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`        A drop of rain with radius 1.5mm falls from a cloud at height 1200m from the ground. The density of water is 1000kg/m3. Assume that the drop was spherical throughout the fall and there is no air drag .the impact speed of drop will be..?`
one year ago

Arun
24739 Points
```							When drag D = 1/2 rho Cd A v^2 = mg = W the raindrop weight, v is at terminal velocity because acceleration a = 0. Let V = 4/3 pi r^3 = 4/3 pi (1.5E-3)^3 = .00000001414 m^3 be the volume of the drop in the clouds. The mass m = Rho V = .00001414 kg where Rho = 1000 kg/m^3. When the drop falls, that volume V = 4/3 pi r^3 = 2/3 pi R^3 = W/2 a hemisphere with the same volume of water. Thus, 2 r^3 = R^3 and R = r (2)^1/3 so the hemisphere cross sectional area = A = pi r^2 (2)^2/3 = pi (1.5E-3)^2 (2)^2/3 = .00001122 m^2 Solve for v = sqrt(2mg/(rho Cd A) = sqrt(2*.00001414*9.81/(rho*.6*.00001122) = 6.419/rho where rho is air density WHICH cannot not possibly be your Pa = 1000kg/m^3 as that's the density of water. Air density near the surface is more like 1.225 kg/m^3; in which case v = 6.419/rho = 6.419/1.225 = 5.25 mps.
```
one year ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions