A student investigated small ball-bearings falling through a liquid to determine the viscosity of the liquid. Complete the free-body force diagram for a ball-bearing when falling at terminal velocity. (2) (ii) The student thought that the liquid she used might be glycerol. She made measurements to determine the terminal velocity of a ball-bearing. The terminal velocity of the ball-bearing was $$\(0.0824 \mathrm{~m} \mathrm{~s}^{-1}\)$$. Deduce whether the liquid she used was glycerol. density of glycerol $$\(=1.26 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{-3}\)$$ radius of ball-bearing $$\(=2.30 \mathrm{~mm}\)$$ weight of ball-bearing $$\(=4.00 \times 10^{-3} \mathrm{~N}\)$$ viscosity of glycerol at room temperature $$\(=0.934 \mathrm{Pas}\)$$ (6) .............................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................. ..............................................................................................................................................................................................................................................

Answer:

Upwards arrow labelled upthrust or U , and downward arrow labelled weight or W or mg. Both arrows start from dot.

Length of weight arrow \(\approx\) length of drag arrow + length of upthrust arrow
Calculates volume of sphere
Use of \(\rho=\frac{m}{V}\)
Use of \(W=m g\)
Use of weight \(=\) upthrust + viscous drag
Use of viscous drag \(=6 \pi \eta r v\)
\(\eta\) is \(0.94(\mathrm{~Pa} \mathrm{~s}) \approx 0.93(\mathrm{~Pa} \mathrm{~s})\) valid conclusion based on comparison of 0 . with value supplied in question.

Example of calculation
\[
\begin{array}{l}
V=\frac{4}{3} \times \pi \times\left(2.3 \times 10^{-3} \mathrm{~m}\right)^{3}=5.10 \times 10^{-8} \mathrm{~m}^{3} \\
m=1.26 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{-3} \times 5.10 \times 10^{-8} \mathrm{~m}^{3}=6.43 \times 10^{-5} \mathrm{~kg}
\end{array}
\]

Weight of liquid displaced \(=6.43 \times 10^{-5} \mathrm{~kg} \times 9.81 \mathrm{~N} \mathrm{~kg}^{-1}\)
\[
=6.31 \times 10^{-4} \mathrm{~N}
\]
viscous drag \(=6.31 \times 10^{-4} \mathrm{~N}-4.00 \times 10^{-3} \mathrm{~N}=3.37 \times 10^{-3} \mathrm{~N}\)
\[
\eta=\frac{3.37 \times 10^{-3} \mathrm{~N}}{6 \pi \times 2.3 \times 10^{-3} \mathrm{~m} \times 0.0824 \mathrm{~m} \mathrm{~s}^{-1}}=0.943 \mathrm{~Pa} \mathrm{~s}
\]

Knowledge points:

Solution: