Two solid spheres of same metal but of mass M and 8M fall simultaneously on a viscous liquid and their terminal velocities are v and nv, then find the value of n.
Question
Two solid spheres of same metal but of mass M and 8M fall simultaneously on a viscous liquid and their terminal velocities are v and nv, then find the value of n.
Solution
The terminal velocity of a sphere falling in a viscous fluid is given by Stokes' Law, which states:
v = (2/9) * (r²) * g * (ρ - σ) / η
where:
- v is the terminal velocity,
- r is the radius of the sphere,
- g is the acceleration due to gravity,
- ρ is the density of the sphere,
- σ is the density of the fluid, and
- η is the viscosity of the fluid.
Given that the spheres are made of the same material and are falling through the same fluid, we can say that g, ρ, σ, and η are constants.
The mass of a sphere is given by M = 4/3 * π * r³ * ρ. Therefore, the radius of the sphere is proportional to the cube root of its mass. So, if the mass of the second sphere is 8M, its radius will be 2r (since the cube root of 8 is 2).
Substituting these values into Stokes' Law gives us:
v = (2/9) * (r²) * g * (ρ - σ) / η nv = (2/9) * ((2r)²) * g * (ρ - σ) / η
Solving for n gives us:
n = (2/9) * (4r²) * g * (ρ - σ) / η / ((2/9) * (r²) * g * (ρ - σ) / η) n = 4
So, the terminal velocity of the second sphere is four times that of the first sphere.
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