Question

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

• A. $1 \times 10^{-1} \, kg \, ms ^{-1}$
• B. $9.9 \times 10^{-1} \, kg \, ms ^{-1}$
• C. $24.3 \times 10^{-2} \, kg \, ms ^{-1}$
• D. $2 \times 10^{-2} \, kg \, ms ^{-1}$

Answer 1

Answer: (B)

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