Eight equal drops of water are falling through air with a steady speed of 10 cm / s. If the drops coalesce, the new velocity is:-

Solution

Physics sir · 3 steps

Terminal speed goes as r², so find the new radius from volume and square the ratio.

1. Terminal velocity. A drop reaches steady speed when Stokes drag plus buoyancy balance its weight. That gives v = 2r²(ρ - σ)g/9η, so for the same liquid and air, v ∝ r².
2. New radius. Volume is conserved when the drops merge: 8 × 4/3π r³ = 4/3π R³. So R³ = 8r³, which gives R = 2r.

   v'/v = (R/r)² = 2² = 4

3. So. v' = 4 × 10 cm s⁻¹ = 40 cm s⁻¹, which is option (a).

The idea to keep. When n drops merge, use volume to get R = n^(1/3)r. Then terminal speed scales as R², so it becomes n^(2/3) times the old speed.

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