61.00*10^-3*0.012

rho = 1219.666 kg/m^3

the density of the spheres is rho = 1219.666 kg /m^3