Small water droplets of radius 0.01 mm are formed in the upper atmosphere and falling with a terminal velocity of 10 cm / s. Due to condensation, if 8 such droplets are coalesced and formed a larger drop, the new terminal velocity will be ________ cm / s.

Solution

Physics sir · 4 steps

Volume is conserved, so 8 drops make one drop of twice the radius, and terminal speed goes as r squared.

1. Terminal velocity. At terminal velocity, weight minus buoyancy is balanced by Stokes drag 6πη r v. Weight and buoyancy both grow as r³, so solving for v leaves v ∝ r².

   vₜ = 2r²(ρ - σ)g/9η

2. New radius. The water in 8 droplets goes into one drop: 4/3π R³ = 8 × 4/3π r³. So R³ = 8r³, which gives R = 2r.
3. Scale the speed. v'/v = (R/r)² = 2² = 4, so v' = 4 × 10 cm/s.
4. So. The new terminal velocity is 40 cm/s.

The common slip: 640. Taking the radius as 8 times larger instead of 8^(1/3) = 2 times gives 64 × 10 = 640 cm/s, not 40.

The idea to keep. For drops merging, first find the new radius from volume conservation, R = n^(1/3)r. Then use vₜ ∝ r², so the speed grows by n^(2/3).

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