Two small metal spheres of densities in the ratio 3:2 and diameters in the ratio 1:2 are released from rest in two vertical liquid columns of coeff. of viscosities in the ratio 4:3. When the viscous force on both of them is same, what will be the ratio of their instantaneous velocities?
To determine the ratio of their instantaneous velocities, we need to consider the balance of forces acting on the spheres. In this case, we have the gravitational force and the viscous drag force.
Let's denote the masses, densities, and diameters of the spheres as follows:
- Sphere 1: Mass m1, Density ρ1, Diameter D1
- Sphere 2: Mass m2, Density ρ2, Diameter D2
Given the ratios:
Density ratio: ρ1/ρ2 = 3/2
Diameter ratio: D1/D2 = 1/2
We know that the mass of each sphere is proportional to its volume (V) and density (ρ).
m1 = ρ1 * V1 and m2 = ρ2 * V2
Since the spheres are made of the same material, the density and volume are related by the formula:
ρ = (3 * mass)/(4 * π * r^3), where r is the radius of the sphere.
Now, we can calculate the ratio of the volumes of the two spheres:
(V1/V2) = (m1/ρ1) / (m2/ρ2) = (ρ2/ρ1) * (V2/V1) = (2/3) * (D1/D2)^3
Given that D1/D2 = 1/2, we can substitute that into the equation:
(V1/V2) = (2/3) * (1/2)^3 = 1/12
Now, let's consider the force balance between gravity and viscous drag:
For each sphere, the viscous drag force (F_drag) is given by Stokes' law:
F_drag = 6 * π * η * v * r,
where η is the coefficient of viscosity of the liquid and v is the instantaneous velocity of the sphere.
Since the ratio of coefficients of viscosity is 4/3, we can write:
(η1/η2) = (4/3)
Given that the viscous drag force is the same for both spheres, we have:
(6 * π * η1 * v1 * r1) = (6 * π * η2 * v2 * r2)
The term (6 * π * r^2) can be replaced by the volume (V) of the sphere:
(6 * π * r^2) = (4/3) * (π * r^3)
(6 * π * r^2) = (4/3) * (π * (3 * mass) / ρ)
Now, we can express the ratio v1/v2 in terms of the masses and densities of the spheres:
(v1/v2) = [(6 * π * r1^2) * ρ1] / [(6 * π * r2^2) * ρ2]
(v1/v2) = [(4/3) * (π * (3 * m1) / ρ1) * ρ1] / [(4/3) * (π * (3 * m2) / ρ2) * ρ2]
(v1/v2) = (3 * m1 * ρ1 / ρ1) / (3 * m2 * ρ2 / ρ2)
(v1/v2) = m1/m2
Therefore, the ratio of the instantaneous velocities (v1/v2) is equal to the ratio of their masses (m1/m2).
In conclusion, the ratio of the instantaneous velocities of the two spheres is equal to the ratio of their masses.
To find the ratio of the instantaneous velocities of the two spheres, we need to consider the forces acting on them.
Let's denote the densities of the two spheres as ρ₁ and ρ₂, and the diameters as D₁ and D₂ respectively.
First, let's find the ratio of the viscous forces acting on the two spheres. The viscous force (F) on a sphere moving through a liquid can be calculated using Stokes' law:
F = 6πηrv
where:
F = viscous force
η = coefficient of viscosity of the liquid
r = radius of the sphere
v = velocity of the sphere
From the question, we know that the ratio of the coefficients of viscosities is 4:3. Let's denote the coefficient of viscosity of the first liquid as η₁ and the second liquid as η₂.
Since the diameters are given in the ratio 1:2, the radius of the first sphere (r₁) will be half the radius of the second sphere (r₂). Hence, r₁ = D₁/2 and r₂ = D₂/2.
We can now calculate the ratio of the viscous forces as follows:
F₁/F₂ = (6πη₁r₁v₁) / (6πη₂r₂v₂)
= (η₁/η₂) * (r₁/r₂) * (v₁/v₂)
= (η₁/η₂) * (D₁/2)/(D₂/2) * (v₁/v₂)
Now, let's find the ratio of the densities of the spheres, which is given as 3:2. Let's denote the density of the first sphere as ρ₁ and the second sphere as ρ₂.
Density is directly proportional to mass, and mass is proportional to the volume of the sphere. The volume of a sphere is given by:
V = (4πr³)/3
Since the radii are given by r₁ = D₁/2 and r₂ = D₂/2, we can write the volumes as:
V₁ = (4π(D₁/2)³)/3 = (πD₁³)/6
V₂ = (4π(D₂/2)³)/3 = (πD₂³)/6
Now, let's find the ratio of densities:
ρ₁/ρ₂ = (V₁ / (4/3)πr₁³) / (V₂ / (4/3)πr₂³)
= (V₁ / r₁³) / (V₂ / r₂³)
= ((πD₁³)/6) / (r₁³) / ((πD₂³)/6) / (r₂³)
= (D₁³ / (6r₁³)) / (D₂³ / (6r₂³))
= (D₁³ / r₁³) / (D₂³ / r₂³)
= (D₁/r₁)³ / (D₂/r₂)³
= (2)³ / (1)³
= 8
Since the viscous force is the same for both spheres, we can write:
F₁/F₂ = 1
Now, using the given ratio of the coefficients of viscosities and densities, we can equate the expressions for the viscous forces:
(η₁/η₂) * (D₁/2)/(D₂/2) * (v₁/v₂) = 1
Substituting the ratio of densities, we get:
(4/3) * (3/2) * (v₁/v₂) = 1
(4/2) * (v₁/v₂) = 1
v₁/v₂ = 1/2
Therefore, the ratio of their instantaneous velocities is 1:2.
In the highly viscous "Stokes Law" Reynolds-number region (Re < 0.1), the viscous force is proportional to velocity*diameter*viscosity.
It does not depend upon density.
If the viscous forces are equal instantaneously,
V2/V1 = (D1/D2)*(Visc1/Visc2)
= (1/2)(4/3) = 2/3
This question is fundamentally flawed in my opinion, for this reason. In the steady-state limiting-velocity regime, the viscous forces cannot be equal in this case, since the force ratio must equal the weight (and mass) ratio, which is not 1 in this case.