A solid steel ball of diameter 3.6 mm acquired terminal velocity 2.45 × 10⁻² m / s while falling under gravity through an oil of density 925 kg m⁻³. Take density of steel as 7825 kg m⁻³ and g as 9.8 m / s². The viscosity of the oil in SI unit is

Solution

Physics sir · 4 steps

At terminal velocity the net force is zero, so use Stokes' formula for vₜ and solve it for η.

1. Balance the forces. At terminal velocity, weight equals buoyancy plus viscous drag 6πη r v. Rearranging gives η in terms of the density difference ρ - σ.

   η = 2r²(ρ - σ)g/9v

2. Put in numbers. The radius is half the diameter: r = 1.8 × 10⁻³ m, so r² = 3.24 × 10⁻⁶ m². The density difference is ρ - σ = 7825 - 925 = 6900 kg m⁻³.
3. Work it out. The numerator is 2 × 3.24 × 10⁻⁶ × 6900 × 9.8 = 0.4382. The denominator is 9 × 2.45 × 10⁻² = 0.2205. So η = 0.4382 / 0.2205 = 1.987 Pa s.
4. So. The viscosity of the oil is about 1.99 Pa s, which is option (d).

The idea to keep. For a sphere at terminal velocity, use vₜ = 2r²(ρ-σ)g/9η. Always halve the diameter, and subtract the fluid density, because buoyancy acts too.

Answer from the NTA final answer key. Solution by PrepWiser.