A drop of water of radius 0.0015 mm is fwlling in air of the coefficient pf viscosity of air is 2 × 10 ^-5kgms the terminal velocity of the drop will be
Answers
Answered by
14
Hey bae,
● Answer -
v = 2.45×10^-4 m/s
● Explanation-
# Given-
r = 1.5×10^-6 m
d = 1000 kg
η = 2×10^-5 kgm/s
# Solution-
Terminal velocity is attained by water drop when
Weight = Viscous force
mg = 6πηrv
4/3 πr^3 dg = 6πηrv
v = 2dr^2g / 9η
v = (2×1000×9.8)×(1.5×10^-6)^2 / (9×2×10^-5)
v = 2.45×10^-4 m/s
Terminal velocity of the water drop is 2.45×10^-4 m/s.
Hope this is helpful...
● Answer -
v = 2.45×10^-4 m/s
● Explanation-
# Given-
r = 1.5×10^-6 m
d = 1000 kg
η = 2×10^-5 kgm/s
# Solution-
Terminal velocity is attained by water drop when
Weight = Viscous force
mg = 6πηrv
4/3 πr^3 dg = 6πηrv
v = 2dr^2g / 9η
v = (2×1000×9.8)×(1.5×10^-6)^2 / (9×2×10^-5)
v = 2.45×10^-4 m/s
Terminal velocity of the water drop is 2.45×10^-4 m/s.
Hope this is helpful...
Similar questions