- Home
- Standard 11
- Physics
The terminal velocity of a copper ball of radius $2.0 \;mm$ falling through a tank of oll at $20\,^{\circ} C$ is $6.5 \;cm s ^{-1} .$ Compute the viscosity of the oil at $20\,^{\circ} C .$ Density of oil is $1.5 \times 10^{3} \;kg m ^{-3},$ density of copper is $8.9 \times 10^{3} \;kg m ^{-3}$
$1.1 \times 10^{-1} kg m ^{-1} s ^{-1}$
$9.9 \times 10^{-1} kg m ^{-1} s ^{-1}$
$6.37 \times 10^{-2} kg m ^{-1} s ^{-1}$
$5.98 \times 10^{-1} kg m ^{-3} s ^{-1}$
Solution
We have $v_{ t }=6.5 \times 10^{-2} ms ^{-1}, a=2 \times 10^{3} m$
$g =9.8 ms ^{-2}, \rho=8.9 \times 10^{3} kg m ^{-3}$
$\sigma =1.5 \times 10^{3} kg m ^{-3}$
$\eta =\frac{2}{9} \times \frac{\left(2 \times 10^{-3}\right)^{2} m ^{2} \times 9.8 m s ^{-2}}{6.5 \times 10^{-2} m s ^{-1}} \times 7.4 \times 10^{3} kg m ^{-3}$
$=9.9 \times 10^{-1}\, kg \,m ^{-1} \,s ^{-1}$
