Two small spherical metal balls, having equal masses, are made from materials of densities ρ₁ and ρ₂ (ρ₁ = 8ρ₂) and have radii of 1 mm and 2 mm, respectively, They are made to fall vertically (from rest) in a viscous medium whose coefficient of viscosity equals η and whose density is 0.1 ρ₂. The ratio of their terminal velocities would be,

CONCEPT:

Terminal Velocity: 

• It is the maximum constant velocity acquired by the body while falling freely in a viscous medium.
• When a body falls freely through a viscous medium, three forces act on it.

1. Weight of the body acting vertically downwards.
2. Upward thrust due to buoyancy equal to the weight of viscous medium displaced.
3. Viscous drag acting in the direction opposite to the motion of the body.

• Mathematically, the terminal velocity for a spherical body is given by the formula:

\(\Rightarrow {\rm{v}} = \frac{{\left( {\frac{2}{9}} \right){{\rm{r}}^2}{\rm{g}}\left( {{\rm{ρ }} - {\rm{σ }}} \right)}}{{\rm{η }}}\)    -----(1)

Where, v = Terminal velocity, 

r = Radius of the spherical body, 

g = Acceleration due to gravity (constant for big and smaller spheres), 

ρ = Density of the body (constant for big and smaller spheres), 

σ = Density of the medium through which the body is falling (constant for big and smaller spheres), 

η = Coefficient of the viscosity of the medium through which the body is falling (constant for big and smaller spheres)

CALCULATION:

Given:

ρ₁ = 8ρ₂ , r₁ = 1 mm, r2 = 2 mm, σ = 0.1 ρ2

Coefficient of viscosity = η 

Now, using equation (1) expression for terminal velocity for ball 1:

v_t1 = \(\frac{{\left( {\frac{2}{9}} \right){{\rm{r_1}}^2}{\rm{g}}\left( {{\rm{ρ_1 }} - {\rm{σ }}} \right)}}{{\rm{η }}}\)= \(\frac{{\left( {\frac{2}{9}} \right){{\rm{1}}^2}{\rm{g}}\left( {{\rm{8ρ_2 }} - {\rm{0.1ρ_2 }}} \right)}}{{\rm{η }}}\) = \(\frac{{\left( {\frac{2}{9}} \right){}{\rm{g}}\left( {{\rm{7.9ρ_2 }}} \right)}}{{\rm{η }}}\)   -----(2)

And  expression for terminal velocity for ball 2:

vt2 = \(\frac{{\left( {\frac{2}{9}} \right){{\rm{r_2}}^2}{\rm{g}}\left( {{\rm{ρ_2 }} - {\rm{σ }}} \right)}}{{\rm{η }}}\) = \(\frac{{\left( {\frac{2}{9}} \right){{\rm{2}}^2}{\rm{g}}\left( {{\rm{ρ_2 }} - {\rm{0.1ρ_2 }}} \right)}}{{\rm{η }}}\) = \(\frac{{\left( {\frac{2}{9}} \right){}{\rm{3.6g}}\left( {{\rm{ρ_2 }}} \right)}}{{\rm{η }}}\)   -----(3)

Dividing equations (2) & (3) we get,

vt1/vt2 = 7.9/3.6 = 79/36

Hence Option 2) is the correct choice.