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JEE Main 202228 JuneEvening Shift
Exam
JEE Main
Year
2022
Date
28 June
Shift
Evening Shift
Subject
Physics
Chapter
Mechanical Properties of Fluids

JEE Main 2022 · 28 June · Evening Shift · Physics · Mechanical Properties of Fluids

A water drop of radius 1 falls in a situation where the effect of...

A water drop of radius 1\mum1\mum falls in a situation where the effect of buoyant force is negligible. Co-efficient of viscosity of air is 1.8×10−5Nsm−21 . 8 \times {10}^{- 5}Ns{m}^{- 2} and its density is negligible as compared to that of water 106gm−3{10}^{6}g{m}^{- 3}. Terminal velocity of the water drop is

(Take acceleration due to gravity =10ms−2= 10{ms}^{- 2})

Options

  1. A145.4×10−6ms−1145 . 4 \times {10}^{- 6}{ms}^{- 1}
  2. B123.4×10−6ms−1123 . 4 \times {10}^{- 6}{ms}^{- 1}
  3. C118.0×10−6ms−1118 . 0 \times {10}^{- 6}{ms}^{- 1}
  4. D132.6×10−6ms−1132 . 6 \times {10}^{- 6}{ms}^{- 1}

Correct answer

B. 123.4×10−6ms−1123 . 4 \times {10}^{- 6}{ms}^{- 1}

Step-by-step solution

Terminal velocity is given by vT=2r2(ρ−σ)g9η{v}_{T} = \dfrac{2{r}^{2}\left(ρ - \sigma\right)g}{9η}, where, rr is radius of falling body, ρρ is density of falling body and σ\sigma is density of fluid.

⇒vT=2×10−12×106×10−3×109×1.8×10−5⇒ {v}_{T} = \dfrac{2 \times {10}^{- 12} \times {10}^{6} \times {10}^{- 3} \times 10}{9 \times 1 . 8 \times {10}^{- 5}}

⇒vT=123.4×10−6ms−1⇒ {v}_{T} = 123 . 4 \times {10}^{- 6}m{s}^{- 1}

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