Q4. A steel ball of radius 1.0 mm (density 7.8 × 10 kg m) is dropped into glycerine (density 1.26 × 10 kg m, viscosity 0.83 Pa s). Find its terminal velocity. Take g = 9.8 m s.
vt = 2 × (1.0 × 10) × (7.8 − 1.26) × 10 × 9.8/(9 × 0.83), about 1.7 × 10 m s, or 1.7 cm s.
Q5. Why does a car engine run more easily once it is warm, and why is a thinner oil often recommended for very cold weather?
The viscosity of a liquid falls as it warms, so warm oil offers less resistance between moving parts. In very cold weather oil becomes much more viscous, so a thinner (lower-viscosity) oil is used to keep it flowing.
Q6. A rigid sphere less dense than a liquid rises steadily in the Stokes regime. Explain the negative sign in its terminal velocity. Can the same drag coefficient automatically be used for a clean air bubble?
Taking downward as positive, ρ < σ gives negative vt, meaning upward motion. Buoyancy exceeds weight, and drag acts downward. A clean bubble has a mobile interface, so its drag differs from that on a no-slip rigid sphere; the sign reasoning applies, but the coefficient need not.