A copper ball of radius `3.0`mm falls in an oil tank of viscosity 1kg/ms. Then, the terminal velocity of the copper ball will be ( Density of oil `=1.5xx10^(3)kg//m^(3)`, Density of copper `=9xx10^(3)kg//m^(3)` and `g=10m//s^(2)`.)

Options

A. `18xx10^(-2)` m/s
B. `25xx10^(-2)` m/s
C. `15xx10^(-2)` m/s
D. `20xx10^(-2)` m/s

Answer

C

Explanation

Given, radius of copper ball, `r=3xx10^(-3)m`,
viscosity of oil, `eta=1`kg/ms,
density of oil, `rho=1.5xx10^(3)kg//m^(3)` and
density of copper, `sigma=9xx10^(3)kg//m^(3)`
Terminal velocity, `v_(T)=2/9 (r^(2)(sigma-rho)g)/(n)`
Putting the given values, we get
`=2/9xx((3xx10^(-3))^(2)(9xx10^(3)-1.5xx10^(3))xx10)/(1)`
`= 2/9xx9xx10^(-6)xx7.5xx10^(3)xx10`
`rArr v_(T)=15xx10^(-2)m//s`
Hence, the correct option is (c).