A water drop of radius 1 μ m falls in a situation where the effect of… | JEE Main 2022 (28 Jun, Shift 2)

Q38JEE Main 2022 · 28 JUN · SHIFT 2Medium

Terminal Velocity

A water drop of radius 1 $\mu$ m falls in a situation where the effect of buoyant force is negligible. Co- efficient of viscosity of air is 1.8 $\times$ 10 $-$ 5 Nsm ${}^{-2}$ and its density is negligible as compared to that of water 106 gm ${}^{-3}$ . Terminal velocity of the water drop is:

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

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Answer: (d)

Buoyancy neglected: $v_t = \dfrac{2r^2\rho g}{9\eta}$ , with $\rho = 10^6\ \text{g/m}^3 = 10^3\ \text{kg/m}^3$ .
$v_t = \dfrac{2 \times (10^{-6})^2 \times 10^3 \times 10}{9 \times 1.8 \times 10^{-5}} = \dfrac{2 \times 10^{-8}}{1.62 \times 10^{-4}}$
$v_t \approx 1.234 \times 10^{-4} = 123.4 \times 10^{-6}\ \text{ms}^{-1}$
So the answer is (d).

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Terminal Velocity question from JEE Main 2022 (28 Jun, Shift 2)

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