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 through 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, and Density of air ρa = 1.2 kg m⁻³, g = 10 m/s², Coefficient of viscosity of air η = 1.8 * 10⁻⁵ Nsm⁻²]

Option 1 - <p>4.94 ms⁻¹<br><!-- [if !supportLineBreakNewLine]--><br><!--[endif]--></p>
Option 2 - <p>14.4 ms⁻¹<br><!-- [if !supportLineBreakNewLine]--><br><!--[endif]--><br><!--[endif]--></p>
Option 3 - <p>43.56 ms⁻¹<br><!-- [if !supportLineBreakNewLine]--><br><!--[endif]--><br><!--[endif]--></p>
Option 4 - <p>250.6 ms⁻¹</p>
7 Views|Posted 5 months ago
Asked by Shiksha User
1 Answer
V
5 months ago
Correct Option - 1
Detailed Solution:

At terminal speed
Mg = Fv = 6πηRv
⇒ V = mg / 6πηR
V = (4/3)πR³ρg / 6πηR
⇒ V = (2/9) * (ρR²g/η)
= (2/9) * (1000 * (0.2 * 10? ³)² * 10) / (1.8 * 10? )
= 4.94 m/s

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Physics Mechanical Properties of Fluids 2025

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