Question

the ratio of terminal velocity of two drops of radii R and R/2 is

Answer 1

From the Concepts of the Fluid Mechanics, Using the Formula,  

Terminal Velocity (or V) = $\frac{2r^2g(sigma - row)}{9 eta}$

or Terminal velocity (or V) = 2r²g(σ - ρ)/9η

where,  σ is the density of the object moving in a fluid , ρ is the density of the fluid, r is the Radius of the spherical object, η is the coefficient of the viscosity, and g is the acceleration due to the gravity.

Now, For the Radius of the molecule R,

V₁ = 2R²g(σ - ρ)/9η

Also, For the Radius of the molecule R/2,

V₂ = 2(R/2)²g(σ - ρ)/9η

V₂ = 2R²g(σ - ρ)/18η

Now, V₁/V₂ = [2R²g(σ - ρ)/9η]/[2R²g(σ - ρ)/18η]

V₁/V₂ = 2/1

∴ V₁ : V₂ = 2 : 1

Hope it helps.

Answer 2

From the Concepts of the Fluid Mechanics, Using the Formula,  

Terminal Velocity (or V) = 2r²g(sigma - row) /9eta

or Terminal velocity (or V) = 2r²g(σ - ρ)/9η

where,  σ is the density of the object moving in a fluid , ρ is the density of the fluid, r is the Radius of the spherical object, η is the coefficient of the viscosity, and g is the acceleration due to the gravity.

Now, For the Radius of the molecule R,

V₁ = 2R²g(σ - ρ)/9η

Also, For the Radius of the molecule R/2,

V₂ = 2(R/2)²g(σ - ρ)/9η

V₂ = 2R²g(σ - ρ)/18η

Now, V₁/V₂ = [2R²g(σ - ρ)/9η]/[2R²g(σ - ρ)/18η]

V₁/V₂ = 2/1

∴ V₁ : V₂ = 2 : 1