Question

The terminal velocity of a copper ball of radius $2.0mm$ falling through a tank of oil at $20^{\circ }C$ is $6.5cm{s}^{-1}$. Compute the viscosity of the oil at $20^{\circ }C$. Density of oil is $1.5\times 10^{3}kg{m}^{-3}$ and density of copper is $8.9\times 10^{3}kg{m}^{-3}$.

• A. $1\times 10^{-1}kgm{s}^{-1}$
• B. $9.9\times 10^{-1}kgm{s}^{-1}$
• C. $24.3\times 10^{-2}kgm{s}^{-1}$
• D. $2\times 10^{-2}kgm{s}^{-1}$

Answer 1

Answer: (B)

Given, $v_T=6.5\times 10^{-2}m{s}^{-1},a=2\times 10^{-3}m$
$g=9.8m{s}^{-2},\rho =8.9\times 10^{3}kg{m}^{-3}$
$\sigma =1.5\times 10^{3}kg{m}^{-3}$
So, terminal velocity, $v_T=\frac{2a^2(\rho -\sigma )g}{9\eta }$
$\Rightarrow \eta =\frac{2}{9}\times \frac{{\left(2\times 10^{-3}\right)}^2\times (8.9-1.5)\times 10^3\times 9.8}{6.5\times 10^{-2}}$
$=9.9\times 10^{-1}kgm{s}^{-1}$