Question

A metal sphere of radius 1 mm and mass 50 mg falls vertically in glycerine. Find (a) the viscous force exerted by the glycerine on the sphere when the speed of the sphere is 1 cm s⁻¹, (b) the hydrostatic force exerted by the glycerine on the sphere and (c) the terminal velocity with which the sphere will move down without acceleration. Density of glycerine = 1260 kg m⁻³ and its coefficient of viscosity at room temperature = 8.0 poise.

Answer 1

Given:
Radius of metallic sphere r = 1 mm = 10⁻³ m
Speed of the sphere v = 10⁻² m/s
Coefficient of viscosity η = 8 poise = 0.8 decapoise
Mass m = 50 mg = 50 × 10⁻³ kg
Density of glycerin σ = 1260 kg/m³

(a)
Viscous force exerted by glycerine on the sphere F = 6πηrv
⇒ F= 6 × (3.14) × (0.8) × 10⁻³ × (10⁻²)
= 1.50 × 10⁻⁴ N

(b)
Let V be the volume of the sphere.
Hydrostatic force exerted by glycerin on the sphere $F\text{'}=V\sigma g$
$\Rightarrow F\text{'}=\frac{4}{3}\pi r^2\sigma g$

$$\begin{gathered} =\left(\frac{4}{3}\right)\times \left(3.14\right)\times \left({10}^{-6}\right)\times 1260\times 10 \\ =5.275\times {10}^{-5}\mathrm{N} \end{gathered}$$

(c)
Let the terminal velocity of the sphere be v'.
The forces acting on the drops are
(i) The weight mg acting downwards
(ii) The force of buoyance, i.e., $\frac{4}{3}\pi r^3\sigma g$ acting upwards
(iii) The force of viscosity, i.e., 6πηrv' acting upwards
From the free body diagram:



$$\begin{gathered} 6\pi \eta rv\text{'}+\frac{4}{3}\pi r^3\sigma g=mg \\ \Rightarrow v=\frac{mg-{\displaystyle \frac{4}{3}}\pi r^2\sigma g}{6\pi \eta r} \\ =\frac{50\times {10}^{-3}-{\displaystyle \frac{4}{3}}\times 3.14\times {10}^{-6}\times 1260\times 10}{6\times 3.14\times 0.8\times {10}^{-3}} \\ =\frac{500-{\displaystyle \frac{4}{3}}\times 3.14\times {10}^{-3}\times 1260\times 10}{6\times 3.14\times 0.8} \\ =2.3\mathrm{cm}/s \end{gathered}$$