Question

Example 2.9: A spherical drop of oil falls
at a constant speed of 4 cm/s in steady air.
Calculate the radius of the drop. The density
of the oil is 0.9 g/cm³, density of air is
1.0 g/cm³ and the coefficient of viscosity of
air is 1.8×10^-4 poise, (g = 980 cm/s²)​

Answer 1

Answer :

The radius of the spherical drop is 0.006 cm

Explanation :

Given,

• A spherical drop of oil falls  at a constant speed of 4 cm/s in steady air.

           i.e., terminal velocity, v = 4 cm/s = 0.04 m/s

• The density  of the oil is 0.9 g/cm³

            ⇒ ρ = 0.9 g/cm³

            ⇒ ρ = 0.9 × 1000 kg/m³

            ⇒ ρ = 900 kg/m³

• density of air is  1.0 g/cm³

            ⇒ σ = 1 g/cm³

            ⇒ σ = 1 × 1000 kg/m³

            ⇒ σ = 1000 kg/m³

• coefficient of viscosity of  air is 1.8×10⁻⁴ poise

            ⇒ η = 1.8 × 10⁻⁴ poise

            ⇒ η = 1.8 × 10⁻⁴ × 0.1 kg/m-s

            ⇒ η = 1.8 × 10⁻⁵ kg/m-s

• acceleration due to gravity = 980 cm/s²

            ⇒ g = 980 cm/s²

            ⇒ g = 9.8 m/s²

To find,

• the radius of the drop

Solution,

terminal velocity is given by,

     $\sf \implies v=\dfrac{2r^2g ( \rho - \sigma)}{9 \eta} \\\\ \implies \boxed{ \bf r^2=\dfrac{9 \eta v}{2g(\rho - \sigma)} }$

Substituting all the values,

    $\sf r^2=\dfrac{9(1.8 \times 10^{-5})(0.04)}{2(9.8)(1000-900)}\\\\\\ r^2=\dfrac{9 \times 1.8 \times 4 \times 10^{-5} \times 10^{-2}}{19.6(100)}\\\\\\ r^2=\dfrac{36 \times 1.8 \times 10^{-7}}{1960}\\\\\\ r^2= \dfrac{9 \times 1.8 \times 10^{-8}}{44} \\\\\\ r^2=\dfrac{16.2 \times 10^{-8}}{44}\\\\ r^2 \simeq 0.36 \times 10^{-8}\\\\ r \simeq \sqrt{0.36 \times 10^{-8} }\\\\ r \simeq 0.6 \times 10^{-4} \ m \\\\\ r \simeq 0.6 \times 10^{-4} \times 100 \ cm\\\\ r \simeq 6 \times 10^{-3} \ cm \\\\ r \simeq 0.006 \ cm$

∴ The radius of the spherical drop is 0.006 cm

Answer 2

Answer :

The radius of the spherical drop is 0.006 cm

Explanation :

Given,

A spherical drop of oil falls  at a constant speed of 4 cm/s in steady air.

          i.e., terminal velocity, v = 4 cm/s = 0.04 m/s

The density  of the oil is 0.9 g/cm³

           ⇒ ρ = 0.9 g/cm³

           ⇒ ρ = 0.9 × 1000 kg/m³

           ⇒ ρ = 900 kg/m³

density of air is  1.0 g/cm³

           ⇒ σ = 1 g/cm³

           ⇒ σ = 1 × 1000 kg/m³

           ⇒ σ = 1000 kg/m³

coefficient of viscosity of  air is 1.8×10⁻⁴ poise

           ⇒ η = 1.8 × 10⁻⁴ poise

           ⇒ η = 1.8 × 10⁻⁴ × 0.1 kg/m-s

           ⇒ η = 1.8 × 10⁻⁵ kg/m-s

acceleration due to gravity = 980 cm/s²

           ⇒ g = 980 cm/s²

           ⇒ g = 9.8 m/s²

To find,

the radius of the drop

∴ The radius of the spherical drop is 0.006 cm