A steel ball of diameter 0.2mm, falls with a
terminal velocity of 0.0029m/s through oil in a
relatively large jar.Calculate the viscosity of
oil, assuming the densities of the oil and the
ball are to be 7200 and 420kg/m^3 respectively
.if the radius is doubled,calculate the new
terminal velocity.[take g=9.8m/s^2]
To calculate the viscosity of the oil, we can use Stokes' Law, which states that the terminal velocity of a small sphere moving through a viscous fluid is given by:
v = (2/9) * ( (ρ_ball - ρ_fluid) * g * r^2 ) / η
Where:
v is the terminal velocity,
ρ_ball is the density of the ball,
ρ_fluid is the density of the fluid,
g is the acceleration due to gravity,
r is the radius of the ball, and
η is the viscosity of the fluid.
We can rearrange this equation to solve for η:
η = (2/9) * ( (ρ_ball - ρ_fluid) * g * r^2 ) / v
Using the given values:
ρ_ball = 420 kg/m^3,
ρ_fluid = 7200 kg/m^3,
g = 9.8 m/s^2, and
v = 0.0029 m/s,
η = (2/9) * ( (420 kg/m^3 - 7200 kg/m^3) * 9.8 m/s^2 * (0.1 mm / 2)^2 ) / 0.0029 m/s
Note that the radius needs to be converted to meters before calculation, hence we divide the radius by 1000.
η = (2/9) * ( (420 kg/m^3 - 7200 kg/m^3) * 9.8 m/s^2 * (0.1 mm / 2 / 1000)^2 ) / 0.0029 m/s
Simplifying this equation will give the value for viscosity η.
Now, to calculate the new terminal velocity when the radius is doubled, we can use the same formula but with the new radius. Let's call the new terminal velocity v_new and the new radius r_new.
v_new = (2/9) * ( (ρ_ball - ρ_fluid) * g * r_new^2 ) / η
The radius is now doubled, so r_new = 2 * r.
v_new = (2/9) * ( (ρ_ball - ρ_fluid) * g * (2r)^2 ) / η
v_new = (2/9) * ( (ρ_ball - ρ_fluid) * g * 4r^2 ) / η
Simplifying this equation will give the value for the new terminal velocity v_new.