A raindrop with radius R = 0.2 mm falls from a cloud at a height h =… | JEE Main 2021 (27 Jul, Shift 2)

Q27JEE Main 2021 · 27 JUL · SHIFT 2Difficult

Viscosity & Stokes' Law

A raindrop with radius R = 0.2 mm falls from a cloud at a height h = 2000 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 fw = 1000 kg m ${}^{-3}$ and Density of air fa = 1.2 kg m ${}^{-3}$ , g = 10 m/s2, Coefficient of viscosity of air = 1.8 $\times$ 10 $-$ 5 Nsm ${}^{-2}$ ]

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Answer: (c)

Terminal speed (buoyancy neglected): $v_t = \frac{2r^2\rho_w g}{9\eta}$ .
$v_t = \frac{2(0.2\times10^{-3})^2(1000)(10)}{9\times1.8\times10^{-5}} = \frac{8\times10^{-4}}{1.62\times10^{-4}}$
$v_t \approx 4.94$ m/s (the height 2000 m is not needed)
So the answer is (c) $4.94\text{ m s}^{-1}$ .

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Viscosity & Stokes' Law question from JEE Main 2021 (27 Jul, Shift 2)

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