What will be the approximate terminal velocity of a rain drop of diameter 1.8 × 10-3 m, when density of rain water ≈ 103 kgm-3 and the co-efficient of viscosity of air ≈ 1.8 × 10-5 Nsm-2 ? (Neglect buoyancy of air).

Question

What will be the approximate terminal velocity of a rain drop of diameter 1.8 × 10-3 m, when density of rain water ≈ 103 kgm-3 and the co-efficient of viscosity of air ≈ 1.8 × 10-5 Nsm-2 ? (Neglect buoyancy of air). (A) 49 ms-1 (B) 98 ms-1 (C) 392 ms-1  (D) 980 ms-1

Tardigrade Admin · Accepted Answer

Terminal velocity, v=(2/9) r2 ((p-σ)/η) g Neglecting buoyancy effect of the fluid, v=(2/9) ⋅ (ρ/η) r2 g Putting the given values, we get v=(2/9) × (103 ×(0.9 × 10-3)2/1.8 × 10-5) × 9.8=98 ms -1

Q. What will be the approximate terminal velocity of a rain drop of diameter $1.8 \times 1 0^{- 3} m$ , when density of rain water $\approx 1 0^{3} k g m^{- 3}$ and the co-efficient of viscosity of air $\approx 1.8 \times 1 0^{- 5} N s m^{- 2}$ ? (Neglect buoyancy of air).

$49 m s^{- 1}$

$98 m s^{- 1}$

$392 m s^{- 1}$

$980 m s^{- 1}$

Q. What will be the approximate terminal velocity of a rain drop of diameter 1.8×10−3m, when density of rain water ≈103kgm−3 and the co-efficient of viscosity of air ≈1.8×10−5Nsm−2 ? (Neglect buoyancy of air).

49ms−1

98ms−1

392ms−1

980ms−1

Terminal velocity, v=92​r2η(p−σ)​g Neglecting buoyancy effect of the fluid, v=92​⋅ηρ​r2g Putting the given values, we get v=92​×1.8×10−5103×(0.9×10−3)2​×9.8=98ms−1

Q. What will be the approximate terminal velocity of a rain drop of diameter $1.8 \times 1 0^{- 3} m$ , when density of rain water $\approx 1 0^{3} k g m^{- 3}$ and the co-efficient of viscosity of air $\approx 1.8 \times 1 0^{- 5} N s m^{- 2}$ ? (Neglect buoyancy of air).

$49 m s^{- 1}$

$98 m s^{- 1}$

$392 m s^{- 1}$

$980 m s^{- 1}$