Terminal velocity is given by vT=9η2r2(ρ−σ)g, where, r is radius of falling body, ρ is density of falling body and σ is density of fluid.
⇒vT=9×1.8×10−52×10−12×106×10−3×10
⇒vT=123.4×10−6ms−1
A water drop of radius 1μ m falls in a situation where the effect of buoyant force is negligible. Co-efficient of viscosity of air is 1.8×…
JEE Main 2022 — Physics Mechanics
2022mcqeasy
A water drop of radius 1μm falls in a situation where the effect of buoyant force is negligible. Co-efficient of viscosity of air is 1.8×10−5Nsm−2 and its density is negligible as compared to that of water 106gm−3. Terminal velocity of the water drop is
(Take acceleration due to gravity =10ms−2)
Official previous-year question
Held on 28 Jun 2022 · Verified 6 Jul 2026.
Options
- A
145.4×10−6ms−1
- B
123.4×10−6ms−1
- C
118.0×10−6ms−1
- D
132.6×10−6ms−1
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