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Terminal velocity of a sphere, or viscosity from it, using Stokes' law

This type of question, asked 4 times: 2021 Paper 17, 28 Jun 2022, 3 Apr 2025, 23 Jan 2026 S2. All 4 questions below, oldest first.

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  1. Question 1

    2021 Paper 17

    3
    A raindrop with radius R=0.2 mmR = 0.2\text{ }mm falls from a cloud at a height h=2000 mh = 2000\text{ }m above the ground. Assume that the drop is spherical throughout its fall and the force of buoyance may be neglected, then the terminal speed attained by the raindrop is : [Density of water ρw=1000 kg m−3\rho_{w}= 1000\text{ }kg{\text{ }m}^{- 3} and Density of air ρa=1.2 kg m−3, g=10 m/s2\rho_{a}= 1.2\text{ }kg{\text{ }m}^{- 3},\text{ }g = 10\text{ }m/s^{2} Coefficient of viscosity of air =1.8×10−5Nsm−2= 1.8 \times10^{- 5}Nsm^{- 2} ]

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  2. Question 2

    28 Jun 2022

    97
    A water drop of radius 1μm1\mu m falls in a situation where the effect of buoyant force is negligible. Coefficient of viscosity of air is 1.8×10−5Nsm−21.8 \times10^{- 5}Nsm^{- 2} and its density is negligible as compared to that of water 106gm−310^{6}gm^{- 3}. Terminal velocity of the water drop is: (Take acceleration due to gravity =10 ms−2= 10{\text{ }ms}^{- 2} )

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  3. Question 3

    3 Apr 2025

    87
    A solid steel ball of diameter 3.6 mm acquired terminal velocity 2.45×10−2 m/s2.45 \times10^{- 2}\text{ }m/s while falling under gravity through an oil of density 925 kg m−3925\text{ }kg{\text{ }m}^{- 3}. Take density of steel as 7825 kg m−37825\text{ }kg{\text{ }m}^{- 3} and g as 9.8 m/s2m/s^{2}. The viscosity of the oil in SI unit is.

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  4. Question 4

    23 Jan 2026 S2

    20
    A small metallic sphere of diameter 2 mm and density 10.5 g/cm310.5\text{ }g/cm^{3} is dropped in glycerine having viscosity 10 Poise and density 1.5 g/cm31.5\text{ }g/cm^{3} respectively. The terminal velocity attained by the sphere is ____\_\_\_\_ cm/scm/s. (π=227\pi =\frac{22}{7} and g=10 m/s2g = 10\text{ }m/s^{2})

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