A spherical ball of radius 1 10-4 m and density 105 kg/m3 falls freely under gravity through a distance h before entering a tank of water. If after entering in water the velocity of the ball does not change, then the value of h is approximately: (The coefficient of viscosity of water is 9.8 10-6 N s/m2)
- 2518 m
- 2396 m
- 2249 m
- 2296 m
Answer
(1) 2518 m
Solution
The velocity does not change in water only if the ball enters with its terminal velocity, so 2gh = vT.
Since water = 103 = 105 kg/m3, buoyancy is neglected and Stokes' law gives
vT = 2r2 g9
= 2 10-8 105 9.89 9.8 10-6
= 20009 222.2 m/s
Hence h = vT22g = (222.2)22 9.8 2.52 103 m, i.e. about 2518 m.
Answer: (1)