Question

a ball of steel (density p= 7.8 g cm-) attains a terminal velocity of 10 cm si when falling in a water (Coefficient
of Viscosity = 8.5 * 10- Pa s) then its terminal velocity in glycerine (p = 1.2 9 cm
n= 132 Pas.)
(A) 6.25 * 10 cm s' (B) 6 45 * 10 cm s (C) 1.5*10% cm s' (D)1.6 *10cms!​

Answer 1

Question:

If a ball of steel (density, ρ =7.8 g cm⁻³) attains a terminal velocity of 10 cm⁻¹ when falling in a tank of water (coefficient of viscosity ηwater  = 8.5 x 10⁻⁴ Pa-s), then its terminal velocity in glycerine (ρ = 12 g cm⁻³,  η = 13.2 Pa-s) would be nearly

(a) 1.6 x 10-5 cm⁻¹

(b) 6.25 x 10-4 cm⁻¹

(c) 6.45 x 10-4 cm⁻¹

(d) 1.5 x 10-5 cm⁻¹

Solution:

Density of ball of steel , ρ₁ = 7.8 g cm⁻³

Terminal velocity of ball of steel when falling in water , v₁ = 10 cm/s

Viscosity of water , ηₐ = 8.5 x 10⁻⁴ PaS

Density of water , ρₐ = 1

Density of glycerine , ρₓ = 1.2 g cm⁻³

Viscosity of glycerine , ηₓ = 13.2 PaS

Terminal velocity of ball of steel in glycerine , v₂ = ?

Now as we know,

Terminal velocity , v = [ 2r²( ρ - ρ₀ )g ] / 9η

Therefore , v ∝ ( ρ - ρ₀ ) / η

v₂ / v₁ = [ ( ρ - ρₓ ) / ηₓ ] * [ ηₐ / ( ρ - ρₐ ) ]

v₂ / v₁ = [ ( 7.8 - 1.2 ) / 13.2 ] * [ 8.5*10⁻⁴ / ( 7.8 - 1 ) ]

v₂ / v₁ = [ 6.6  / 13.2 ] * [ 8.5*10⁻⁴ / 6.8 ]

v₂ / v₁ = [ (6.6  * 8.5 * 10⁻⁴) / (13.2 * 6.8) ]

v₂ / v₁ = [ (8.5 * 10⁻⁴) / (2 * 6.8) ]

v₂ / v₁ = [ (4.25 * 10⁻⁴) / 6.8 ]

v₂ / v₁ = 0.625 * 10⁻⁴

v₂ = 0.625 * 10⁻⁴ * v₁

v₂ = 0.625 * 10⁻⁴ * 10

v₂ = 6.25 * 10⁻⁴ cm s⁻¹

The terminal velocity in glycerine would be 6.25 * 10⁻⁴ cm s⁻¹