Question

A metal sphere of radius 1 mm and mass 50 mg falls verticcally 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 = $1260kg{m}^{-3}$ and its coefficient of viscosity at room temperature = 8.0 poise.

Answer 1

r = 1 mm =${10}^{-3}m$

v = ${10}^{-2}m/s$

$\eta$ = 8 poise = 0.8 decapoise

m = 50 mg = $50\times {10}^{-3}kg$

$\sigma =1260kg/m^3$

(a) Viscous force = $6\pi \eta rv$

= $6\times \left(3.14\right)\times \left(0.8\right)\times {10}^{-3}\times \left({10}^{-2}\right)$

(b) Hydrostatic force = B = $\left(\frac{4}{3}\right)\pi r^2\sigma g$

= $=\left(\frac{4}{3}\right)\times \left(3.14\right)\times \left({10}^{-9}\right)\times 1260\times 10$

= $5.275\times {10}^{-5}N$

(c) $6\pi \eta rv+\left(\frac{4}{3}\right)\pi r^3\rho g=mg$

$\Rightarrow v=\frac{mg-\frac{4}{3}\pi r^2\sigma g}{6\pi \eta r}$

$=\frac{50\times {10}^{-3}-\frac{4}{3}\times 3.14\times {10}^{-6}\times 1260\times 10}{6\times 3.14\times 0.8\times {10}^{-3}}$

$=\frac{500-\frac{4}{3}\times 3.14\times {10}^{-3}\times 1260\times 10}{6\times 3.14\times 0.8}$

= 2.3 cm/sec.