A small metallic sphere of diameter 2 mm and density $10.5 \mathrm{~g} / \mathrm{cm}^3$ is dropped i

Question

A small metallic sphere of diameter 2 mm and density $10.5 \mathrm{~g} / \mathrm{cm}^3$ is dropped in glycerine having viscosity 10 Poise and density $1.5 \mathrm{~g} / \mathrm{cm}^3$ respectively. The terminal velocity attained by the sphere is $\_\_\_\_$ $\mathrm{cm} / \mathrm{s}$.
$\left(\pi=\frac{22}{7}\right.$ and $\left.g=10 \mathrm{~m} / \mathrm{s}^2\right)$

Professor PC Thomas Classes · Answer

The terminal velocity of a sphere in a viscous fluid is $v_t = \frac{2}{9} \frac{r^2 (ho - \sigma) g}{\eta}$. Working in CGS units, the radius is $r = 1 	ext{ mm} = 0.1 	ext{ cm}$. The viscosity is $\eta = 10 	ext{ Poise}$. The densities are $ho = 10.5 	ext{ g/cm}^3$ and $\sigma = 1.5 	ext{ g/cm}^3$. The acceleration due to gravity is $g = 1000 	ext{ cm/s}^2$. Substituting these values gives $v_t = \frac{2}{9} \frac{(0.1)^2 (10.5 - 1.5) 	imes 1000}{10} = \frac{2}{9} \frac{0.01 	imes 9 	imes 1000}{10} = \frac{2}{9} 	imes \frac{90}{10} = 2 	ext{ cm/s}$.