A raindrop with radius R = 0.2 mm falls from a cloud at a height h = 2000 m above the ground. Assume that the drop is spherical throughout its fall and the force of buoyance may be neglected, then the terminal speed attained by the raindrop is :
[Density of water f w = 1000 kg m⁻³ and Density of air fₐ = 1.2 kg m⁻³, g = 10 m/s², Coefficient of viscosity of air = 1.8 × 10⁻⁵ Nsm⁻²]

Solution

Physics sir · 4 steps

At terminal speed, Stokes drag just balances the weight. The 2000 m height plays no part at all.

1. Balance the forces. At terminal speed the drop stops accelerating, so the viscous drag 6πη R v equals the weight 4/3π R³ ρ_w g. Buoyancy is neglected, so there is no upthrust term.

   vₜ = (2R² ρ_w g)/9η

2. Put in numbers. With R = 0.2 mm = 2 × 10⁻⁴ m, we get R² = 4 × 10⁻⁸ m². The numerator is 2 × 4 × 10⁻⁸ × 1000 × 10 = 8 × 10⁻⁴ and the denominator is 9 × 1.8 × 10⁻⁵ = 1.62 × 10⁻⁴.
3. Height does not matter. Terminal speed depends only on size, density and viscosity. The drop reaches it early in the fall, so h = 2000 m is extra information. The air density only matters if you include buoyancy, and even then it changes the answer by about 0.1%.
4. So. vₜ = (8 × 10⁻⁴)/(1.62 × 10⁻⁴) ≈ 4.94 m s⁻¹, which is option (c).

The idea to keep. Terminal speed comes from drag equals weight, giving vₜ = 2R²(ρ - σ)g/9η. It goes as R², so convert mm to m before squaring, and ignore the fall height.

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