A water drop of radius 1 μm falls in a situation where the effect of buoyant force is negligible. Co-efficient of viscosity of air is 1.8 × 10⁻⁵ Nsm⁻² and its density is negligible as compared to that of water 10⁶ gm⁻³. Terminal velocity of the water drop is :
(Take acceleration due to gravity = 10 ms⁻²)

Solution

Physics sir · 4 steps

At terminal velocity, weight is exactly balanced by Stokes drag, so set mg = 6πη r v and solve.

1. Balance the forces. Buoyancy is negligible, so only weight acts down and viscous drag acts up. At terminal velocity they are equal: 4/3π r³ ρ g = 6πη r v.

   v = (2 r² ρ g)/9η

2. Convert units. r = 1 μm = 10⁻⁶ m. The density 10⁶ g m⁻³ is 10³ kg m⁻³, since 1000 g make 1 kg.
3. Put in numbers. The numerator is 2 × 10⁻¹² × 10³ × 10 = 2 × 10⁻⁸. The denominator is 9 × 1.8 × 10⁻⁵ = 1.62 × 10⁻⁴. So v = (2 × 10⁻⁸)/(1.62 × 10⁻⁴) = 1.234 × 10⁻⁴ m s⁻¹.
4. So. v = 123.4 × 10⁻⁶ m s⁻¹, which is option (d).

The idea to keep. Terminal velocity in Stokes' regime is v = 2r²(ρ - σ)g/9η. Drop σ when buoyancy is negligible, and put everything in SI units first.

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