Question

The terminal velocity of copper ball of radius 2 mm falling through a tank of oil at 20°C is 6.5 cm/s. If viscosity of the oil at 20°C is 9.9 × 10–1 kg m–1 s–1 and density of copper is 8.9 × 103 kg m–3. Then density of the oil will be​

Answer 1

density of oil is 1.7 × 10³ kg/m³

it is given that the terminal velocity of copper ball of radius, r = 2 mm falling through a tank of oil at 20°C is $v_t$ = 6.5 cm/s. If viscosity of the oil at 20°C is $\eta$= 9.9 × 10–1 kg m–1 s–1 and density of copper is ρ =  8.9 × 103 kg m–3.

we have to find the density of fluid (σ )​

  using formula, coefficient of viscosity η = $\frac{2}{9} \frac{r^{2}g(\rho-\sigma) }{v_t}$

     ⇒ 9.9 × 10⁻¹ = 2/9 (2 × 10⁻³)² × 10 (8.9 × 10³ - σ)/(6.5 × 10⁻²)

    ⇒ 9.9 × 9 × 6.5 × 10⁻³ = 8 × 10⁻⁵(8.9 × 10³ - σ)

   ⇒ 7.2 × 10³ = (8.9 × 10³ - σ)

   ⇒ σ = 1.7 × 10³ kg/m³

therefore, density of oil is 1.7 × 10³ kg/m³

Answer 2

Answer:

we have to find the density of fluid (σ )​

  using formula, coefficient of viscosity η =

     ⇒ 9.9 × 10⁻¹ = 2/9 (2 × 10⁻³)² × 10 (8.9 × 10³ - σ)/(6.5 × 10⁻²)

    ⇒ 9.9 × 9 × 6.5 × 10⁻³ = 8 × 10⁻⁵(8.9 × 10³ - σ)

   ⇒ 7.2 × 10³ = (8.9 × 10³ - σ)

   ⇒ σ = 1.7 × 10³ kg/m³