A spherical ball of radius `3xx10^(-4)`m and density `10^(4)kg//m^(3)` falls freely under gravity through a distance h before entering a tank of water. If after entering the water the velocity of the ball does not change, find h the viscosity of water is `9.8xx10^(-6)N-s//m^(2)`

Explanation

After falling a height h velocity of the ball will become `v=sqrt(2gh)`. After entering into the water as this velocity does not change, this velocity is equal to the terminal velocity.
`sqrt(2gh)=2/9r^(2)[(rho-sigma)/(eta)]g`
`2gh=[2/9xx(10^(-4))^(2)xx((10^(4)-10^(3))xx9.8)/(9.8xx10^(-6))]^(2)`
`impliesh=(20xx20),(2xx9.8)=20.41m`