Q4. In the same liquid, a ball of twice the radius is used. How does the time to fall between the marks change?
If both balls obey Stokes’ law and wall effects are negligible, vt ∝ a, so terminal speed is four times larger and the time a quarter as long.
Q5. For steel balls (density 7.8 × 10 kg m) in castor oil (density 0.96 × 10 kg m) at about 20 °C, the graph of vt against a has a slope of 1.5 × 10 m s. Find the viscosity of the oil. Take g = 9.8 m s.
η = 2(ρ − σ)g/(9 × slope) = 2 × 6.84 × 10 × 9.8/(9 × 1.5 × 10), about 0.99 Pa s.
Q6. Why is this method not used to measure the viscosity of water?
For the millimetre-scale steel balls in this classroom setup, water gives a sphere Reynolds number too large for Stokes’ law and a fall too fast for convenient stopwatch timing. Much smaller particles can settle in water in the Stokes regime, so the limitation belongs to this apparatus, not to water in principle.