Question

The terminal velocity of a steel ball 2 mm in diameter falling through glycerin is $44\times {10}^{-2}cm/s$ (Given that specific gravity of steel = 8, specific gravity of steel =8, specific gravity of glycerin a 1.3, viscosity of glycerine 8.3 poise.)

Answer 1

$d=2$mm

$=2\times {10}^{-3}$m

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

$\eta =8.3$ poise$=8.3\times \frac{1}{10}$ deca poise $=0.83$

$g=9.8$

$s={10}^3\times 8$kg

$6g=1.3\times {10}^3$ $kg/m^3$

${\theta }_t=\frac{2r^2(s-6g)g}{9\eta }$

$=\frac{2\times {10}^{-3}\times {10}^{-3}\times 9.8\times 8\times {10}^3-1.3\times {10}^3}{9\times 0.83}$

$=\frac{19.6\times {10}^{-6}\times {10}^3\times 6.7\times {10}^2}{242}=0.17\times {10}^{-1}$

${\theta }_t=1.7\times {10}^{-2}$ m/s.