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)`

Options

A. `1.65xx10^3m`
B. `2.65xx10^2m`
C. `3.65xx19^4m`
D. `1.45xx10^2m`

Answer

A

Explanation

Before entering the water the velocity of ball is `sqrt(2gh)` . If after entering the water this velocity does not change, then this value should be equal to the terminal velocity . Therefore

`sqrt(2gh) = 2/9 (r^2(rho-sigma)g)/(eta)`
`thereforeh=({2/9(r^2(rho-sigma)g)/(eta)}^2)/(2g) = 2/81xx(r^4(rho-sigma)^2g)/(eta^2)`
`=2/81xx((3xx10^(-4))^4(10^4-10^3)^2xx9.8)/((9.8xx10^(-6))^2)=1.65xx10^3 m`